Welcome to SRM GatewayDear Student, Welcome to SRM Gateway. We thank you very much for joining SRM Gateway. We are all committed to your satisfaction and fulfill your expectations in training, quality in our services, which is our primary goal. We are sure that you will become very skilled and will be positioning yourself well in the IT industry. We shall always ensure that you are guided well during the course. We wish you all the best for the complete learning of the technology Stay ahead in the technology. Way ahead of others. Warm Regards, Management SRM Gateway Document Number : SRMGW - IIT-JEE - MATH-01 Course Code : IIT-JEE SRM Gateway This document is the property of SRM Gateway. This document has been prepared exclusively for the use of the students. No part of this document shall be copied or transferred in any form or by any means which would initiate legal proceedings, if found. SRM Gateway No. 9, 3rd Avenue, Ashok Nagar, Chennai – 600 083 Preface About the book This book will be an effective training supplement for students to master the subject. This material has in-depth coverage and review questions to give you a better understanding. A panel of experts have complied the material, which has been rigorously reviewed by industry experts and we hope that this material will be a value addition. We also would value your feedback, which will be useful for us to fine-tune it better. Wishing you all success. SRM Gateway IIT - MATHS SET - 1 INDEX 1. BASIC MATHEMATICS ................................................................................... 2 2. ALGEBRA PROGRESSION ............................................................................. 54 3. BASIC TRIGONOMETRY .............................................................................. 132 4. TRIGONOMETRY EQUATIONS ................................................................... 164 5. INVERSE TRIGONOMETRY FUNCTION ................................................... 186 6. PROPERTIES OF TRIANGLE ....................................................................... 210 7. CO - ORDINATE GEOMENTRY ................................................................... 232 IIT- MATHS 1 BASIC MATHEMATICS 2 e. 0. . –1. 1. represented by R. The set of non positive integers is {…. 151 / 3 . . Zero is neither positive nor negative but it is non–positive as well as non–negative. Rational numbers when represented in decimal form are either ‘terminating’ or ‘non– terminating’ but repeating. q have no common factor except 1. } (iii) Integers : Z or I = {. 3. .25 (terminating) 5/3 = 1. All integers are rational numbers with q = 1 When q 1 and p. Note : p 22/7. –3. . p. 5/4 = 1. Z– or I– = {. . . . . . . . . . (i) Odd and even natural numbers. . Also note that Integers which give an integer on division by 2 are called even integers otherwise they are called odd integers. . . –1.g. 4. . which cannot be represented in q form. the rational numbers are called fractions. . 2. . . 1.6666 . 0}. (non terminating but repeating) p (v) Irrational numbers: Numbers. 3. . 2.. .14159 .BASIC TRIGONOMETRY Number System (i) Natural numbers : N = {1. The set of negative integers. –3. 2 . Zero is considered as even number. This is the largest set in the real world of numbers. . In decimal representation. . –1}. 22/7 is only an approximate value of p in terms of rational numbers. . . q Î Z and q 0 (because division by zero is not defined). (ii) Prime numbers (which are not divisible by any number except 1 and themselves) and composite numbers (which have some other factor apart from 1 and themselves). 2. 3 . –2. . (iv)Rational numbers: Numbers of the form p/q where p. .. .} Natural numbers are also called positive integers (denoted by Z+ or I+) Whole numbers are also called non–negative integers. . etc. 3. ‘Q’ represents their set. taken for convenience Actually p = 3. –2.. they are neither terminating nor repeating all surds fall into this category e. (vi)Real numbers: All rational and irrational numbers taken together form the set of real numbers. –2. 4. } (ii) Whole numbers : W = {0. –3.g. The set of natural numbers can be divided in two ways. e. 8}. 7. y. 2. A. and the elements are enclosed between brackets {}. A set may contain no elements and such a set is called Void set or Null set or empty set and is denoted by (phi).g. c. . AB = Union of set A and set B = {x : x Î A or x Î B(or both)} e. 4. 8}.e.g. 3. U.IIT. V etc. set of all English alphabets) or continuous (e. 5. If A = {1. Each element in a set is unique. . 4. 2.. A – B = {x : x A and xB} Similarly B – A = {x : x B and x A}. Usually but not necessarily a set is denoted by a capital letter e. b. 8} Intersection of sets It is the set of all the elements. 4. 6. 2. AB = {x : xA and xB} e. Union of sets Union of two or more sets is the set of all elements that belong to any of these sets. . b. then ABC = {1. If A = {a. 2. x. . The symbol used for intersection of sets is ‘’ i. d} and B – A = {e. . 2. 6. . The symbol used for union of sets is ‘’ i. d} and B = {b. y etc. 2.. then ABC = {2}. 4} and B = {2. .g. z} B = Set of all positive integers less than or equal to 10 = {1.g. Remember that n(AB) = n(A) + n(B) – n (AB) Difference of two sets The difference of set A to B denoted as A – B is the set of those elements that are in the set A but not in the set B i. e. set of real numbers). ..e. . 6} and C {1. If A = {1. denoted by small letters a. 6. 9. 6}and C = {1. . .MATHS 1 is neither prime nor composite 2 is the only even number which is prime Set Theory Basic Concept Set: A set is a well–defined collection of objects or elements. c. b. . In general A – B B – A e. B. 4} & B = {2. . 3. 3. The set may contain finite or infinite number of elements. f}. 10} R = Set of real numbers = {x : – ¥ < x < ¥} The elements of a set can be discrete (e. f} 4 . 5. . 3. 5. 5. x. The number of elements of a set A is denoted as n(A) and hence n () = 0 as it contains no element. 6. . c..g. For example: A = Set of all small English alphabets = {a. then A – B = {a. which are common to all the sets.g. 4. 8. x 0 x 0 defined as |x| = x. then we write logac = b obviously c is positive. For example log3 81 = 4 34 = 81 Note The expression logb a is meaningful for a > 0 and for either 0 < b < 1. Thus we can define |x| as |x| = x2 e. |x| in fact represents the distance of number ‘x’ from the origin. then the magnitude of x is called it’s absolute value and in general.g if x = –2. or b > 1 a = b log b a log c b logab = log a c a1 a 2 0 if b 1 logba1 ³ logba2 0 a a if 0 b 1 1 2 Brain Teaser 1 : log x2 = 2logx. . Thus it’s distance from the origin is x 2 Hence |x| = x 2 .5 then |x| = 2. or not defined. Thus |x| ³ 0 secondly. 0). Note that x = 0 can be included either with positive values of x or with negative values of x.5. denoted by |x| and x . is it true or false? Formulae (i) loga|mn| = loga|m| + loga |n| (ii) loga m = loga|m| – loga|n| n (iii) loga|mn| = n log a|m| (iv) logab = logcb × logac (v) log a k N = 1 logaN k Modulus Function Let x Î R. if x = 3. measured along the number–line. As we know all real numbers can be plotted on the real number line.8 Brain Teaser 2 : If x R– then x 4x 2 is 1 1 1 .BASIC TRIGONOMETRY LOGARITHM If a is a positive real number other then 1 and ab = c. any point ‘x’ lying on the real number line will have it’s coordinate as (x. 2 2 2 Basic Properties : 5 ||x|| = |x| |x| > a x > a or x < –a if a R+ and x R if a R– |x| < a – a < x < a if a R+ and no solution if a R––{0} |x + y| |x| + |y| |xy| = |x||y| .8 then |x| = 3. This is possible only when both a and b are finite.e.y0 y | y| Intervals Intervals are basically subsets of R and are of very much importance in calculus as you will get to know shortly.. . 0 < a < b Þ ar < br if r > 0 and ar > br if r < 0. . inequality sign reverses if both sides are multiplied by a negative number. and a1a2a3 . a < b and b < c a < c. 6 . b] = {x : x b} intervals are particularly important in solving inequalities or in finding domains etc. . > b1b2b3 . a If a1 > b1. . Inequalities The following are some very useful points to remember a b Þ either a < b or a = b. . a < b a + c < b + c cR a < b –a > –b i. If a > b. Closed interval: [a. . b] = {x : a < x b} Closed–open interval:[a.… Then a1 + a2 + a3 + . If there are two numbers a. > b1 + b2 + b3 + .e. 1 1 a ³ 2 a > 0 and equality holds for a = 1. b R such that a < b. b) = {x : a x < b} The infinite intervals are defined as follows (a.. . b] = {x : a x b} i. 2. then following results are evident. i = 1. . Open–closed Interval: (a. end points are also included. .IIT.MATHS x |x| .e. a < b and c < d a + c < b + d and a – d < b – c. we can define four types of intervals as follows: Open interval : (a. ) = {x : x > a} [a. .. b) = {x : a < x < b} i. a a –2 a < 0 and equality holds for a = –1. . a3 > b3 . . ) = {x : x ³ a} (–. . p and q are some positive integers. where ai > 0. end points are not included. a < b ma < mb if m > 0 and ma > mb if m < 0. bi > 0. a2 > b2. The intervals in which the curve is above the real line will be the intervals for which f(x) is positive and intervals in which the curve is below the real line will be the intervals in which f(x) is negative. b2. . an and f(x) is not defined for x = b1. we use the following method: Qx Qx If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation. a2. . . . DETERMINANTS Consider the equations a1x + b1y = 0 and a2x + b2y = 0. bm in an increasing order say c1. If p(x) is divided by (x–a). if (x–a) is a factor of p(x). . . Factor Theorem Let p(x) be a polynomial of degree greater than or equal to 1 and a be a real number such that p(a) = 0. b2. . . where P(x) and Q(x) are polynomials. then within this interval the polynomial itself takes on values having the same sign. . . . . Then f(x) = 0 for x = a1. a2. Plot them on the real line. x m . . This curve is known as the wavy curve. . Qx Where a1. . an. bm apart from these (m + n) real numbers f(x) is either positive or negative. . . b2. . . And draw a curve starting from right of cm+n along the real line which alternately changes its position at these points. c5. . . bm are distinct real numbers. c3. 2 2 7 . then (x – a) is a factor of p(x). . . b1. . . . . . Now arrange a1. . . . . . then p(a) = 0. c4. a2. c2. . . . an. . x n Px f x x 1 x 2 x 3 . . Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. . . These give a1 y a 2 a a 1 2 Þab –ab =0 1 2 2 1 b1 x b2 b1 b 2 a1 b1 We express this eliminant as a b = 0. then the remainder is equal to p(a). . . cm+n. Remainder Theorem Let p(x) be any polynomial of degree greater than or equal to one and a be any real number. Conversely. . . Ignore the common roots and write x 1 x 2 x 3 . . b1. . 0 . . . .BASIC TRIGONOMETRY a > b an > bn Þ a–n < b–n where n Î N a > b al/q > bl/q Þ ap/q > bp/q Wavy Curve Method In order to solve inequalities of the form Px Px 0. Thus the figure (a) would represent the graph of a function. Here we say that the distance traversed is a function of time. the area assumes different numerical values. in functional notation y = f(x). the path traversed is regarded as a variable. In the figure any line drawn parallel to yaxis would meet the curve at only one point. f(x) f(x) y = f(x) L y = f(x) C y3 y2 B y2 y1 y1 A x2 x3 x1 Fig (a) x x0 x Fig (b) These figures show the graph of two arbitrary curves. 3) are called the elements of the determinant. If R takes on various numerical values.IIT. bi. The area of a circle. Thus this curve will not represent a function. The relation between the variable x and y is called a functional relation. For example. in terms of its radius R.MATHS a1 b1 The expression a b is called a determinant of order two. it is necessary to consider the variation of one quantity as dependent on the variation of another. is pR2. Function In the study of natural phenomena and the solution of technical and mathematical problems. The variable x is called the independent variable or argument. 2 2 A determinant of order three consisting of 3 rows and 3 columns is written as a1 b1 c1 a2 b2 c2 a3 b3 b 2c 2 a 2c 2 a 2b2 and is equal to a1 b c – b1 a c + c1 a b 3 3 3 3 3 3 c3 = a1 (b2c3 – c2b3) – b1 (a2c3 – c2a3) + c1(a2b3 – b2a3) The numbers ai. 2. If to each value of variable x (within a certain range) there corresponds a unique value of another variable y. line L) would meet the curve in more than one points (A. The letter f in the functional notation y = f(x) indicates that some kind of operation must be performed on the value of x in order to obtain the values of y. In the figure (b) certain line (e. Hence area of the circle is a function of the radius R. and equals a1b2 – a2b1.g. B and C). or. So the variation of one variable brings about a variation in the other. The set of all possible values which the independent variable (here ‘x’) is permitted to take for a given functional dependence to be defined is called the domain of definition or simply the domain of the 8 . And the variable y is called the dependent variable. ci (i = 1. then we say that y is a function of x. That means each element of X would have one and only one image. Thus element x0 of X would have three distinct images. which varies with time. in studies of motion. a 0 x n a 1 x n 1 . b 0 1 m e. e. This function is defined for all values of x. y = cosecx. All these function have a very important property that is Periodicity.g. defined for R – (2n + 1) .g. Therefore its domain of definition is the infinite interval – < x < .BASIC TRIGONOMETRY function. y = ax + b. defined for R – n . (iii) General exponential function: y = ax. (c) Irrational function e. (b) a is negative integer. a 0 (a linear function) y = ax2 + bx + c.. Elementary Functions: (i) Constant function: y = c where c is a constant.. where nl It must be noted that in all these function the variable x is expressed in radians. a n (b) Rational Function y = b x m b x m 1 . (v) Trigonometric function: y = sinx. defined for all real x. where a0. ` Is sec2 θ – tan2 θ = 1 valid for all θ R (real) ? (vi) Algebraic function: (a) Polynomial function: y = a0xn + a1xn–1 + … + an. a 0 (a quadratic function) A polynomial function is defined for all real values of x. y = 9 2x 2 x 1 5x 2 . ). The function y = 1 x 1 is defined for all x > 1 its domain is (1. The function is defined for all values of x except for x = 0.. y = cosx defined for all real x y = tanx. The function y = sin x is defined for all values of x. y = secx. (ii) Power function: y x (a) a is positive integer. e. The function is defined in the infinite interval – < x < .. called the polynomial of degree n. This function is defined for all x > 0. a > 0 but a 1. 2 y = cotx. (iv)Logarithmic function: y = logax. a1 … an are real constants (a0 0) and n is a positive integer.g:y = a/x (inverse variation) The rational function is defined for all values of x except for those where the denominator becomes zero. where a is positive not equal to unity.g. x2] . it just provides an overall information.r. This sufficiently close is the key word here. Thus..t. x in the interval [x1. the quantity x x will tell us the average rate of change of y w. or they may even have a mixed trend.t. 2 1 dy we say that x2 nearly coincides with x1 and represent it as x2 ® x1. x at x = x1. depending upon the type of function. To know the rate of change of y w. For example if y2 = y1 it does not necessarily mean that y is same for all x in the interval [x1. Let us take two values of x: x1 and x2(x1 < x2). In the limiting case. Suppose we start putting some values of ‘x’ in increasing order. Putting the values of ‘x’ in this relation. i. 10 . 2 1 y 2 y1 Let y2 > y1 x x is positive Function is increasing on an average. or in decreasing order. x2].e. We call dy the derivative or the differential coefficient of y w.t. we take x2 y 2 y1 very near to x1 (as much as possible). x.MATHS Differential Calculus Let y = f(x) be a function. the quantity x x can not give the exact 2 1 idea of the variation of y w.r. dx y y2 – y1 x2 – x 1 x1 x2 x (You can understand it physically by taking x as time and y as displacement of a body. We use the notation dx y 2 y1 x x1 for x x 2 1 as x2 ® x1. if x1 and x2 are sufficiently far apart. The respective values of ‘y’ that we obtain may turn out to be in increasing order. 2 1 y 2 y1 If y2 = y1 x x is zero Function is constant on an average. y1 = f(x1) and y2 = f(x2) y y 2 1 Then. x 2 tends to x1 and then calculate x x . 2 1 y y 2 1 As you can see. x in the interval [x1. 2 1 y 2 y1 if y2 < y1 x x is negative Function is decreasing on an average.r.r. dx means small change in x (near x = x1) and dy means the corresponding change in y.IIT. to obtain a sufficiently accurate information. So.t. x2]. we obtain the corresponding values of ‘y’. or they may remain constant. we have to choose x1 and x2 sufficiently close to each other. where k is a constant. Basic Differentiation Formulae y = constant dy 0 dx y = tan–1x dy 1 dx 1 x 2 y = xn dy nx n 1 dx y = cot–1x dy 1 dx 1 x 2 y = sinx dy cos x dx y = cosec–1x dx | x | x 2 1 y = cosx dy sin x dx y = sec–1 x dx | x | x 2 1 y = tanx dy sec 2 x dx y = ax dy a x ln a dx y = cotx dy cos ec 2 x dx y = ex dy x e dx y = sin–1x dx 1 x2 y = logax dx x ln a y = cos–1x dx 1 x2 y = ln x dy dy 1 dy 1 dy 1 dy 1 1 dy 1 dx x Some Important Theorems The following are very important theorems. dx x x1 (i.. Theorem 1: If a function is of the form y = k f(x).BASIC TRIGONOMETRY Then dy denotes the magnitude of velocity). then dy df ( x ) k dx dx Theorem 2: The derivative of the sum or difference of a finite number of differentiable functions is equal to the sum or difference of the derivatives of these functions. dx dy df ( x ) is also represented as f ¢ (x) or dx dx dy Graphically. dy computed at x = x1) denotes the slope of the tangent to the curve y = dx f(x) at x = x1 We will not here derive the formulae for ¢(x) of various functions. 11 .e. but we give the results of the derivations here. which can be applied directly. e. We take the co-ordinate system as shown and assume that the load is dropped at the instant the aeroplane cuts the y-axis.e.. where u and v are functions of x. if y = f(u(v)). if y = u (x) + v (x) + w(x) then y = u(x) + v(x) + w(x).IIT. Since the horizontal translation is uniform. then y = dy df du dv . then y = dy dF du dx du dx This is called the chain rule. u = f (x) or y = F[f(x)].g. y = (t) 12 . dx du dv dx Parametric Representation of a Function and it’s Derivatives We find the trajectory of a load dropped from an aeroplane moving horizontally with uniform velocity v0 at an altitude y0. y = y0 – gt 2 2 Y v0 (x... y) X Those two equations are called the parametric equations of the trajectory because the two variables x and y have been expressed in terms of the third variable t (parameter) i. a function represented by y = F(u).MATHS i. then y¢ = dx Theorem 5 u v uv v2 If y = uv. Theorem 3 The derivative of the product of two differentiable functions is equal to the product of the derivative of the first function with the second function plus the product of the first function with the derivative of the second function: i.v Derivative of a Composite Function Given a composite function y = f(x). This formula can be extended for the derivatives of the product of any (finite) number of functions. then y = vuv-1 u + uv ln u.e. if y = uv. the position of the load at any time t. two equations x = (t). Theorem 4 u(x) dy If y = v( x ) . i. then y = uv + uv.e. is given x = v0t.e. The rule can be extended to any number of composite function. b].r. b]. then y is said to be an increasing function of x. f(x) decreases in [a. If y = x then dx dx d 2 y d dy d = (5x 4 ) = 5. (ii) Decreasing Function If x2 > x1 y2 < y1 for any x belonging to [a. b) For a constant function.r. b] f (x) < 0 x in (a.t. i. f (x) = 0 For a non-decreasing function f (x) 0 For a non increasing function. then y is said to be a decreasing function of x.BASIC TRIGONOMETRY where t assumes values that lie in a given interval (t1. (x) increases in [a. t2) dy dy / dt (d / dt ) ( t ) Then dx dx / dt (d / dt ) ( t ) Second Derivative of a Function The second derivative of y w. a function ‘f’ is said to have a minima at x = x0 if f(x) > f(x0). if s = f(t) then v = ds d 2s dv = f (t) and a = 2 = = f (t) dt dt dt THE BEHAVIOUR OF FUNCTIONS The following behaviours of a function are important to study Increasing and Decreasing Functions (i) Increasing Functions If y = f(x) and x2 > x1 implies y2 > y1 for any x belonging to the interval [a. dx d2 y dy 5 = 5x4 2 or y or f (x). b).t.g. x in the immediate neighbourhood of x0. f (x) 0 Maxima and Minima of Functions A function ‘f ’ is said to have a maxima at x = x0 if f(x) < f(x0). e.4 x3 = 20x3 2 dx dx dx dx The acceleration ‘a’ of a particle is the second derivative of the distance ‘s’ (given as a function of time). dy w. x is the function obtained by differentiating It is represented as So. Similarly. b] f (x) > 0 x in (a. x in the immediate neighbourhood of x0 We have used the word immediate here because a given function may have any number of high 13 . x.e. e. Integral Calculus: The Antiderivative of Function A function F(x) is called the antiderivative of the function f(x) on the interval [a. the antiderivative of the function f(x) = x is . b]. as 3 = x2. is 3 3 3 2 the antiderivative of x2. dy or f ¢ (x) dx (a) Find (b) Find the points at which it becomes zero.MATHS and low points. The highest of all the bumps is the global maxima and the lowest of all the depressions is the global minima. (a) First Derivative Test Suppose x = x0 is a critical point i..e. These points are called critical points.. If f (x) changes sign from positive to negative in the neighbourhood of x = x0 Maxima at x0 If f ¢ (x) changes sign from negative to positive in the neighbourhood of x = x0 Minima at ‘x0’ (b) Second Derivative Test (i) d2y Find 2 or f (x) dx (ii) Compute the value of f (x) at the critical points If it is positive Minima at those values of x. Mathematically.e. all of them being contained in the expression F(x) + C. Infact.IIT. Then the largest among them gives the global maximum values and smallest gives the global minimum values. It is just like moving on an uneven surface (which has many bumps and depressions). then it possesses infinitely many antiderivatives. y = f(a) and f(b)). b] if. y) at all the critical points and also the end points (i. then to find the maxima and minimum values i. To find the points of maxima and minima we resort to either of the following tests. these bumps are called the points of local maxima and the depressions are called the points of local minima. at all points of the interval f(x) = F(x).e.. f (x0) = 0. C . We state here the preliminary methods only to find the maxima and minima of functions. where C is a constant. where C is an arbitrary constant. 14 . If it is negative maxima at those values of x. global maxima and global minimum of the function in that interval we com pare the values of the function (i. x3 ' x3 For example. The function 3 3 x3 x3 x 2 and 1 are also antiderivatives of f(x) = x2.. If the function is defined in an interval [a. So if a function f(x) possesses an anti-derivative F(x). In fact sin–1x – (–cos– 1 x) = sin–1x + cos–1x = /2. Thus. (i) 1 x2 1 x2 2 dx + c = . ( n 1) n 1 (f ( x )) n f ( x ) dx (f ( x )) n 1 C (n 1 ) n 1 sec x dx = ln |sec x + tan x| + c dx a2 x2 dx 1 1 x tan–1 c a a = x dx ln | x | c 1 x 2 tan x x e dx e c sin x dx = – cosx + c cos x dx = sinx + c x dx a2 x2 1 x c sin 1 x x c or –cos–1 + C a a dx –1 1 x 2 = sin x + c dx x 2 1 = sec–1x + c or – cosec–1 x + c sec2 x dx = tanx + c cosec2 x dx = –cotx + c tan x dx = – ln|cosx| + c = ln |secx| + c cot x dx = ln |sin x| + c = – ln |cosecx| + c 1 e. The only legitimate conclusion is that they differ by some constant. if F(x) = f(x). by definition òf(x) dx = F(x) + C. then the expression F(x) + C is called the indefinite integral of the function f(x) and is denoted by the symbol òf(x) dx. If a function f(x) is continuous on an interval [a. n x dx x n 1 c. b]. Two different integrals of a function differ by a constant. Standard Elementary Integrals In the following integrals. then this function has an antiderivative. It does not mean that sin–1 x = –cos–1x.g.BASIC TRIGONOMETRY If the function F(x) is an antiderivative of f(x). x3/2 + c 1 3 1 2 (ii) 1 x 21 1 2 dx x dx c c x2 2 1 x The following points are to be noted: 1 x dx lnx + c if x is positive = ln (–x) + c if x is negative because 1) = 1 x 1 1 x2 d 1 (ln (–x)) = (– dx x 1 x dx ln | x | C dx = sin–1x or –cos–1x. then a f(x) dx = a f ( x ) dx . C stands for an arbitrary constant. The process of finding the antiderivative of a function f(x) is called integration. 15 If a is a constant. x and t being connected by the relation x =f (t). we write b b f (x ) dx | F( x) dx |a F(b) F(a ).. a Note: b a f (x ) dx f (x) dx a b b c b f (x ) dx f ( x) dx f ( x) dx a a c where c is any point inside or outside the interval (a. is called the definite integral of f(x) over the interval (a. a is called the lower limit and b the upper limit of integration.e. = f(x) dx g(x) dx Methods of Integration (i) Integration by Substitution This method consists of expressing the integral ò f(x) dx. is the independent variable.IIT. b. i. 16 . in terms of another integral where some other.MATHS [f(x) g(x)] dx. say ‘t’. where x is the independent variable. b). This method is useful only when a relation x = (t) can be so selected that the new integrand f(x) dx is of a form whose integral is known dt (ii) Integration by parts f (x) (x) dx (i ) ( ii ) df ( x ) dx dx dx = f (x) ( x ) dx Integral of the product of two function = first function × integral of second–integral of (derivative of first × integral of second). b) and is denoted by b b f ( x) dx. f(x)dx = òf[j (t)] j¢ (t) dt. Geometrically definite integral represents area under curve. a a where F(x) is the antiderivative of f(x). Or. say a. Definite Integral The difference in the values of an integral of a function f(x) for two assigned values of the independent variable x. Thus f ( x) dx F(b) F (a ). (DDBC). GC = 2 FG GC = 2 × 4 = 8 Example 2 : Triangles ABC and DBC are on the same base BC with A. Therefore the altitudes corresponding to BC are equal i. what is BE ? (b) If FG = 4. what is GC ? Solution: (a) We have. Show that BC bisects AD. (DABC) = ar. We have 1 = Ð2 (vertically opposite angles) AEO = ÐDFO (90° each) and AE = DF A 1 B DAEO @ DDFO (by A.A. such that ar.. D on opposite sides of lne BC.S) AO = OD 17 E O F C 2 D .AE = DF Now. (a) If BG = 6. BE and CF pass through G.e. Solution: Since D’s ABC and DBC are equal in area and have a common side BC.BASIC TRIGONOMETRY SOLVED SUBJECTIVE EXAMPLES Example 1 : In a DABC. the medians AD. in D’s AEO and DFO. BG = 6= 2 BE 3 2 BE BE = 9 3 (b) We have. we get (4x – 3)log 18 = (3x – 4) log(18. … (1) c c ra2 = a a b2 c [Using (1)] 2 2 a ( r 1) a 18 .3 2 ) (since 3 2 = 18 ) or. 4x – 3 = (3x – 4) 3 2 or. x2 – 3x – 4 ³ 0 (x – 4) (x + 1) ³ 0 x ³ 4 or x –1. (4x – 3) log 18 = (3x – 4) log (18)3/2 or. 8x – 6 = 9x – 12. a × ra = r. Example 4 : Solve 184x–3 = (54 2 )3x–4 Solution: Given equation is 184x–3 = (54 2 )3x–4 Taking log on both the sides.MATHS BC bisects AD Example 3 : Solve |x2 – 3x – 4| = x2–3x – 4 Solution: We know |x| = x when x ³ 0 So. a + ra = b b b a(1 + r) = a = a (r 1) a a Also. show that (r 1) 2 b 2 r ac Solution: Let a and ra be the roots of the equation ax2 + bx + c = 0 So. or x = 6 Example 5 : Solve for x if log3x + log9(x2) + log27 (x3) = 3 Solution: log3x + log9(x2) + log27(x3) = 3 log x 2 log x 3 log x 3 log x log 3 2 log 3 3 log 3 3 log 3 3 logx = log3 x=3 Example 6 : If r be the ratio of the roots of the equation ax2 + bx + c = 0.IIT. D 7 14 7 c. radius of circle = 2 7 A Now the area of circle = r 2 = (7) 2 = 49 c. Solution: BC = 14 c. 1) Example 10 : If x = 2 ln cot t and y = tan t + cot t.m2 Example 9 : x2 If f(x) = .m. find the range of f(x) 1 x2 Solution: f(x) = 1 x2 x2 11 = = 1 2 2 1 x2 1 x 1 x Clearly f(x) [0.m. find Solution: Since 19 dy dy / dt dx dx / dt C dy dx B .m.BASIC TRIGONOMETRY (r 1) 2 b 2 r ac Example 7 : If x is real. prove that 3x2 – 5x + 4 is always positive Solution: 5 4 2 3x2 – 5x + 4 = 3 x 3 x 3 2 2 5 48 25 5 23 = 3 x 3 x 0 since square of real number is always non-negative. x R. 6 36 6 36 Example 8 : Find the area of the largest circle that can be inscribed in a square of side 14 c. tan d (Integrating by parts) = 0 1 ln 2 4 2 Example 13 : Solve : log| x | | x | = 0 Solution: We have log| x | | x | = 0 |x| = 1 but | x | 1 (being in base of logarithm) x Example 14 : 20 .IIT. when x = 0. sinq = 0 When x = 1 / 2 .MATHS dx cos ec 2 t 2 4 2 Now dt cot t cos t sin t sin 2t dy sin 2 t cos 2 t cos 2 t 2 2 sec t cos ec t – 4 Also 2 2 dt sin t cos t sin 2 2 t dy 4 cos 2t sin 2 t cos 2 t Hence dx = cot 2t 2 sin 2t 4 sin 2 t Example 11 : sin 3 x cos 3 x sin 2 x cos 2 x dx Solution: sin 3 x cos 3 x sin 2 x cos 2 x dx = sin 3 x cos 3 x dx sin 2 x cos 2 x sin 2 x cos 2 x dx = tanx secx dx + cotx cosecx dx = secx – cosec x + c Example 12 : 1/ 2 Evaluate: 0 sin 1 x dx (1 x 2 ) 3 / 2 Solution: Let x = sinq dx = cos q dq and q = sin–1x. sinq = p/4 1 2 4 1 sin x 1 x 2 3/ 2 dx 0 0 4 cos d sec 2 d cos 3 0 / 4 /4 = tan 0 1. y-axis and x = Solution: /2 The represented area = cos x dx 0 /2 sin x 0 1 0 = 1 sp. Now. 3]. xcos . (1) = 2 – 14 + 107 = 85. unit. f (x) = 6x2– 24 Now. Solution: We have f(x) = 2x3 – 24x + 107 So. f (x) = 0 Þ 6x2– 24 = 0 x = ± 2 But x = –2 Ï [1. Example 15 : Find area bounded by y = cosx. 21 . (3) = 2 (3)3 – 24 × 3 + 107 = 89. the maximum value of f (x) is 89 which attains at x = 3 and the minimum value is 75 which is attained at x = 2. Hence. f(2) = 2(2)3 – 24 (2) + 107 = 75 And. 2 . 3] So x = 2 is the only stationary point.BASIC TRIGONOMETRY The maximum and minimum value of f (x) = 2x3 – 24x + 107 in the interval [1. MATHS SECTION . If the roots of (1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0 are equal. 10.IIT. For every x R. then prove that 2 y 0 dt 8.I SUBJECTIVE LEVEL . then find the value of m. 5. d2 y 2 If y = acos t + sin t . prove that 2x2 – 6x + 9 is always positive. 2 (ii) (2xe x ) dx /2 (ii) (sin x cos x) dx 0 22 . Evaluate : (i) (x 2 2x 1) dx . Find the intervals of increase and decrease of (x – 3) (x + 1). Differentiate the following with respect to ‘x’. ( x 3) 2 ( x 3) 5 ( x 1) Find the solution to the inequality >0 ( x 5) ( x 4) 2.I REVIEW YOUR CONCEPTS 1. Solve for x (a) |x –4| > 7 3. 6. (b) |x| > x (a) Solve for x: log2x > 3 (b) Which is greater: log23 or log1/25 4. (i) sinx + cosx (ii) xlogx (iii) Differentiate sin2x + cos2x with respect to x 7. (a) Find domain of the definition of the following: (i) (b) (ii) x log e x Draw the graph of the following: (i) f(x) = x3 (ii) f(x) = logex 9. Solve for x ( x 4 x 4) (i) |x – 3| + |x + 2| = 3 (ii) (a) Solve for x. 6. loge ( x 1) ( x 3) . h (5) = 97. – x (i) sin x3 x 1 (ii) Integrate the following sin x . Differentiate cos (4x3–3x) with respect to x. Find the intervals of increase and decrease of the function y = cosx. Find Domain of definition 5. 2 3. given by h(x) = kx2 + 7x – 4. then prove that log (a + 2b) = 4. In the figure given below. cos 2 x dx .II BRUSH YOUR CONCEPTS 1. 7.BASIC TRIGONOMETRY LEVEL . 0 9. find k. If for the function h. If a2 + 4b2 = 12 ab. 3 2 5 Find the solution common to both the inequalities ( x 1) ( x2 3x 2)7 | x 4 | 0 & 1 < |x – 3| < 5 2. find the value of angle P A 24° R P 36° B 10. 54° C D In the figure given below. find the value of x T 5 P 23 4 B x A ( x 2) ( x 4) 2 2 8. x2 5 2x 7 1 (log a log b 4 log 2) . In the figure given below. Find the intervals of decrease and increase of (x + 2) e–x .1 4. log10 0. (i) In the figure given below. Evaluate: 7. 2. Evaluate: 7 log3 5 3log 5 7 5log 3 7 7 log 5 3 3 5 1 log 7 5 1 .3 (x – 1) < log0.MATHS LEVEL . Find the set of all solution of the equation 2|y| = 2 y1 1 3.IIT. ABCD is square and triangles BCX and DYC are equilateral triangle. 2 B A y x C D 9. Find QSR Q S O T 50° P R 24 .III CHECK YOUR SKILLS 1. 8. Evaluate: 5. Solve: log0.09 (x – 1). x 6. Find domain of definition (i) f(x) = log 2 ( x 3) x 2 3x 2 (ii) f(x) = 5x x 2 y log 10 (iii) 4 q 5 cos x 3sin x cos (iv) y | x | 1 2 | x | x2 1 x x2 1 x dx . Find the value of y. 25 D ABC is a triangle in which BC is produced to D.BASIC TRIGONOMETRY (ii) In the given figure. O is the centre of the circle. DCA = 108° and EAB = 124°. If OCA = 26°. CA is produced to E. then find ODB C B O A 10. Then find ABC. . x x is (b) always decreases (d) some times increases and some times decreases The function f(x) = x3 – 3x is (a) increasing on (–. 1) 26 . then x = (a) 16 (c) 64 2. 0) 6. –4 (d) none of these 1 5. (b ) 6 (d) 36 The value of p for which (x – 1) is a factor of x3 + (p + 1)2 x2 – 10 is given by (a) 4.) (d) none of these The differential coefficient of f(x) = log (logx) with respect to x is x 7. 7 3 6 2 (b) . ) and decreasing on (–1.II OBJECTIVE LEVEL . 2 (c) –2. (b) (0. 4 (b) 2. 6 2 7 3 (c) .MATHS SECTION . If | 4 – 3x | £ (b) 32 (d) none 1 then x is equal to 2 7 3 (a) .I 1.|x| – 6 = 0 is (a) – 9 (c) 9 4. 6 2 3. –1] [1. (a) always increases (c) never decreases 8. If log16x + log4x + log2x = 14. log x x (a) log x (b) (c) (xlogx)–1 (d) xlogx The function f(x) = tanx – x.IIT. The domain of definition of the function f(x) = (a) R (c) (–. (d) none of these The product of all the roots of the equation x 2 . (b) 110° (d) 90° In the figure PAQ is a tangent of the circle with centre O at a point A if ÐOBA = 32°. the number of digits in 340 is (a) 18 (c) 20 13. What is the sum of the other two angles of the triangle? (a) 112° (c) 120° 14. Evaluate 0 12. sin x dx cos 3 x 3 cos x (a) 1 2 (b) 1 6 (c) 1 8 (d) 1 12 If log10 3 = 0. (b) 2 (d) none of these The maximum value of x3 – 3x in the interval [0. ) and decreasing on (–. The minimum value of 2( x 2 3)2 27 is (a) 227 (c) 1 10. 1) (c) increasing on (0. 2] is (a) – 2 (c) 2 (b) 0 (d) 1 x/4 11.BASIC TRIGONOMETRY (b) decreasing on (–. (b) 19 (d) 21 The interior and its adjacent exterior angle of a triangle are in the ratio 1 : 2. ) and increasing on (–1. The value of x and y is P C y A O x B Q 27 . 0) (d) decreasing on (0. –1] [1. 0) 9.477. ) and increasing on (– . The number of sides of the polygon is (a) 12 (c) 15 (b) 8 (d) 20 28 .IIT. (b) both 40° (d) 30°. 60° The difference between the interior and exterior angles of a regular polygon is 132°. 50° (c) both 58° 15.MATHS (a) 30°. The greatest value of f(x) = cos (xex + 7x2 – 3x). 1+ = dx (a) tan2 (b) sec2 (c) sec (d) |sec| 3 3 8. (b) 3 (d) 7 If logkA . 2 7. x [– 1. (b) k3 (d) 243 If the product of the roots of the equation x 2 5 x 4 log 2 0 is 8. ) (a) R (c) (0. then A = (a) 53 (c) 12 3. log5k = 3. x2 9 The domain of definition of the function f (x) = log x (b) (1. 1) (1. ) is (a) –1 (b) 1 (c) 0 (d) none of these 29 .BASIC TRIGONOMETRY LEVEL . If a and b are roots of the equation x2 + x + 1 = 0 then a2 + b2 = (a) 1 (b) 2 (c) –1 (d) 3 6. y = a sin . If A = log2log2log4256 + 2 log 2 2. then l is (a) 2 2 (c) 3 (b) 2 2 (d) none of these 4. The largest value of 2x3 – 3x2 – 12x + 5 for – 2 x 4 occurs at x = (a) – 2 (b) – 1 (c) 2 (d) 4 10.II 1. then A equals to (a) 2 (c) 5 2. The number of real roots of the equation (x –1)2 + (x – 2)2 + (x – 3)2 = 0 (a) 3 (b) 2 (c) 1 (d) 0 5. The function y = x3 – 3x2 + 6x – 17 (a) increases everywhere (b) decreases everywhere (c) increases for positive x and decreases for negative x (d) increases for negative x and decreases for positive x 9. dy If x = a cos . ) (d) [1. (a) 3 2 (b) 2 (c) 1– 4 (d) 6 5 2 If the area of parallelogram ABCD is 32 sq.IIT.5 cm If A. CX = 10 cm find BX B A x C D O P T (a) 3 cm (c) 4 cm 15.MATHS 11.5 cm (d) 4. Then the value of angle BAC is equal to (a) 5° (c) 20° (b) 10° (d) 30° 30 . B. cm. (b) 2 cm2 (d) none of these In the given figure if AX = 5 cm. ex 1 e 2 x dx (a) cot–1ex+c (c) tanex + c (b) tan–1ex + c (d) sin–1ex + c x/4 12. XD = 7 cm. (b) 3. Area of DAMN is equal to A B N D C M (a) 8 cm2 (c) 4 cm2 14. Eva luate tan 2 x dx 0 13. C are three consecutive points on the arc of a semicircle such that the angles subtended by the chords AB and AC at the centre O is 90° and 100° respectively. + – + –3 –5 – – 1 3 x ( .I (CBSE Level) CHECK YOUR SKILLS 1. x > x which is not possible. 0) or R– 3. 5) ( 3 1) (4 ) 2. Solve for x (a) |x – 4| > 7 Solution: (a) (b) |x| > x |x – 4| > 7 x – 4 > 7 or x – 4 < – 7 x > 11 or x < – 3 –3 x (– . ) (b) |x| > x Case I : x > 0.BASIC TRIGONOMETRY SUBJECTIVE SOLUTIONS LEVEL . Case II: x < 0 x < – x 2x < 0 x < 0 x ( – . – 3) (11. ( x 3)2 ( x 3)5 ( x 1) 0 Find the solution to the inequality ( x 5)( x 4) Solution: (x 3)2 (x 3)5 (x 1) 0 (x 5) (x 4) By wavy curve method. (a) (b) Solution: (a) Solve for x: log2x > 3 Which is greater: log23 or log1/25 log2x > 3 x > 23 [ base is greater than 1] 31 11 + 4 . ) log23 or log1/2 5 log1/2 5 = – log25 < 0. [ log1/a x = – logax] log23 is greater than log1/2 5. Solution: Let f(x) = 2x2 – 6x + 9 = 2(x2 – 3x + 9/2) = 2( x – 3/2)2 + 9/4 > 0 [ square of real number is always non–negative] Hence f(x) is always positive. For every x R.MATHS (b) x > 8 x (8. prove that 2x2 – 6x + 9 is always positive. If the roots of (1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0 are equal then find the value of m.IIT. Solution: Given equation is (1 + m) x2 – 2(1 + 3m) x + (1 + 8m) = 0 roots are equal . Differentiate with respect to ‘x’ (i) sinx + cosx (ii) xlogx Solution: (i) (ii) Let y = sinx + cosx dy cos x sin x dx d (sinx + cosx) = cosx – sinx dx Let y = x logx dy d d 1 x (log x) log x (x) = x log x1 dx dx dx x d (x log x) = 1 + logx dx 8. 6. (a) Find domain (i) (b) x Draw graph (i) f(x) = x3 (ii) log e x (ii) f(x) = logex 32 . 5. 4. 3. then D = 0 4 (1 + 3m)2 – 4 (1 + m) (1 + 8m) = 0 1 + 9m2 + 6m – 1 – 9m – 8m2 = 0 m2 – 3m = 0 m (m – 3) = 0 m = 0. x 0 f (x) 6x ve. y = a sint. f (x) = – 2 x x (ii) Df R {0} (iii) Rf R 9. Differentiate with respect to x (i) sin2x + cos2x (ii) Solution: (i) x = a (cost + log tan t ). 0) x 1 1 (i) f (x) = . ) (b) Let f(x) = x3 It is odd function so graph will be symmetric about the origin (ii) Df R (iii) Rf R (iv) y x f (x) = 3x2 f (x) = 0 x = 0 (v) f (x) = 6x (vi) ve. 2 Let y = sin2x + cos2x dy d d d d (sin 2 x) (cos 2 x) (cos x) (sinx) + dx dx dx dx dx = 2sinx cosx + 2cosx (–sinx) = 2sinx cosx – 2sinx cosx = 0 33 d (sin 2 x cos 2 x) 0 dx d d (1) 0 [ (constant) = 0] dx dx (ii) x = a (cost + log tan t ). x 0 (ii) Let f(x) = logx y y = logx (1.BASIC TRIGONOMETRY Solution: (a) (i) Let f(x) = x f(x) is real for all x 0 Df = [0. 2 y = asint . ] (ii) Let f(x) = log e x f(x) is real for all x 1 Df = [1. f (x) 0 2(x – 1) 0 x 1 x [1. f (x) = 0 x 3 2 f (x) = 2 > 0 –3/2 f(x) is local minima at x (ii) f(x) = logx + x f (x) 3 2 1 1 x for local maxima or local minima.to t 1 t 1 sin t sec 2 dy dx t 2 2 .sin t = = tant 2 1 sin t cos 2 t 10.r. ) for decreasing. w. for local maxima or local minima we must have. Solution: Let f(x) = (x – 3) (x + 1) = x2 – 2x – 3 f (x) 2x 2 = 2(x – 1) for increasing. Find the point of local maxima and minima of the following (i) x2 + 3x (ii) logex + x Solution: Let f(x) = x2 + 3x = x(x + 3) f (x) = 2x + 3. f (x) 0 x ( .sin t cos t. we must have f (x) 0 1 1 0 x = –1 x 1 0 x2 f(x) is local maxima at x = –1. Find the intervals of increase and decrease of (x – 3) (x + 1).IIT.1] 11.MATHS Diff. f (x) 34 . a a cos t tan dt dt 2 = a[– sint + cosect] dy cos t dx cos ect sin t cos t. Find QSR . Integrate the following (i) (ii) 2 xe x2 x2 + 2x + 1 x/2 (iii) 2 (sin x cos x ) dx (iv) x dx 1 0 Solution: (i) Let I = (x 2 2x 1) dx x 2 dx 2 x dx 1. 35 5 2 In the figure given below. /2 (iii) I (sin x cos x) dx 0 /2 = /2 sin x dx 0 cos x dx 0 = cos x 0 / 2 [sin x]0 / 2 = – cos cos 0 sin sin 0 = –[0 –1] + [1 –0] = 2 2 2 2 (iv) Let I | x |dx 1 x. 3 Let I = 2x e x2 dx Put x2 = t 2xdx = dt 2 x t I e dt = et + C = e C. x 0 Let f(x) = |x| = x. x 0 0 2 I f (x)dx f (x) dx 1 0 0 2 0 2 2 2 xdx xdx x x 2 1 0 1 2 0 1 1 [0 1] (4 0) 2 2 13.BASIC TRIGONOMETRY 12. where C is integral constant.dx (ii) x3 x2 2 x C 3 2 = x3 x 2 x C. where C is integral constant. 36 . then find ODB C B O A D Solution: OCA = 26° = DBA OB = OD OBD = ODB = 26° C B 26° 26° O A D ABC is a triangle in which BC is produced to D. Then find ABC. DCA = 108° and EAB = 124°. CA is produced to E.MATHS Q S O 50° T P R Solution: QSR + PRQ + QPR + QOR = 360° 90° + 90° + 50° + QOR = 360° QOR = 130° 1 QSR = QOR = 65° 2 14. Solution: DCA = 108° E ACB = 72° 124° A 56° EAB = 124° 108° BAC = 56° D C B ABC = 180° – (72° + 56°) = 62° 15.IIT. O is the centre of the circle. If OCA = 26°. In the given figure. 4) Case II : x > –4 – + –4 –2 + – + 1 –1 Similarly.. x (4 2) ( 1.1) . (i) x ( ..I BRUSH UP YOUR CONCEPTS 1. Find the solution common to both the inequalities ( x 1)3 ( x 2 3x 2) 5 | x 4 | 0 & 1 < |x – 3| < 5 ( x 2 4 x 4) 7 Solution: Case I : x < – 4 (x 1)3 (x 2)5 (x 1)5 (x 4) 0 (x 2)14 (x 1)3 (x 2)5 (x 1)5 (x 4) 0 (x 2)14 – + –2 –4 + – + 1 –1 x ( .. (ii) x ( 2. –1 x ( 1. 4) ( 4.BASIC TRIGONOMETRY LEVEL . 2) Case II: x > 3 1<x–3<5 4<x<8 x (4.1) Solve for x (i) |x – 3| + |x + 2| = 3 Solution: (i) |x – 3| + |x + 2| = 3 Case I: 37 –2 (ii) x2 5 2x 7 2 4 8 .1) And 1 < | x – 3| < 5 Case I : x < 3 1 < – x + 3 < 5 – 2 < –x < 2 – 2 < x < 2 x (–2..8) from (i) & (ii) –4 2. 2) ( 1.8) . 2) (4. – x ( . which is not possible Case III: x3 x–3+x+2=3 2x = 4 x = 2. then prove that log (a + 2b) = 1 (log a log b 4log 2) 2 Solution: a2 + 4b2 = 12ab a2 + 2a2b + 4b2 = 16 ab (a + 2b)2 = 16ab taking log on both the sides. + –7/2 37 7 ) ( . which is impossible x (ii) x2 5 2x 7 x 2 5(2x 7) x 2 10x 35 0 0 2x 7 2x 7 9x 37 0 (9x + 37) (2x + 7) > 0 2x 7 + –37/9 3. which is impossible Case II: –2 x < 3 –x + 3 + x + 2 = 3.IIT. 2 log (a + 2b) = log16 + log a + logb 4. log (a + 2b) = Find Domain 1 (loga + logb + 4log2) 2 (i) sin x3 x 1 (ii) ( x 1)( x 3) loge ( x 2)( x 4) Solution: x 3 x 1 f(x) is defined for (i) Let f(x) = sin x 3 0 (x – 1) (x + 3) 0 x 1 38 . ) 9 2 If a2 + 4b2 = 12 ab.MATHS x<–2 –x + 3 – x – 2 = 3 –2x = 2 x = – 1. 3) ( 2. given by h(x) = kx2 + 7x – 4. log sin x = x cotx + log sinx y dx sin x dy = (sinx)x [x cotx + log sinx] dx d {(sin x) x } (sin x) x[x cot x log sin x] dx Let y = cos(4x3 – 3x) (iii) dy = –sin (4x3 – 3x) [12x2 – 3] dx d (cos 4x3 – 3x) = 3 (1 – 4x2) sin (4x3 – 3x) dx 6.3] (1.BASIC TRIGONOMETRY x + – –3 1 x ( . If for the1 function h. Taking log on both the sides. ) (ii) Let f(x) = log e (x 1) (x 3) (x 2) (x 4) f(x) is defined for (x 1) (x 3) 0 (x 2) (x 4) x ( . find k. we get logy = x log sinx 1 dy x cos x .1) (4. ) 5. + – –3 + –2 – + 4 1 Differentiate with respect to x (i) (ii) (iii) Solution: (i) x sin x + log (1 x 2 ) (1 x 2 ) (sinx)x cos (4x3–3x) Let y x sin x log 1 x 2 2 1 x dy (1 x 2 ) (sin x x cos x) (2x)x sin x 1 (x 2x) 2 2 dx (1 x ) 1 x2 2 1 x2 (ii) (1 x 2 ) x cos x (1 x 2 ) sin x 2x 2 sin x 1 2 (1 x )2 1 x2 (1 x 2 ) x cos x (1 x 2 ) sin x 1 2 2 (1 x ) 1 x2 Let y = (sinx)x. Solution: 39 . h (5) = 97. f (x) 1 0 1– Hence local minima at x = 1 at x = – 1.MATHS 7. if f (x) 0 – sinx 0 sinx 0 x [0. 2 8.IIT. f (x) 1 0 Hence local maxima at x = – 1. 1 Solution: / 2 (i) Let I = sin x. Integrate the following (i) sin x .cos 2 x dx 0 0 put cosx = t = t 2 dt –sinx dx = dt when x = 0. x 2 (ii) Find the point of local maxima & minima of the function. 0 2 /2 f(x) decreases. ] (ii) 1 f (x) x x f (x) 1 f (x) O O /2 1 x2 2 x3 For local maximum or local minimum we must have f (x) 0 1 0 x = 1 x2 at x = 1. h(x) = kx2 + 7x – 4 h(x) 2kx + 7 h(5) = 10k + 7 97 = 10k + 7 k = 9 [ h (5) = 97] (i) Find the intervals of increase and decrease of the function y = cosx. t 0 2 40 . if f (x) 0 – sinx 0 sinx 0 x . f (x) = x 1 . when x = 1 . cos 0 8 2 x dx 3 e x (ii) x 2 / 3 dx . x 2 f (x) = –sinx f(x) increases. x Solution: (i) Let f(x) = cosx. t = 1. 4 4 B x A . 2 / 3 dx dt when x = 8. In the figure given below. x 3 x 1 2 t t 2 x = 1. find the value of angle P A 24° R P 36° B 54° C D Solution: ACD = ABC + BAC = 36 + 24° = 60° P = RCD + RDC = 60 + 52 = 104°. find the value of x T 5 P Solution: PT2 = PA. In the figure given below. t = 1 = 3 e dt 3(e ) .BASIC TRIGONOMETRY 1 1 t3 1 t dt = 3 3 0 0 2 8 (ii) 3 e x 1 1 Let I = 2 / 3 dx put x1/3 = t .PB 5 5 25 4 4 AB = PA – PB PA = x= 25 9 4 units 4 4 Hence x = 41 9 units. = 3(e2 – e) = 3e (e – 1) 1 9. 10. t = 2 . III CHECK YOUR SKILLS 1. ) Find the set of all solution of the equation 2 y 2 y 1 1 Solution: 2|y| = 2y–1 + 1 Case I: y < 0 2–y = 2y–1 + 1 2.3 (x – 1) < 0 2 x–1>1 x>2 x (2. which is not possible 3 1 3 1 y = log2 ( 3 1 ) Case II: y < 0 2y = 2y–1 + 1 2. log0.3 (x – 1) < log0.3 (x – 1) < log0. 2 2y 2 y (2y)2 + 2.2y – 2 = 0 2 a2 + 2a – 2 = 0 .3 (x – 1) 2 1 log0.MATHS LEVEL . where a = 2y 1= a 2 4 8 2 2 3 2 2 a = 1 3 . log2 ( 3 1) 3.3 (x – 1) < log0. – 1 + a=–1– a= 2y = log22y = log2 ( 3 1 ) 3 3 .IIT.09 (x – 1) Solution: log0. Evaluate: 7 log 3 5 3log 5 7 5log 3 7 7 log 5 3 Solution: 5 7 7 Let y = 7log3 3log5 5log3 7 log5 3 = 7log3 5 3log5 7 7 log3 5 3log 5 7 [ alogb c clogb a ] 42 . Solve.2y = 2y + 2 2y = 2 y=1 set of solution 1.09 (x – 1) 1 log0. – 1} Df = Dg Dh = (–3.1 1/ 3 1 log 7 5 5 log10 10 Hence the result. Evaluate: 3 2 log 75 1 log 100.BASIC TRIGONOMETRY =0 Hence the result. = (7 + 1)1/3 = (8)1/3 = 2 Find domain (i) f(x) = (ii) f(x) x 1 2 x Solution: log 2 (x 3) x 2 3x 2 Let g(x) = log2 (x + 3) and h(x) = x2 + 3x + 2 g(x) is defined for x+3>0 x>–3 Dg = (–3. ) {–2.1 Solution: 3 Let y 5 1 1 log 7 5 log10 0. (2 – |x|) (|x| – 1) 0 2 – |x| > 0 or |x| – 1 0 |x| < 2 or |x| 1 – 2 < x < 2 or x 1 or x – 1 Df = (–2. 2) Find the domain 5x x 2 4 (i) y log10 (iii) y x2 1 x x2 1 x Solution: (i) 43 2 y log10 5x x f (x) 4 (ii) y cos ecx (sin x )1 / 3 . –1} (i) f(x) = (ii) | x | 1 f(x) is defined for 2 | x | f(x) = | x | 1 (| x | 1) (2 | x |) 0 0 2 | x | (2 | x |) 2 6. ) R – {–2. 5. ) And h(x) is defined for x2 + 3x + 2 0 (x + 2) (x + 1) 0 Dh = R – {–2. 1 4. – 1} = (–3. –1] [1. – 2) [2. (2n + 1) ) (iii) Let f (x) x2 1 x x2 1 x x2 and h(x) = x2 g(x) is defined for x 2 0 either x – 2 0 x2 or x + 2 < 0 x 2 or x + 2 < 0 Dg = (– . 4] Let f(x) = cos ecx + (sinx)1/3 Let g(x) = cos ecx and h(x) = (sinx)1/3 g(x) is defined for cosecx > 0 Dg = (2n .IIT. 1] Df = Dg Dh = 7. Integrate the following (a) q 5 cos x – 3 sinx + cos 2 x 4 (b) 2 3 sin x dx cos 2 x 0 Solution: (a) Let I = (5 cos x 3sin x a )dx cos 2 x 5 cos x dx 3 sin x dx 9 sec 2 x dx 44 . (2n + 1) ). And h(x) is defined for R Df = Dg Dh = (2n .MATHS f(x) is defined for 5x x 2 log10 0 4 (ii) 5x x 2 1 5x x 2 4 0 4 x2 – 5x + 4 0 (x – 1) (x – 4) 0 either x 1 or x 4 Df = [1. ) h(x) is defined for Let g(x) = 1 x 1 x 1 x x 1 0 0 1 x x 1 either 1 – x 0 or 1 + x > 0 x 1 or x > – 1 Dn = (–1. ‘x’. 1] (b) (i) y = –3x4 + 6x2 – 1. ) for increasing . at x = 0 . 1 2 dx least value y = 2. Find the greatest and least values of the following functions on the intervals y = – 3x4 + 6x2 – 1 (–2 x 2) y= x3 2 x 2 3x 1 3 (–1 x 5) Solution: (a) Let f(x) = (x + 2) e–x f (x) = –e–x (x + 2) + e–x = – e–x (x + 1) for decreasing f (x) 0 – e–x (x + 1) 0 e–x (x + 1) 0 ( e–x 0] x+1>0 x > – 1 x [ 1. (a) (b) x 2 1 w. where C is integral constant.e x 2 1 x2 1 Find the intervals of decrease and increase of (x + 2) e–x . – 2 x 2 dy 12 x 3 12x dx d2 y 36x 2 12 12(x 2 1) 0 2 dx 45 dy 0 – 12x (x2 – 1) = 0 x = 0. f (x) 0 –e–x (x + 1) 0 e–x (x + 1) 0 x+1 0 [ e–x 0] x 1 x (. Differentiate e Solution: Let y = e dy e dx 9.BASIC TRIGONOMETRY = 5 sinx + 3cosx + 9 tanx + C.t. /4 (b) Let I 0 2 3sin x dx cos 2 x /4 2 = 2 sec xdx 3 tan x sec x dx 0 = 2[tan x]0 / 4 3[sec x]0 / 4 = 2[1 – 0] + 3 [ 2 1] = 3 2 1 8. x 2 1 x 2 1 1 × 2 x2 1 × (2x – 0) x.r. – 1 x 5 3 dy d2 y x 2 4x 3 2x 4 2(x 2) dx dx 2 dy 0 x2 – 4x + 3 = 0 dx (x – 1) (x – 3) = 0 x = 1. 3. 7 3 d2 y 20 at x = 3. Find the value of y. dx 2 minimum at x = 5 10.MATHS greatest value y = –25 at x = 2 (ii) y x3 –– 2x2 + 3x + 1 . ABCD is square and triangles BCX and DYC are equilateral triangle. B A y x D Solution: C BCX and DYC are equilateral AB = BC = CD = DA = CX = DY = CY = BX CBX = 60° ABX = 150° BAX = 15° A [ AD DY] DAY = 75° = DYA 46 .IIT. minimum value of y 23 3 In the figure given below. 5 d2 y 20 dx 2 maximum at x = 1 maximum value of y = at x = 1. c R+ Solution: Equation is a2x.0625 = or 1 1 4 16 2 2x = 2–4 or x = –4 Illustration 3: Solve the equation a2x b3x = c5. Illustration 5: Solve |2x – 1| < 3. AB. d} then evaluate AB. A – B and B – A Solution: AB = {x : x A or x B} = {a. d} AB = {x : x A and x B} = {b. b. b.0625 to the base ‘2’. b. where a. Solution: Suppose 2x = 0. log1/2 (x – 2) > 2. we have 2x log a + 3x logb = 5 logc or x (2loga + 3 logb) = 5logc 5 log c x = 2 log a 3 log b Illustration 4: Solve for x.BASIC TRIGONOMETRY Illustration 1: If A = {a. Solution: |2x – 1| < 3 – 3 < 2x – 1 < 3 – 2 < 2x < 4 – 1 < x < 2. c. Illustration 6: 47 .b3x = c5 Taking log on both sides. c. c} and B = {b. Solution: 1 4 1 4 9 4 log1/2(x – 2) > log1/2 0 < x – 2 < 2 < x < So 9 4 x 2. c} A – B = {x : x A and x B} = {a} B – A = {x : x Î B and x Ï A} = {d} Illustration 2: Find the logarithms of 0. x x = 24 x2 – 5x = 24 x2 – 8x + 3x – 24 = 0 (x – 8) (x + 3) = 0 x = – 3. x . Solution: 48 . x 3x 2 x 5 x 1x 7 Illustration 7: x 5 If x x = 24. we get x = 5 . y = –3. Solution: Given f(x) = (ii) f(x) < 0.MATHS x 3x 2 x 5 x 1x 7 then find x such that If f(x) = (i) f(x) > 0.x 11 dx 2 3 2 3 1 1 1 2 2/3 x 3 (x) x Illustration 10: Find the derivative of y = (a + x) ex w..r. y x 4 2x 3y 4 x y 7 … (1) 7 and 2 1 2 x 2 y 2 … (2) Solving (1) and (2).t. v x 1 / 3 and w = 1 x Hence we can use theorem 2 1 1 dy 1 1 1 1 1 3 x 2 x 3 1. where u = 3x .IIT. Solution: x 5 Here. 2 Illustration 9: Find the derivative of y = 3x ( x )1 / 3 1 x Solution: Here we have y = u + v + w. 8 Illustration 8: 2 3 x y 7 Find the value of x and y when y x = 4 and 2 1 2 Solution: 2 3 Since. then find the value of x. 2x.r. x (i) y = sinx2 (ii) y = (lnx)3 (iii) y = sin (lnx)3 (iv) y = cos–1 (lnx) Solution: (i) y = sin (x)2.t.2 x 2x cosx2 dx du dx dy 1 du 1 1 1 dx 1 u 2 dx 1 u 2 x x 1 (ln x ) 2 .BASIC TRIGONOMETRY Using theorem 3 dy d d (a x ) e x e x (a x ) dx dx dx = (a x ) e x e x .t x. (i) y = x x (ii) y = (sin x ) x Solution: (i) y = xx 2 dy x. v = u3 y = sin v (iv) dy dy dv du d(sin v) d( u ) 3 d 1 3(ln x ) 2 (ln x ) = cos v.r.1 e x (a x 1) Illustration 11: Find the derivative of y = ax w.1 x x ln x. Illustration 13: Find the derivative of the following functions w.t.1 x x (1 ln x ) dx (ii) y (sin x ) x 2 2 2 dy d dx 2 x 2 (sin x ) x 1 (sin x ) (sin x ) x ln sin x dx dx dx = x2 (sin x ) x 2 1 2 cos x (sin x ) x ln sin x. 3u 2 cos [(ln x ) 3 ] dx dv du dx dv du dx x x Let u = ln x = cos–1 u 49 dy d d 2 (sin u ) ( x ) cos u. x ax Solution: Illustration 12: Differentiate the following w.r. Let u = ln x y = u3 (iii) y = sin (lnx)3 Let u = ln x.x x 1 . Let u = x2 y = sin u dy d( u 3 ) du 1 3 3u 2 (ln x ) 2 = dx du dx x x (ii) y = (ln x)3. y = a sin t. Solution: dy (a sin t ) cos t – cot t dx (a cos t ) sin t If we want to compute dy dy cot 1 at a particular t. y > 0 the function increases in (0. Thus we investigate the character of the given function in an interval containing point x = 0. 3) Is tanx always increasing x R ? Illustration 18: Test y = 1–x4 for maximum and minimum Solution: Here y = –4x3 = 0 for x = 0 x = 0 is the critical point.t. at x = 0. y > 0 the function is increasing for x < 0 For x 0.e. 2) and (3. For x < 0. f (x) > 0 Hence the given function is increasing in (–.. Solution: y = x4 y = 4x3 For x > 0. y < 0 the function is decreasing for x > 0. ). f (x) > 0 over 2 < x < 3. ). Now y = –12 x2 = 0 at x = 0 It is thus impossible to determine the character of the critical point by means of the sign of the second derivative.IIT. yx = 0 = 1 50 . where y = sin2x dx 2 Solution: dy = 2sinxcosx = sin2x dx Illustration 16: Find the interval of increase and decrease of the function y = x4. 0) Illustration 17: Separate the intervals in which f(x) = 2x3 – 15x2 + 36x + 1 is increasing or decreasing Solution: We have f (x) = 6x2 – 30x + 36 = 6 (x – 2) (x – 3) Thus for x < 2. f (x) < 0 and for x > 3. then dx 4 x / 4 dx 4 Illustration 15: d2y Find . Find the derivative of y w. x = a cos t.r. y < 0 the function decreases in (–. Consequently. say t = . x. and decreasing in (2.MATHS Illustration 14: The function y of x is given by. For x < 0. the function has a maximum i. y = 3x2 – 3 For the critical points. It should be noted that the values are actually the largest and smallest values of the function in the given interval. y’’ < 0 at x = – 1 Hence there is maximum at x = –1 at which y = –1 + 3 + 3 = 5 Also there is minimum at x = 1 at which y = 1–3 + 3 = 1 Now at x = –3. Illustration 20: Evaluate: 2 (i) (a0 + a1x + a2x2 ) dx (ii) cos x x e (iii) x2 1 x 2 dx (iv) x4 x 2 1 dx Solution: (i) x dx (a0 + a1x + a2x2) dx = a0 dx + a1 x dx + a2 x2 dx x2 x3 a c = a0 x + a1 2 2 3 2 x 1 x dx cos x dx 2 dx e dx = sin x + 2 log |x| –ex + c x (ii) cos x x e (iii) x2 x2 11 1 dx dx 1 x 2 1 x 2 dx 1 1 x 2 dx dx 1 x 2 = x – tan–1 x + c (iv) x4 x4 11 1 2 dx x 2 1 1 x 2 dx x 1 1 x 2 dx 2 = x dx dx dx x3 x tan 1 x c 1 x2 3 Illustration 21: Integrate the following w.t. y = –27 + 9 + 3 = –15 and at x = 3/2. x.BASIC TRIGONOMETRY Illustration 19: Determine the maximum and minimum of the function y = x3 – 3x + 3 on the interval [–3.r. (i) sin2x cosx (ii) Solution: (i) 51 x3 1 x8 Let sin x = t cos x dx = dt . y = 15/8 Hence the minimum value of the given function is –15 at x = –3 and the maximum value is 5 at x = –1. 3 ]. 2 Solution: For the given function. 3x2 – 3 = 0 x = 1 Then y = 6x > 0 at x = 1. (x) = 1 1 1 1 sin x dx sin x. x = /2. (x) = ex xe x dx x.IIT.e x 1. t = 1 /2 0 1 cos xdx dt (tan 1 t )10 = tan–1 1– tan–1 0 = – 0 = /4 2 2 1 sin x 0 1 t 4 52 . d sin 1 c a cos a 1 bx c sin 1 a a 2 ( bx c) 2 b dx Illustration 23: Evaluate (i) xe x (ii) dx sin 1 x dx Solution: (i) let f(x) = x. t = 0.x 1 1 x2 x dx = x sin–1 x – x (1 x 2 ) 1/ 2 dx x sin 1 x 1 x 2 c Note: e x (f ( x ) f ( x )) dx e x f ( x ) dx e x f ( x ) dx = f(x) ex – f ( x ) e x dx e x f ( x ) dx f ( x ) e x c Illustration 24: Evaluate : / 2 cos x 1 sin 2 x dx 0 Solution: Let sin x = t cos x dx = dt When x = 0.e x dx xe x e x c (ii) Let f(x) = sin–1x. 1 dx sin x.MATHS 2 2 sin x cos x dx t dt (ii) t3 sin 3 x c c 3 3 Let x4 = t = 4 x3 dx = dt x3 dx = ¼ dt x3 1 1 dt 1 1 dx dt = tan 1 t c tan 1 x 4 c 8 2 2 1 x 4(1 t ) 4 1 t 4 4 Illustration 22: 1 Evaluate 2 a x2 dx Solution: Let x = a sin dx = a cos d a2 – x2 = a2 cos2 Note: dx 2 a x 2 a cos d x 1. BASIC TRIGONOMETRY 53 . IIT.MATHS 2 ARITHMETIC PROGESSION 54 . .D..P. (e) Sum of squares of first n natural numbers ( n 2 ) n 2 n( n 1) (2n 1) 6 (f) Sum of cubes of first n natural numbers (n 3 ) n(n 1) n 3 2 2 (g) Middle term: If the number of terms is n. if there is a number d. l = tn = a + (n – 1) d. d is called the common difference (C. we choose them as (a – 3d).. (a + d) For four terms in A. and their sum is known.) A sequence of numbers {an} is called an arithmetic progression. then (a) nth term is denoted by tn and is given by tn = a + (n – 1) d. d = common difference and n is the number of terms..e. then and 1 terms are middle terms.P.. . (a + d). formed. . where l = last term in the series i. (i) Useful Formulae If a = first term. we choose them as (a – d). an. such that d = an-an–1 for all n. ARITHMETIC PROGESSION (A.) of the A. 2 (d) Sum of first n natural numebrs ( n ) n(n 1) n 2 . and th n 1 n is odd. a. 2 (c) Arithmetic mean A of any two numbers a and b a b A . is called a sequence. then the terms must be picked up in following way For three terms in A. then term is the middle terms 2 th th n n n is even.P. a2. 2 2 (h) If terms are given in A.. (a – d). (a + 3d) 55 . (b) Sum of first n terms is denoted by Sn and is given by n Sn [2a (n 1)d] 2 n or S n (a l ) ..ARTHMETIC PROGESSION SQUENCE AND SERIES A succession of numbers a1. according to some definite rule. where n N .P. a3..P. .P. the result ing terms will also be in A..P. r = common ratio and n is the number of terms.R. (c) Sum of infinite terms (S ) a S (for | r | 1) 1 r Note: When |r| 1.’s will form an A. (g) If terms a1. the sum of terms equidistant from the beginning and end is constant and equal to sum of first and last terms.P.P. .P. 1 r 1 r r 1 where l is the last term (the nth term) in the series.P. (h) If terms a1. then sum of these terms will be equal to (2n + 1)an+1. (a – d). (b) If Sn = an2 + bn + c. a2n–1.P.) A sequence of the numbers {an}. ` (c) If every term of an A... . a.. a2. denoted by tn . and their product is known. an. the series is divergent and so its sum is not possible.. the resulting terms will also be in A.) G ab is the geometric mean of two positive numbers a and b..P. is increased or decreased by some quantity.M.P. r 1 In case r = 1. is called a geometric progression. is given by tn = arn–1 (b) Sum of first n terms denoted by Sn is given by a rl a(1 r n ) a(r n 1) Sn or or .MATHS For five terms in A. a2n+1 are in A. n 1 (i) Useful Formulae If a = first term..P.P. 56 . (d) Geometric mean (G.P.IIT.r is called the common ratio (C.P.. then (a) nth term.) of the G. 2 GEOMETRIC PROGRESSION (G. (e) If terms are given in G.. (a + 2d) etc. is multiplied or divided by some non-zero quantity. a2. then the terms must be picked up in the following way. if there is a an number r 0 such that a r for all n. an+1. (ii) Useful Properties (a) If tn = an + b. a2n are in A. (d) If every term of an A. (f) Sum and difference of corresponding terms of two A.P. we choose them as (a – 2d).. in which a1 0 . Sn = na.P. (e) In an A. The sum of these terms will be equal to a n a n 1 (2n) .. . then the series so formed is an A. (a + d). then series so formed is an A. and a1 a 2 a 3 r . we choose them as 3 . a d a a d (ii) For four terms in H. HARMONIC PROGRESSION (H. and it is equal to the product of the first and the last term. a2.’s of common ratio r1 and r2 respectively.) (a) A sequence is said to be in harmonic progression. 4.. (b) If every term of a G.. are in A. r r (ii) Useful Properties (a) The product of the terms equidistant from the beginning and end is constant. and conversely. a2b2 .ARTHMETIC PROGESSION a For three terms in G. be two G.P. will be an A. 6.P.P.P. the resulting progression is a G.P..P.P. . b are first two terms of an H. . .. (c) If a1. a2. . if and only if the reciprocal of its terms form an arithmetic progression. a3.. a.. of positive terms. then the terms could be picked up in the following way (i) For three terms in H. then a1b1 . we choose them as 2 . . ab (d) If terms are given in H. For example 1 1 1 . then tn 1 1 1 1 (n 1) a b a (c) Harmonic mean H of any two numbers a and b H 2 1 1 a b 2ab . loga3. we choose them as 1 1 1 .. . Common ratio will be r1r2 and 1 b1 b 2 b3 r2 respectively..P... then loga1. ar etc. ar.P... is multiplied or divided by the some non-zero quantity. because 2. and b1... .P. form an H. 2 4 6 (b) If a. be a G. ar. . where a. b are two non-zero numbers..P.. ar r a a 3 For four terms in G. loga2.P. we choose them as 57 .P.... ar r r a a 2 For five terms in G.P.. b2. a3 . (d) If a1. . a. b3..P.P. . we choose them as . . will also form a G. IIT- MATHS 1 1 1 1 , , , a 3d a d a d a 3d (iii) For five terms in H.P, we choose them as 1 1 1 1 1 , , , , a 2d a d a a d a 2d Insertion of Means Between two numbers Let a and b be two given numbers. (i) Arithmetic Means If a, A1, A2 , ... An, b are in A.P., then A1, A2, ... An are called n A.M.’s between a and b. If d is the common difference, then b = a + (n + 2 – 1) d d = Ai = a + id = a + i ba n 1 b a a(n 1 i) ib , i = 1, 2, 3, ..., n n 1 n 1 Note: The sum of n-A. M s, i.e., A1 + A2 + ... + An = n (a b) 2 (ii) Geometric means If a, G1, G2 ... Gn, b are in G.P., then G1, G2 ... Gn are called n G.M.s between a and b. If r is the common ratio, then 1 b = a.r n+1 (n 1) r = b a i b n 1 Gi = ar = a a a i n 1 i n 1 .b i n 1 , i = 1, 2, ..., n Note: The product of n-G. M s i.e., G1 G2 ... Gn = ab n (iii)Harmonic Means If a, H1, H2... Hn, b are in H.P., then H1, H2 ... Hn are called n H.M.s between a and b. If d is the common difference of the corresponding A.P., then 1 1 a b (n 2 1) d d b a ab(n 1) 1 1 1 ab ab(n 1) id i , i 1, 2, 3, ..., n Hi a a ab(n 1) b(n i 1) ia 58 ARTHMETIC PROGESSION MEANS OF NUMBERS Let a1, a2, ... an be n given numbers (i) Arithmetic Means (A.M.) = a1 a 2 ... a n n (ii) Geometric Means (G.M.) = (a1 a2 ... an)1/n n (iii) Harmonic Mean (H.M.) = 1/ a 1/ a ... 1/ a 1 2 n If weights of a1, a2, ... , an are w1, w2, ..., wn respectively, then their weighted arithmetic mean, weighted geometric mean and weighted harmonic mean are respectively defined by 1 a1w1 a 2 w 2 ... a n w n (a1w1 .a 2 w 2 ...a n w n ) w1 w 2 ...w n w1 w 2 ... w n and w1 w 2 ... w n . w1 w 2 wn ... a1 a 2 an RELATION BETWEEN A, G AND H If A, G and H are A.M., G.M. and H.M. of positive numbers a1, a2 ... an ( a1 a 2 ... a n ) then a1 H G A a n ...(1) Note: (i) The equality at any place in (1) holds if and only if the numbers a1, a2, ..., an are all equal (ii) (1) is true for weighted means also (iii) G2 = AH, if n = 2. ARITHMETIC MEAN OF mth POWER Let a 1 , a 2 a1m a m2 ... a mn n ... , a n be n positive real numbers and let m be a real number, then m a a ... a n 1 2 , if m R [0,1]. n a1m a m2 ... a mn a1 a 2 ... a n However if m (0, 1), then n n m a m a m2 ... a mn a1 a 2 ... a n Obviously if m {0,1}, then 1 n n m ARITHMETIC-GEOMETRIC SERIES A series whose each term is formed, by multiplying corresponding terms of an A.P. and a G.P., is called an Arithmetic-geometric series. 59 IIT- MATHS e.g. 1 + 2x + 4x2 + 6x3 + ..... ; a + (a + d) r + (a + 2d)r2 + ..... (i) Summation of n terms of an Arithmetic-Geometric Series Let Sn = a + (a + d) r + (a + 2d)r2 + ... + [a + (n – 1)d] rn–1, d 0 , r 1 Multiply by ‘r’ and rewrite the series in the following way rSn = ar + (a + d)r2 + (a + 2d)r3 + ... + [a + (n – 2)d]rn–1 + [a + (n – 1)d ]rn on subtraction, Sn (1 – r) = a + d(r + r2 + ... + rn–1) – [a + (n– 1)d]rn or, Sn (1 r) a or, Sn dr(1 r n 1 ) [a (n 1)d].r n 1 r a dr(1 r n 1 ) [a (n 1)d] n .r 1 r (1 r)2 1 r (ii) Summation of Infinite Series If |r| < 1, then (n –1)rn, rn–1 0, as n . Thus S = S = a dr 1 r (1 r)2 SUM OF MISCLENIOUS SERIES (i) Defference Method Let T1, T2, T3 ... Tn be the trms of a sequence and let (T2 – T1) = T1 , (T3 – T2) = T2 ... , (Tn – Tn–1) = T n 1 . Case I: If T1, T2 ,....Tn1 are in A.P. then Tn is a quadratic in ‘n’. If T1 – T2 , T2 T3 , ... are in A.P., then Tn is a cubic in n. Case II: If T1, T2 ,....Tn1 are not in A.P., but if T1 , T2 ,..., Tn1 are in G.P., then Tn = arn + b, where r is the C.R. of the G.P. T1 , T2 , T3 .....and a, b R. Again if T1,T2,...Tn1 are not in G.P. but T2 T1, T3 T2 ,...Tn1 Tn2 are in G.P., then Tn is of the form arn + bn + c; r is the C.R. of the G.P. T2 T1, T3 T2 T4 T3 ... and a, b, c R. (ii) Vn – Vn–1 Method Let T1, T2, T3 , ... be the terms of a sequence. If there exists a sequence V1, V2, V3 ... satisfying Tk = Vk – Vk–1, k 1, n n k 1 k 1 then S T (V V ) V V . k k k 1 n 0 n 60 ARTHMETIC PROGESSION OBJECTIVE SECTION - I LEVEL - I Multiple Choice Questions with Single Answer: 1. 2. If a1, a2, a3 ... are in AP, then ap, aq, ar are in AP, if p, q, r are in (a) AP (b) GP (c) HP (d) none of these Let tr denote the rth term of an AP. If tm = (a) 1 mn (c) 1 3. 1 1 and t n = , then tmn equal to m n (b) 1 1 –1 m n (d) 0 If p, q, r, s N and they are four consecutive terms of an AP, then the pth, qth, rth, sth terms of a GP are in 4. 5. 6. 61 (a) AP (b) GP (c) HP (d) none of these Let a1, a2, a3, ... be in AP and ap, aq, ar be in GP. Then aq : ap is equal to (a) rp qp (b) qp rq (c) rq qp (d) none of these If a, b, c are in G.P., then a + b, 2b, b + c are in (a) A.P. (b) G.P. (c) H.P. (d) none of these If a, b, c, d are nonzero real numbers such that (a2 + b2 + c2) (b2 + c2 + d2) (ab + bc + cd)2, then a, b, c, d are in (a) AP (b) GP (c) HP (d) none of these IIT- MATHS 7. 8 9. 10. 11. 12. 13. If 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca), where a, b, c are nonzero numbers, then a, b, c are in (a) AP (b) GP (c) HP (d) none of these If a, b, c are in H.P., then c, c – a, c – b are in : (a) A.P. (b) G.P. (c) H.P. (d) none of these Let S be the sum, P be the product and R be the sum of the reciprocals of n terms of a GP. Then P2Rn: Sn is equal to (a) 1 : 1 (b) (common ratio)n : 1 (c) (first term)2 : (common ratio)n (d) none of these If a1, a2, a3 are in AP, a2, a3, a4 are in GP and a3, a4, a5 are in HP, then a1, a3, a5 are in (a) AP (b) GP (c) HP (d) none of these 1 1 1 If x > 1, y > 1, z > 1 are three numbers in GP then 1+ ln x , 1+ ln y , 1+ ln z are in (a) AP (b) HP (c) GP (d) none of these If a, a1, a2, a3, ... a2n–1, b are in AP, a, b1 , b2 , b3 ..., b2n–1, b are in GP and a, c1, c2, c3, ... , c2n–1, b are in HP, where a, b are positive, then the equation anx2 – bnx + cn = 0 has its roots (a) real and unequal (b) real and equal (c) imaginary (d) none of these The product of n positive numbers is unity. Then their sum is (a) a positive integer (c) equal to n + 14. 1 n (b) divisible by n (d) never less than n If x > 0 and a is known positive number, then the least values of ax + (a) a2 (b) a (c) 2a (d) none of these a is x 62 ARTHMETIC PROGESSION 15. 16. 17. If p, q, r be three positive real numbers, then the value of (p + q) (q + r) (r + p) is (a) > 8 pqr (b) < 8 pqr (c) 8 pqr (d) none of these a1 a2 an If a1, a2, a3 ….. an are in H.P., then a 2 a 3 ....an , a 2 a 3 ....an ,.... a 2 a 3 ....an 1 are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 1 1 1 1 Sn = 1 + 2 3 4 ..... 2 n 1 , then (a) S100 < 100 (b) S200 < 200 (c) S200 > 100 (d) S50 > 25 n 18. Given Sn r0 19. 1 1 1 ,s . , then least value of ‘n’ is r r If S – Sn 2 1000 r 0 2 (a) 8 (b) 9 (c) 10 (d) 11 If x15 – x13 + x11 – x9 + x7 – x5 + x3 – x = 7,(x > 0), then (a) x16 is equal to 15 (b) x16 is less than 15 (c) x16 greater than 15 (d) none of these 20. 1 (n 1) (n 2) (n 3) ......(n k) is equal to n 1 1 (a) (k 1) k 1 1 (b) k k 1 (c) (k 1) k 1 (d) k n 21. 63 n 1 1 Maximum value of n for which n is 2 14 1 (a) 4 (b) 5 (c) 6 (d) 7 MATHS 22.7 7 217. 1 7 1 20 . c are in a a-b c c-b The minimum value of x4 1 is x2 1/ 3 1 (a) 3 4 (b) 1 2 1/ 3 1 (c) 2 3 (d) 2 64 . 1 1 1 1 + + + = 0 and a c b then a.9 (d) 2285 (c) 25.P. (c) H. 24. is The nth term of the series 2 1 1 2 13 9 23 20 (5n 3) (b) 20 (5n 3) (b) 20 (5n + 3) (d) 20 (5n 2 3) (a) 23... If (a) A. the maximum value of p3q5 r7 will be (a) 2180 54.P.P. (b) G..IIT.. (d) None of these If 2p + 3q + 4r = 15. b.35 (b) 15 2 55. II Multiple Choice Questions with one or more than one correct Answers: 1. 8. (b) 3n (d) independent of n a.P. b2. 8.. z are in GP then logc b is equal to (a) logba (c) z/y (b) logab (d) none of these n 7. . then x r x r+1 is equal to r=1 (a) (n – 1)x1xn (c) (n + 1) x1 xn 6. b.P. c are distinct real numbers such that a. where n consecutive terms have the value n. .. 8.. x2. The value of r=1 8.. then (a) 2b2 = – ac (c) 2b2 = ac (b) 4b2 = – ac (d) 4b2 = ac n-1 5. [x].ARTHMETIC PROGESSION LEVEL . 2. (b) 1 (d) none of these In the sequence 1. 8. If x1. 4. c of a triangle are in G.P. If ln . Total number of positive real values of x such that x. .P. {x} are in H.. 8. (b) nx1 xn (d) none of these If ax = by = cz and x. ln a must be 3b a and ln are in A. 8. then triangle 5c 3b . b. y. b.] denotes the greatest integer function and {.P.P.} denotes fraction part. and a2. c2 are in H. 65 (a) n a a nx (c) n( a nx a) x 1 a + rx + a + (r -1)x is (b) a nx a x (d) none of these 5c Sides a. the 1025th term is (a) 210 (c) 29 3. 8. (b) 211 (d) 28 Sr denotes the sum of the first r terms of an AP. 4. xn are in H. 4. Then S3n : (S2n – Sn) is (a) n (c) 3 4.. 2. where [. 4.. c are in A. 8. is equal to (a) zero (c) 2 2. b c d e (c) 15. then the minimum value of a1 + a2 + . i = 1. (b) Equilateral (d) None of these G13 + G 32 If A1 be the A.. nP1. 1 (a) .IIT. then G1G 2 A1 is equal to (a) 1 (c) 3 10. (b) in HP (d) none of these In a GP the product of the first four terms is 4 and the second term is the reciprocal of the fourth term. an are positive real numbers whose product is a fixed number c. a2. .. (b) -8 (d) –8/3 If a1. c. 2 2 (b) [–1. nP3 are three consecutive terms of an AP then they are (a) in GP (c) equal 12..MATHS (a) Isosceles (c) Obtuse angled 9. (b) (n + 1)c1/n (d) (n + 1) (2c)1/n a b c d e 5 b c d e a a e a c b d (b) 4. b.M. If a. . then 24 b i is equal to i=1 (a) 600 (c) 300 11. The sum of the GP up to infinite terms is (a) 8 (c) 8/3 13... n is an arithmetic sequence. 2.Ms between two positive numbers a and b. 2] 1 (c) . and G1. (b) 2 (d) none of these {bi}. 2 66 . an–1 + 2an is (a) n(2c)1/n (c) 2nc1/n 14. G2 be two G. nP2. b dc e (d) a e b c d e a 1 a b c d e 5 If a2 + b2 + c2 = 1 then ab + bc + ca lies in the interval... d are four positive numbers then a b c d (a) 4.1 2 1 (d) 1. If b1 + b5 + b10 + b15 + b20 + b24 = 255. (b) 900 (d) none of these If 2. d are distinct integer in A. If the middle number is increased by 8.P. when n is even 2 (d) (a) 1 an . the resulting numbers form a GP again. Then (a) common ratio = 3 (c) common ratio = –5 21. q are in AP such that 2a.1) (b . when n is even 6 (b) n (c) – (n 1) ... c. p. be squares such that for n 1. the three numbers form an AP.ARTHMETIC PROGESSION 16. Let f(x) = 2 3 n +1 1. 67 n(n 1) . (b) 1 (d) none of these Three positive numbers form a GP.a) . 1 a 1 a (d) 1 an . n(n 1) . If the last number is also increased by 64 along with the previous increase in the middle number. 1 a 1 a Let S1. S3. c are in GP and a. where A & B are respectively If tn = n 1 n n (1 a x) (1 a x) 1 x 1 an (a) 7 (c) 9 19. the length of a side of Sn equals the length of a diagonal of Sn+1. then a + b + c + d is (a) 0 (c) 2 20.. 1 a 1 a (c) 1 an . Then the constant term in x x xn 1.x n +1 and g(x) = 1 – + 2 – . S2. . c + q are in GP then the common difference of the AP is (a) 2a (b) ( 2 + 1) (a – b) (c) 2 (a + b) (d) ( 2 . n(n 2 1) (a) .. when n is odd 2 a n 1 A B . b + p. If the length of a side of S1 is 10 cm then for which of the following values of n is the area of Sn less than 1 cm2? (b) 8 (d) 10 If a. such that d = a4 + b4 + c4. b. 18. b. then S = .x f (x) × g(x) is equal to 17. when n is odd 2 (b) first number = 4/9 (d) first number = 4 If a. 1 a 1 a (b) 1 an . + (–1)n . a2 are two AMs.. (b) y 2 xz xy yz xy yz 4 has the minimum value 2(d) 2y x 2y z 2y x 2y z n 24. 23. if a1. 1 1 1 Value of 1 + 1 2 1 2 3 .MATHS 22. 1 2 3 .g2 is equal to (a) a1h1 (c) a2h2 (b) a1h2 (d) a2h1 68 .... g2 are two GMs and h1. h2 are two HMs then g1.n is equal to (a) 2n n 1 (b) (c) 4n 3n 1 (d) none of these If x. y... 3n 2n 1 r 1 is equal to (b) 3n2 (d) none of these Between two unequal numbers. z are positive numbers in AP then (a) y2 xz (c) 1 r 1 n r 1 (a) 2n2 (c) n2 25.IIT. g1. The n numbers A1. the A. If ‘d’ is the common difference of this A. ….M. 3.’s are inserted between 1 and . b.P.P.. The sum sn of the first n terms of such an A. A1. 2 A2.…… The nth term an is given by an = a + (n – 1)d.An.M.P. can be written as a...P.. is a and qth term is b..e. a 2n 1 a 2n is equal to (a) 69 13 5 2n 1 2 a1 a 2n2 n (b) n a12 a 2n2 2n 1 ba n 1 . A2. the nth 2 2 a c is the A. then the 4th arithmetic mean is equal to 2 (a) 3 2 (b)3 (c) 2 3 (d) The pth term of an A.…. b are in A.M. of a and c. is a sequence whose terms increase or decrease by a fixed number. c are in A.P.P.II COMPREHENSIVE PASSAGE I An A... then d Ar a r 1. (b)log25 (d)log53 2 2 If am be the mth term of an A.P. If ‘a is the first term and ‘d’ the common difference. An are said to be A.P. a + 2d.P. then sum of its (p + q) terns is (a) pq 2 pq (b) a b pq (c) pq a b ab 2 p q (d)none of these If log 2..P.ARTHMETIC PROGESSION SECTION .. ). then the value of x is (a)5/2 (c)log35 4. is given by : sn = n n (2a + (n – 1)d) = (a + l) where l is the last term (i... a+d.P. called the common difference of the A. (b a) th n 1 . log (2x – 1) and log (2x + 3) are in A.... then b a. then a12 a 22 a 32 a 42 . 2..’s between the numbers a & b if term of the A. If a. where Ar is the r mean 9 If 6 A. . c > 0 and the minimum value of a(b2 + c2) + b(c2 + a2)+ c(a2 + b2) is labc.. e.M. ... x1 x 2 ..xn ) 1/n 1 1 1 n .. ³ H. 33 44 77 (b) [-3. then Pn = Vn 1 9. d.3] (c) [-3 2 . V3.xn n (x1 x 2. xn are ‘n’ positive real numbers. ... (b) 2 (d) 6 If a.MATHS (c) II n a12 a 22n 2n 1 (d) 2n 1 2 2 a1 a 2n n If x1... 3 2 ] III. If a > 0. If a and b are two positive real numbers... where V1.. . then Sn T k Vn V 0 . equality occurs when numbers are same using x1 x 2 xn this concept. 8. V2. 5. and a + b = 1... then the greatest value of a3b4 is (a) 32 43 75 (b) (c) 77 33 44 (d) none of these If x2 + y2 = 9. c. then A. .3 2 ] If nth term Tn of a given sequence is of the form Vn – Vn – 1.M.. then l is (a) 1 (c) 3 6. then x = (a + f)(b + e)(c + d) satisfies the relation (a) 0 < x 1 (b) 1 x 2 (d) 3 x 4 (c) 2 x 3 7. .. 0 n 4 Lt tan -1 2 = n®¥ 4n + 3 n =1 (a) 1 (c) 3 (b) 2 (d) 4 70 . n Vn T V n k 1 . b > 0... b..3] (d) [0. ³ G.. . is some other n sequence...IIT. f are positive real numbers such that a + b + c + d + e + f = 3.. .M.. Similarly if the nth term Tn of a given sequence is of the form K 1 Vn . then the value of x + y lies in the interval (a) [0. x2. 5. then sum of first n terms of the sequence is (b) less than n 1 n 1 (d) can not be determined n 1 If cos 1 + cos 2 + cos 3 + .3. IV.5. 1 n 1 n .7.4 2.11 1 1 (a) 840 8(2n 1) (2n 3) (2n 5) 71 1 1 (b) 840 8(2n 1) (2n 3) (2n 5) (2n 7) .3. then lim n Sn n (a) equals 1 (b) equals sin 1 (c) does not exist (d) none of these 1 1 1 Sum of the infinitely many terms of the series 2 3 4 .7. 12. . is equal to (a) 2 3 (b) 1 (c) 3 2 (d) does not exist as the sum tends to infinity The sum of n terms of a series each term of which is composed of the reciprocal of the product of r factors in arithmetical progression..6 .9 3.. 14.2.. affix the next factor at the end. divide by the number of factors so diminished and by the common difference. .. strike off a factor from the beginning. is 1. can be done by following rule: Write down the nth term. The sum of n term of a series each term of which is composed of r factors in arithmetical progression..ARTHMETIC PROGESSION 10.3. change the sign and add a constant..5 3.. If nth term of a sequence is (a) more than (c) equal to 11. 13. is 1 1 (a) 18 3(n 1) (n 2) (n 3) 1 1 (b) 18 3(n 1) (n 2) (n 3) 1 1 (c) 18 3(n 1) (n 2) (n 3) 1 1 (d) 18 3(n 1) (n 2) (n 3) The sum of n terms of the series 1 1 .9..4. 1 1 1 The sum of n terms of the series 1. + cos n = Sn. divide by the number of factors thus increased and by the common difference and add a constant.4. Write down the nth term..5... b .4.c 36 2 3 (c) a 29 3 4 .c 36 2 3 (b) a 29 3 4 . If 1 b c 3 4 5 .2.IIT.b . = a n 3 (n 2) (n 3) (n 1) (n 2) (n 3) .5 3..b .3.c 36 2 3 (d) a 29 3 4 ..6 (a) a 29 3 4 .4 2.c 36 2 3 72 .b .MATHS 1 1 1 1 (c) 840 8(2n 1) (2n 3) (2n 5) (2n 7) (d) 840 8(2n 1) (2n 3) (2n 5) (2n 7) 15. then 1. P. between a & b. 4. If the reciprocal of its common ratio is an integer. Find the number of sides of the polygon.ARTHMETIC PROGESSION SUBJECTIVE SECTION . The sum of an infinite geometric series is 162 and the sum of its first n terms is 160. Let x = 1 + 3a + 6a2 + 10a3 + . St–r.. If H be the H. 6. rth terms of an A. find all possible values of the common ratio. S and T respectively.’s is 7n + 1 : 4n + 27. 7. Tr–s = 1. then show that G2 = (2p – q) (2q – p).III LEVEL . the resulting numbers form an A.. If the first two numbers are increased by 2 and third is decreased by 4. then show that (H – 2a) (H – 2b) = H2 5.. Find S = 1 + 3 (ab) + 5 (ab)2 . are R.. y = 1 + 4b + 10b2 + 20b3 + .. The smallest angle is 120° and the common difference is 5. c are the sides of a triangle . the rth term of which is (2r + 1) 2r. 9. Evaluate: 1 + 2..M.. Find the numbers of G. (i) (ii) If one G. then prove that a2 + b2 + c2 > ab + bc + ca. then prove that p(b – c) + q (c – a) + r (a – b) = 0. + 100. 8.P. The interior angles of a polygon are in arithmetic progression.2 + 3. n and the first term of the series.M.. in terms of x and y. (i) (ii) The sum of three numbers in G.. If a.P. The r th .2 99 . b. Prove that Rs–t ..I 1. Find the sum of n terms of the series. 10. Find the ratio between their nth terms. 2. qth. |a| < 1.P.23 + . 3. c respectively. The ratio between the sum of n terms of two A.P.22 + 4. is 42. b..P. If pth. 73 . be a. G and two arithmetic means p and q be inserted between any given numbers. s th and t th terms of a certain G. |b| < 1. If the (m + 1)th.IIT. are in A. with common difference d. prove that 1 1 1 8 x y z 10.P. a2. Evaluate: 6..M.. Find the sum Sn of the cubes of the first n terms of an A. 1 1 3 1 3 5 5. show 2 that the ratio of the common difference to the first term in the A..P.. (n + 1)th and (r + 1)th terms of an A. and show that the sum of first n terms of the A. 7. 91 digits 3.. 2. 8.16 terms. n. tan 1 tan 1 1 a 1a 2 1 a 2a 3 1 a n a n 1 1 a 1a n 1 (ii) Sum to n terms the series 12 – 22 + 32 – 42 + 52 – 62 + . z are postive and x + y + z = 1.P.II BRUSH UP YOUR CONCEPT 1. is greater than or equal to the sum of the G. (i) If a1. r are in H. are in G..P..P. then show that the ratio S3n/Sn is equal to 6.P... Show that the number 1111.P.... In a triangle ABC prove that 3 a b c 2 2 bc ca ab 74 . d d d nd tan 1 . An A. If S2n = 3Sn . .. Show that the sum of the A. of a and b is t wice as great as their G.. If the roots of 10x3 – cx2 – 54x – 27 = 0 are in harmonic progression.MATHS LEVEL . Let Sn denote the sum of first n terms of an A. a3. then prove that tan 1 4. is .P. is a factor of Sn. y. 9.M. then find c and all the roots. and a G. 1 1 1 If x.P.. m. with positive terms have the same number of terms and their first terms as well as last terms are equal. n 13 13 23 13 23 33 . then show t hat (ii) a : b = (2 3):(2 3) .P.P.1 is a composite number.P. (i) If the A. x2. prove that the value of q cannot be between p and p.P.. The real numbers x1. The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. b2. 4 4 4 4 Find the smallest natural number n0 such that Bn > An for all n n0. c are in G. then find the value of 2 3 4 n 1 2 S12 S 22 S32 .. b are in arithmetic progression. A2.. then prove that either –a/2..3. a2. b.P.. G1 . b. b are in geometric progression and a. or a = b = c. c are in A. geometric mean and harmonic mean of a1. For each n. [1986] If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means 2 n 1 between the same two numbers.. 3. Prove that (a + 1)7 (b + 1)7 (c + 1)7 > 77 a4 b4 c4.. (1) n 1 . [1996] 5. 3) . S 2n 1 . [2002] 7.. S2. H1. the arithmetic mean. … ... Sn are the sums of infinite geometric series whose first terms are 1.. then find the exhaustive range of x ? [2004] 11.P. [1991] 4.III 1. An infinite G. A2. . c2 are in H. G2. H1. n and whose common ratios are 1 1 1 1 ..x2 + x + = 0 are in A.. be positive real numbers in geometric progression. . H2. . Hn. c.P. x3 satisfying the equation x3 . and a2... be positive real number’s. H2 . Gn in terms of A1. [2003] 9. 3 2. .. 75 [2006] . S3. R [2004] 10. Let a1. . where a. show that G1G 2 A1 A 2 (2a b) (a 2b) . b are in harmonic progression. If their sum is S. an. Gn. b.ARTHMETIC PROGESSION LEVEL . . n 1 [1991] 3. a2. a. respectively.1 (1. If a.P. .P has first term x and sum 5.. 2.. and Bn = 1 – An . . 1 2 show that . Find an expression for the geometric mean of G1. 2. [2001] 6..... is S2. Find the intervals in which and lie. The sum of the squares of three distinct real numbers. H1H 2 H1 H 2 9ab [2002] 8. . let A n 2 3 n 3 3 3 3 . which are in G. G2. b.. Prove that the resulting sum is the square of an integer. let An. An. If a. ... If S1.. Hn be respectively. For n = 1. Let a.. A1. 15.d c b. 4.d a a. c a 76 . c 12. 7. 4. 3. 13.d c. c 17. 23.b 16.c a. 5. b a c. 11. 5. 15. 3. a. 2. 9. c d 14. 8. 11. 9. 10.d 12. c a 21. 7. 17.d b. 19. 11. 25.c. 6.d a a a. 13. b.II 1. 24. 13.MATHS ANSWER OBJECTIVE SECTION . b a c c a LEVEL .b c a.d a. 22.I LEVEL . 14. 22. a c 21. 9.IIT. 5. 12. 3.b c.d SECTION .I 1. 18. 19. 18. 2. b c c 20. a b b c c b c c a b b c d 14.b c. 8. 2. c b b a. 8. 23. a b b c c b c c a b b 16. 7. 24. 15. 6. 20. 4. 10.II 1. 25. 6.c c b. 10. 1 1 and 3 27 (A1 A2 . 11. 0 < x < 10. 8. (ii) 5.ARTHMETIC PROGESSION SUBJECTIVE SECTION . 99. 2 5 LEVEL .2100 + 1 n2n+2 – 2n+1 + 2 9. 12.I 1. .. 6. 1 n (2n 1) (4n 1) 1 3 5. when n be even & . Hn)1/2n 10. 2 or 1 and a = 108. 6 (ii) n = 4.III LEVEL-III 3.II n(n 1) n(n 1) .III LEVEL . 77 . 446 6. 144 or 160 7. An . 12. H1 H2 .. 3 3 c = 9... 2. S 1 ab where a = 1 – x–1/3 & b = 1 – y –1/4 (1 ab)2 LEVEL . 2a (n 1)d [2a 2 2ad (n 1) nd 2 (n 1)] 2 2 7. 24 OR 24. least value of n0 = 2 4. when n be odd 2 2 3.. roots are 3. 9 14n 6 8n 23 (i) 6.. 1 n Sn .. ’s is 7n + 1 : 4n + 27..IIT.P. a4 are om AP b1b2 . Find the sum of first 24 terms of the A. Given a1 + a5 + a10 + a15 + a20 + a24 = 225 First term = a1 commoun ratio = d a1 + a1 + 4d + a1 + ad + a1 + 14d + a1 + 19d + a1 + 23d = 225 6a1 + 69d = 225 2a1 + 23d = 75 = t24 Sn = n/2 (a + l) 524 = 2. Solution : First term a = 120° common ratio = 5 Given : sum of enterior angles of a polygon ‘n’ is (2n . if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225.. a24 are in A.... a 1 .P. The ratio between the sum of n terms of two A. Solution : Let s = a1 a2 . a 2 .25n + 144 = 0 n = 16 n = a 3.2)180° 2 n2 . . Solution : Let a1 a2 .MATHS SUBJECTIVE SOLUTIONS LEVEL . Find the number of sides of the polygon.P..I (CBSE LEVEL) REVIEW YOUR CONCEPTS 1. n n Sn = {2a1 + (n – 1) d1} Sn = {2b1 + (n – 1)d2} 2 2 78 . 24 (75) = 900 2 The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5...P. Find the ratio between their nth terms. bn are in A.4) = (n – 2) (n – 2) 180° 2 n Sn = [2a (4 1) d] = (n – 2) 180° 2 n [240 5n 5] = (n . a 3 . a1 = 4 b1 = ratio of nth term is 4. ar2 are in G. ar. Prove that Rs–t .r Tr–s = 5. is 42.. Tr – s = r° = 1 The sum of three numbers in G.P. the resulting numbers form an A.r tr S Ts a s .P. sth and tth terms of a certain G. are R.P.r ts s RS–t . T r a r .r = t tr t .r rs s St r S a . 79 .P. Solution : Let a. Find the numbers of G. If the first two numbers are increased by 2 and third is decreased by 4. S and T respectively.P. St – r .r rs t Rt a . Tr–s = 1. St–r. 31 d = 7 d2 = 4 2 1 14n 6 a1 (n 1)d1 4 (n 1)7 = 31 23 8n b1 (n 1)d 2 (n 1) 4 2 The rth. b1 and d. Sr a r . Solution : Given Tr = arr – 1 = R TS = ars–1 = S Tt = ar t – 1 = T S–t then R t t St t RS a s .ARTHMETIC PROGESSION 7n 1 2a1 (n 1)d ratio is 4n 27 2b (n 1)d 1 2a1 (n 1)d 2 4 7(n 1) 31 2 4(n 1) 2b1 (n 1)d 2 2a1 (n 1)d 8 – 1 (7 – 7n) = 2b (n 1)d 1 2 4 7(n 1) 2a1 (n 1)d 31 2 4(n 1) 2b1 (n 1)d 2 comparing the values a1. y in these A. The sum of an infinite geometric series is 162 and the sum of its first n terms is 160.P. y 2q = p + y y = 2q – p G2 = 2y (G . G and two arithmetic means p and q be inserted between any given numbers. 2 a = 6. then show that G2 = (2p – q) (2q – p). Solution : x. p. n and the first term of the series.MATHS 2 Given a + ar + av = 42 a(1 + r + v2) = 42 . (1) and a + 2. If the reciprocal of its common ratio is an integer. 6 6.M.M. find all possible values of the common ratio.q and p. r = 1/2 24. q.Geometric mean) G2 = (2p – q) (2q – p) 7. r = 2 6. q. 24 If a = 24. ar2 – 4 are in A. 12. 24 If a = 6.15r – 6 = 0 1 r = 2. of x. p. If one G. 12... ar + 2.IIT. q 2p = x + q x = 2p . 2(ar + 2) = a + 2 + ar2 – 4 a (1 + r2 – 2r) = 6 6 a= 2 1 r 2r Put ‘a’ in 0 6 (1 + r + r2) = 42 2 (1 r 2r) 6r2 . Solution : Sum of an infinite geometric series 80 . 22 + .25 = (1 + 2 + .2 + 3 .. a a = 108.2100 –8 = 1 + 2 (299 ..1) . r = then a = 160 81 1 n = 2..299 + 100. Evaluate: 1 + 2. + 100. .. Solution : Let s = 1 + 2.P.2 99 .22 + 3. .1) d1 = b Tr = a1 + (r .23 + .1) d1 = c ab a . r = 1/3.2100 s = –99. + 100. then prove that p(b – c) + q (c – a) + r (a – b) = 0.. 144 8. .2100 – 1 = 99.1) d1 = a Tq = a1 + (q .. r then a = 144. + 299) + 100. . b. r th terms of an A. If p th .q) d1 = = d1 pq 81 .. q th . c respectively. 160. be a.. Solution : Given Tp = a1 + (p .23 + .299 2.2100 + 1 9. 99..22 + 4.. 162 1 rn 3 4 if n = 4..2100 s .ARTHMETIC PROGESSION a Sn = = 162 a = 162 (1 – r) 1 r sum of first in terms is Sn = a(1 r n ) 160 1 r 162 (1 r) (1 r n ) 160 (1 r) 1 160 n r r is an integer.2 + 3.b = (p . then a = 108 1 if n = 1.100.5 = 2 + 2. P. then show that (H – 2a) (H – 2b) = H2 Solution : a.r) + qd1 (r.2a) b a H2 = (H – 2a) (H – 2b) 12... 10. Let x = 1 + 3a + 6a2 + 10a3 + . H.. 49 Solution : First term a = 28 tH = 4 ar 3 r = 1/7 49 S = 11.M.. whose first term is 28 and the fourth term is 4 . Solution : 13.MATHS Similarly bc ca d1 . 82 .IIT. Find S of the G.. Solution : Series can be written as s = 1 + 3z + 5z2 + . the rth term of which is (2r + 1) 2r. |b| < 1..(–H) = a b (H 2b) (H.p) + rd1 (p – q) 0..P. Find the sum of n terms of the series.a) + r (a .b) pd. in terms of x and y.. Find S = 1 + 3 (ab) + 5 (ab)2 . a 1 r If H be the H.. between a & b. and H b a (–H).. y = 1 + 4b + 10b2 + 20b3 + . = d1 qr rp p(b .c) + q (c . (q . H 2ab ab Ha + Hb = 2ab Ha – 2ab = –Hb a(H 2b) b(H 2a) H. |a| < 1. b are in H.. 1 + 2z [1 + z + z2 + ....ARTHMETIC PROGESSION sz = z + 3z2 + .b)-1/4 1 – y–1/4 = b 1 b 1 1 x = (1 a)3 = (1.b)2 = 1 + 2b + 3b2 + ...b)y = b + 3b2 + ...b)3 y = (1 – a)2a = 1 + a + a2 + .. ax = a + 3a2 + ....... y = 1 + 4b + 10b2 + 20b3 + . c are the sides of a triangle . s(1 – z) = 1 + 2 z + 2z2 + . 1 b b2 .. b(1. |b| < 1 by = b + 4b2 + 10b3 + ... ] 1 z (1 ab) 1 1 z s (1 – z) = 1 + 2 s= = 2 (1 z) (1 ab)2 1 z 1 z where a = 1 – x–1/3 and b = 1 – y–1/4 x = 1 + 3a + 6a2 + .. 14... |a| < 1 b(1 ... then prove that a2 + b2 + c2 > ab + bc + ca. x(1 .. b. ax(1 – a) = a + 2a2 + . = 1 y = (1 .. y (1 ...... (1 ... b. c are sides of a triangle (a 2 b 2 c 2 ) a b c 3 3 2 a 2 b 2 c 2 a 2 b 2 c 2 2(ab bc ca) 3 9 9 3a 2 3b 2 3c 2 a 2 b 2 c2 2(ab bc ca) 9 9 a2 + b2 + c2 > ab + bc + ca 83 .b) = 1 + 3b + 6b2 + 10b3 + . If a....a)–3 1 a a = 1 – x–1/3 .a) = 1 + 2a + 3a2 + . Solution : a.. y(1 ...b) y = b 2b 2 .. . x n (x1.... (i = 1. + xn > 0 x1 x 2 .. ... xn)n n (x1 x 2 . + xn) x + x + .... x n ) n n 1 1 1 ...IIT. n).... xn x1 x 2 1 1 1 (x1 + x2 + . then prove that (x1 + x2 + ... + x ³n . 1 2 n Solution : xi > 0 (Given) x1 + x2 + .. 1 1 1 2 If xi > 0..... x 2 n 1 2 n 84 . x2. + xn) x x .. 2. x3 .MATHS 15.. 1)d} {(n + 1)d + 3nd – d} = 6n2d 2 2 S3n = 6Sn S3n = 6 Sn 2. Let Sn denote the sum of first n terms of an A..I BRUSH UP YOUR CONCEPTS 1. If S2n = 3Sn . then find the common difference. tan 1 tan 1 1 a 1a 2 1 a 2a 3 1 a n a n 1 1 a 1a n 1 Solution : 85 . 4. are in A.P. 1..P. 7 are in A. 4. Solution : Let f(x) = x3 – 12x2 + 39x – 28 = 0 if x = 1 f(x) = 0 (x – 1) is a factor x2 – 11x + 28 = 0 (x – 7) (x – 4) = 0 x = 7.. If the roots of the equation x3 – 12x2 + 39x – 28 = 0 are in A.. (or) 7.P..P. common ratio is d = 3 3. a2. Solution : Given S2n = 3Sn 2n n {2a (2n 1)d} 3 {2a (n 1)d} 2 2 2a = (n + 1)d n Sn {2a (n 1)d} {(n 1)d (n 1)d} n 2 d 2 3n 3n S3n = {2a + (3n .. 4 series 1. If a1. .. then prove that tan 1 d d d nd tan 1 .ARTHMETIC PROGESSION LEVEL . a3. . with common difference d. then show that the ratio S3n/Sn is equal to 6. ......IIT. + 1) (106 + 105 + ... are in AP tan-1 a2 – tan–1 a1 tan-1 a3 – tan–1 a2 tan-1 an + 1 – tan–1 an 1 (a n 1 a 1 ) tan–1 an+1 – tan–1 a1 tan (1 a a ) = tan–1 1 n 1 d = n (an + 1 – an) nd 1 a1 a n 1 n=1 d = a2 – a1 nd = an+1 – an nd = an + 1 – a1 4....1 is a composite number...... then prove that 1 + ab + a2b2 + a3b3 + .. + 10 + 1) 86 . y 1 y 1 b 1 b y S = 1 + ab + (ab)2 + . to (|b| < 1). to = xy . x= 1 x 1 a 1 a x y = 1 + b + b2 + . xy 1 = x y 1 1 ab Show that the number 1111... are in AP also tan–1a1 tan–1a2 . S = 5.. x + y -1 Solution : x = 1 + a + a2 + ... to (|a| < 1) and y = 1 + b + b2 + b3 + . 91 digits Solution : We have 111 .......MATHS a1 a2 .......... 1 (91 digits) = 1090 + 1089 + . + 102 + 10 + 1 1091 1 1091 1 107 1 10 1 107 1 1 0 1 = (1084 + 1077 + 1070 + . If x = 1 + a + a2 + a3 + ... . (n 1)n n(n 1) n2 2 2 If the A.. Sum to n terms the series 12 – 22 + 32 – 42 + 52 – 62 + ... then show that a : b = (2 + 3 ):(2 ...ARTHMETIC PROGESSION [ 91 = 13 × 7.3 ) . + [(n – 2)2 – (n –1)2 ] + n2 = – 1 [1 + 2 + ..17 16 446 4 1 4 6 2 1 1 7. n 3 n 2 (n 1)2 1 2 (n 2n 1) 1 3 5 . + n) = n(n 1) 2 Case II n is odd (12 – 22) + (32 – 42) + .. 111 ... (91 divide and multiply by (10–1 – 1) or (1013 – 1] Thus....n 2 4 16 16 1 16 1 1 1 S16 Tn n 2 1 (16.16 terms.16... + [(n – 1) – n] [(n – 1) + n] = – 1 (1 + 2 + ..33) 2. 1 (91 digits) is a compossite number. Solution : Am = ab GM = 2 ab and A = 2G ab 2 ab a b 4 ab 2 a a 1 4 0 put b b a =x b x2 + 1 – 4x = 0 x = 2 3 87 ..n terms 4...17..M. Evaluate: + 1 1+ 3 1+ 3 + 5 Solution : Tn 13 23 33 .. 13 13 + 23 13 + 23 + 33 + +. ... of a and b is twice as great as their G. 6. we consider two cases (i) Let n be even (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + ... Solution : 12 – 22 + 32 – 42 + . [(n – 1) – n] [(n – 1) + n] (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + . + (n – 1)] + n2 = 8...M. b. b/r 12 = Diving .br. 24.. 2 b +c c+a a+b Solution : Ist part adding 3 to both sides a bc a bc a b c 3 3 bc ca ab 2 9 (b + c)–1 + (c + a) –1 + (a + b)–1 2(a b c) Applying A. 3/2 3 a. 1 2 (a b c)3 3 3 2 9 3(a b c) 2(a b c) 2nd part b + c > a in a triangle add (b + c) to both sides 2 (b + c) > (a + b + c) 1 1 2(b c) a b c 88 .b / r 2 b 1 1 r b(1 ) r 1 and hence from any b = 24 3 five numbers are 8.IIT.b 2 br b(1 r) 1 r 36 = 2b.N of nth powers (A. c are in G. b. 2. 72.M)m (b c) (c a) (a b) then LHS 3 3 2 3 (a b c)3 3 1 3. 648 In a triangle ABC prove that 3 a b c £ + + <2.P. r = 10.MATHS taking positive sign a 2 3 (2 3)2 b 2 3 taking negative sign a : b = 2 – 9. find the first five terms of the series. c are the first three terms of a geometric series. Let choose them to br. 216. Solution : a. . b. If the harmonic mean of a and b is 12 and that of b and c is 36. ARTHMETIC PROGESSION a 2a write similar inequalities and add ac abc a a bc a c 2 a b c 2 89 . is S2... (2) and ( 2 1) ( 2 3) <0 ( 2 1) 2 ..IIT. 3) . α S.MATHS LEVEL . which are in G.. a1 a/r 1 a(r 1 ) S r 1 2 2 2 and a (r 1 2 ) S r 1 1 2 2 put r t.1 È (1...II CHECK UP YOUR SKILLS 1. If their sum is 1 α 2 Î . (1) r2 – r t + 1 = 0 for t to be real t2 – 4 > 0 t < – 2 or t > 2 from (1) 2 1 2 1 2 or >2 2 1 2 1 In an enequality we can multiply only by a +ve quantity. r 2 t 2 r r a (t + 1) = S and a2 (t2 – 1) = S2 Eliminating S. and 3 90 . The sum of the squares of three distinct real numbers. (3) (3) implies that 2 lies between 1. a2 (t2 – 1) = (t – 1) 2 = (t + 1) t 2 1 d2 1 Now t = r 1 r a 2 (t 1) 2 2 . show that 3 Solution : Let the numbers are ar. 2 1 is we whether it is +ve or –ve 3 2 1 3 2 2 3 0 0 0 2 1 2 1 2 1 1 3( 2 ) ( 2 1) 3 0 ( 2 1)2 .P. P. .. and show that the sum of first n terms of the A.1) (1.P. If the roots of 10x3 – cx2 – 54x – 27 = 0 are in harmonic progression. then find c and all the roots. (1) 2q = p + r or 3q = p + q = r = 54 2 27 q = –2/3 is roots of (1) 27x2 + 54x2 + 9x – 10 = 0 has a root (3x + 2) (9x2 + 12x – 5) = 0 x = 1/3. r be the roots of the equation 10 c 54 27 0 x3 x2 x 27x3 + 54x2 + (x – 10 = 0 are in AP .P 2 or 3 3x + 2 is its factor 3.. 4. 91 If the (m + 1)th.. are in G. + (a + nd) n n [2a (n 1)d] Sn = [2a + (n – nd] 2 2 Sn = (a + d)3 + (a + 2d)3 + . + (a + nd)3 = na3 + 3a2 d n + 3ad2 n2 + d2 n3 = na3 + 3a2 d n(4 1) n(n 1) (2n 1) 3 n 2 (n 1) 2 3ad 2 d 2 6 4 1 n .P. be the roots in H. Solution : Let A. show that .P..P. –3/2. Solution : . (2) implies that 2 lies between 1/3 and 1. [2a + (n + 1)d] [2a2 + 2ad + (n + 1 + d2 n (n + 1)] 2 2 3.P.ARTHMETIC PROGESSION 2. (n + 1)th and (r + 1)th terms of an A.. –2/3. n... –5/3 are in A. r are in H.P. m. 1 2 ( . is a factor of Sn..P... –3/5 are in H. be (a + d) + (a + d) + .3) 3 Find the sum Sn of the cubes of the first n terms of an A. then p. q. . y. prove that .. (iii) (i)..M G.. (ii) & (iii) (x y) (z x) (y z) xyz 8 (1 – x) (1 – y) (1 – z) 8 xyz 6.P (a + nd)2 = (a + md) (a + rd) a (2n – m – r) = d (mr – n2) d 2n (m r) a mr r 2 .IIT. (1) but m. .P. (ii) xy 2 xy . z are positive and x + y + z = 1. (2) 2 2n (m r) 2 by () & (2) (m r) 2n n n 1 1 1 If x. y numbers each equal to y and z numbers each equal to z.. n Solution : a + md. 2mr mr n d 2n (m r) a (m r) n n 2 2 ..M. (i) zn 2 zx . then A... a + rd are in G. r are in H..P.MATHS 2 the ratio of the common difference to the first term in the A. is .1 ³ 8 x y z Solution : 5. On these x + y + z numbers 92 . A.1 . a + nd. Prove that x2 + y 2 + z 2 x+y+z x+ y + z x+y +z > xx y y zz > 3 x+ y + z Solution : Consider x numbers each equal to x. n.NM yz 2 yz .. > G.1 .M. . (A) 2nd part : consider x number each equal to 1/x y is 1/y 1 z is z AM > GM 1 1 1 1 1 1 . z 3 1 x y z 1 1 1 xyz x y z xyz x ..... y times) (z z . 93 ab > aa...M > G..M (a a . y factors 1 1 x yz . (3) Both (A) & (B) prove the required results... z factors]1/x + y + z But x + x ...x . a 2 aa. ...P.... x factors) (x..z .. y .ARTHMETIC PROGESSION x x x....a ..z factors . z times) x yz [x. y . z x yz xyz x y z x yz 3 x y z x x y y z z or xyz x y z > 3 x y z x y z ..x times . z times y y x x z z x yz 1 1 1 1 1 . is equal to 155. .a factors a 2 a 2 b2 ab 8. x times) (y y .x factors x x y y z z 1 1 1 1 xyz x .x = x2 and x.. y factors) (z.. and the sum of the first two terms of a ... a 2 + b2 Prove that a+b Solution : a+b > aa b b Apply A... 7. x times = x.. x factors = xx x 2 y2 z 2 x 2 y2 z 2 x y z 1/ x y z (x y z ) or xyz xyz x y z x x y y z z ... a times a 2 . . bb The sum of first ten terms of an A...x . y times ... P.. and so on.P.P. is 3 3 6 9.P = 155 S2 of G.P 25 79 83 . (2. 2 25 625 .P.P. and the first term of G. 3..P is d + da + da2 + . Hence 1st term of nth group is one more than the last term of (n – 1) th group A = (n – 1)2 + 1 = n2 – 2n – 2 Also terms in each group are in A. find these progressions. is equal to common ratio of G. (5.P. and a G. 3. An A..P. S1o of A. whose D = 1. 6... . Solution : The number of terms in successive groups are 1. d = 3.P.P is 2 + 5 + 8 + ... 5.P.. ‘b’ the last term and n the numbers of terms of A. Show that the sum of the A. 9). or a = A. is 9.IIT. 7.P...P = [2A (n 1)D] 2 2n 1 2 = [2n – 4n + 4 + (2n – 2)] 2 = 2n3 – 3n2 + 3n – 1 n3 + (n – 1)3 .. Solution : Let ‘a’ be the first terms. 4)..P = 2n –n = N. The C. hence nth groups is n2 and of (n – 1)th group is (n – 1)2 . with positive terms have the same number of terms and their first terms as well as last terms are equal. 10.. G. if the first term of A. is equal to common difference of A. Solution : Let the AP be a + (a + d) + (a + 2d) + . G.. n sum of terms is nth group is sum of an A.d of 94 ..MATHS G.P.and 2 6 6 The series of natural numbers is divided into groups (1). . The last terms of successive groups are 12...P. By given condition G. is greater than or equal to the sum of the G.P = 9 2a + ad = 31 d + da = 9 Solving a – 2. and hence nth group will be nth terms of this A.. 8. 25 2 d 2 3 or A.. Show that the sum of the numbers in the nth group is (n – 1)3 + n3.P and G.P. 22 . is 3 + 6 + 12 + 24 + . P and S the sum of n terms G.P. n – 1 r a b ab an (1 r n 1 ) = n 1 = n S [ from (1)] 2 a 2 2 S S . + 1) a S [1 + rn–1) + (r + rn–1) + .r of G. (rx + rn-k ) + .... . + rn–1) S = a(rn–1 +rn–2 + .... (2) 2 k (r + rn–k–1) – (rn–1 + 1) = (rk – 1) + (rn–1 (r–x – 1) r n 1 k n k 1 )0 = (r – 1) 1 k (r 1) (1 r r k [for 0 r 1 and also for 1 < r < ] rk + rn–k–1 from (2) S 95 n–1 + 1 for k = 0.. n Now S (a b) 2 S = a (1 + r + r2 . + (rn–1 + 1)] ... 1...P = and C. . . 2.P = n 1 a Let S be the sum of n terms of A......ARTHMETIC PROGESSION 1/ n 1 ba b A. JEE 1.P d1 = ab ab(n 1) q = first H..M. (A) (bn a) (n 1) 2 2 n 1 p ab(n 1) n 1 an b (n 1) (a b) n Again q – – = = (n 1)2 (bn a) ..MATHS PROBLEMS ASKED IN IIT . (B) bn a n 1 m 1 n 1 n 1 2 n 2 (a b) 2 (a b) 2 Now (q – p) q p = (n 1)2 (bn a) 2 0 n 1 2 n 1 p Hence q can not lie between p and n 1 96 .P = a + d = a + ba n 1 Since n + 1 m’s have been inserted a and b C. If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means between the same two numbers..’s have been inserted between a and b common difference d ba h 1 Now P = first A.IIT.. n -1 [1991] Solution : Let two numbers be a and b Since n A.M.. (iii) n 1 x not will between and is ( x ) ( x ) > 0 n(a b)2 Now p – q = .. (ii) bx a an b . = 2nd terms of A.M = 2nd term of H.d of A. prove that the value of q cannot be between 2 n +1 p and p...P 1 1 1 ab bn a q a d1 a ab(n 1) ab(n 1) q = p ab(n 1) . . an = a1rn–1 An = .. an )1/n = a1(1.. + (24)2 = S1 12 + s2 . a2.. (2) n r 1 Gn = (a1 a2 ... 2n (24 1) (4n 1) n(2n 1) (4n 1) 1 6 3 Let a1.. S3. If S1.r2 .r.. A2. (4) Hn = r n 1 from (2) (3) & 4) 97 ... ..a r 1 r n 1 H n x a1 = n a1 r r 2 r n 1 1 na1 r n 1 (r 1) ..+ S 2n-12 ... a2 = a1r : a3 = a1r2 .. S2..P. H2... Hn be respectively. geometric mean and harmonic mean of a1. + (24)2}– 12 = 3. H1.. ....... 2... Hn.. + S22 + .. An.. [1 r 2 . Gn in terms of A1.. S2 1 1 3 3. . . . G2. respectively. be positive real numbers in geometric progression. the arithmetic mean.... an. .. Gn. = b. then find the value of 2 3 4 n +1 S12 +S 22 +S 32 + ... a4 are in G..r n 1 ) by (1) n n or An a1 r 1 ... 1 .3.. . ... (3) 1 1 1 1 1 1 1 . [1991] Solution : 1 S1 1 1 2 2 2. let An..(i) Now S12 S22 + .... Find an expression for the geometric mean of G1. Sn are the sums of infinite geometric series whose first terms are 1. n 1 ] = H n n a1 n a1 r r 1 1 1 1 1 1 1 1 r n 1 1 1 .. . . rn–1)1/n orGn = a1 [r(1/2) (n – 1) ]1/4 = a1r1/2(n– 1) .ARTHMETIC PROGESSION 2.. (1) a 1 a1 (1 r r 2 .. a2. n and whose common ratios are 1 1 1 1 . .... [2001] Solution : a1 a2 ..n) ..... S22 (n – 1) = 22 + 32 + . For each n.. of G1. Gx whose value we have determined in terms of A1A2 . 5. A4) (H1 . b. H ... a. H2. H1... Gn2 = (–A1 A2 .P. An and H1 H2 . hence G12 ..... An) (H1 H2 .. H1 H 2 H1 + H 2 9ab [2002] Solution : a....P 1 1 1 1 a . Hn)]1/2x LHS is G. G2...P be taken as (a – 3d).MATHS 2 n 2 1 G = Hn An = a n–1 i(r ) above is true for each n. G22 . show that G1G 2 A1 + A2 (2a + b)(a + 2b) = = . G3. [2002] Solution : Any four consecutive integers in A. be positive real number’s If a.IIT... G1 . b are in geometric progression and a. G4)1/4 = [(A1A2 . b are in arithmetic progression. G2. b are in H. (a + 3d) so that common difference is 2d and their product is (a2 – 9d2) (a2 – d2) we have to evaluate (2d)4 + (a2 – 9d2) (a2 – d2) a4 – 10a2 d2 + 25d2 = (a2 – 5d2)2 (a2 – 5d2)2 is also an enteger.. b are A. H1. (a – d). H . Hn . A1. . Prove that the resulting sum is the square of an integer.. The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. H2 . Let a. A2.M/. b are in harmonic progression. 4. (a + d). H4) (G1 . 1 2 1 1 3D b a 11 1 D 33 a 1 1 1 1 1 2b a D 3 H a a b a 3ab 1 1 1 1 1 1 1 2b a D 3 3 H1 a a a b a 3ab 98 .. c. c2 are in A. [2003] Solution : 2b = a + c ab. 2 7.P.P....ARTHMETIC PROGESSION G1G 2 ab (ab a) (2a b) (2b a) (2a b) = 2 2 HH 9a b 9ab 1 2 6.P . c are in G. and a2.P. c are in A. c2 are in H. where a. a 2 c2 . b. b. b2. Prove that (a + 1)7 (b + 1)7 (c + 1)7 > 77 a4 b4 c4..P. b2. c are in H. b.. If a. or a = b = c. c are in G. b. R Solution : (a + 1) (b + 1) (c + 1) = abc + ab + bc + ca + a + b + c + 1 > abc + ab + bc + (a + a + b + c = 7 4 4 4 1/7 7 (a b c ) 99 abc ab bc ca a c 7 [2004] . then prove that either –a/2. b. (2) b2 [(a + c)2 – 2ac] = 2a2c2 b2 [(4b2 – 2ac)] = 2a2 c2 2b4 – b2ac – a2 c2 = 0 factorize b2 – ac = 0 or 2b2 + ac = 0 2 a c ac if b = ac then by 2 2 (a + c)2 – 4ac = 0 or (a – c)2 = 0 a = c and 2b = a + c = a + a = 2a b=a b=a a=b=c if 2b2 + ac = 0 then b2 = 1 c 2 a ..P. (1) b2 2a 2 c 2 a2. An infinite G.IIT.MATHS Thus (a + 1) (b + 1) (c + 1) > 7 (a4 b4 c4)1/7 (a + 1)7 (b + 1)7 (c + 1)7 > 77 (a4 b4 c4) 8.P has first term x and sum 5. then find the exhaustive range of x ? [2004] 100 . . d = –1 and a = m + n – 1 Tp = a + (p –1)d = m + n – 1 + (p – 1) (–1) = m + n – p Now. . ..ARTHMETIC PROGESSION Illustration 1: The mth term of an A. T26.P.. Also show that its (m + n) th term is zero. find the series. is n and its nth term is m.. Illustration 2: 1 2 Find the number of terms in the series 20. . Illustration 3: The first term of an infinite G. Solution: Given Tm = a + (m – 1) d = n and Tn = a + (n – 1) d = m Solving we get. Solution: Given that Tp = (Tp+1 + Tp+2 + . 18 . Tm + n = a + (m + n – 1)d = (m + n –1) + (m + n–1) (–1) = 0.. ) or... ..e. 19 . Prove that its pth term is m + n – p... T27 .. 2 4 8 Illustration 4: 101 ..] 1 r 1 2 1 1 1 Hence the series is 1.P.. rp–1 = 1–r=r r= rp [sum of an infinite G. . . . n2 – 61n + 900 = 0 n = 25 or 36. of which the sum is 300. d 2 and Sn = 300. arp–1 = arp + arp+1 + arp+2 + ... Explain the 3 3 double answer... T36 is zero. Since common ratio is negative and S25 = S36 = 300. it shows that the sum of the eleven terms i.. Solution: Clearly here a = 20. 3 n 2 2 20 (n 1) 300 2 3 Simplifying.P is 1 and any term is equal to the sum of all the succeeding terms.. IIT- MATHS If a1, a2, a3, ... an are in harmonic pregression, prove that a1a2 + a2a3 + ... + an–1 an = (n – 1) a1an . Solution: Since a1, a2 , ... , an are in H.P., 1 1 1 1 , , ,..., a1 a 2 a 3 a n are in A.P. having common difference d (say) . 1 1 1 1 1 1 d, d, ... d a 2 a1 a3 a 2 a n a n 1 or a1 – a2 = d(a1a2), a2 – a3 = d (a2a3), ... , (an–1 – an) = d(an–1 an ) Adding the above relations, we get a1 – an = d (a1a2 + a2a3 +... + an–1 an) ... (1) 1 1 Now a a (n 1)d n 1 1 1 (n 1)d a n a1 or (a1 – an) = (n – 1) d an a1 ... (2) Putting the value of a1 – an from (2) in (1), we get (n – 1) an a1d = d (a1a2 + a2a3 + ... + an–1 an) (n – 1) ana1 = a1a2 + a2a3 + ... + an–1 an. Illustration 5: If A1, A2; G1 , G2 and H1 , H2 be two A.M.s, G.M.s and H.M.s between two quantities ‘a’ and ‘b’ then show that A1H2 = A2 H1 = G1G2 = ab Solution: a, A1 , A2, b be are in A.P. ... (1) a, H1 , H2 , b are in H.P. 1 1 1 1 , , , a H1 H 2 b are in A.P.. Multiply by ab. b, ab ab , , a are in A.P. H1 H 2 take in reverse order. or a, ab ab , , b are in A.P.. H 2 H1 ... (2) Compare (1) and (2) ab ab and A 2 H2 H1 A1 A1H2 = A2H1 = ab = G1G2 102 ARTHMETIC PROGESSION Illustration 6: ab a b Prove that 2 ab.ba , a,b N;a b. Solution: Let us consider b quantities each equal to a and a quantities each equal to b. Then since A.M. > G.M. (a a a ...b times) (b b b ...a times) [(a.a.a...b times) (b.b.b. ... a times)]1/(a+b) ab ab ab (a b ba )1/(a b) ab 2ab (a b ba )1/(a b) ab Now a b 2ab (A.M. > H.M.) 2 ab ab 2 a b a b .b a . Illustration 7: Prove that a4+ b4 + c4 abc (a + b + c), [a, b, c > 0] Solution: Using mth power inequality, we get a 4 b4 c4 a b c 3 3 a b c a b c 3 3 or 4 3 abc 1/ 3 3 [(abc) ] 3 ( A.M G.M) .M) a 4 b 4 c4 a b c abc 3 3 4 4 4 a + b + c abc (a + b + c). Illustration 8: Prove that s s s 9 , if s = a + b + c, [a, b, c > 0] sa sb sc 2 Illustration 9: Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ... . Solution: Clearly here the differences between the successive terms are 7 – 3, 14 – 7, 24 – 14, ... i.e., 4, 7, 10, ... which are in A.P. Tn = an2 + bn + c Thus we have 3 = a + b + c, 7 = 4a + 2b + c and 14 = 9a + 3b + c 103 IIT- MATHS 3 2 Solving we get, a , b 1 ,c 2 . 2 1 Hence Tn (3n 2 n 4) 2 Sn = 1 [3n 2 n 4n] 2 1 n(n 1) (2n 1) n(n 1) n 3 4n (n 2 n 4) 2 6 2 2 Illustration 10: Find the sum of n terms of the series 3 + 8 + 22 + 72 + 266 + 1036 + ..... Solution: 1st difference 5, 14, 50, 194, 770, ... 2nd difference 9, 36, 144, 576, ..... They are in G.P. whose nth term is arn–1 = a4n–1 Tn of the given series will be of the form Tn = a4n–1 + bn + c T1 = a + b + c = 3 T2 = 4a + 2b + c = 8 T3 = 16a + 3b + c = 22 Solving we have a = 1, b = 2, c = 0. Tn = 4n–1 + 2n 1 Sn = 4n 1 2n (4 n 1) n(n 1) . 3 104 ARTHMETIC PROGESSION PROGRESSION AND SERIES ARITHMETIC PROGRESSION (A.P.) If a is the first term and d the common difference, the A.P. can be written as a, a+d, a+2d,... The nth term a n is given by a n=a+(n-1)d. The sum S n of the first n terms of such an A.P. is given by Sn n n ( 2a ( n 1)d) ( a l) where l is the last term (i.e., the nth term of the A.P.). 2 2 nth Term and Sum of n Terms : • If a fixed number is added (subtracted) to each term of given A.P. then the resulting sequence is also an A.P. with the same common difference as that of the given A.P. • If each term of an A.P. is multiplied by a fixed number (say k) (or divided by a nonzero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k. REMARKS : • If a 1, a 2, a 3,.... and b1, b2, b3,..... are two A.P.’s with common differences d and d' respectively then a 1+b1, a 2+b2, a 3+b3 ,.... is also an A.P. with common difference d+d'. • If we have to take three terms in an A.P., it is convenient to take them as a-d, a, a+d. In general, we take a-rd, a-(r-1)d,...... a-d, a, a+d,.... a+rd in case we have to take (2r+1) terms in an A.P. • If we have to take four terms, we take a-3d, a-d, a+d, a+3d. In general, we take a-(2r-1)d, a-(2r-3)d,..... a-d, a+d,....a+(2r-1)d, in case we have to take 2r terms in an A.P. • If a 1, a 2, a 3,.... a n are in A.P. then a 1+a n= a 2+a n-1= a 3+a n-2 = ..... and so on. • ar • If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) a r k a r k k , k n r, k 0 2 between the other two i.e., if a, b, c are in A.P., then b ac is the A.M. of a and c. 2 Arithmetic Mean(s): • If a 1, a 2,.... a n are n numbers, then the arithmetic mean (A) of these numbers is A . 1 (a 1 a 2 a 3 .....a n ) n The n numbers A1, A2,...... An are said to be A.M.’s between the numbers a and b if a, A1, A2,....., Anb are in A.P. A n a 105 n ( b a ) a nb . n 1 n 1 IIT- MATHS GEOMETRIC PROGRESSION (G.P.) nth Term and Sum of n Terms : If a is the first term and r the common ratio, then G.P. can be written as a, ar, ar 2,.... The nth term, a n, is given by a n = ar n-1. The sum S n of the first n terms of the G.P. is a (r n 1) Sn , r 1 r 1 na , r 1 If -1<r<1, then the sum of the infinite G.P. is a +ar + ar 2+..... a 1 r REMARKS : • If each term of a G.P. is multiplied (divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same ratio as that of the given G.P. • If each term of a G.P. (with common ratio r) is raised to the power k, then the resulting sequence is also a G.P. with common ratio r k . • If a 1, a 2, a 3, .... and b1, b2 , b3,.... are two G.P.’s with common ratios r and r respectively, then the sequence a 1b1, a 2b2, a 3b3.... is also a G.P. with common ratio rr'. • If we have to take three terms in G.P., it is convenient to take them as a/r, a, ar. In general, we take • a a , k 1 ,..., a , ar ,..., ar k in case we have to take (2k+1) terms in a G.P.. k r r If we have to take four terms in a G.P., it is convenient to take them as a/r 3, a/r, ar, ar 3. In general, we take a r 2 k 1 , a r 2 k 3 a ,....., , ar ,.....ar 2 k 1 , in case we have to take 2k terms r in a G.P. • If a 1, a 2,...., a n are in G.P., then a 1a n = a 2a n-1 = a 3a n-2 = ........ • If a 1, a 2, a 3,...... is a G.P. (each a 1 > 0), then loga 1, loga 2 , loga 3...... is an A.P. The converse is also true. Geometric Means : • If three terms are in G.P., then the middle term is called the geometric mean (G.M) between the two. So if a, b, c are in G.P., then b ac is the geometric mean of a and c. • If a 1, a 2,...... a n are non-zero positive numbers, then their GM. (G) is given G = (a 1a 2a 3 ......a n) 1/n • If G 1, G 2,..... G n are n geometric means, between a and b, then a, G 1, G 2,...., G n, b 106 ARTHMETIC PROGESSION b will be a G.P. G n a n 1 a n HARMONIC PROGRESSION (H.P.) 1 1 1 1 The sequence a 1, a 2, a 3,.....a n (a i 0) is said to be an H.P. if the sequence a , a , a ,.... a ,.... 1 2 3 n is an A.P. 1 1 1 1 The nth term, a n of the H.P. is a n a (n 1)d , where a a , and d a a . 1 2 1 NOTE : There is no formula for the sum of n terms of an H.P. • If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have 1 11 1 2ab H . H 2a b ab • If a 1, a 2,..... a n are ‘n’ non-zero numbers, then the harmonic mean H of these numbers is given by 1 1 1 1 ..... 1 H n a 1 a 2 a n • The n numbers H 1, H 2,....., H n are said to be n-harmonic means between a and b, 1 1 1 1 1 if a, H 1, H 2......, H n, b are in H.P. i.e. if a , H , H ..... H , b are in A.P.. 1 2 n 1 1 n (a b ) b na H n a (n 1)ab ab(n 1) ARITHMETIC - GEOMETRIC PROGRESSION : The sum of S n of first n terms of an A.G.P. is obtained in the following way : Sn = ab + (a + d)br + (a + 2d)br 2 +.....+ (a + (n - 2)d)br n-2 + (a + (n - 1)d)br n-1 Multiply both sides by r, so that rS n = abr + (a+d)br 2 +....+ (a + (n - 3)d)br n-2 + (a + (n - 2)d)br n-1 + (a + (n - 1)d)br n Subtracting, we get (1-r)S n=ab+dbr+dbr 2+.....+dbr n-2+dbr n-1-(a+(n-1)d)br n dbr (1 r n 1 ) ab (a (n 1)d )br n (1 r ) Sn 107 ab dbr (1 r n 1 ) (a (n 1)d )br n 1 r (1 r ) 2 1 r . n an a1 a 2 It can be shown that A* > G* > H*.a 2 2 ........ weighted Geometric mean (G*) and Weighted mean (H*) as A* m1a 1 ..a n n m1 m 2 . geometric mean H (G) and harmonic n 1 1 1 1 ... if -1 < x < 1 INEQUALITIES : Let a 1..x) -2 .. n ) 2 (sum of cubes of first n natural 4 numbers) • 1 + x + x2 + x3 + .....MATHS If -1 < r < 1. m n a n m m m .. m n 1 ( m1 m 2 . Then we defined weighted Arithmetic mean (A*). = (1 . n 3 n ( n 1)(2n 1) (sum of squares of the first n natural numbers) 6 n 2 (n 1) 2 (1 2 3 ...a n ) 1/n and It can be shown that A>G>H.IIT... a n be n positive real numbers and m1.... an a1 a 2 a 3 mean (H) as A a 1 a 2 ... n • 12 2 2 32 ....x) -1 ...a n be n positive real numbers. a 2....=a n . mn be n positive rational numbers.... Moreover equality holds at either place if and only if a 1=a 2=..... a 2...... n G=(a 1 a 2 .... Moreover equality holds at either place if and only if a1 = a 2 = .. Arithmetic Mean of mth Power : 108 . n 2 • 13 23 33 .. if -1 < x < 1 • 1 + 2x + 3x2 + . = (1 .... m n m1 m 2 m ......... Weighted Means : Let a 1... = a n ...... m n ) and H* m1 m 2 .. the sum of the infinite number of terms of the progression lim n Sn S ab dbr 1 r (1 r ) 2 MISCELLANEOUS PROGRESSION : Some Important Results : n ( n 1) (sum of the first n natural numbers) 2 • 1 2 3 ... a n ... m2. then we define their arithmetic mean (A)...... G* a 1 1 . 1}.ARTHMETIC PROGESSION Let a 1... c=x+y Substitute a=y+z.. + bn2) > (a 1b1+a 2b2+.. a 2. a n a a 2 . z=s-c then.. then n n m Obviously if m {0.. Proving Inequalities : (i) Any inequality has to be solved using a clever manipulation of the previous results.. a n 1 if m R [0..e... a + c > b Do the following : put x = s-a. b=x+z.... then m m m a 1 a 2 .. z > 0 Similar b + c > a = x > 0. b. which is generally easier to prove... (involving only positive reals without any other constraints). (ii) Any inequality involving the sides of a triangle can be reduced to an inequality involving only positive reals. a n a a 2 .... a n be n positive real numbers (not all equal) and let m be a real number. 2s > 2c OR s .... a n 1 n n m Cauchy’s Inequality : If a 1's and bis are reals... c to get an inequality involving x.... .. + a n2) (b12 + . 1] n n m m m a 1 a 2 . a n 1 However if m (0. a + c > b = y > 0 109 The inequality obtained after the substitution is easier to prove.... a n a a 2 .....c > 0. then (a 12 + .+a nbn) 2 . then m m m m a 1 a 2 ... x+y+z=3s-2s=s and a=y+z...1). y=s-b.. b + c > a... i. b=x+z. For the triangle we have the constraints a + b > c. c=x+y in the inequality involving a.e. y... z. Also note that the condition a+b>c is equivalent to a + b + c > 2c i. ......IIT. Let 5x k ==> a 5k 5 1 k2 2 k k 1 2 1 ==> a 5 k k 2 k k As the sum of a positive number and its reciprocal is always greater than or equal to 2.e........P. are in the ratio 5n +4 : 9n + 6. (A) 129 131 (B) 127 132 (C) 125 134 (D) 121 139 Solution : Ans : (A) Let be the first terms of two A.MATHS WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 The sum of n terms of two series in A..P. Find the ratio of their 13th terms.. n = 25 in equation (1) ==> a 12d 9 25 6 231 2 2 ILLUSTRATION : 02 At what values of parameter 'a' are there values of n such that the numbers : 51 x 51 x . n 1 d n a1 2a1 n 1 d1 5n 4 1 5n 4 2 2 ==> n ==> n 1 d 9n 6 2a2 n 1 d 2 9n 6 a2 2 2 2 . a x .. k k Hence a 5 2 2 ==> a 12 110 ... x 2 25 25 form an A. ? (A) a £ 8 (B) a £ 8 (C) a £ 12 (D) a £ 12 Ans : (C) Solution : For the given numbers to be in A.s and are their respective common differences. k 1 1 2 and k 2 2 2 . (1) a1 12d1 Now the ratio of 13th terms = a 12d 2 2 ==> put n 1 12 2 a1 12d1 5 25 4 129 i.P.P. .6 Find the sum of first n terms of the series : 1 1 (A) 2 2 n 1 n 2 1 1 (B) 4 n 1 n 2 1 1 (C) 2 n 1 n 2 1 1 (D) 4 2 n 1 n 2 Ans : (D) Solution : 1 1 1 1 Let S = 1..3 2...3...f(n+1).. Now if we increase the second digit of the required number by 2.. ILLUSTRATION : 04 Find the three digit number whose consecutive number form a G.4...2 2..2 n 1 n 2 ==> 1 1 S = 4 2 n 1 n 2 Note : You should observe that here........3 3... 1...5 ....... n n 1 n 1 n 2 1 1 2S = 1.. (A) 901 (B)931 (C)981 Ans : (B) Solution : Let the three digit be a...........5 4.. ar........4.3....... ar2 then according to hypothesis 100a + 10ar + ar2 +792 = 100ar2 + 10ar + a ==> a r 2 1 8 .3 2... n n 1 n 2 1 1 1 1 1 1 2S = 1..(1) 111 (D)991 .5.......3.4.... the resulting number will form an A.3 2.4 3.....5 .4 3.3 2..... n n 1 n 2 n 2 n 3 1 4 2 5 3 2S = 1....P...4 3. we get a number consisting of the same digits written in the reverse order...2. Tn 1 1 1 1 n n 1 n 2 2 n n 1 n 1 n 2 It is in the form f(n) ...ARTHMETIC PROGESSION ILLUSTRATION : 03 1 1 1 1 ....2...P.4 .2.... If we subtract 792 from this number.. then 2 ar 2 a ar 2 ==> a r 2 2r 1 4 ........ sum to n terms of the given series.... ar + 2.... = tan 1 2 tan 1 1 tan 1 3 tan 1 2 + tan 1 4 tan 1 3 ...... n cos ec 1 10 cos ec 1 50 cos ec 1 170 ..MATHS and a.. cos ec 1 1 (A) tan n 1 4 1 (B) tan n 1 4 2 1 n 2 2n 2 is.P... 3........................(2) Dividing (1) by (2)..........IIT. tan 1 n 1 tan 1 n 112 .... a r 2 1 8 Then a r 2 2r 1 4 r 1 r 1 2 2 r 1 ==> r=3 r 1 2 r 1 ==> from (1)..... 1 (C) tan n 1 2 1 (D) tan n 1 2 Ans : (A) Solution : n Let cos ec1 ==> 2 1 n 2 2n 2 n cosec2 = = n 2 2 1 n 2 2n 2 = 2 1 2n n 2 1 n 2 1 ==> cot2 = ==> tan n 2 n 1 = n 2 n 2 1 n 2 1 2n 1 2 n 1 1 2 n 1 n 1 n n 1 1 n 1 n 2 n 1 n 1 1 tan 1 tan n 1 tan n 1 n 1 n ==> Thus.. ILLUSTRATION : 05 Sum to n terms of the series. ar2 are in A..... 9 and so the required number is 931.. a =1 Thus digits are 1......... . log2b .log2b.P.log3c. a x / 2 and log b x are in G. log b a log x log b log b ==> x log a log b a log a log e a log a log e b ILLUSTRATION : 08 113 . b. as a is the greatest side. Also.angled (D) equilateral Ans : (B) Solution : We have b 2 ac and 2 log 2b log 3c = log a log 2b log 3c log a 2a 4a and 2b 3c ==> b and c 3 9 ==> b 2 ac since a b 5a 10a 13a b . ABC is an obtuse-angled triangle.log b x x ==> a log a log x log a .and loga . therefore are the sides of a > c. and sides of a triangle which is (A) acute .ARTHMETIC PROGESSION = tan 1 n 1 tan 1 1 tan 1 n 1 4 ILLUSTRATION : 06 If a. c are in G. then are the lengths of the (C) right .P. a x / 2 . then is equal to (A) log a log b a (B) log a log e a log a log e b (C) log a log a b (D) log a log e b log a log e a Ans : (a) and (b) Solution : As log x a. log b x are in G.angles are in A. cos A b 2 c 2 a 2 29 0 2bc 48 Hence. . ILLUSTRATION : 07 If log x a. let us find angle A of ABC. x/2 2 a log x a. b c >a and c a 3 9 9 triangle.P..angled (B) obtuse .P. then the value of a b c 1is independent of (A) a (B) b (C) c (D) None of these Ans : (D) Solution : As are in G.P. a b are in H.P.. a b are in H. According to the given condition Sn k n 1 S 2n S n S1 S2 ==> S S S S 2 1 4 2 ==> S1 S4 . b and c. ILLUSTRATION : 09 An A. ac 2 1 1 = b c a b c a b c a b ==> 2 b c a b a c ==> 2 ab ac b 2 bc = ==> 2 ab 2b 2 bc ==> 2b a c 2 2 2 a c a c a c 2 2 a c a c a c 2 2 2 a c 0 ==> 2b a c 2 ac a c 2b ==> a b c 3 ac Which is not independent of a.S1S2 = -S1S2 114 . c a. the common difference d is given by (a) 1 (b) 2 (c) 3 (d) 4 Ans: (B) Solution : Let denote the sum to n terms of the A. c a.MATHS If there unequal positive real numbers a.P.P. c are in G. b.P.P. and b c.. b 2 ac and b c.IIT. whose first term is unity and in which the sum of the first half of any even number of terms to that of the second half of the same number of terms is in constant ratio. for n 1. ILLUSTRATION : 10 n If t r 1 r n 1 1 n n 1 n 2 n 1 . t r r 1 2 r r 1 r n n 1 1 1 1 2 ==> t r r 1 2 1 n 1 r 1 r r 1 n 1 1 lim 2 1 ==> lim 2 1 0 2 z n n n 1 r 1 t r 115 (c) 3/2 (d) 6 . for r 1 . n 1 n t n t r t r r 1 r 1 1 1 n n 1 n 2 n 1 n n 1 6 6 1 1 n n 1 n 2 n 1 n n 1 6 2 1 2 1 1 Now. we get d = 2.ARTHMETIC PROGESSION ==> S1S 4 S22 2 4 ==> a 2a 4 1 d a a d 2 ==> a 4a 6d 2a d 2 ==> 4a 2 6ac 4a 2 4ad d 2 ==> 2ad d 2 ==> 2a d As a =1. Then lim is n 6 r 1 t r (a) 2 (b) 3 Ans : (A) Solution : We have. D) z2 = x 1 1 1 .P D) none of these n1 .care in 2 2 2 B) G.S3 B) 9 C) 16 D) 13 If the arithmetic mean of two positive numbers a and b (a > b) is twice their geometric mean then a : b is A) 2 + 3 :2- C) 6+ 7 :6- 3 7 B) 5 + 6 : 5- D) 1 + 2 :1.r2 are in A.P A) 2 6.S2.1/r in A. C) S1.n 2 is equal to n1 n2 n3 A) 0 8. C) z-3 = y The first term of an A.r3 are in A. Its pth term is A) m +n+p 2. B) 1 10.n 3 + 2 n 3 .P then a A) A. 2b-x. C) (p+1)3 B) p2.r are in A.(a2+b2)x . are in A. .b.q3. n3 t erms respectively of an A.q2.a B) ab( b .P If S 1.P C) 1/p.P A.S 2 . D) n1n2n3 If the interior angles of a polygon are in A.P 7. The mth term of an A.a ) C) a (b-a) D) ab(a-b) 116 . B) m+n-p B) x = y -1 If B) (2p+1)(p+1)2 D) p3 + (p+1)3 D) p3. If the sum of all the terms is 5 times the sum of the terms occupying odd places.P consists of an even number of terms. D) m-n-p If 1. the common ratio will be B) 3 If b-c.IIT.P 5.P is n and its nth term is m. log y x .S 3 denote the sums of C) H. C) m-n+p C) 4 D) 5 x x x .2 6 If m is a root of the equation (1-ab)x2 .q.P then A) z3 = x 3.P of consecutive integers is p2 + 1.n1 + 3 n1 . log z y -15log x z are in A.MATHS SECTION A SINGLE ANSWER TYPE QUESTIONS 1.(1+ab) = 0 and m harmonic means are inserted between a and b then the difference between the last and the first of the means equals A) b .G.P then q+r r+p p+q A) p. 1/q.P then s1 s s n 2 . n2 . The sum of (2p +1) terms of this series can be expressed as A) (p+1)2 4. b-a are in H.P with common difference 50 and smallest angle is 1200 then the number of sides of the polygon is A) 9 or 16 9. B) 1 If (0.b.2100+1 C)99.P C) G. 3 3 2 D) . x2.c are in A) A. 18.299 is A) 100.c. bcd are B) A.P then a1a 2 + a 2 a 3 + a 3a 4 + . C) 2n 1 mn B) 1 1 m n C) 1 D) 0 If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the square of their reciprocals then a b c .d be in A. . then Tmn is A) 13. 3 3 Sum of the series 1+2.P If a. then the minimum value of is A) 1/4 16.P. + a n-1a n is equal to A) n 12. B) (n-1) B) Arithmetic progression D) Harmonic progression Suppose a.P/H. are in c a b A) Arithmetic ..Geometric progression C) Geometric progression 14.P 15.P 19.2 + 3..2100-1 . to = 43 then S = z 2 B) . a n are in H. abd.d are +ive real no such that a+b+c+d =2 then M = (a+b) (c+d) satisfies the relation B) 1 M 2 C) 2 M 3 D) 3 M 4 Let S (0. . ) denote the set of values of x satisfying the equations 81 |cosx|+ cos2x + |cos3x| +.P D) H. C) H.b >0 and x1 . x3 (x1 > x2 > x3) are the roots of x-a x-b b a + = + b a x -a x -b and then a. D) (n-2) Let Tr be te rth term of an A.. D) none of these An infinite G. B) G.299-1 D)100..P has first term 'x' and sum then x belongs to A) x < . acd. ..P/G.b.P B)99. 3 3 2 C) .c.. If a1 . n we have Tm= 1/n and Tn = 1/m.22 + 4. C) 3/4 x2 + x - A) 0 M 1 20.10 17. .+100.. . a 2 a 3 . .b.ARTHMETIC PROGESSION 11.. B) 1 D) does not exist tan 2 α x2 + x C) 0 < x < 10 D) x > 10 is always greater than equal to D) sec2 C) 2 Let the positive numbers a. π /2) then A) Not in A.3.23 +.2. Then abc.10 < x < 0 A) 2 tan A) 3 21.. If for some positive integers m.P If is the nth term of a G.2100+1 117 B) ..P with first term 1 and common ratio r.P for r =1. P D) are in H. an n log n A) 2 b 27. 28.be y b . If the sum of first n terms of a series be 5n 2 + 2n . The roots of the equation A) form an A.to If 2 = 1 22 32 6 2 A) 8 31. 2cotq. a n 1 n log n B) 2 b If 2 B) 12 C) are in G..4 | x -1 | +3 = 0 C) 1/2 B) are in A.. sec .P 1 1 1 π2 + + + . to =8.P.c are in arithmetic progression then A)A.IIT.d are in a ... iff ? is equal to 2 2 A) /6 25.P. then its second term is A) 16 24. B) -1 x -1 D) do not form any progression 2 . . B) G.P.b.MATHS 22. C) Form a H. ..P. 23. B) G.c..P. 0 < θ < are in G. 2 D) None of these c A 3b If in a Δ ABC acos2 +ccos2 = then the sides a. then n is equal to a a +b b A) 0 29.P..P.de y A) A.b. If a.ce y c .P. n+1 n+1 . C) H. 1 1 1 .P then equals : 2 C) 3 2 D) 2 a + be y b + ce y c + de y = = then. x -1 1 a n + bn 1 If ..c 2 2 2 A) satisfy a+b=c 30.. If B) 17 D) None of these C) 27/14 D) 56/15 1 π cosec 2 q. are in : a+ b a+ c b+ c If 3 + B) /4 C) /3 D) None of these 1 1 3 + d + 2 3 + 2d + . then the value of is 4 4 A) 9 B) 5 C) 1 D) None of these a n 1 n log n1 C) 2 b a n 1 n log n 1 D) 2 b ar log r-1 = The value of r=1 b n 26. C) H. are in A.P.4 | x -1 | +3 = 0 B) form a G.. a.b. D) None of the above 118 .P.P. C) a=2 b=32 1 D) ab (a .(a2 + b2) x . b.(1+ab) = 0 and m harmonic means are inserted between a and b then the difference between the last and the first of the means equals A) b .b) 1 1 {an} and {bn} are two sequences given by an = x 2n + y 2n and bn = x 2n .cosx 3 C) 2 .. .d be the root of x2-12x + q= 0 where a. B) a=12 b=3 B) 11 : 10 C) 5 1 .P then the sines of the actute angles are A) 3/5.P only when x > 0 D) G. 1/ 3 D) 3 1 . B) 5 C) 6 The sum of all the products of first n positive integers taken two at a time is A) 1 n 1 n n 1 3n 2 24 B) C) 1 n n 1 n 2 n 5 6 D) None of the above B) G.P then A) a=3 b=12 37.3 B) ab (b -a) D) 1 + C) a (b ..8x + 7 = 0 α .P C) G. be roots of x2-3x+p=0 and Let c. D) 1 : 2 If for 0 < x < /2 y = exp [ sin 2 x sin 4 x . If f x is a function satisfying f x + y = f x f y for all x. then the value of n is and x =1 A) 4 33. D) a=4 b=16 Let a..16x + 5 = 0 C) 5x2-16x + 7 =0 D) 5x2 . 4/5 39.a) 1 119 3 1 2 3 If m is a root of the equation (1-ab)x2 . c are in A. form an increasing G.P.P only when x < 0 If the A. Then the ratio of p+q : q . 1 n 2 n n 1 n 2 48 If a.d form an increasing G.ARTHMETIC PROGESSION 32.a 41. β be roots of the equation x2 . 34. 2 B) 2+ sinx + cosx is sinx .P for all x 35. D) None of these B) 7x2 . . C) 17 : 15 If the sides of a right angles triangle are in A.b.p is equal to A) 8 : 7 38.b.c.12x + b = 0 and numbers . y N ..P then 10ax+10. then the quadratic equation is A) 7x2 + 16x +5 = 0 36. such that f 1 = 3 n f x = 120.M of their reciprocals is 8/7. ] is a zero of the quadratic equation x2 -9x + 8 =0 then the value of 40. 10bx+10.M of the roots of a quadratic equation is 8/5 and the A. 2 5 1 2 A) 0 B) 3 . y 2n for all n = .3x + a =0 and λ .10cx+10 are in A) A. δ be roots of x2 . a2 . If a < b < c and a + b + c = 3/2 then the value of a is A) 44.c2 are in G.r 2 r +1 x + r 5 =0 The value of 3α r + 2β r is r=1 120 . .3 C) -6.3 D) -6.P.d.M of two numbers is twice of their geometric mean. .P D) none of these If the A.e are in G.[a] are in A. a 20 .c are in A. ..f ax is G.h20[a] are in H. β be the roots of x2 . D) none of these B) G.b) (c. xy D) b n B) 20/9 C) 9/20 x2 - D) 11/23 Suppose a.h1.f) are the vertices of a triangle such that a.P and a2. δ be the roots of x2 .P then the value of the determinant is A) -2 45.} loge16 0 < x < /4..4x + q =0. 46.c. B) 12 B) 3 / 2 C) 4 D) 2/ 3 Let α . δ are in G. -32 B) -2.h2.P C) H. x y B) b n 1 2 2 B) 1 C) 2 3 1 1 2 3 D) 1 1 2 2 If .b..P with common ratio 's' then area of the triangle is A) ab r 1 ( s 2)( s r ) 2 B) ab r 1 s 1 s r 2 C) ab r 1 s 1 s r 2 D) r 1 s 1 s r The roots of equation x2 + 2(a-3) + 9= 0 lie between -6 and 1 and 2. β r α r < β r are the roots of x 2 .tan4x+tan6x . .P. satisfies the quadratic equation 3x + 2 = 0 the value of cos2x + cos4x is equal to A) 21/16 43. C) 18 If a.an is equal to A) x . . where [a] denotes the integral part of a and 2.MATHS N. x y C) b n If exp {tan2x . then ratio of sum of numbers to the difference of numbers equals A) 2 49.b. B) 1 C) 2 D) 0 If (a.IIT. b. -32 n 50.y 42.2(ab+bc+cd)p + (b2+c2+d2) 0 then a. are in G.x + p = 0 and λ. c. α r .P. ..d and p are distinct real numbers such that (a2+b2+c2)p2 .c.tan8 x +. β .b2.P 48. then the integral values of p and q respectively are A) -2. . λ. The value of a1.. If α .d) (e.P then a3h18 = A) 6 47. d are in A) A.P with common ratio 'r' and b. . a1a 2 . are in A.P.. ... . and when 1 1 1 1 1 1 1 1 5 .c. 3 n n 1 n 2 n 1 2 2 1 bc b 2 If for 0 < x < A) 0 C) 12 B) 1 3 2 1 4 c 2 ca a 2 C) 3 2 2 b ab π 2 4 6 . is x x x x 20 1 x 22 1 A) x 2 1 x 20 20 x18 1 x11 1 B) x 2 1 x 9 20 x18 1 x11 1 C) x 2 1 x9 20 D) None of these When are in A.. D) None of the above If a. . 1 n n 1 n 2 n 3 2 B) 10 equation then the value of the value of 54.100) is A) 5050 57.cosx B) 2 3 C) 2 3 D) none of these B) a b c a C) 3ac D) 3bd B) bx 2 c C) cx 2 d B) 5000 C) -5050 2 58...b.c are in H... y = exp is a zero of the quadratic sin x + sin x + sin x + . + + = then the numbers a and b a x y z b x y z 3 are : A) 8. 25 + 25 are three consecutive terms of an A. + ..ARTHMETIC PROGESSION 1 2 A) n n 1 n 3n 1 2 C) 51.. to log2 2 sinx + cosx is : sinx .3 D) none of these .d are in G.c. D) None of these If a. B) D) -5000 2 2 1 2 1 3 1 The sum of the first 10 terms of the series x + + x + 2 + x + 3 + . . D) x a The coefficient of x99in the expansion of (x .d are in H.P..1 C) 7..P.. D) 1 1 1 1 1 1 If a ...2 121 B) 9. then a factor of ax 3 + bx 2 + cx + d is A) ax 2 c 56.P. (x . 1 n n 1 3n 2 n 1 2 1+x 1-x a x -x The least value of 'a' for which 5 + 5 . is b c a c a b A) 53.... then the value of + .2) .P. then ab + bc + cd is A) 3ad 55. is 2 : A) 5 52.b..1) (x . b.P. then the common difference will be A) 1 62.t are in A) A. B) 11:10 The sum of the cubes of first n natural numbers does not exceed 1360.G2 be two G. .b. B) 8 B) 2 D) 4 C) 3a2 D) 4a2 r r 1 B) 2 x r r 1 C) 2 x xy r r 1 D) 2 x rx B) second term D) last term If the numbers a.IIT.c.. then a1 + a 2 + .α3 .P. B) 75 B) (a+b)(c+d) A1 +A2 If A1.d.P then the value of a .d form in increasing G. C) 750 D) 900 C) 3ac D) 3cd If a.Ms between a and b then G G is equal to 1 2 A) a b 2ab B) 2ab a b C) a b ab D) a b ab 122 .. A second square is formed by joining the middle points of this square.MATHS 59.z..P C) H.. C) 9 n A series whose nth terms is +y then sum of r terms will be x r r 1 A) 2 x ry 64. C) 17 :15 If the roots of the equation are in A. 2 B) G.P then x.P D) none of these If α1 .b.P then ab+bc+cd is equal to A) 3ad 69. D) 10 In an A.e form an A.P 67. D) none of these The length of a side of a square is 'a' metre. .c.P. A2 be two A.y. 2 Let a and b be the roots of x . .3x + p = 0 and let c and d be the roots of x -12x + q = 0.4b + 6c .b.Π such that a1 + a 5 + a10 + a15 + a 20 + a 24 = 225. the maximum value of n is A) 7 61.P the sum of terms equidistant from the beginning and end is equal to A) 1 66.c. Then the common ratio of q + p : q .c. Then a third square is formed by joining the middle points of the second square and so on.d are in G.b.d are in H. C) 3 B) 2a2 A) first term C) sum of 1st and last terms 65.Α.4d + e is B) 2 C) 0 D) 3 If ax = by = cz = dt and a. Then the sum of the area of the squares which carried upto infinity is A) a2 63.α2 . ..p is A) 8:7 60. where a. a 23 + a 24 equal to A) 909 68.Ms and G1. P B) G.z then x . . . .an are in H.h10 be in H..P C) H. D) 6 If the system of linear equations x + 2ay + az = 0 .x . then the minimum value of a1 + a 2 + a 3 + .ARTHMETIC PROGESSION 70.. C) H.c A) satisfy a+2b+3c =0 C) are in G. D) none of these 10 a j=1 79. .P. + a +. a ..a.1 " n³1 then lim x®¥ A) 3 73. C) 5 If B) 2 C) 3 D) 4 are the +ve real numbers where product is a fixed number c. . If a1 = h1 = 2 and a10 = h10 = 3 then a4 h7 is equal to A) 2 76.a . If Skx/Sx is independent of x then A) a = 2d 72. a10 be in A... a n -1 +2an is 1/ n A) n 2c 123 B) n 1 c1/ n C) 2nc1/ n D) 1/ n n 1 2c . y . + a + a + a +.a are in A) A. .. then 1 2 21 A) 361 B) 396 a 2i+1 is i=0 C) 363 D) data insufficient If the roots of the equation x3-12x2+39x-28=0 are in A.. ..a are in A.P B) a = 2.P with first term 'a' and common difference 'd' Let Sk denotes the sum of the first k terms. C) 2a = d n A) 8/5 74..P 71. D) none of these consider an A.P then a + a +.P 21 78.P is 2 and its common difference is greater than one. r = 3/7 75.P... .M between y . fourth and fifth terms is greatest is B) 2/3 C) 5/8 D) 3/2 Consider an infinite geometric series with first term 'a' and common ratio 'r'. .a2. B) a = d t r = 2 3 n . B) G.. If j = 693 where a . then their common difference will be A) 1 80.P D) are in H... are in 2 3 n 1 3 n 1 2 n A) A. r =1/2 D) a=3.. x + 4cy + cz = 0 has a non-zero solution then a.. + a + a + a +. The value of the common difference of the progrenion so that the product of the first. r = 3/8 C) a=3/2. . y . B) 3 B) are in A. a1 a2 a3 If a1.a3.z then x . x + 3by + bz = 0 .P If 2(y-a) is the H..P.b.. .. If its sum is 4 and the second term is 3/4 then A) a = 7/4. ..P 77.P and h1 h2 . y . r = 1/4 If a1a 2 .a. z ..P n If r=1 r=1 B) 3/2 D) a = 3d 1 is equal to tr C) 3/4 D) 3/8 The sixth term of an A. . ...P. = 2 4 8 16 C) n + 2-n . is equal to the sum of the first n terms of the A. c. C)15 B) G.2-n D) p3 + (p + 1)3 1 3 7 15 + + + + . The sum of A) x 2 n2 x + 2 n-1 x 1 n + x + 2 C) H.81.MATHS If the sum of the first 2n terms of the A..a1b1 = 0. whose last term is l. 2b and 3c form an A..+ cosec an-1 cosec an ) is ..a n are in Arithmetic series with common difference 'd' ..P.59.. of common ratio r with 0 < r < 1. b. D) 13 If the lines a1y +b1x .. 2.61.... of consecutive integers is p2 +1.P. a2 y + b2 x = a2 b2 cut the co-ordinate axes in co cyclic points then a1. B) 10 l 2 a2 B) 2S a l l 2 a2 C) 2 S a l D) None of these If ax3 + bx2 + cx + d is divisible by ax2 + c then a. C) (p + 1)3 B) 1/3 C) 2/3 D) None of these The geometric and harmonic means of two numbers x1 and x2 are 18 and 16 8 respectively..P. + x +1 x 1 n-1 is equal to : n 1 124 .6 D) 4.P C) G. D) 20 If S denotes the sum of first n terms of the A..5... B) 12 B) H... b2 a2 may be in A) A.The value of sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + . Their arithmetic mean is A and geometric mean is G.. n-2 n-3 D) None of these x +1 + x + 2 x +1 B) x 2 n 1 2 + ..P..... B) 1 . a 2 . 91... is la A) n 90.P 83. If G satisfies 2A + G2 = 27. IIT.... A) sec a1 sec an 85. 57. Then n equals A) 10 82.P.P. b.... then r equals : A) 1/2 88...P. The 13 value of | x1 .8...1 Let a. d are in A) A...P. C) 11 B) 9.1 D) 2n .13 84... If a..P D) none of these The harmonic mean of 2 numbers is 4.n + 1 87. b1. the numbers are A) 1.... then the common difference 'd' of the A. B) cos eca1 cos ecan C) tan a1 tan an D) cot a1 cot an The first term of an A..8 If a1 .. a 3 . B) (2p + 1) (p + 1)2 The sum of the first n terms of the series A) 2n . a + (a + d) + (a + 2d) +.. The sum of (2p+1) term of this series can be expressed as: A) (p+1)2 86..x 2 | is A) 5 89.12 C) 3.P. c form a G. I3 . ..P C) H. are in G. .P 94.M. .P.25 1 . 6n n 1 B) 9n n 1 If In = B) log 34 π k cot 2k y= q then k=0 C) p + q + pq D) p+q+pq 3 5 7 +. then A) n. . D) n 1 1 1 If x > 1 and . c are in x x x A) A. are in 1 .P A) n digits 4 10n 1 9 B) 4 102n 1 9 C) 2 4 10n 1 9 Given that 0 < x < ?/4. a.8) is equal to n digits 95.x + 2 and log9 (4. a c b B) G. log 9 3 A) log 4 3 98..(n-1)3 + .. then I1 .M. -1 D) is 1 1 1 A) p q pq 96. Let A denote the single A. then a. and S denote the sum of n A.P 99...+ (-1)n-1 13 is equal to 2 A) 125 k=0 -1 A) A. k tan 2k x= P 4 n 10 2 9 1. b a 93. .P C) H..ARTHMETIC PROGESSION n n C) x 2 x 1 92. Given two numbers a and b. I 2 . 2 + 2 2 + 2 1 1 +2 1 + 22 + 32 C) 12n n 1 D) 15n n 1 C) 1 .cos2x 0 B) G. . D)None of the above n 1 2n 1 4 n 2 2n 1 B) 4 2 C) n 1 2n 1 4 D) n 1 2n 1 4 2 . 6)2 + (88. D) none of these (666.. b C) n.3x -1 ) are in A. . ?/4 < y < ?/2 and tan 2k x cot 2k y k=0 1 B) 1 1 1 p q pq The sum of the first n terms of the series A) 97.'s between a and b. then x is equal to If 1.P.log3 4 D) log 3 0. S depends on : A B) n..sin2nx dx.P D) none of these For any odd integer n 1 n3 . b.. 3 C) -6.b. The sum of first n terms of the series 1 +2. .P C) D) then B) G. -32 is C) 0 D) 1 are in : C) H.P.P. If a. d are in H.6 + .P and f(k) = A) A. b.16 =0 B) x2 . Let B) nbe the roots of x2 . If in H. -32 B) -2.c and d are in G. then B) G.. If cos (x . .P D) none of these 104. D) -6.n + 1 D) 2-n + n -1 101.1 B) 1 ..P. is 2 2 2 2 2 2 2 2 when n is even.P then cosx sec y/2 is equal to A) ± B) C) ±2 D) none of these 105. .18x .IIT.P.y). 0. The sum of first n terms of the series A) 2n .222 + .2 + 0. are in G.4 + 5 + 2. A)-1 108. are in A.22 + 0. then : A) a + b > c +d B) a +c > b +d 126 .16 =0 D) x2 + 18x + 16 =0 n n +1 102. If a. 3 107.c. .P. . is 2 4 8 16 100. If B) 2 . to n terms is equal to A) 106. .2 + 3 + 2.x + p = 0 and C) D) be the roots of x2 .18x + 16 =0 C) x2 + 18x . cosx. cos (x +y) are in H. then integral values of p and q respectively are : A) -2.4x + q = 0 If .P are in C) H. .b. . Let two numbers have arithmetic mean 9 and geometric mean 4. A) A. Then these numbers are the roots of the quadratic equation A) x2 .P.P.MATHS 1 3 7 15 + + + + . D) None of these C) a + d > b +c D) none of the above 109. When n is odd the sum is n n 1 A) 2 2 B) 103. .2-n C) 2-n . the common ratio will be equal to A) 2 B) 3 C)4 112. P D) H.P then f--1(a). consists of an even number of terms. for which A) 113. then A) a = 2d B) a =d C) 2a = d denote the sum of first k D) None of these 111.P. Let f(x) be a polynomial function of second degree if f(1)=f(-1) and a. If is independent of x. with common ratio r. A G. if the sum of all the terms is 5 times the sum of the terms occupying odd places. the value of r for holds is given by A) 1< r < 3 B) -3 < r < -1 C) r >3 or r <1 D) none of the above 115. c are in A.P 127 B) A.Geometric progression C) G.ARTHMETIC PROGESSION 110. f1(b).P.P . The coefficient of D)5 and 1 are in A.P with some common ratio then (x1y1) (x2y2) (x3y3) A) lie on a straight line B) lie on an ellipse C) lie on circle D) are vertices of triangle 116.. Let terms. If x1 x2 x3 as well as y1 y2 y3 are in G. with first term 'a' and common difference 'd'.P.P. b. is B) C) D) in the polynomial given by is A) 5511 114. f1(c) are in A) Arithmetic . Consider an A. The value of x. If which B) 5151 C) 1515 D) 1155 are three consecutive terms of a G. d are distinct positive numbers then the inequality an + an > bn + cn holds for every positive integer n provided a. r are in H. 4.P. then A) x. then the ratio p : q : r is A) A.B MULTIPLE ANSWER TYPE QUESTIONS 1. e.P. B) C) 2dbf = aef + cde 3. b. If tan-1x. are equal and their nth terms are a... f are in G. c. B) x.P.P.P. q >1 then A) Tp+q = pq 6.P. then A) 8. then A) a = b = c 7. . D) (x-y)2 + (y-z)2 + (z-x)2 = 0 If d.. c be three unequal positive quantities in H.P is q (p+q) and qth term Tq is p (p+q) when p > 1.P. and their squares are in A. y.P. z are also in A.. D) b2 df = ace2 If three unequal numbers p. are in G. then A) are in H.P. B) G. B) C) The sum of n terms of the series is A) B) C) D) Given that 0 < x < /4 and /4 < y < D) /2 and = a..P and x.P.P 5. D) increasing progression B) Tpq = p+q C) Tp+q > Tpq D) Tpq > Tp+q B) a + c = b C) a b c D) ac = b2 If a. and a H.. and the two quadratic equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root.. z are in H. b.P If the first and (2n-1) th terms of an A. z are in G. c. C) x = y = 1 2. d are in 9. y. (y being not equal to 0. y. a G. b and c respectively. 1 or -1). A) 1 - : -2 : 1 + B) 1 : - :- C) 1 - : D) 1 + :-2:1- If a. q.P The pth term Tp of an H.IIT. b.MATHS SECTION .P. tan-1z are in A.. then 128 . C) H.P. tan-1y. P. If 1.P.3b1 holds is given by A) r > 3 C) r = 3.2 If logx a. ..loga (logea) If a. ax/2 and logb x are in GP. A) B) a+b-ab C) D) If a.. B) r < 1 D) r = 5. B) C) aD) 129 are in G. b.. then A) are in H.. b3 (b1 > 0) are three successive terms of a G.loge (logeb) D) loga (loge b) . then A) z3 = x C) z -3 = y 13.. c are in H.ARTHMETIC PROGESSION 10. then x is equal to A) loga (logba) C) -loga (loga b) 15. c are in H. logyx. are in H..P.5 14. n2 B) x = y-1 D) x = y-1 = z3 I f b1. then the value of is A) B) C) D) none of these A) B) C) D) 11. the value of r for which the inequality b3 > 4b2 .. b2.P. logzy. with common ratio r.15 logxz are in A. b.P. B) loga (logea) . 12.P.P. . log9 (31-x+2) and log3(4.3x-1) are in A.75) 130 .. then B) b = 8/3 C) c = 9/2 D) e = 0 I f 1.MATHS 16.log3 4 D) log3 (0. If A) a = 1/2 17. then x is equal to A) log43 B) log3 4 C) 1 .IIT.P. ARTHMETIC PROGESSION 131 . IIT.MATHS 3 BASIC TRIGONOMETERY 132 . . . thus tracing out the angle MOP. i. Perp. OM Base Hyp.e. MP FUNDAMENTAL RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS OF AN ANGLE It is clear from the definitions that if one of the trigonometric ratios of an angle is known. . OM Base P O M Base OM . . of previous article) be . OP Base OM .e. Perp. i.e.. MP OP Hyp. is called the Cotangent of the angle AOP. The trigonometrical ratios. i. . . the numerical magnitude of each of the others is also known. . PM is the perpendicular. cot = cos . then 133 sin2 + cos2 = 1 or sin2 =1 –cos2 or cos2 = 1 – sin2 1 + tan2 = sec2 or sec2 – tan2 = 1 1 + cot2 = cosec2 or cosec2 – cot2 = 1 tan = sin . MP . .. sec = 1 It is possible to express a trigonometrical ratio in terms of any one of the other ratios: sin cos and cot cos sin . OP is the hypotenuse. Hyp. of the angle MOP are defined as follows. and OM is the base.e. is called the Tangent of the angle AOP.. OP MP Perp. or functions.e. i. is called the Sine of the angle AOP. OP . is called the Secant of the angle AOP. Perp. Let the angle MOP (fig. Draw PM perpendicular to the initial line OM. is called the Cosine of the angle AOP.e.. i. is called the Cosecant of the angle AOP. i. In the triangle MOP.. Hyp.BASIC TRIGONOMETRY DEFINITION OF TRIGONOMETRIC RATIOS Let a revolving line OP starts from OM and revolves into the position OP. cosec = tan . . g. when it coincides with OB. and. the distance OM2 is negative and increases numerically from 0 to a [i. Let the revolving line OP be of constant length a. the distance OM4 increases from 0 to a. cos 1 cot cos ec 1 cot 2 . in the second quadrant. whilst in the fourth quadrant M4 P4 increases algebraically from –a to 0. and. the point M1 coincides with O and OM1 vanishes. Therefore it is clear that 134 ..e. sec cot 2 . as the angle increases from 0° to 360°. all trigonometrical functions have been expressed in terms of cot .e. This is left as an exercise for you to derive other results of this type. In the first quadrant.MATHS 1 e. the length OM1 is equal to a. the distance OM3 increases algebraically from –a to 0. Also. M2P2 decreases from a to 0..IIT. in the fourth quadrant. in the third quadrant. the length M1P1 increases from 0 to a. as the revolving line turns from OA to OB. M3P3 decreases algebraically from 0 to –a. When it coincides with OA. B P1 P2 A M2 M1 M3 O P3 M4 A P4 B In the third quadrant. SIGNS OF TRIGONOMETRIC RATIOS Tracing of the changes in the sign and magnitude of the trigonometrical ratios of an angle. sin 2 . Whilst the revolving line is in the second quadrant and is revolving from OB to OA . the distance OM1 decreases from a to zero. tan 1 cot 1 cot 1 cot 2 cot i. it decreases algebraically from 0 to –a]. . i. and cosecant go through all their changes as the angle increases by 2 . and thus the sine goes through all its changes. the period of the tangent and cotangent is radians. GRAPHS OF THE TRIGONOMETRIC RATIOS The variations in the values of the trigonometric ratios may be graphically represented to the eye by means of curves constructed in the following manner. however. cosine. are same.e. the cosine. Since the values of the trigonometrical functions repeat over and over again as the angle increases. goes through all its changes as the angle increases from 0 to radians.BASIC TRIGONOMETRY B i i A A O i i B PERIODS OF THE TRIGONOMETRICAL FUNCTIONS As an angle increases from 0 to 2 radians. returning to its original value. secant. The period of the sine. as the angle increases from 2 radians to 4 radians. whilst the revolving line makes a complete revolution.e. i. Similarly. Also. i. whilst the revolving line turns through two right angles. Similarly.. its sine first increases from 0 to 1. the sine goes through the same series of changes. 135 . they are called periodic functions. Similarly for the cotangent. 2 radians. The tangent. This is expressed by saying that the period of the sine function is 2 . and finally increases from –1 to 0. then decreases from 1 to – 1. the sines of any two angles which differ by four right angles.. secant and cosecant is therefore 2 radians.e. 136 .IIT.MATHS Y Sine-Graph: / 2 1 O /2 2 X –1 Y 1 O / 2 Cosine-Graph : / 2 3 / 2 2 X –1 Y Tangent-Graph: / 2 O 3 / 2X /2 Cosecant-Graph: Y 2 X 3 2 1 O –1 2 2 X The secant-graph and the cotangent-graph are left as an exercise to the students. the supplement of any angle is 180° – . 2 2 The trigonometric functions of these angles can be expressed as trigonometric functions of . where n is any integer. 137 . ] TRIGONOMETRIC RATIOS FOR AN ANGLE OF ANY MAGNITUDE Complementary Angles Two angles are said to be complementary when their sum is equal to a right angle. i. Allied or Related Angles 1 1 n and n .e. 1. i. are known as allied or related angles. sine becomes cosine. Supplementary angles Two angles are said to be supplementary when their sum is equal to two right angles. note the quadrant in which the given angle lies. tangent becomes cotangent. But if n is odd. The angles with either a plus or a minus sign.. If n is even. 2. Assuming that 0 < < 90°. the result contains the corresponding cofunction..BASIC TRIGONOMETRY TRIGONOMETRIC RATIOS OF SOME ANGLES Angle 0° 30° 45° 60° 1 2 1 2 1 2 3 2 1 2 3 2 1 2 3 3 1 3 2 3 1 3 2 3 3 2 1 3 3 2 2 3 2 90° 120° 135° 1 2 1 2 150° 180° 1 2 3 2 1 3 3 2 2 2 3 Note: Later on we shall learn that infact tan(90°–) = and tan(90°+) = – etc. secant becomes cosecant and vice-versa.e. the result contains the same trigonometric function as the given function. The result has a plus or minus sign according as the given function is positive or negative in that quadrant. The following working rules can be used in determining these functions. Thus any angle and the angle 90° – are complementary.. tan (45° + A) = 1 tan A tan A tan B tan(A–B) = 1 tan A tan B . tan A tan B 1 tan A tan(A + B) = 1 tan A tan B .MATHS P2 P1 90°– +ve A –ve O +ve +ve A –ve –ve P3 P4 Here angle AOP1 = angle AOP4 (measured clockwise) = – angle AOP2 = 180° – angle OP1A = 90° – and similarly the other angles. cot A cot B 1 cot A cot B 1 cot (A + B) = cot A cot B . Some formulae and results regarding compound angles: sin (A + B) = sin A cosB + cosA sinB sin(A – B) = sinA cosB – cos A sinB cos (A + B) = cosA cos B – sinA sin B cos(A – B) = cosA cosB + sin A sin B. cot (A – B) = cot B cot A sin(A + B) sin(A – B) = sin2A – sin2B = cos2B – cos2A 1 tan A tan (45° – A) = 1 tan A 138 .IIT. It is clear from the figure that when equals sin cos tan cot sec cosec – –sin cos –tan –cot sec –cosec 90°– cos sin cot tan cosec sec 90° + cos –sin –cot –tan –cosec sec 180° – sin –cos –tan –cot –sec cosec 180° + –sin –cos tan cot –sec –cosec 360° – –sin cos –tan –cot sec –cosec 360° + sin cos tan cot sec cosec TRIGONOMETRIC RATIOS OF COMPOUND ANGLES An angle made up of the algebraic sum of two or more angles is called compound angle. 1 tan 2 A 2 2 2 2 1 + cos2A = 2cos2A. tan A tan B tan C tan A tan B tan C tan (A + B + C) = 1 tan A tan B tan B tan C tan C tan A . acos + bsin = r(cos sin + sin cos ) = rsin ( ) Now the maximum and minimum values of sin( ) are 1 & – 1 respectively.. 1 – cos2A = 2sin2A 2 tan A 1 tan 2 A tan2A sin3A = 3sinA – 4sin3A = 4sin(60° – A) sinA sin(60° + A) cos3A = 4 cos3A – 3cosA = 4cos(60°–A) cosA(cos60°+A) 3 tan A tan 3 A tan 3A = tan(60°–A) tanA tan(60°+A) 1 3 tan 2 A PRODUCT OF SINES/COSINES IN TERM OF SUMS 2 sinA cosB = sin (A + B) – sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos (A + B) SUM OF SINES/COSINES IN TERM OF PRODUCTS 139 sinC + sinD = 2sin CD CD cos 2 2 . TRIGONOMETRIC RATIOS OF MULTIPLES OF AN ANGLE 2 tan A 1 tan 2 A sin2A = 2sinA cosA = 1 tan 2 A cos2A = cos A –sin A = 1 – 2 sin A = 2 cos A–1 = .BASIC TRIGONOMETRY cos(A + B) cos(A – B) = cos2A – sin2B = cos2B –sin2A. Hence – r rsin( ) r – a 2 b 2 a cos b sin a 2 b 2 Hence the maximum value = a2 b2 and minimum value is a 2 b 2 .MAXIMUM AND MINIMUM VALUES OF acos + bsin Let a = rsin . b = rcos so that r = a2 b2 also. otherwise A A cos | 1 sin A 2 2 | sin or A A sin cos 2 2 tan A 5 2n ve.MATHS sinC – sinD = 2 cos CD CD sin 2 2 cosC + cosD = 2cos CD CD cos 2 2 cosC – cosD = –2sin CD CD sin 2 2 tanA + tanB = sin ( A B) sin ( A B) . tanA – tanB = cos A cos B cos A cos B TRIGONOMETRIC RATIO OF SUBMULTIPLE OF AN ANGLE | sin A A cos | 1 sin A 2 2 or sin A 3 A A ve.IIT. if 2n 1 sin A 4 2 4 ve. if 2 n 2n cos 1 sin A 4 2 4 2 2 ve. otherwise A tan 2 A 1 1 2 tan A The ambiguities of signs are removed by locating the quadrant in which A lies or one can use the 2 following figure. sin A A cos is ve 2 2 sin sin sin A A cos is ve 2 2 4 A A cos is ve 2 2 sin sin A A cos is ve 2 2 |a cosA + bsinA| Also cosA sinA = sin sin A A cos is ve 2 2 A A cos is ve 2 2 A A cos is ve 2 2 A A cos is ve 2 2 a2 b2 2 sin 4 A = 2 cos A 4 140 . A given identity may be established by (i) reducing either side to the other one. then sin (B + C) = sinA. or (iii) any convenient modification of the methods given in (i) & (ii). sin cos 2 2 2 2 tan BC A B CA cot . CONDITIONAL IDENTITIES When the angles A. B and C satisfy a given relation. if A + B + C = . C are angles of a triangle (or A + B + C = ): 141 tanA + tanB + tanC = tanA tanB tanC cotA cotB + cotB cotC + cotC cotA = 1 tan A B B C C A tan + tan tan + tan tan 1 2 2 2 2 2 2 cot A B C A B C cot cot cot cot cot 2 2 2 2 2 2 sin2A + sin2B + sin2C = 4sinA sinB sinC cos2A + cos2B + cos2C = –1 – 4cosA cosB cosC sinA + sinB + sinC = 4cos cosA + cosB + cosC = 1 + 4 sin A B C cos cos 2 2 2 A B C sin sin 2 2 2 .BASIC TRIGONOMETRY IDENTITIES A trigonometric equation is an identity if it is true for all values of the angle or angles involved. many interesting identities can be established connecting the trigonometric functions of these angles. we require the properties of complementary and supplementary angles. or (ii) reducing each side to the same expression. cotA = –cot(B + C) cos sin CA B A BC cos . tan cot 2 2 2 2 AB C C AB sin . B. In providing these identities. cos sin 2 2 2 2 Some important identities: If A. For example. cosB = –cos (C + A) cos (A + B) = –cosC. sinC = sin(A + B) tan (C + A) = tanB. .. +cos{ ( n 1) }= sin 2 142 .. + sin{ ( n 1) }= sin 2 n 2 (n 1) cos sin 2 2 cos +cos( )+cos( 2 )+ .MATHS TWO SIMPLE TRIGONOMETRICAL SERIES n 2 (n 1) sin sin 2 2 sin + sin( )+sin ( 2 ) + ..IIT. . then tan 2 cos 2 (b) 2 (c) 3 (d) tan is equal to 2 1 Z 3 Ans : (C) Solution : From the given relation we have 2cos 1 2 cos 2 cos 1 1 cos = 1 2 cos 2 cos or 2cos 2 1 cos 2 2cos 2 2 2 cos 1 2sin 2 2 or cos 2 2 cos 2 2 1 2sin 2 ..BASIC TRIGONOMETRY WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 1 sin 3 cos3 is equal to sin cos sin cos (a) sin 2 (b) cos 2 (c) sin cos (d) sin 2 Ans : (C) Solution : We can write the given expression as sin cos sin 2 cos 2 sin cos sin 3 cos3 1 1 sin cos sin cos =1. (1) 2 cos 2 1 2sin 2 cos2 2 2 2 1 cos 2 1 ==> 2 1 2sin 2 1 2sin 2 2 2 143 ... .......( 1 sin cos ) = sin cos ILLUSTRATION : 02 If cos (a) 1 2cos 1 0 . .. then + is equal to a2 b2 a3 b3 1 a b 1 (b) a b 2 1 (c) 3 a b (d) a b Ans : (C) Solution : sin 4 cos 4 k Let a2 b2 144 ...IIT. then Pn Pn 2 kPn 4 where (a) k 1 (b) k sin 2 cos 2 (c) k sin 2 (d) k cos 2 Ans : (B) Solution : We have Pn Pn 2 cos n sin n cos n 2 sin n 2 = cos n 2 cos 2 1 sin n 2 sin 2 1 = sin 2 cos n 2 cos 2 sin n 2 = sin 2 cos 2 cos n 4 sin n 4 = sin 2 cos 2 Pn 4 kPn 4 where k = sin 2 cos 2 ILLUSTRATION : 04 If (a) sin 4 cos4 sin 8 cos8 ... (2) From (1) and (2) we get tan 2 3 2 2 tan 3 tan ==> 2 2 tan 2 ILLUSTRATION : 03 If Pn cos n sin n ..MATHS 3sin 2 2 sin 2 or 2 1 2sin 2 2 ..... tan .(3/4) sin3A (d) 0 Ans : (C) Solution : The given expression is equal to 1 0 0 0 0 3sin A sin 3 A 3sin 120 A sin 360 3 A 3sin 240 A sin 720 3 A 4 3 1 0 0 0 0 = sin A sin 180 60 A sin 180 60 A sin 3 A sin 3 A sin 3 A 4 4 = 3 3 0 0 sin A sin 60 A sin 60 A sin 3 A 4 4 3 3 0 = sin A 2 cos 60 sin A sin 3 A 4 4 3 = sin 3 A 4 ILLUSTRATION : 06 If tan .BASIC TRIGONOMETRY k then 1 sin 2 cos 2 a k b k ==> 1 a b 2 sin 8 cos8 k 2 a 4 k 2b 4 1 2 so that a 3 b3 a3 b3 k a b a b 3 ILLUSTRATION : 05 sin 3 A sin 3 1200 A sin 3 240 0 A is equal to (a) (9/4) sinA (b) (3/4) sin3A(c) . tan are the roots of the equation x 3 px 2 r 0 then the value of (1 + tan?) (1 + tan2 ) (1 + tan2 ) is equal to 2 (a) p r (b) 1 p r 2 (c) 1 p r 2 Ans : (B) Solution : From the given equation we have tan tan 0 and tan p tan tan tan r so that 1 tan 2 1 tan 2 1 tan 2 145 (d) None of these . MATHS = 1 tan 2 tan 2 tan 2 tan 2 tan 2 tan 2 =1 + tan 2 2 tan tan tan tan 2 2 tan tan tan tan tan 2 tan 2 tan 2 = 1 + p 2 2 pr r 2 = 1 p r 2 ILLUSTRATION : 07 If A and B be acute positive angles satisfying 3sin 2 A 2sin 2 B 1.IIT.tan2BtanA =0 3sin 2 A ==> A+2B = A ==> B 2 4 2 ILLUSTRATION : 08 If (a) a c cos x cos x cos x 2 cos x 3 then is equal to b d a b c d a d (b) c d (c) b c (d) d a Ans : (C) Solution : 146 . 3sin 2 A 2sin 2B 0 then (a) B A 4 2 (b) B A 4 2 (c) B A 2 4 (d) A B 4 2 Ans : (A) Solution : From the given relations we have 3 Sin2B = sin 2 A 2 and 3sin 2 A 1 2sin 2 cos 2 B 3 sin 2 A so that tan 2 B 2 cot A or 1. then is equals (a) tan A B tan A B tan C (b) tan A B tan A B tan C (c) tan A B tan A B sin C (d) tan A B tan A B cos C 2 2 2 2 2 2 Ans : (B) Solution : cos A sin C cos B sin C cos A cos B sin C sin C ==> cos A cos B sin C sin C ==> tan A B AB tan cot C tan 2 2 ==> tan tan 147 A B A B tan tan C 2 2 2 2 2 2 2 . sec 2 becomes indeterminant. ILLUSTRATION : 10 cos A sin C If cos B sin C .BASIC TRIGONOMETRY cos x cos x 2 2 cos x cos a c b = b d cos x cos x 3 = 2 cos x 2 cos c ILLUSTRATION : 09 sec 2 4 xy x y (a) x y 0 2 is true if and only if (c) x y (b) x y . y 0 Ans : (C) Solution : 2 2 Since sec 2 1 so 4 xy x y x y 0 x y Which is true if and only if x y 0 . x 0 (d) x 0. because for x y 0 . ...cos2x is A) 2 8..... C) 1/8 cos2A If f(x) = cos2x + sec2x its value always is A) f(x) < 1 6...p 2. 2 .. n A) 1/2n/2 and cot 1.. C) p 2 p 2 B) 1/3 and 3 The maximum value of cos 1..MATHS SECTION A SINGLE ANSWER TYPE QUESTIONS 1....sin 20° = P the cos40° is equal to A) .. D) p2 =q2 (q2 .. 3 .. n2x ) is 4 C) n n 1 2n 1 3 D) n n 1 2n 1 C) 1/4 and 4 D) 1/6 and 6 The value of tan3 cot can't lie between A) 1/2 and 2 11. D) 4cos2A C) p2 + q2 = 2q B) f(x) = 1 The value of cos A) 0 7.. If cos20° . cos 2 cos 3..cos n under the restriction 0 1 . Cos n = 1 is 2 B) 1/2n C) -1/2n D) 1 148 .. C) 2 < f(x) < 1 π 3π 5π 7π 9π + cos + cos + cos + cos 11 11 11 11 11 12 6 A) n 2n 1 n 1 B) n 2n 1 n 1 10.IIT.. D) 1/2 If sec and cosec are the roots of x2-px+q=0 then A) p2 = q (q-2) 5. D) none of these If tan2 .....2 tan2 +1 then cos2 + sin2 is equal to A) -1 3. B) -1/2 D) f(x) 2 is C) 1/2 D) 1 B) 1 C) 3 D) 1 C) /2 D) /4 sin2x is 3cos4x B) The period of sin (x + 4x + 9x+ . C) 1 sin 2 3A cos 2 3A is equal to sin 2 A cos 2 A A) cos2A 4.4) Maximum value of 2sin2x . cot 2 .. 2 p2 B) p 2 p 2 B) 0 B) 8 cos2A B) p2 = q (q+2) The period of A) 2 9... x and tan 9 18 A) 2x = y 13. of roots of equation x sinx = 1 in the interval 0 < x < 2 π is A) 32 21. then (A + 2B) is equal to B) /4 C) /2 D) /6 The number of solutions of the equation |sinx| = |cos3x| in [-2 .tan2x = 1 then the value of tan4x . y and tan 9 18 π 3π 5π 7π 1 + cos 1 + cos 1 + cos 1+ cos is equal to 8 8 8 8 A) 1/2 14. 1/3] 17. D) x = 2y 2π π 2π π . D) The no.3 tan tan . are in A.tan2x + 2tanx + 1 is A) n 23. 1 2 2 2 Let ABC be a acute angled triangle such that A = π /3 and cot B cos C = P. C) ± 3/4 B) 1 A) 0 22. B) x > y tan D) ± 1/5 B) (0.P.1] C) [1/3 ) D) [1 ) C) 2 D) 4 B) 2 C) 3 D) 4 If 3sin2A + 2sin2B = 1 and 3sin2A . The possible values of P will be A) /3 20. C) 1 B) ± 1/2 A) 1 19. If tan .2sin2B = 0 where A and B are acute angles. 2 ] is B) 28 1 1 2 + + 4 4 2 sec α cosec α sec α + cosec 2 α B) 1 C) 24 D) 30 C) sin2 D) cos2 = If sin α + cosec α = 2 then sinn α + cosecn α = B) 2n C) 0 D) 2 If x2 = 1 . C) 1/8 B) 1/ 3 A) (0.tan2 θ then tan3 θ cosec θ + sin θ = A) 149 3 If tan x .tan is equal to 5 15 5 15 A) ± 1/4 16. π 7π . and tan 2 1/ 2 2 x B) x 2 3/ 2 2 3/ 2 C) 2 x 2 D) 2 5/ 2 2 x . D) If sin ( π cos θ ) = cos ( π sin θ ). then C) x = y B) cos /8 A) 3 15. are also in A.P.BASIC TRIGONOMETRY π 5π 12.2tan2x . then sin2 θ is equal to A) 0 18. D) /2 2sinθ 1 . the value of the expression tanθ -1 + tanθ 1 . then 1 1 A) m . B) n-1 For 0 < < π if . D) None 3(sinx . then cos θ 1 + cos θ 2 + cos θ 3 +.tanθ .cosA sinA B) tanA C) cosA D) sinA n 25.. Z = n =0 n=0 n =0 B) xyz = xy+z cos 2n φsin 2n φ . C) 2n B) m 0.. If x = B) /6 B) cos B) zero C) x D) 13 B) a2 + 2ab . C) x tan 2 θ 1 If 0 < θ < π /2.cosθ + sinθ then = 1 + cosθ + sinθ 1 + sinθ A) sin 30. x = 2 A) xyz = xz + y 27. 2] 1 D) m ... If sinθ i=1 i = n . b.. The value of the expression 1 A) 1 1. 1 The smallest value of θ in the first quadrant which satisfies the equation exp {(1+cos2 θ +cos4 θ +cos6 θ +.+cos θ n = A) n 26..MATHS 24.secθcosecθ is A) 1 31. 1 4 If sin6 α + cos6 a = m. then C) xyz = x+y+z D) xy2 = y2+x C) m [-2. )loge16} = 256 is A) /4 29.b2=0 If (1-sinA)(1-sinB)(1-sinC)=(1+sinA)(1+sinB)(1+sinC) then each equal side is equal to B) 1 C) sinAsinBsinC D) cosAcosBcosC If cosec θ + cot θ = 11/2.tanB C) 44/117 2 1/ 2 D) 117/44 is equal to 150 . 6 6 28.cos 2 A sinA 1 + cosA + is equal to 1 + cosA 1 . D) zero ¥ ¥ ¥ cos 2n φ. c is A) zero 34.cosx)4 + 6(sinx + cosx)2 + 4 sin 6 x + cos 6 x is A) sinx 32. then tan θ is equal to A) 21/22 35. C) /3 1 + tanAtanB B) 15/16 2 + tanA ..b2 =0 C) a2 + 2ac + b2=0 D) a2 + 2ac ... C) -1 B) cosx A) a2 + 2ab + b2 =0 33.IIT. y = sin 2n φ. D) None If sin and cos are roots of equation ax2 + bx + c =0 then the relation-ship between a. 4 4 3 m 2 1 The number of integral values of k for which the equation 7cosx + 5sinx = 2k+1 has a solution is A) ( /6. D) 1/2 If A.P. C) -1 If sin α + cos α = m then sin6 α + cos6 α is equal to A) 38.θ )/ sin (c + θ ) then tan tan is 2 2 A) 4 43.m 1+ m C) 1/64 π π then tan . 4 2 If cos6 α + sin6 α + k sin22 α = 1 then k is equal to A) cot c cot 41.BASIC TRIGONOMETRY A) tan2A + tan2B 36.α tan . C are acute positive angles such that A +B + C = and cot A cot B cot C = k then A) k 39. /6) C) ( /6 /4) D) ( /2.β 4 4 B) 1 + m 13π is equal to 14 D) 1/32 is equal to C) 2m D) 2 + m If A and B are acute angles such that A+B and A-B satisfy the equation tan2 . B) cos2Acos2B C) secAsecB D) tanAtanB If tan2 α tan2 β + tan2 β tan2 γ + tan2 γ tan2 α + 2tan2 α tan2 β tan2 γ = 1 then the value of sin2 α + sin2 β + sin2 γ is A) 0 37. B. 4 3 m 2 1 2 B) 4 1 3 3 151 D) C) k < 1 9 2 4 D) k > 1/3 C) 1/2 D) 1 C) tan c tan D) tan c cot B) 2 C) 1/ 2 D) 2 B) 8 C) 10 D) 12 The value of sin π /14 sin 3 π /14 sin 5 π /14 sin 7 π /4 sin 9 π /14 sin11 1 π /4 sin If sin α + β cos α . then cos α sec β /2 be A) m 45. sec ( α + β ) are in A.sin4x + cos2x + α 2 + a = 0 will have at least one solution if A) -2 2 47.β ). 3 3 B) cot c tan A) 1 44. 2 If sec( α .4tan +1=0 then (A. sec α . /6) 46. C) 3 4 m 2 1 A+B A -B If cos A/ cos B = sin (c . B) 1 B) -3 1 C) 2 1 D) 1 2 The number of points inside or on the circle x2+y2=4 satisfying tan4x + cot4x + 1 = 3sin2y . 4 3 m 2 1 B) k A) 1/4 40.β B) 1/16 = 1. /6) The equation cos4x . B) be B) ( /4. 1 B) 3/4 A) 2 42. β satisfies both the equations cos2x + acosx + b= 0 and sin2x + p sinx + q = 0 then relation between a.IIT.MATHS A) one 48.b 57. tan θ 2. D) a2+b2 = p+q π sin 2π .A) 2 A) cosA 52. then a2/b2 = A) 1 55.sin = 0 then tan θ1 + θ 2 + θ3 + θ 4 is equal to A) sin 49. tan θ 3.3x + a =0. then the minimum value of tan2 α + tan2 β + tan2 γ = A) zero 54. p and q is A) 1+b+a2 = p2-q-1 50. D) 3 If a sin2 α + b cos2 β = C.sin θ ) (sec θ . tan θ 4 are roots of the equation x4-x3 sin 2 + x2 cos2 . B) 2 C) 3 B) cosec A D) 4 C) sin A D) tan A (cosec θ . b.x cos .cos θ ) (tan θ + cot θ ) is equal to A) 4 cos sin 53. C) 2 B) b + a C) a b 2 D) a b 2 4 4 sin 8 θ cos 8 θ + = If sin θ + cos θ = 1 .A cos 2π + A sin(π . B) cos C) tan D) cot If x = α . B) a2+b2 = p2+q2 B) 4 sec tan C) 4 cosec cot D) 1 If α.A 2 is equal to π sin . bsin2 β + acos2 β = d and atan α . B) two C) four D) infinite I f tan θ 1. C) If 0 < A < /2 and sinA+cosA+tanA+cotA+secA+cosecA=7. then a3 b3 a b a+b 1 A) 1 8 a b B) a b 1 4 C) 3 a b D) 1 a b 152 .b tan β . then A) a=1 56. B) 1 d a a c B) b d c b B) a=0 If p = a b D) None bc C) a > 1 D) a > 0 a 2 cos 2 θ + b 2 sin 2 θ + a 2 sin 2 θ + b 2 cos 2 θ then the maximum value of P is A) a . C) 2(b+q) = a2+p2-2 The number of solutions of the equation 1 + x2 + 2x sin (cos-1y) = 0 is A) 1 51.Α cos π + A tan . β and λ are variables subject to the relation 2tan α + 3tan β + 6 tan γ = 7. and sinA and cosA are the roots of the equation 4x2 . D) 12 If 3cosx = 2cos (x .y) tan y is equal to A) 1/5 67. cos (120-x) cos (120+x) is A) -1/4 65.. then for all values of A) 1 A 2 68. a d c a b c d b B) B) 2 d a c a b c d b C) C) B) -1/2 2 B) n n 1 B) . sin α = and tan β = the value of sin ( β . B) /3 21 221 B) C) tan + 2tan D) 2tan + tan 3π 15 12 .13/18 If A + B = π /4 then (tan A +1) (tan B +1) is equal to A) 1 64... D) /2 The minimum value of 9tan2 θ + 4cot2 θ is A) 13 61. C) 6 The period of tan (x + 2x + 3x + .α ) is 2 17 5 21 221 C) 171 221 D) 171 221 . B) 9 The minimum value of cosx.. C) /6 If α is the root of 25cos2 θ + 5cos θ .5 B) 13 A1 16 B) tan + tan If /2 < α < π .. B) tan A tan B < 1 If A = cos2 +sin4 .2y) then tan (x . C) 13/18 If a sin2 x + b cos2x = c . C > 0) and the angle C is obtuse then A) tan A tan B > 1 62.24/25 If A+B+C = (A. bsin2y + a cos2y = d and a tanx = b tany then b c d b A) a d c a 63. π /2 < α < π then sin 2 α is equal to A) 24/25 60. D) . B) . + nx) is A) n n 1 66.12 = 0. B.1/5 C) 3 13 A 4 16 D) 3 A 1 4 If + = /2 and + γ = then tan equals A) 2 (tan + tan ) 69. π < β < A) 153 C) tan A + tan B = 1 D) tan A tan B = 0 a2 is equal to b2 D) b c b d a c a d 3 D) 1/ 3 C) 1 D) -1 C) 2 n 1 D) 2n n 1 C) 5 D) ..BASIC TRIGONOMETRY 58. If A and B are positive acute angles satisfying the equations 3cos2A + 2cos2B = 4 and 3sinA 2cosB = sinB cosA then A + 2B is equal to A) /4 59... 4cos4 +8cos2 = A) 3 83. then value of x satisfying f(x) where xε 0. cos B = tan C. III could lie in quadrant C) III.9 sin 4 θ + cos 4 θ + 15 is equal to B) 10 C) 12 D) . 2 2 2 A) 2 cos 80. /2 A) 1 81. then cos6 . C) 2cos18° If sin x + sin2x = 1 then the value of cos12 x + 3cos10 x + 3cos8x + cos6x . B) 5 76. then |sinx-cosx| = 154 . D) 2cos36° The value of y for which the equation 4sinx+3cosx=y2 . IV B) 8 C) 10 D) 12 If cosec .1 is equal to A) 0 74.70. C) [ /2.12 If sin + sin2 +sin3 =1. B) 2sin18° If A and B are acute positive angles satisfying the equation 3sin2A + 2sin2B = 1 and 3sin2A 2sin2B = 0 then A +2B = A) /4 73. IV D) I.sinA = sinA / 2 . C) /3 B) 5 A) 6 78.MATHS The number of values of x in the interval [0. cos C = tan A. D) If [cosx] + [sinx + 1] = 0.3 B) (0. sec .6y + 14 has a real solution is/are A) I.sin = m. 2 ] If A π 1.cosA / 2 then 2 . C) .5 D) None 6 sin 6 θ + cos 6 θ .4 D) 1 B) II. .1 2 + 2 + 2 + 2cos4θ B) 2cos 2 C) 2cos /2 D) 2sin /2 If sinx + sin2x = 1. B) /2 B) 1 A) 3 75. then the value of cos2x + cos4x is B) 2 C) 1. 2 ] are A) -10 82. then cos2 A is equal to A) sin 18° 72. C) 6 B) -3 C) 4 D) -4 If sinx+cosx = a. D) 2 Total number of roots of the equation 3cosx = |sinx| belongs to [-2 . IIT. C) -1 3 A) x . 2 ] D) x [ . II 77. 5 ] satisfying the equation 3sin x-7sinx+2=0 2 A) 0 71. 2π A) 0 79. D) 10 If cos A = tan B.cos = n then (m2n) 2/3 + (mn2) 2/3 = If < /8 the value of B) 1 C) 2 D) . ] U [3 /2. 16x + 15 < 0 and cos β equals to the slope of the bisector of the first and quadrant. D) 4/5 B) a2 + (b .3 m 2 -1 if 4 B) m R A) /7 88. C) m If f(x) = 3(sinx-cosx)4 + 6(sinx+cosx)2 + 4(sin6x+cos6x) and g(13) = π /7 then (gof)x = A) x =1 89. D) None If x = sin2 θ + cos2 θ +tan2 θ +cosec2 θ +sec2 θ +cot2 θ . C) x = y 0 B) sin +cos D) None 2 cos If tan α equals to the integral solution of the inequality 4x2 . z = γ cos then x2 + y2 + z2 = A) sin2 B) sin2 D) 2 C) cos2 4xy 85. 2 For all real values of x and y.β ) is equal to A) 3/5 93. /4) 155 C) 2 / 5 1 If maximum value of |a sin2 θ + b sin θ cos θ +c cos2 θ (a+c)| is k/2 then k2 is equal to 2 A) b2 + (a -c)2 94. y = bsin3 θ .BASIC TRIGONOMETRY A) 84. /4 /4) . /2) D) (. then sin ( α + β ) sin ( α .sin . then A) a + b 90.cosα .b C) a/b D) 2 ab If x = a cos3. then sinα + cosα A) sin -cos = 2 sin C) sin = 2 (sin +cos ) 92. B) .3/5 B) (0. B) 2 a2 C) 2a D) 2 a2 2a If x = γ sin . B) x y sin θ + cos θ = m. then a C) x 2/3 2/3 If tan θ = y b b y 2/3 x B) b 1 2/3 2/3 b D) x 1 y a 2/3 2/3 a y 1 2/3 1 sinα . then sin6 θ +cos6 θ = A) m R 87. B) 2 /7 D) -2 /7 B) x ? 4 C) x ? 6 D) None B) a .b )2 D) b2 + (a +c)2 For what value of α lying between 0 and π is the inequality sin α cos3 α > sin3 α cos α valid A) (0. C) .c )2 C) c2 + (a . D) x = y = 0 2 1 4 . 0. /7 The minimum value of a2tan2? + b2cot2 θ is x A) a 91.cos . the equation sec θ = x + y 2 is possible when A) x = y 86. /2) C) ( /4. y = γ sin . 4 .7 has a solution A) 2 a 6 96. cosec + θ + x cos θ cot + θ = sin + θ then x = 2 2 2 156 .cot2A is equal to A) 1 B) -1 C) 1/2 D) -1/2 C) -1/2 D) 0 100. then cot6A . 3 B) 3.P. cos A.16x . then tan and tan ( π /4.e -cos2 x + 4 = 0 in [0. D) 1 a 6 If sinx cos y = 1/4 and 3 tanx = 4 tany then sin(x+y) A) 5/16 97.MATHS 95. Then tan and tan ( π /4 . B) 1 a 5 B) 7/16 C) ( B) ( / 2. then the quadratic equation whose root are cosC. a straight line drawn from the mid point of one equal sides to the opposite angle. C) 1/16 Given both θ and are acute angles and sin = 1/2 cos = 1/3 then the value of + belongs to A) ( / 3 / 2] 98.5 / 6] 3 D) ( 5 .y + a will have no solution in x and y if a belongs to A) 0.4x |sin | . tan A are in G.cos2 is A) -2 B) -1 101. The equation of sinx (sinx + cosx) = k has real solutions then A) 0 k 1 2 2 C) 0 k 2 3 B) 2 3 k 2 3 D) 1 2 1 2 k 2 2 π π π 105.5 = 0 and 5cos B + 3 = 0. C) a 5 B) 39x2 + 88x + 48 = 0 D) 39x2 + 16x + 48 = 01 If sinA. 2 / 3] D) 4/15 2 . ] 6 If ABCD is a cyclic quadrilateral such that 12 tan A .48 = 0 C) 39x2 . 3 4 B) 1 1 . It divides the angle into two parts and ( π /4 . 4 5 C) 1 1 . 2 3 104. ). ). 5 6 D) 1 1 . In an isosceles right angled triangle. ) are equal to A) 1 1 .88x + 48 = 0 99. The values of a for which the equation cos2x + a sin x = 2a . tanD is A) 39x2 . 0 C) 1 D) 2 . sinx + siny = y2 . The number of solution of equation e cos 2 x . Minimum value of 4x2 . 2 π ] A) 1 B) 2 C) 3 D) none of these 102. 3 103.IIT. If x = acos2 θ sin θ . then A) a sin B) a2sin x 2 B) -2cos 4a 2 x 2 x2 y 2 3 x 2 y2 = C) a2sin2 3π 110. 0 < α.cosθ 1 + cosθ + 1+ cosθ 1. 3 sin 2 2 A) 0 B) 1 C) 3 D) 2 107.α + sin 4 3π + α . β < then 2 A) cos + cos = C) sin + sin = 4ax 2 x y2 B) cos cos = 4ay 2 x y2 D) sin sin = 109.BASIC TRIGONOMETRY A) cot B) sin C) tan D) cos 4 3π π . If cosecA = cosecBcosecC + cotBcotC. then cosecB = A) cosecAcosecC+cotAcotC C) cosecAcosecC ? cotAcotC 157 D) a2(sin2 +cos2 ) B) cosecAcosecC . then the expression is equal to 2 A) +2sin + y2 4a 2 y 2 x2 y 2 1 . If π < θ < .2 sin 6 + α + sin 6 5π . If x cos α + y sin β = x cos β + y sin β = 2a .α = 106.cotAcotC D) None D) -2cosec . (sin α + cosec α )2 + (cos α + sec α )2 = k + tan2 α + cot2 α then k = A) 9 B) 7 C) 5 D) 3 π 108. y = a sin2 θ cos θ .cosθ C) +2sec 111. 2 1 1 C) a . Let Pn (u) be a polynomial in u of degree n. then 5 13 A) cos ( + ) = 33 65 1 C) sin2 2 65 6.3 cot (2 )] = 6 B) sin = cos ( + ) sin C) 2 sin = sin ( + ) co s D) tan ( + ) = 2 sin 4. ) = 63 65 The equation sin6x + cos6x = a2 has real solutions if A) a (-1.MATHS SECTION . If sin + sin = a and cos θ + cos = b. If 3 sin β = sin (2 α + β ) then A) [cot + cot ( + )] [cot .tan2 B) cos2 .1 2 If tan α and tan β are the roots of the equation x2 + px + q = 0 (p 0).sin2 C) 3(sin2 + cos4 ) . sin 2nx is expressible is A) P2n (sinx) B) P2n (cos x) C) cos x P2n-1 (sinx) D) sin x P2n-1 (cos x) If cos α = 3 5 and cos β = . 2 2 1 D) a . 5.IIT. B) sin ( + ) = 1 B) a 1. then 158 . 56 65 D) cos ( . For any real x one has A) cos (cos x) > sin (sin x) C) cos (sin (cos x)) > sin (cos (sin x)) B) cos (sin x) > sin (cos x) D) cos (cos (cosx)) > sin (sin (sinx)) 7 2. for every positive integer n.2(sin6 + cos6 ) D) none of these 3. 1) 7. then A) sin2 ( + ) + p sin ( + ) cos ( + ) + q cos2 ( + ) = q B) tan ( + ) = p/q -1 C) cos ( + ) = 1-q D) sin ( + )=-p 8. 2 0 2 0 2 0 I f sec 20 + sec 40 + sec 80 = a and tan k =1 2 kπ = b then the value of a-b can be expressed as 16 A) sec2 . Then.B SINGLE ANSWER TYPE QUESTIONS 1. then the values of tanA. Then 2 Let fn ( θ ) = tan A) f2 =1 16 B) f3 =1 32 C) f4 =1 64 D) f5 =1 128 If (secA + tanA) (secB + tanB) (secC + tanC) = (secA-tanA) (secB-tanB) (secC-tanC) B)-1 C) 0 ¥ For 0 < /2 if A) xyz = xz + y If tanx = D) none of these ¥ 2n x = cos φ. (a c) a -c y = a cos2x + 2b sinx cos x + c sin2x z = a sin2x . 3 C) 1. 3 159 (n.BASIC TRIGONOMETRY 1 A) cos 2 2 a 2 B) cos 2 b2 4 a2 b2 2 2 C) tan 2 a b 9. 2. B) (a2+b2 ) cos = 2ab D) sin = 2. 2. then A) y = z B) y + z = a+c n 14. (1+sec2n θ ). 1. a ¥ sin 2n φ z= n =0 B) xyz = xy + z cos 2n φsin 2n φ .sinB cosA . 11..tanB and tanC are A) 1. then n =0 C) xyz = x+y+z D) xyz = yz+x C)y-z = a-c D) y-z = (a-c)2 + 4b2 2b . b. AB B) 2cotn 2 C) 0 D) none of these B) cot 760 C) tan 460 D) cot 440 In a triangle tanA + tanB + tanC = 6 and tanA tanB = 2. b2 Which of the following statements are possible a. n B) 2.2b sinx cosx + c cos2x. m and n being non-zero real numbers ? A) 1 12.375 θ (1+sec θ ) (1+sec2 θ ) (1+sec 4 θ ) .cosB AB A) 2tann 2 15. 2 a2 b2 2 D) cos 2 A) 4 sin2 = 5 C) m 2 n 2 cos ec m 2 n 2 10. y = n =0 13. 0 D) none of these . cosA + cosB sinA + sinB + sinA ... even or odd) = 3 + cot76 0 cot16 0 = cot76 0 + cot16 0 A) tan 160 16. IIT. 19.MATHS 17. B) sin tan = y 2 sin B) sin + cos = 2 cos D) sin2 + cos2 = 0 Let 0 /2 and x = X cos + Y sin . b = 3 B) = /4 C) x = 3.cosα then sinα + cosα A) sin -cos = C) cos2 = sin2 20.(xy ) = 1 A) sinq cosq = 18. Then A) a = -1. then 1 x 2 2/3 2 2/3 C) (x y) . where a. b = -1 D) θ = π /3 160 .cos θ = y. If cot θ + tan θ = x and sec θ . y = X sin -Y cos such that x2 + 4xy + y2 = aX2 + bY2. If D) (x2y)1/3 + (xy2)1/3 = 1 x cosA = where A B then y cosB A B x tan A y tan B A) tan x y 2 A B x tan A y tan B B) tan x y 2 sin A B y sin A x sin B C) sin A B y sin A x sin B D) x cosA + y cosB = 0 If tan θ = sinα . b are constants. 8. D C 29. 10. 23 24.BASIC TRIGONOMETRY SECTION . 22. 11. 3. 30. 12. 26. 18.A SINGLE ANSWER TYPE QUESTIONS 1. A C 27. 25. 9. A C D C C B D C C D B. 17.C D A C A 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 D D D C C B A A B B B B C A B 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 C B D C A B D B C C B C D B D 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 B B B A B A D C D C B B A A 161 A . 28. 20. 13. 4. 6. 15. 7. B B B B D C C B A B A 16. 2. 19. 5. 14. 21. MATHS 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 A B B C A C C C D A C A D A 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 A. 14. AB AD ABC CD BCD BD AB ACD BD 16 17 18 4. 2. 11.IIT.B. 13. 10.B. A B AB CD BC BC BC CD 162 . 3.C D A A A B B A A 106 107 108 109 B B A. 19 20 AB ABC ABC AB CD BC 6. 7. 9. 8.C A B D D C D D C 110 111 D C SECTION . 12. 5.B SINGLE ANSWER TYPE QUESTIONS 1. 15. BASIC TRIGONOMETRY 163 . IIT.MATHS 4 TRIGONOMETRIC EQUATIONS 164 . cos2x – 4 sinx = 1 It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle whereas. SOLUTION OF A TRIGONOMETRIC EQUATION A value of the unknown angle which satisfies the given equation is called a solution of the equation e.TRIGONOMETRY EQUATIONS An equation involving one or more trigonometrical ratios of unknown angle is called trigonometric equation e. sin2x + cos2x = 1 is a trigonometrical identity as it is satisfied for every value of x R. We use the following formulae for solving the trigonometric equations: ( n I) sin = 0 = n.g. / 6 is a solution of sin = 1 . trigonometric equation is satisfied only for some values (finite or infinite in number) of unknown angle. 2 GENERAL SOLUTION Since trigonometrical functions are periodic functions. The general solution should be given unless the solution is required in a specified nterval or range. sin = 1 = (4n + 1) sin = –1 = (4n – 1) 2 2 cos = 1 = 2n cos = –1 = (2n + 1) sin = sin and cos = cos = 2n + Note: Everywhere in this chapter n is taken as an integer. If squaring is necessary. cos = 0 = (2n + 1) .g.g. check the solution for extraneous values. sin = sin = n + (– 1)n cos = cos = 2n tan = tan = n + sin2 = sin2 or cos2 = cos2 or tan2 = tan2 = n . e. 2 tan = 0 = n. squaring the equation at any step should be avoided as far as possible. if not stated otherwise. SOME IMPORTANT POINTS TO REMEMBER While solving a trigonometric equation. solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. 165 . necessary corrections must be incorporated. Domain should not be changed. SOLVING SIMULTANEOUS EQUATIONS Here we will discuss problems related to the solution of two equations satisfied simultaneously. (i) Two equations in one unknown (ii) Two equations in two unknowns. which make any of the terms undefined. n + of .g. This may be due to the different methods of solving the same problem. The answer should not contain such values of angles.IIT. k I the second part of the second set of solution (you can check by putting 5 10 k = 5 m + 2 (mI). e. While solving trigonometric equations you may get same set of solution repeated in your answer.MATHS Never cancel terms containing unknown terms on the two sides. Some times you may find that your answers differ from those in the package in their notations. 5 10 Some times the two solution set consist partly of common values. Which will highlight the importance of above mentioned points. TRIGNOMETRIC INEQUATIONS y /6 5/6 y = 1/2 x 166 . It may cause loss of genuine solution. ( n I) forms a part 2 k . If it is changed. Hence the final answer is k . In all such cases the common part must be presented only once. We may divide the problems in two categories.k I . This will ensure that your answer is correct. Now we present some illustrations for solving the different forms of trigonometric equations. you must check their authenticity. Check that the denominator is not zero at any stage while solving equations. which are in product. Whenever you come across such situation. It is necessary for you to exclude these repetitions. | cosec x | 1. |sec x| 1. 2 n The required solution set = n I 6 6 PROBLEMS BASED ON BOUNDARY CONDITIONS If the problem involves only one equation consisting of more than one variable or equation involves variable of different natures then the boundary conditions of trigonometric functions is generally used. it is obvous that. |cot x| 0 167 . the graph of y = sin x.TRIGONOMETRY EQUATIONS From. | cosx | 1. between 0 and 2p sinx > 1/2 for 5 x . |tan x| 0. It must be noted that |sinx| 1 . 6 6 Hence sin x > 1/ 2 Þ 2np + p/6 < x < 2np+ 5p/6 5 2n . 2 so that required number of solutions is 2. then = 2 (b) 3 . 6 6 ILLUSTRATION : 02 2 If 6 cos 2 2 cos (a) 3 2 sin 2 0.MATHS WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 Number of solutions of the equation tan x sec x 2cos x lying in the interval [0. .IIT. cos 1 3 5 Ans : (D) Solution : The given equation can be written as ==> 10 cos 2 cos 3 0 ==> 5cos 3 2cos 1 0 ==> cos 1 3 or cos 2 5 ==> 1 3 or cos as 3 5 ==> 3 . cos 1 3 5 1 3 (c) cos 5 (d) 3 . in 0. 2 ] is (a) 0 (b) 1 (c) 2 (d) 3 Ans : (A) Solution : The given equation can be written as sin x 1 2 cos x cos x == sin x 1 2cos 2 x cos x 0 ==> 2sin 2 x sin x 1 0 ==> 2sin x 1 sin x 1 0 ==> x ==> sin x 2 sin x 1 0 as cosx 0 5 . cos 1 3 5 ILLUSTRATION : 03 The set of values of x for which tan 3x tan 2 x 1 is 1 tan 3x tan 2 x 168 . . 0 < k < 1.TRIGONOMETRY EQUATIONS (b) 4 (a) (c) n / n 1. 3B < as 0 < k < 1 169 . the angle A is greater than the angle B.4sin3x ... ILLUSTRATION : 04 The number of all possible triplets such than a1 a2 cos 2 x a3 sin 2 x 0 for all is (a) 0 (b) 1 (c) 3 (d) infinite Ans : (D) Solution : The given equation can be written as a1 a2 cos 2 x a3 1 cos 2 x 0 2 a3 a3 ==> a1 a2 cos 2 x 0 2 2 which is zero for all values of x.3..3.. If a1 a3 a2 2 k k or a1 .. If the values of the angles A and B satisfy the equation 3sinx ... tan2x is not defined so the given equation has no solutions. 4 Ans : (C) Solution : The given equation can be written as tan 3x 2 x 1 ==> tanx = 1 ==> x n 4 But for these values of x.k =0. then the measure of angle C is (a) 3 (b) 2 (c) 2 3 (d) 5 6 Ans :(C) Solution : The given equation can be written as sin 3 x k .. 2. ILLUSTRATION : 05 In a triangle ABC.. 2. a2 . a3 k for any 2 2 Hence the required number of triplets is infinite. 4 (d) 2n / n 1. 0 k 1 Since A and B satisfy this equation 0 3 A. 2 (b) . 2 (d) None of these Solution : 1 sin x cos x 1 ==> sin x 4 2 sin 4 n n ==> x n 1 ==> x n 1 . (b). 2 3 ILLUSTRATION : 06 The equation sinx + cosx =1 has a solution in the open interval (a) 0. n I 4 4 4 4 ==> x 2n or x 2n 2 so that x does not belong to the intervals given by (a). 3B < so cos 0 2 2 ==> cos 3 C 0 2 ==> sin 3C 2 0 ==> C .IIT. ILLUSTRATION : 07 3 7 1 1 sin The principal value of 2 cos cos 6 is (a) 5 6 (b) 2 (C) 3 2 (d) None of these Ans : (B) Solution : 3 3 7 1 1 sin 1 sin and cos cos 3 6 2 2 170 .MATHS Also sin 3 A k sin 3B ==> sin 3 A sin 3B 0 ==> 2cos 3 A B 3 A B sin 0 2 2 ==> either cos But sin 3 A B 3 A B 0 or sin 0 2 2 3 A B 3 A B 0 as A > B and 0 < 3A. 2 3 (c) . or (c) for any value of x. and 0 x... sin n x ............ 1 cos 2 x (a) 1 (c) 1 n n 3 n 1 n (b) 1 n 6 n 6 (d) 1 n 1 n ........ n I 3 Ans : Solution : n 1 sin x ... y (a) -2 (b) 0 (c) 2 2 then sinx + cosy is equal to (d) none of these Ans : (C) Solution : The given equation can be written as sin 4 x cos 4 y 2 4sin x cos y 0 2 2 2 2 ==> sin 2 x 1 cos 2 y 1 2sin 2 x 2 cos2 y 4sin x cos y 0 ==> sin 2 x 1 cos 2 y 1 2(sin x cos y )2 0 which is true if sin 2 x 1 ... 1 cos 2 x The general solution of the equation 1 sin x . y we get sinx=cosy=1 2 ==> sin x cos y 2 ILLUSTRATION : 09 n 1 sin x ... sin n x .. cos 2 y 1 and sin x cos y as 0 x.TRIGONOMETRY EQUATIONS 5 5 1 1 = cos cos 2 cos cos 6 6 3 7 1 sin 1 cos cos 6 2 so that 5 6 5 a 3 6 2 ILLUSTRATION : 08 If sin 4 x cos 4 y 2 4sin x cos y..... 1 sin n x ......... 1 cos 2 x The equation 1 sin x . 1 sin n x ......... 1 cos 2 x ==> 171 1 1-sinx 2sin 2 x X 1 sin x 1 2 cos 2 x is .. IIT. which is not possible for any . ILLUSTRATION : 11 cos 3 1 if 2 cos 2 1 2 (a) n 3 (b) 2n 3 (c) 2n 6 (d) 2n 6 Ans :(B) 172 .MATHS ==> cos 2 x cos 2 x sin x sin 2 x sin 3 x (sinx +1 0) ==> cos 2 x sin 2 x sin x cos 2 x sin 2 x ==> 2sin 2 x sin x 1 0 ==> sin x 1 1 1 8 1 3 ==> or sin x 2 4 4 ILLUSTRATION : 10 5 5 The number of solutions of the equation sin x cos x (a) 0 (b) 1 (c) infinite 1 1 sin x cos x is cos x sin x (d) none of these Ans : Solution : The given equation can be written as sin 5 x cos5 x sin x cos x sin x cos x sin x cos x ==> sin x cos x 1 sin x cos x 5 5 1 4 3 2 2 3 4 ==> sin 2 x sin x sin x cos x sin x cos x sin x cos x cos x =1 2 2 2 2 2 2 2 2 2 2 ==> sin 2 x sin x cos x 2sin x cos x sin x cos x sin x cos x sin x cos x 2 ==> sin 2 x 1 sin 2 x cos 2 x sin x cos x 2 1 1 2 ==> sin 2 x 1 sin 2 x sin 2 x 2 2 4 ==> sin 3 2 x 2 sin 2 2 x 4 sin 2 x 8 0 2 ==> sin 2 x 2 sin 2 x 2 0 ==> sin 2 x 2 . as for this value of L. of the given equation is not defined. cos 2 173 .S.TRIGONOMETRY EQUATIONS Solution : 4 cos3 3cos 1 The given equality can be written as 2 2 cos 2 1 1 2 4cos 3 cos 1 2 ==> 4 cos 2 3 ==> cos 2 1 2 ==> 2n 3 3 .H. ..} 3. = 2 .. The most general value of which satisfies sin = -1/2 and tan = 1/ 3 A) 2n /6 4.. then the number of different solutions of 3cos + 4sin = K is sin 4 x + cos 4 x = A) x = B) two C) one D) infinite 7 sinxcosx then x equal to 2 n n 1 2 2 C) x = n 1 10...2. sin2x . n = 1..12 = 0 π /2 < α < π A) 24/25 8. D) n ± /3 n B) x = n 1 24 n D) x = n 1 6 n 12 If tan x + tan 4x + tan 7x = tan x..1) D) x = 5 (n+ 1 ) 2 C) 13/18 D) -13/18 If |K| = 5 and 0 £ θ £3 6 0° ...tan2x = 1 is 1 + tan3xtan2x A) C) {n /4 n = 1. 1 ) 2 2 C) x = 5 (n + B) -24/25 A) zero 9. D) / 8 If α is the root of 25cos2 θ + 5cos θ . is Solution of the equation 4cos2 sin . C) / 6 6x x If sin = 0 and cos = 0 then 5 5 A) x = (n . D) 2n + /4 B) = n + (-1)n (3 /10) C) = 2n ± /6 6.A SINGLE ANSWER TYPE QUESTIONS 1..5) 7. B) /4 D) 2n /4 .. B) 2n + 11 1 /6 C) 2n + 7 /6 D) ? = n + 3 /6 If 1 + cos α + cos2 α +... 3 . B) 2n ± /4 The set of values of x for which C) 2n ± /6 tan3x .2cos x + 1 = 0 then x equal to 4 A) 2n ± /3 2.2 then α (0 < α < π ) is A) 3 /4 B) / 4 B) x = 6 (n .2sin2 = 3sin is 3 A) n + (-1)n 10 5. tan 4x tan 7x then x is equal to A) n /6 B) n /12 C) /12 D) n /2 174 .IIT.MATHS SECTION .... 2. D) n + /12 The number of distinct solutions of sin 5 θ cos 3 θ = sin9 θ . B) (2n-1) /24 B) 5 C) 8 D) 9 The most general solutions of the equation secx . If sin 4 θ = cos θ . n 4 4 4 D) None If [x] denotes the greatest integer less than or equal to x and let f(x) = sinx + 3 cosx. C) 2n + 7 /6 1 . then 8 3 A) = n + /3 18. 2n + /4 D) n /4 C) 2n The most general values of ? satisfying the equation (1+2sin )2 + 3sinθ -1 2 = 0 are given by B) n + (-1)n A) n /6 16.cos 7 θ A) n n 1 / 18 3 B) n n 1 / 9 3 C) n n 1 / 6 3 D) n n 1 / 12 2 If tan 7 θ = cot 5 θ then general solution for θ is A) 2n ± /24 13. cos7 θ in [0.1 = ( 2 1) tanx are given by A) n /8 15. n D) 2n + 11 1 /6 6 B) x = D) x n n 1 3 18 n n 1 3 9 cos3θ 1 = .sin(600-x) sin(600+x) = A) x = n + (-(A)n C) x n 1 17.TRIGONOMETRY EQUATIONS 11. C) n + /24 B) 2n . then the π most general solution of f(x) = f are 10 A) n /4 20. sinx. n I. 175 B) 2n /3 C) n + (-1)n /6 . /3 D) None The smallest positive value of x for which tanxtannx = 1(n N) is . 7 6 B) n . if 2cos2θ -1 2 B) = 2n /3 C) =2n /6 D) =2n /6 General solution of the equation 1+sin2x = (sin3x-cos3x)2 is/are given by A) n 2 19. 12.2 ] is A) 4 14. n C) n . 5/6] C) [5/6. 5/6] The equation cos4x – (a+2) cos2x – (a+3)=0 possesses a solution if B) a < -2 C) –3 a 2 D) If sin8x + cos8y + 2 = y sin2x cos2y and 0 x. /2 C) /6. then is equal to A) = /4 B) = /3 C) = /2 D) /6 22. /6] 30. /4} D) tan x = 2 0 <q < p solution for the equation cos2 = ( 2 1) COS A) {/4. C) infinite If max {5sin + 3sin () = 7 then the set of possible values of is A) [0. D) 2 If sin 4sin2x – 8sinx + 3 0 0 / 2 then the solution set for x is A) a >-3 31. D) {/4. The equation 3sin2x+2cos2 x + 31-sin2x+2sin 2x = 28 is satisfied for the values of x given by A) tan x = 1 24. p/2] B) [0. B) {/3. 2] D) [/6. e. /6} 25. y /2 then sinx + cosy is equal to A) –2 32. 1 is 2 x 1 2 2 . C) tan x = B) 0 C) 2 D) None {2.y + z = 0 (cos2 ) x + 4y +3z = 0 and 2x + 7y + 7z = 0 have non .trivial solution. C) {/3} 2sin 2 cos 2 x = 1-cos sin 2x if 2 A) /6. B) 2 n 1 C) 3 2 n 1 D) None If 5 cos2 θ + 2cos2 θ /2 +1 = 0 . /3 B) {x|x = 2np ± 2p/3 n e z} D) [p/3./4 D) /4. . 2 2 A) 0 26. then is equal to A) = n B) = (2n+1) C) 2n 1 D) 4n 1 / 2 2 2 23. If the system of equations (sin3 ) x . B) cosx = 0 B) 1 If sin = B) x = n 2 C) tan x = 1 / 4 D) x n / 4 B) /3.MATHS A) 2 n 1 21. 2 4 7} are given by The most general values of x for which 3 sin x cos x min R 176 . 2p/3 ] 29. /3} 3 .IIT. /3 28.sin x x The number of solutions of 2cos 2 x 2 0 x / 2 is A) x = 2n 1 / 2 27. + 4sin = 2 ( 3 1) 0 / 2 then is equal to A) {x|x = 2np ± p/3 n e z } C) [x/3. 3/2) D) (3/2. 4 The solution set of (2cosx –1) (3+2 cosx) = 0 in the interval 0 x 2 is A) {/3} 35.P. = n n z B) = n n z C) = n + (-1)n+1 /3 n z D) = n nz 2 The value of x between 0 and 2 which satisfy the equation sin x 8 cos 2 x 1 are in A. then is equal to 2 2 A) (0. 5/3} A) n +7 /4 36. D) All The most general value of which satisfies both the equations tan = -1 and cos = 1 / 2 will be A) n +/4 37. C) (2n±1) B) {/3. with common difference A) /4 177 B) /8 C) 3/8 D) 5/8 . 5 . B) 2n ± /4 The general solution of the equation 42./4 x / 4 is sin x cos x B) 2 C) 1 D) 3 The smallest positive root of the equation tan x – x = 0 lies in B) (/2.TRIGONOMETRY EQUATIONS A) 2n B) 2n + 2/3 33. 2) 7 cos3x + sin 2 x = -2 then x is equal to 6 A) 41. ) C) (./4 3 1 sin D) 2n ± /4 3 1 cos 2 n A) n 1 / 12 4 B) 2n / 4 / 12 n C) n 1 / 12 4 D) 2n / 4 / 12 sin x cos x The number of distinct real roots of cos x cos x sin x cos x cos x 0 in the interval . C) {/3. C) n + (-1)n/4 +/6 D) None 6k 1 3 B) 6k 1 3 The general solution of sin2 sec + C) 6k 1 3 D) 6k 1 2 3 tan = 0 is A) = n + (-1)n+1 /3./2) 40. /6} 3 If sin cot cos tan . cos-1 (-3/2} D) {/3. B) n + (-1)n 7/4 A) 0 39. 4 B) (2n-1) 38. D) n + (-1)n /4 C) 2n + 7/4 C) n . If 2 cos 2 X 1 2 sin 2 X 1 2 2 then x = A) (2n+1) 34. /(B) D) (0./3 C) 8 D) 10 B) 2n + /3 n z C) 2n ± n z D) n + (-1)n /3 n z B) 4 C) 2 D) 1 B) 1 B) n + /4 C) 2 D) 3 sin x i cos x is purely imaginary are given by 1 i C) n D) 2n + /2 The equation (cosp-1)x2 + (cosp)x+sinp=0 where x is a variable has real roots if p lies in the interval A) (0.MATHS 43. /2) . B) /6 If sec + tan = 2 3 then general solution for is A) n + /3 46. . 6 3 2 4 D) (0. where x e [0. B) 2n + /3 The number of solutions of |cosx|= sinx 4 is A) 8 49. C) 2n- . 2) 52.] A) /3 or 2 /3 45. /3) . D) n ./4 51. A) (0. ] C) no solution in [. ) n If 2tan2x –3secx is equal to 0 for exactly 7 distinct values of x 0.IIT. 2] such that A) n . The inequality 2sin2x – 3sinx + 1 > 0 holds for all values of x 5 .. B) 10 C) 13 D) 15 If [x] denotes the greatest integer less than or equal to x and sinx = [1+sinx]+[1-cosx] has A) no solution in [-/2. 0) C) (-/2. /2] B) no solution in [/2. B) (-. 1 + sin x + sin2x +………. /6) 5 . C) /4 The number of values of x in [0. 44. D) /2 3 = 2 are The most general values of satisfying tan + tan 4 A) n ± /3 n z 48. n N then the greatest 2 value of n A) 4 53.¥ = 4 + 2 3 .5] satisfying the equation 3cos2x – 10cosx + 7 = 0 is A) 5 47. 6 B) (0. B) 6 The values of x [-2./3 The number of values of x for which sin2x + cos4x = 2 is A) 0 50. /3) 3 C) (0. 3/2] D) no solution for x 178 . /4} 3 2 n /8 2 1 D) 2n cos then q is equal to 2 D) {/3. 3 3 B) x : x 2n 2 . D) None The general solution of sinx .3cos2x + cos3x is A) np + p/8 56. 6p] is 4 D) 15/3 If max {5 sin+ 3sin(-a)}=7. n I 63. C) both B) {/6. C) a = ½ 2 19 3 sinpx + cos px = x2 . B) a = 1 B) 1/3 B) 2 a 2 D) 1/ 2 a 1/ 2 C) 110 1 x e [0. D) –1 The equations sin4x – 2cos2x + a2 = 0 is solvable if A) 3 a 3 61. C) 2/3 The sum of all the solutions of the equation cosx. /6} C) {/4} The value of lying between = 0 and = /2 and satisfying the equation 1 cos 2 sin 2 4 sin 4 2 2 4 sin 4 0 cos 1 sin cos 2 sin 2 1 4 sin 4 A) 10/24 58. B) 7/24 The value of ‘a’ for which the equation a2 – 2a + sec2 { (a+ x) } = 0 has solution is A) a = 2 59.TRIGONOMETRY EQUATIONS 54.3 x + 9 . If cos2 = 1 2 1 cos then = 2 A) 2n + /4 55. then the set of possible values of is . 3 2 C) . cos ( / 3 x) cos ( / 3 x ) A) 15 62. /4. B) 2n ± /3 B) n /8 2 n C) 1 If cos6 + cos4 + cos2 + 1 = 0 0 A) {/6. n I 3 D) None 1 1 The number of solutions of the equation sin5x – cos5x = cos x sin x (sinx cosx) is A) zero 179 D) ½ C) 1 a 1 B) 30 A) x : x 2n . then x is equal to If A) –1/3 60.3sin2x + sin3x = cosx . /2} 57. D) /24 C) 5/24 B) 5 C) ¥ D) None . MATHS 64.IIT. If the system of equations (sin3)x – y +z=0. then = A) n. n + (-1)n 3 D) None 180 . n+ (-1)n 6 C) both B) n ± a. 2x+7y+7z=0 have non-trivial solutions. (cos2)x +4y+3z=0. n.nI 4 D) none of these B) x = np + The solutions of the system of equation sinx siny = 3 /4.. 4.. 2 sin x cos2x = sinx if A) x = n + /6 (n l) B) x = n . n I . 2 2 6. C) cosx = 1/2 B) ± /4 C) ± 3/4 D) none of these . 2... 1 3 cos = 6x ./6 (nÎ l) C) x = n (n Î l) D) x = n + /2 (n Î l) X X cos 2 X 2sin sin 2 X cos2 X sin 2 X has a root for which 2 2 The equation 2 sin A) sin 2x=1 3. D) cos (x-p/4) = 2 2 C) two values of x and two values of D) two pairs of values of (x. 2 A) no value of x and 7. D) cos2x = -1/2 sinx + cosx = 1 + sinx cosx...B MULTIPLE ANSWER TYPE QUESTIONS 1. if A) sin (x+p/4) = 4. then x = A) ±/2 181 1 B) one value of x and two values of C) no solution in . cosx cosy = A) x1 = 2n k 3 2 B) y1 = k 2n 6 2 C) x2 = 2n k 6 2 D) y2 = k 2n 3 2 3 /4 are .TRIGONOMETRY EQUATIONS SECTION . holds for sin + A) no solution in .p/4) = 2 1 C) cos (x + p/4) = 5. 8. k I 7 14 C) x = n + /2. x. 2sin2x + sin22x = 2. 2 D) no solution for x Solutions of the equation sin7x + cos2x = -2 are A) x = 2 k 3 ...x2-11. B) sin2x=-1 1 B) sin (x . ) The equation sinx = [1+sinx] + [1-cosx] has (where [x] is the greatest integer less than or equal to x) 3 2 B) no solution in . y are respectively 6 A) 10.IIT. B) - 7 11 . 24 24 If cosx = A) p 11.MATHS 9. then a value of x is B) 0 C) tan-12 D) none of these The solution of the equation cos103x . 0 x. then x. 5 . 1 tan x If 1 tan x = tany and x-y . 24 24 D) none of these 1 sin 2 x .sin103x = 1 are A) - 2 B) 0 C) 2 D) p 182 . 24 24 C) 115 119 . A A A A A D B B A B A 16. 11.D B A B B B 61 62 63 64 B A A A 183 B . 30. A. 12. 13. 28.B. 15.B C. B D 27. B C 29. 20.TRIGONOMETRY EQUATIONS SECTION . 22. 8. 14. 9. 21. 7. 19. 3.A SINGLE ANSWER TYPE QUESTIONS 1. 26. 6. 10. 4. 25. 18. 2. 23 24. B B A C A B A B A A A A A D C 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 C B C B C A D C C A B C B A B 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 C A B A A D D A. 17. 5. A B AC AB AB CD C C 9. 4. 6. 8.B MULTIPLE ANSWER TYPE QUESTIONS 1. 11. AB C 10. 7. 3.MATHS SECTION . BC AB 184 .IIT. AB C 2. A B AD BD CD 5. TRIGONOMETRY EQUATIONS 185 . IIT.MATHS 5 INVERSE TRIGNOMETRIC FUNCTIONS 186 . If x is negative. If x is positive. [0. Similar argument also works for cos– 1 x. 6 The functions sin–1 x. Thus. tan–1 x. 2 – {0} Note. Remark 1. tan–1x and cosec–1 x lie between and 0. the principal values of all the inverse circular functions lie between 0 and . 2 2 Function Domain sin–1x [ –1. cos–1 x. ] tan–1x R . whose sine is .INVERSE TRIGONOMETRY FUNCTIONS DEFINITION If a function is one to one and onto from A to B. i. while that of sin–1x. sec–1 x and cot–1x lie 2 between and . we denote g = f-1 {Read as f inverse} –1 x = f (y). cosec–1 and sec–1 x are called inverse circular or inverse trigonometric functions. arc cos x etc. sin–1 where. The sin–1 x is merely a symbol denoting a certain angle sin x whose sine is x. then g is called the inverse function of f denoted by x = g(y). The ‘–1’ used in sin–1 x is not an exponent. It means that is an angle whose cosine is x. then function g which associates each element y B to one and only one element x A. = sin–1 = n + (–1)n 2 2 2 6 is the least positive value of . such that y = f(x). tan–1x etc. 2 cos–1 x [–1. 187 . the principal values of cos–1x. we define principal value. To make each inverse circular function single valued. 1] [0. is an angle. 1) 2 . then may be any angle whose cosine is x. If cos = x.e. 2 2 cot–1 x R (0. Usually. and we write = cos–1 x. 1] 2 . 1). ] – { /2} cosec–1 x R – (–1. The inverse trigonometric functions are also written arc sin x. ) sec–1 x R – (–1. Each of the inverse circular function is multivalued. cot–1 x. sin–1 x is not to be interpreted as Range (Principal Values) 1 . ] and x [–1.IIT. –1] [1. ) y –1 y = cos x cos = x = cos–1 x where [0. 1] – /2 y y = cosec x y=x 2 cosec = x = cosec–1 x 1 –1 y = cosec x .MATHS GRAPHS OF INVERSE TRIGONOMETRIC FUNCTIONS sin = x = sin–1 x. 2 2 y = sin–1x y y=x /2 y = sin x –1 x 1 and x [– 1. 2 2 and x (– . . where . 0 0. 1] 2 y=x 1 2 1 x y = cos x y y=x sec = x = sec–1x where 0. 2 2 2 1 1 – 2 and x (– . –1] [1. ) –1 1 1 y = sec–1x x –1 y = sec x 188 .2 –1 x where . cot (cot–1 x) = x. for all [– /2. sec (sec–1 x) = x. tan (tan–1 x) = x. ] tan–1 (tan ) = . –1] [1. ]. ) for all x R Property III: (i) (ii) (iii) (iv) 189 sin–1 (– x) = – sin–1 ( x). 1] for all x [–1. 1] for all x R for all x [– . ) Property II: (i) (ii) (iii) (iv) (v) (vi) sin (sin–1x) = x. and x (– . 1] for all x R for all x [– . –1] [1. tan–1 (– x) = – tan–1 x. /2) cosec–1 (cosec ) = . ) and x (– . /2]. –1] [1. for all [0. 1] for all x [–1. for all [0. ) for all x (– . cosec (cosec–1 x) = x. for all x [–1. /2 cot–1 (cot ) = . ) x PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS Property I: (i) (ii) (iii) (iv) (v) (vi) sin–1 (sin ) = . cos–1 (– x) = – cos–1 (x).INVERSE TRIGONOMETRY FUNCTIONS y y = tan x y=x 2 –1 y = tan x – 2 tan = x = tan–1 x x 2 where . for all (– /2. for all [– /2. cosec–1 (– x) = – cosec–1 x. for all x [–1. ) 2 2 – 2 y y = cot x y=x –1 y = cot x –1 cot = x = cot x where (0. 0 sec–1 (sec ) = . for all (0. /2] cos–1 (cos ) = . cos (cos–1 x) = x. ) . 1] [1. if 0 x. ) 1 x for all x (– . y 1 and x 2 y2 1 or if xy 0 and x 2 y 2 1. ) for all x R Property IV: 1 x for all x (– . y 1 and x 2 y 2 1. π sin π sin 1 x 1 y2 y 1 x 2 . xy Property VII: (i) sin–1 x + sin–1y = sin 1 x 1 y2 y 1 x 2 . = 1 x cot x . 190 . if xy 1 1 xy (ii) 1 tan–1x – tan–1y = tan 1 xy .MATHS (v) (vi) sec–1 (– x) = – sec–1 x. cot–1 (– x) = – cot–1 x. 1 x 1 y 2 y 1 x 2 . 1] (ii) tan–1 x + cot–1 x = . (ii) cos–1 = sec–1 x. if xy 1 1 xy π . ) Property VI: If x. if xy 1 tan–1x + tan–1 y = 2 1 x y π tan . 2 for all x [–1. y 0. if 1 x. 1] [1. – 1] [1. y 0 and x 2 y 2 1. for x 0 for x 0 (i) sin–1 = cosec–1 x. –1] [1. if 1 x. ) 1 1 cot x. 2 for all x R (iii) sec–1 x + cosec–1 x = . 2 for all x (– . then (i) 1 x y tan .IIT. for all x (– . (iii) tan –1 Property V: (i) sin–1 x + cos–1x = . x (0. x (0.0 x y 1 and x 2 y 2 1. if 1 x. if π sin 1 x. y 1 . 1 y 0 and x 2 y 2 1. if and xy0 1 x. 2π cos 1 xy – 1 x 2 1 y2 (ii) if 1 x. if or if 1 x 1 y 2 y 1 x 2 . if π sin 1 x 1 y 2 y 1 x 2 . cos1 xy 1 x 2 1 y 2 . 0 x 1 and x y. y 1 and x y 0 cos–1 x – cos–1 y = cos1 xy 1 x 2 1 y 2 . x>0 . y 1. Property IX: (i) x 2 1 x sin–1 x = cos–1 1 x 2 tan 1 = cot (ii) 1 x2 x sec 1 2 x 1 x –1 cos ec 1 1 . Property VIII: (i) cos–1x + cos–1y = cos1 xy 1 x 2 1 y2 . if 1 y 0. y 1 and x y. 1) 1 x2 1 1 2 1 sec 1 x cos ec x x .INVERSE TRIGONOMETRY FUNCTIONS (ii) sin –1 x – sin–1 y = sin 1 x 1 y 2 y 1 x 2 . 1) x 1 x2 cos–1 x = sin–1 1 x 2 tan 1 x 1 1 1 cosec 1 sec 2 2 x 1 x 1 x = cot –1 (iii) 1 cos 2 1 x tan–1 x = sin–1 = cot 191 x –1 x 1 2 1 x . 1 x 0. y 1 and x 2 y 2 1 xy 0 and x 2 y2 1. 0 x. π sin 1 2 2sin–1 x = sin 2x 1 x . π tan 2 1 x 1 2x 2tan–1x = tan 1 x 2 .IIT. π sin –1 3sin x = sin 1 3x 4x 3 . if 1 x 2π cos 2 1 1 1 3 3 cos–1 x = 2π cos 4x 3x . 1 2x 1 x 2 . if x 1 cos 2 (ii) if 1 x 0 Property XII: (i) 1 2x . π sin (ii) if 1 x Property XI: (i) 2π cos 1 2x 2 1 . if x 2 2 1 1 4x 3 3x . 2 cos x = cos1 2x 2 1 . 1 x2 if x 1 if 1 x 1 if x 1 192 . π sin (i) if 2 1 2 x 2 1 2 x 1 if 1 x if 1 1 1 2 1 1 if x 2 2 1 if x 1 2 1 3x 4x 3 . 1 3x 4x 3 . 1 2x π tan . –1 if 0 x 1 1 1 4x 3 3x .MATHS Property X: 1 2x 1 x 2 . 2 1 x if x 1 if 1 x 1 if x 1 if 0 x if x 0 . 1 3x 2 if x if 1 3 1 1 x 3 3 if x 1 3 Property XIII (i) (ii) 193 1 2x . 1 x 2 2 tan–1 x = 1 x2 cos . 1 x2 1 2x π sin . π tan 2 1 3x 3 1 3x x –1 tan . π sin 2 1 x 1 2x 2 tan–1 x = sin . 3 tan x = 1 3x 2 3 π tan 1 3x x .INVERSE TRIGONOMETRY FUNCTIONS (ii) 3x x 3 1 . 1 x2 1 1 x 2 cos . then (a) /2 (b) /4 (c) = /4 (d) /4 /2 Ans : (D) Solution : sin 1 x cos 1 x tan 1 x tan 1 x and 0 tan 1 x since 0 x 1 when we fined 2 4 4 2 ILLUSTRATION : 02 If x (a) 1 1 1 .IIT.MATHS WORKED OUT ILLUSTRATIONS 1ILLUSTRATION : 01 If 0 x 1 and sin 1 x cos 1 x tan 1 x . the value of cos cos x 2sin x is 5 24 25 (b) 24 25 (c) 1 5 (d) 1 5 Ans :(C) Solution : The given expression is equal to cos cos 1 x sin 1 x cos sin 1 x 2 1 = sin sin x x 1 5 ILLUSTRATION : 03 2 tan 1 cos ec tan 1 3 tan cot 1 3 is equal too (a) /16 (b) /6 (c) /3 (d) /2 Ans :(C) Solution : The given expression is equal to 2 tan 1 cos ec tan 3 6 194 . tan 1 = tan 1 1 x2 1 sec 1 tan 1 x tan 1 cos 1 1 tan 1 tan tan 1 x sin 2 2 2 So that according to the given condition 1 1 tan x 4 tan 1 x 8 or x tan 8 2 ILLUSTRATION : 06 sec 2 tan 1 2 cos ec 2 cot 1 3 is equal too (a) 1 Ans : 195 (b) 5 (D) (c) 10 (d) 15 (d) x tan 8 .INVERSE TRIGONOMETRY FUNCTIONS 1 1 2 1 1 2x 2 tan 6 3 3 3 3 = 2 tan ILLUSTRATION : 04 2 tan 1 1 tan 1 2 tan 1 3 is equal too (a) /4 (b) /2 (c) (d) 2 Ans : (D) Solution : 1 2 1 tan 1 3 The given expression is equal to 2 tan 1 2 1 1 = 2 tan 3 tan 3 2 ILLUSTRATION : 05 1 x2 1 4 . then x 1 If tan (a) x tan 2 (b) x tan 4 (c)x = tan(1/4) Ans : (D) Solution : Taking x tan . (2) From (1) and (2) we find 0 2 y 0 y 2 4 1 0 cos x which holds if 4 1 x 1 2 ILLUSTRATION : 08 u If u cot 1 tan tan 1 tan .IIT. then tan is equal to o 4 2 196 . then Since 0 cos 1 x 0 2 y 2 Also since ………(1) sin 1 2 x 1 x 2 2 2 sin 1 sin 2 y 2 2 2y 2 2 ………..MATHS Solution : The given expression is equal to 2 1 tan tan 1 2 1 cot cot 1 3 2 = 1+ 4 +1 9 = 15 ILLUSTRATION : 07 1 1 2 The equation 2 cos x sin 2 x 1 x (a) 1 x 1 (b) 0 x 1 is valid for all values of satisfying (c) 0 x 1 2 (d) 1 x 1 2 Ans : (D) Solution : If we denote cos 1 x by y. INVERSE TRIGONOMETRY FUNCTIONS (a) (b) cot tan (c) tan (d) cot Ans : Solution : Let 1 1 tan tan x, then u cot tan x tan tan x x x 2x 2 2 = u 2 2x u 4 2 x u 4 2 tan x tan tan x tan 4 2 ILLUSTRATION : 09 1 1 2 1 1 2 1 The value of cos 2sin 3cos (a) 7/4 (b) 11/4 (c) /12 1 1 4 tan 1 is equal too 2 (d) 25/12 Ans : (D) Solution : 1 1 2 1 1 2 1 We have - cos sin 3cos = 1 2 2 3 25 2 x 3 x 4 3 6 4 4 12 ILLUSTRATION : 10 3 5 1 1 1 If cos ec x 2 cos 7 cos then the value of is (a) 44/117 Ans : (B) 197 (b) 125/117 (c) 24/7 (d) 5/3 IIT- MATHS Solution : 2 3 3 1 7 1 2cot 7 cos cot cot 1 2x7 4 5 1 1 = cot 1 24 3 cot 1 7 4 24 3 x 1 cot 1 7 4 = 24 3 7 4 cot 1 44 117 cos ec 1 125 117 198 INVERSE TRIGONOMETRY FUNCTIONS SECTION - A SINGLE ANSWER TYPE QUESTIONS 1. cos 1 (cos 5 / 4) is given by A) 5/4 2. B) 3/4 cos 1 B) /2 140 221 tan 1 1 If cot x sin A) 2 9. 199 C) /2 B) 0 D) B) 0, /4 C) - /4, /4 D) /4, /2 B) /2 C) D) 3/2 B) 1/ 3 C) 1 D) C) –2 D) ½ 1 / 4 then x equal to 5 B) 3 Number of solutions of sin -1 x + sin -1 2x = /3 is A) 2 10. D) 2 2 tan 3 tan tan is equal to 4 15 5 15 A) - 3 8. 2 If x + 1 / x = 2, the principle value of sin 1 x is A) /4 7. C) 1 1 x The smallest and the largest values of tan ; 0 x 1 are 1 x A) 0, 6. 171 B) cos 1 221 2x If x 1 then 2tan-1 x + sin-1 1 x 2 is equal to A) 4tan-1 x 5. D) /4 C) 0 15 1 2 tan 1 is equal to 17 5 1 A) cos 4. D) - 5/4 yz xz 1 xy tan 1 tan 1 If x 2 y 2 z 2 y 2 then tan is equal to zr xr yr A) 3. C) -/4 B) 3 C) 1 4 x2 x3 x6 -1 -1 2 x sin x + ........ + cos x + ....... = for 0 <|x| < If 2 4 2 4 D) 4 2 3 IIT- MATHS A) 11. 1 2 B) 1 C) – ½ If a and b are roots of the equation 6X 2 +11X + 3= 0 then A) both cos 1 and cos 1 are real C) both cot 1 and cot 1 are real 2n 12. i 1 X i 1 B) x tan 3 3 D) x 2 , 2 C) x 1,0 C) D) 2 B) cot C) tan2a D) cot a C) 16/7 D) 17/6 -1 3 -1 3 The value of tan sin + cot is 5 2 B) 7/16 3 If sin-1x + sin-1y + sin-1z = then the value of 2 B) 1 x x 101 y 101 303 303 y x x 202 y 202 404 404 y is equal to C) 2 D) 3 C) a = x = b D) a > b, x ax x b 1 cos-1 a b sin a b is possible if B) a < x < b If sin-1x = cot--1x , then A) x2 = 20. 2 1 2 u 1 tan a then tan is equal to If u = cot-1 tan a tan 4 2 A) a > x > b 19. , D) 2n-1 holds for B) /2 A) zero 18. 1 n n 1 2 2(tan-11 + tan-12+tan-13) is equal to A) 6/17 17. C) A) 16. is equal to 1 1 2 The formula 2 sin x sin 2 x. 1 x A) /4 15. i B) 2n A) x0,1 14. B) both cos ec 1 and cos ec 1 are real D) none of these 2n If X i n then A) n 13. D) –1 5 1 2 B) x2 = 5 1 2 C) sin(cos-1x)= 5 1 2 D) x = 5 1 2 If x1, x2, x3, x4 are roots of the equation x 4 x 3 sin 2 x 2 cos 2 x cos sin 0 then tan 1 x 1 tan 1 x 2 tan 1 x 3 tan 1 x 4 is equal to 200 INVERSE TRIGONOMETRY FUNCTIONS A) B) /2 - 21. x2 2 x2 1 If minimum value of -1 If cos 201 D) 1, ½ C) 0 D) none of these C) (a + b) (a2 + b2) D) (a - b) (a2 - b2) C) ¼ 2 sin x sin x y 1 2 D) 3/2 2 then the value of K is K C) 8 D) 16 B) 5/4 C) 1 D) ½ C) 2 / 8 D) 2 / 32 Minimum value of (sec-1 x)2 + (cosec-1 x)2 is B) /32 The number of real solutions of tan 1 x( x 1 ) sin 1 x 2 x 1 / 2 is B) one C) two D) infinite 1 3 1 1 1 3 sin 1 sin 1 =‡ is then The two angles are A = 2 tan 2 2 1 and B = 3 sin 3 3 5 B) A < B C) A = B D) A + B = 0 An integral solution of the equation tan-1x + tan-1y = tan-13 is equal to A) (2, 7) 32. C) 0, 1/2 x y 5π y x 2 y2 1 x + cot -1 = sin 1 and sin then value of 2 2 is 2 b 12 2 b 12 a b A) A > B 31. B) (a + b) (a2 - b2) B) 6 A) zero 30. B) 0,1/2 B) ½ A) /8 29. x2 2 x 2 xy y 2 x 1 y cos / 6 then the value of If cos is 4 2 3 9 2 3 A) ¾ 28. x2 2 1 D) 1 A) 4 27. x2 2 x2 1 3 21 1 cos ec 2 tan 1 sec tan 2 2 á is equal to 2 2 A) ¾ 26. C) If sin 1 (1 x ) 2 sin 1 x / 2, then x equals A) (a - b) (a2 + b2) 25. x2 1 B) 1, 0 A) 0, -1/2 24. B) If sin 1 x sin 1 (1 x ) cos 1 x , then x equals A) 1, -1 23. D) The value of sin (cot-1 (cos (tan-1x)) is A) 22. C) B) (4, -13) The number of real solutions of (x, y) where C) (5, -8) D) (1, 2) cos 1 (cos X ) tan x ,0 x 2 is 2 A) 2 33. B) 1 IIT- MATHS D) None C) 4 Sum of infinite series 3 3 2 3 1 cot 1 22 cot 1 32 .... cot is 4 4 4 -1 B) tan-12 A) /4 34. tan 1 m 1 B) -/4 1 2 x2 2 x 1 x 1 cos-1 2 4 A) |x| 1 36. B) x R 2x C) 0 x 1 1 x2 if 0 x 1 2a The solution of sin 1 a 2 ab 1 ab B) f(x) = 1 x2 if x < 1 B) 2 1 1 b cos 2 1 b 1 2x tan 2 is 1 x 1 ab ab C) ab 1 ab D) ab 1 ab B) /2 C) 1 D) none of these C) p/3 D) 4p/3 2 1 The principal value of sin sin is 3 A) 2p/3 B) - 2p/3 3 1 1 1 The equation sin x cos x cos 2 has A) no solution C) infinite no.of solution 41. 2x If xy + yz + zx = 1 then tan-1x + tan-1 y + tan-1z is equal to A) 40. D) –1 x < 0 D) None -1 A) 39. D) None x cos 1 2 - cos--1 x holds for C) not finite if x > 1 38. C) ± /4 2 1 1 1 2x 1 1 x sin cos The value of f(x) = tan 2 1 x2 2 1 x 2 is A) f(x) = 37. D) None 2m 4 is equal to 2 m m 2 zA) /4 35. C) tan-13 B) unique solution D) none of the these If tan + tan ( + /3) + tan ( - /3) = k tan 3 then the value of k is 202 1] 49.1] D) [-1. then x 2xy cos y a 2 ab b2 -1 A) sin2 B) cos2 C) tan2 D) cot2 1 1 1 If tan-1 ( x 1 ) tan x tan ( x 1 ) tan ( 3 ) then x= A) zero 203 B) + 3 tan-1 x B) (-1. 1) x 1 x2 sin 1 x holds is C) [0. D) . C) x 1 2x 1 23 tan 1 tan 1 then x equal to x 1 2x 1 36 A) x[-1. D) 3 -1 A) ¾ 45. 3 7 3 1 1 3x x tan . C) 2 If sin -1x + sin -1 2x = π / 3 then x equal to A) 44.3tan-1 x D) -2tan-1 x 4 3 5 5 4 3 3 8 B) . C) 3 B) –1 C) 3 D) 13 .1) A) 17 52.0] C) –1/2 D) 3 / 2 C) 25 D) 27 If sin 1 x cos 1 x / 6 then x is equal to B) 3 /2 sec2(tan-12) + cosec2(cot-13)+2 is equal to B) 15 x y 2 2 -1 If cos a +cos b = a. 1 2 B) ½ A) . 3 If x e the value of 2 is 3 1 3x 4 3 5 8 48. 4 3 5 7 C) .3tan-1 x 46. C) ¼ 1 The set of values of x for which tan A) 1/2 51. B) 1 1 3 2 7 1 If tan D) 1 7 2 3 D) 3/2 C) . 47. B) xR C) x(--1] [1] D) x(-1.INVERSE TRIGONOMETRY FUNCTIONS A) 1 B) 1/3 42. B) x 2 xy y 2 x -1 y + + sin = π / 6 hen value of If sin is 9 4 3 16 3 4 A) . 1 7 2 3 Let f(x) = sec 1 x tan 1 x then f(x) is real for A) R 50. D) 1/6 9 100 100 100 If sin -1x + sin -1 y + sin -1z = 3π / 2 the value of x y z x101 y101 z101 is A) 0 43. /4 60. then x equals 8 1 1 2 62. C) /3 2 3 2 /3 3 1 3 1 3 If a.IIT. B) . 2 1 If < 1/32 then the number of solutions of (sin-1x)3 + (cot-1 x)3 = ap3 is A) n . D) 2 y then 1 B) 45° A) –1 59. 1 If l is a root of x2 + 3x+1=x0. g are the roots of the equation x 3 mx 2 3x m 0. then the general value of 204 . C) 0 x B) y = 3 / 2 A) 0° 57. x1 5 2 If (tan x ) (cot x ) .MATHS 53. b./4 3 5 cos x cos 1 is equal to 5 3 cos x x 2 x 2 1 B) 2 tan 2 tan C) 1 x tan 1 2 tan 2 2 1 2 x 2 1 D) 2 tan tan Minimum value of (tan–1 x)2 + (cot–1 x)2 is A)2/4 B) 2/8 C) 2/2 D) 2/32 1 1 1 x 3 3x 2 then If f ( x ) cos x cos 2 2 2 3 1 B) f 2 cos 1 3 1 /3 D) f 2 cos A) f / 3 C) f / 3 63./4 D) 2n . then tan-1l + tan1 is equal to A) /2 54. D) None of these 1 x 2 ) is valid for all values of x satisfying B) 0 x 1 A) y = 4/5 56./2 The equation 2cos-1x = sin-1(2x A) –1 x 1 55. 1 2x k x 3 1 -1 tan and B = tan k 3 then the value of A – B is If A = 2k x A) 0 58. 2 B) 1 1 A) tan tan 61. If cos tan 1 sin cot 1 2 C) y = -1 D) None C) 60° D) 30° C) 2 D) infinite C) 0 D) –1/2 2 B) 1 If 2tan-1 (cosq) = tan-2 (2 cosecq) then q equal to B) n + /4 C) . INVERSE TRIGONOMETRY FUNCTIONS tan -1α + tan -1β + tan -1 γ is A) (2n+1) /2 64.. then the number of values of (x..6 . C) n/2 The number of positive integral solution of the equation tan–1x + tan–1 1/y = tan–1 3 A) 2 65.. cos-1x – cos-1y = /3. . B) n B) 1 C) 2 D) The sum of the infinite series 2 1 3 2 n n 1 1 1 1 sin 1 sin .. D) B) 4 C) 6 D) 8 If sin1x + sin-1y = 2/3. sin 2 6 12 n ( n 1 ) -1 A) /8 67... D) /4 + n B) 1 C) 3 D) 4 1 x 2 x 1 /2 is The number of real solutions of tan-1 x( x 1 ) sin A) zero 66. y) is A) two 205 C) /2 n 2 10n 21.. then the maximum value of n is If cot-1 6 A) 2 68. n N.. B) /4 B) four C) zero D) None . sin . 1 8 = 5 / 18 9 e B) f 7 = /12 4 e D) f 8 = 13 / 8 9 e 7 = 11 / 12 9 e If a £ sin-1 x + cos-1x + sin-1 x £ b.75 C) x = 3 D) x = 4 B) x = 2 5 2 Let q = tan-1 tan and f = tan-1 tan then 4 3 A) > 6. If B) 1/2 D) cos-1x B) x (0.3 = 0 A) f 7.3/8 C) 73/8 D) /2 a( a b c ) b( a b c ) c( a b c ) tan 1 tan 1 is bc ca ab B) /2 C) D) 0 206 . 1) D) x = 0. The x satisfying sin-1x + sin-1 (1-x) = cos-1 x are A) 0 2. 1/ 2 ) C) x (1/ 2 . D) none of these Let f(x) = e cos sin( x / 3 ) then C) f 8.IIT.5) = . then A) =0 B) = /2 C) = /4 D) = The greatest and least values of (sin-1x)3 + (cos-1x)3 are A) 3/32 9.MATHS SECTION . if A) x = 1 5. B) tan-1x sin-1x > cos-1x holds for A) all values of x 4. C) sec-1x 6 sin-1 (x2-6x+8. D) 2 1 X 1 then which of the following are real ? 2 A) sin-1x 3.B MULTIPLE ANSWER TYPE QUESTION 1. C) + = 7/12 B) 4 . C) 1 tan-1 A) /4 B) . 28. 8. B B A D B B D B C B C 16. 23 24. D D B A. 9.INVERSE TRIGONOMETRY FUNCTIONS SECTION . 6. 20.A SINGLE ANSWER TYPE QUESTION 1.C D D C B D A A C C 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 B C B B A A B B D B D A B D 61 62 63 64 65 66 67 68 B A B A C C C D 207 A . 2. 13. 30.B C. 11. 3. 22. D A 29. 14. 12. 17. 21.C B B C C C C C B C C A 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 A. 7. 5. 18. 4. 26. 25. B B 27.D C D B C A. 19. 15. 10. 8. CD BD 5. CD 208 .B MULTIPLE ANSWER TYPE QUESTIONS 1. 4. BC BC AD AC 9. 6. 7. 2.IIT.MATHS SECTION . AB AB D 3. INVERSE TRIGONOMETRY FUNCTIONS 209 . IIT.MATHS 6 PROPERTIES OF TRIANGLE 210 . b and c. a b c sin A sin B sin C COSINE RULE In a triangle ABC. b2 c2 a 2 (i) cosA = 2bc (iii)cosC = c2 a 2 b2 (ii) cosB = 2ca a 2 b2 c2 2ab PROJECTION FORMULAE (i) a = b cos C + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos C NAPIER’S ANALOGY (TANGENT RULE) A BC bc cot 2 2 bc (i) tan B CA ca cot 2 2 ca (ii) tan C A B a b cot 2 2 ab (iii) tan HALF ANGLE FORMULAE (a) (b) (i) sin A 2 (iii) sin C 2 (i) cos A = 2 (iii) cos 211 s bs c bc (ii) sin B 2 s c s a ca s a s b ab ss a bc s s c C 2 ab (ii) cos B = 2 ss b ca . B and C and the length of the sides opposite to these angles are denoted by small letters a.PROPERTIES OF TRIANGLE In a triangle ABC the angles are denoted by capital letters A. Semi perimeter of the triangle is given by s = abc and its area is denoted by . 2 SINE RULE In a triangle ABC. BAD = and DAC = . The centroid divides the median in the ratio 2 : 1 a sin B sin = 2b 2 2c 2 a 2 CIRCUM CIRCLE The circle which passes through the angular points of a ABC.. is called the circumcentre. The three medians of a triangle are concurrent and the point of concurrency of the medians of any triangle is called the centroid of the triangle.MATHS (c) (i) tan s b s c s s a A = 2 2 2Δ s s a s b s c bc bc (iii) sin C = B = 2 s cs a ss b s a s b ss c C (iii) tan 2 (i) sin A = (ii) tan (ii) sinB= 2 2Δ s s a s b s c ca ca 2 2Δ s s a s b s c .e. The centre of this circle i. is called its circumcircle. Then (i) (m + n) cot = m cot – n cot (ii) (m + n) cot = n cot B – m cot C CENTROID AND MEDIANS OF A TRIANGLE The line joining any vertex of a triangle to the mid point of the opposite side of the triangle is called the median of the triangle. A E F O A A B a/2 D a/2 C 212 . the point of concurrency of the perpendicular bisectors of the sides of the ABC.IIT. ab ab m-n THEOREM Let D be a point on the side BC of a ABC such that BD : DC = m : n and ADC = . e. = tan A or MD b a 2 = tan A. tan B.. The triangle formed by joining the feet of these perpendiculars is called the pedal triangle i. 2 y = tan B. 2z a b B y x D C c But in a triangle ABC. DEF is the pedal triangle of ABC. In a triangle the altitudes drawn from the three vertices to the opposite sides are concurrent and the point of cuncurrency of the altitudes of the triangle is called the orthocentre of the triangle. A E r I 900– B/2 B/2 B 213 r D F r C/2 C . A F E P 0 90 – C B C D INCIRCLE The circle which can be inscribed within the triangle so as to touch each of the sides of the triangle is called its incircle.. tan B. The centre of this circle i.e. 2 y . i. ORTHOCENTRE AND PEDAL TRIANGLE OF A TRIANGLE.PROPERTIES OF TRIANGLE Radius of the circumcircle is given by the following formulae R= a b c abc 2sin A 2 sin B 2 sin C 4 BD In BDM. tan C = 2x . the point of concurrency of angle bisectors of the triangle is called the incentre of the ABC. a = tan A.tan A + tan B + tan C = tan A. tan C a b c abc x y z 4xyz . 2x x c A F Similarly.e. = tan C 2z E z M A a b c tan A + tan B + tan C = 2x 2y 2z and tan A. B and C can be obtained in 2bc the similar way. the triangle is known completely. 2 bc 2 2 214 . (i) (ii) b 2 c2 a 2 If the sides a. then using tan BC b c A BC cot . 2 2 2 2 2 i. s 2 2 2 2 2 2 ESCRIBED CIRCLES The circle which touches the side BC and the two sides AB and AC produced is called the escribed circle opposite the angle A. If two sides b and c and the included angle A are given. r3 = sc 2 2 2 2 M BISECTORS OF THE ANGLES If AD bisects the angle A and divide the base into portions x and y. bc sin A 2bc A cos b c sin A b c 2 2 A B x D y C SOLUTION OF TRIANGLES When any three of the six elements (except all the three angles) of a triangle are given. we get .IIT. Radii of the excircles are given by the following formuale A (i) (ii) r1 = A A B C s tan 4R sin cos cos sa 2 2 2 2 B C F1 B A B C s tan 4R cos sin cos r2 = sb 2 2 2 2 E1 L (iii) D1 I1 C A B C s tan 4R cos cos sin .MATHS Radius of the Incircle is given by the following formulae r= A B C A B C = (s – a) tan = (s – b) tan = (s – c) tan = 4R sin sin sin . by Geometry. This process is called the solution of triangles. Its centre and radius will be denoted by I1 and r1 respectively. b and c are given. then cos A = . we have.. x AB c y AC b x y xy a c b bc bc ac ab and y = bc bc Also let be the length of AD we have ABD + ACD = ABC x= 1 A 1 A 1 c sin b sin bc sin A.e. Let OL be parallel to BC. If b < c sin B. If c < b and B is an acute angle. 2 2 The third side is given by a = (iii) b sin A . If c sin B < b < c and B is an acute angle. Let I be the in-centre of O be the circumcentre of the triangle ABC. NOC A tan IL IM LM IM ON OL BM BN BM NC r R cos A B r cot R sin A 2 A B C sin sin R cos A 2 2 2 A B C B 4R sin sin sin . then there are two values of angle C (fig 3). b b sin A give the remaining elements. there is sin B no triangle possible (fig 1). then there is only one A = 1800 – (B + C) and a = triangle possible (fig 2).PROPERTIES OF TRIANGLE BC A Also = 900 – .cot R sin A 2 2 2 2 A 4R sin 215 cos A cos B cos C 1 cos A cos B cos C 1 sin A sin C sin B sin A sin C sin B cos B cos C 1 tan 1 . then there is only one triangle (fig 4). A A c c b B B D (Fig 1) b c sinB c sinB D (Fig 2) A A b c b b D B C2 c sinB b c c sinB C2 C1 B C1 (Fig 4) (Fig 3) This case is. sin B c If two sides b and c and the angle B (opposite to side b) are given. so that B and C can be evaluated. sometimes. Let IOL . called an ambiguous case. If b = c sin B and B is an acute angle. IM = r OC = R. then sin C = sin B. sin C sin B I O B L NM C . C are in Arithmetical progression 2 B A C also A + B +C = 1800 so that B = 600 In a triangle ABC. B. B. and the included angle is 600. C of a triangle ABC are in arithmetical progression then (a) tan A tan C 3 tan A tan C 3 (b) tan A tan C 3 tan A tan C 3 (c) tan A tan C 3 tan A tan C 3 (d) tan A tan C 3 tan A tan C 3 Ans : (C) Solution : Since A.3 tanAtanC = 3 ILLUSTRATION : 03 216 . then the third side is (d) 2 3 Ans : (B) Solution : If a represents the third side then 2 2 1 1 a 2 6 2 6 2 cos 60 0 1 1 2x x 6 2 6 2 1 26 2 a 2 6 22 2 26 2 2 a 3 3 a 4 2 ILLUSTRATION : 02 If the angles A.MATHS WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 1 If two sides of a triangle are (a) (b) 3 1 6 2 3 2 (c) and 6 2 1 3 . we know TanA + tanB + tanC = tanAtanBtanC tanA+tanC = 3 (tanAtanC-(A) tanA+tanC .IIT. d denote the length of the sides AB. B 60 0 . then the ratio of the largest side to the smallest side of the triangle is 3 sin cos 2 sin (a) (b) 3 cos sin 2 sin (c) cos 3 sin 2 sin (d) 3 cos sin 2 sin Ans : (D) Solution : Since the angles of the triangle are in the ration 1: 3: 5 Let A 20 0 . a. C 100 0 We have 3 5 180 0 20 0 . CD and DA respectively. b. BC. then cosA is equal to (a) a 2 b2 c2 d 2 2ab cd 217 (b) b2 c2 d 2 a2 2bc da (c) c2 d 2 a 2 b 2 2cd ab (d) d 2 a 2 b 2 c2 2da bc . c. a b c then from sin A sin B sin C c sin C sin 120 0 We get the required ratio = a sin A sin 3 cos 1 sin 2 2 = sin ILLUSTRATION : 04 The expression a b cb c a c a b a b c is equal to 2 A (a) cos 2 2 A (b) sin 2 2 A (c) cot 2 2 A (d) tan 2 Ans : (C) Solution : The given expression is equal to b c2 a 2 a 2 b c 2 b 2 c 2 a 2 2bc 2bc b 2 c 2 a 2 2b cos A 2bc 1 cos A A cot 2 2bc 2bc cos A 1 cos A 2 ILLUSTRATION : 05 In a cyclic quadrilateral ABCD.PROPERTIES OF TRIANGLE If the angles of a triangle are in the ratio 1: 3: 5 and denotes the smallest angle. MATHS Ans : (D) Solution : D We have from ABD c d BD2 = a 2 d 2 2ad cos A C from BDC A a BD2 = b 2 c 2 2bc cos C B = b 2 c 2 2bc cos A Equating the two values we get b cos A d a 2 b2 c 2 2 da bc ILLUSTRATION : 06 2 p q 1 pq A B tan p .IIT. then A is a b c bc (b) an obtuse angle (c) a right angle (d) equal to B-C Ans : Solution : We have cos A cos B cos C a b c 218 . if then 1 p 2 1 q 2 is equal to 2 2 Ans : (C) Solution : The given expression is equal to A B A B 2 tan tan 1 tan tan 2 2 2 2 A B 1 tan 2 1 tan 2 2 2 C C A B A B = 2 sin cos 2 cos sin sin C 2 2 2 2 2 2 ILLUSTRATION : 07 In a triangle ABC if (a) an acute angle cos A cos B cos C a . In a triangle ABC. tan q . B. D are the angles of a quadrilateral.PROPERTIES OF TRIANGLE b 2 c 2 a 2 c 2 a 2 b 2 a 2 b 2 c 2 2abc a 2 b2 c2 a (given) 2abc bc a 2 b 2 c 2 2a 2 b 2 c 2 a 2 b 2 c 2 a 2 0 cosA = 0 A = 2 ILLUSTRATION : 08 If A. C. if sin A (a) k cos 2 Ans : (D) Solution : sin 219 A B acb cos k 2 2 2c C (d) k cos 2 . then 2 2 2c B C (b) k sin (c) k sin 2 2 In a triangle ABC. then tan A tan B tan C tan D is equal to cot A cot B cot C cot D (a) tanAtanBtanCtanD (b) cotAcotBcotCcotD (c) tan 2 A tan 2 B tan 2 C tan 2 D (d) Ans: tan A tan B tan C (A) Solution : We have tanA B tan 360 0 C D tanC D tan A tan B tan C tan D 1 tan A tan B 1 tan C tan D tan A tan B1 tan C tan D 1 tan A tan Btan C tan D 0 tan A tan B tan C tamD tan A tan B tan C tan A tan B tan C tan D 1 1 1 1 tan A tan B tan C tan D tan A tan B tan C tan D = tan A tan B tan C tan D tan A tan B tan C tan D cot A cot B cot C cot D ILLUSTRATION : 09 A B acb cos k. b a c b k bc ca 2c s b ss c s b k c ab c cos C k 2 220 .MATHS s b s c x ss .IIT. P In a triangle ABC if tan A/2.P 7.P 9.P B) H.P 5. cot C/2 are in A.c are in A) A. C of a triangle are in A.P C) H.P D) none of these B) G.P D) none of these B) 1 A C) A 1 1 1 A1 A2 A3 D) 1 A2 B) 12 cm C) 16 cm D) 18 cm If orthocentre H of a DABC bisect the altitude AD of the triangle ABC.P 3.P C) G.b.B. then value of tanB tan C is A) 3 221 D) none of these If length of the side BC of a DABC is 6cm and BAC = 120° then the distance between in centre and excentre of the circle touching the side BC internally is A) 10 cm 11.P D) none of these If twice the squares of the diameter of a circle is equal to half the sum of the squares of the sides of inscribed triangle ABC then sin 2 A sin 2 B sin 2 C is equal to A) 1 4. 2 then the triangle is a b 2 sin ( A B ) A) right angled or isosceles C) equilateral 6.c are in H. c are in A) A. A3 are the areas of incircle and the ex-circles of a triangle.P. A2. c2 are in H.P D) a2.P. B) 2 C) 4 sin A sin( A B ) If in a triangle ABC sin C sin( B C ) then B) a2.P 8. c2 are in A) A. b2.P and sides a. then a.PROPERTIES OF TRIANGLE SECTION . B) H. B) G.c are in A) A. b. cosA +2cosB + cosC = 2 then a.P C) a. a 2 b 2 sin( A B ) If in a triangle ABC.A SINGLE ANSWER TYPE QUESTIONS 1.P then a. cot B/2.P A). tan B/2. D) 8 B) 2 C) 1 D) 0 . C) H.P D) none of these If the triangles A. C) G. c are in A) A. c2 are in A. then A) 10. B) H.b.P If A.P If r 1 2r 2 3r 3 then a.b.b. c in A. 1 A C) G. then a2. a. b2. tan C/2 are in H. B) right angled and isosceles D) none of these If in a DABC.P 2. A1. b. In any DABC if cot A/2. b2. b.c are in G. B) a.. sin B 13. such that BD = DE = EC. (a+b+c) (b+c-a) = l bc if A) 3b2 = a2 – c2 18. D) 4 In a triangle ABC.P.3 If the sides of a triangle are in A. c C) a.P. 20. B) l > 6 In a triangle ABC a : b : c = 4 : 5 : 6 then the ratio of the circumcircle to that of incircle is A) 3+ 3 22. the length of the perpendicular from A on BC produced is A) 23. R In a triangle ABC.P C) G. B) b + c B) isosceles C) obtuse angled D) None of these Points D. IIT. then 2(r + R) is equal to A) a + b 14. If Tn +1 – Tn = 21 then n equals A) 5 16.P. Let C / 2 . b. D) l > 4 B) 3a2 = b2 – 3c2 C) b2 = a2 – c2 D) a2+b2 = 5c2 B) tanB = b/a D) sin2A+sin2B+sin2C=0 C) cosC =0 B) H. and the greatest angle of the triangle exceeds the least by 900 then the sine of the third angle is 7 /4 B) 7 /2 C) 7 /8 D) 7 If one angle of a triangle is 300 and the length of the sides adjacent to it are 40 and 40 3 then the triangle is A) right angled 24. c are in A) 16/7 21. C) c + a In a triangle ABC 3sinA = 6sinB = 2 3 sin C then angle A is A) 30° 15. b.If r is the inradius and R is the circumradius of the triangle. C) 6 If the tangents of the angles A and B of a triangle ABC. sin A. D) a + b + c Let Tn denotes the number of triangles which can be formed using the vertices of a regular polygon of n sides.E are taken on the side BC of a triangle ABC. sinA. sin B.12. C) 90° In a triangle ABC. R D) a. if tan A/2 = 5/6. D) None of these B) 16/9 C) 25/27 D) 27/25 B) 20 3 3 C) 20 3 3 D) 3. B) 60° B) 7 A) l < 0 17. then A) tanA = a/b 19. D) 120° If D is the midpoint BC of a triangle ABC and AD is perpendicular to AC. and tan B/2 = 20/37 and the sides a. If 222 . C=1200 and a=40.MATHS Which of the following pieces of data do not uniquely determine an acute angle triangle ABC (R being the radius of the circumcircle) A) a. C) 0 < l < 4 If ABC is a triangle in which B=450. satisfying the equation abx2 -c2x+ab=0 then A) A. I3 are the centres of the excircles of the triangle ABC then radius of the circumcircle of I1 I2 I3 is A) 4.P 29.5 cm 32. 2s Radius of the circumcircle of ABC is 3cm. A1. B) 2 sin( x y ) sin ( y z ) is equal to sin x sin z C) 4 D) None of these If A.B. A3 denote the respectively the areas of an inscribed polygon of 2n sides. If I1.5 cm D) 6 cm C) abc D) 4 abc B) sin /n : cos /n D) tan /n : cos /n P Q In a triangle PQR R / 2 If tan and tan are roots of the equation ax 2 bx c 0 2 2 a 0 then A) a + b = c 223 b 3 then angle A is equal to c 2 The ratio of the radius of circumcircle and incircle of a regular polygon of side n is A) cosec /n : cot /n C) tan /n : cot /n 34. sa B) B) G. C are angles of a triangle.P B) /4 2 B) 2 :1:1 B) 5 cm B) 3 abc C) 5/12 D) /2 C) 1 : 2 : 1 D) 1 : 1 : 1 C) 5. I2. C) If A1. s C) H. B and C of a triangle ABC are in A.PROPERTIES OF TRIANGLE BAD X . then the minimum value of tan a 2 A / 2 tan 2 B / 2 tan 2 C / 2 is equal to A) 0 26. D) If in a DABC cos B cos C + sin A sin B sin C = 1 then a : b : c is equal to A) 1 : 1 : 31. A3 are in A) A.P Angles A. DAE Y EAC X then value of A) 1 25. B) 1 If in a triangle ABC. sb B) b + c = a C) a + c = b D) b = c . P3 are the altitudes of a triangle ABC from vertices A. C) ½ D) none of these bc ca a b then cosA is equal to 11 12 13 B) 5/7 C) 19/35 D) 20/35 If P1. B. A) 1/5 27. A2. If A) /6 30. D) none of these The minimum value of bc (b+c) cos A + ca(c+a) cos B + ab (a +b) cos C is equal to A) 2abc 33.P. inscribed polygon of n sides and circumscribed polygon of n sides then A2. C and D is the area of the triangle then p1 1 p 2 1 p 3 1 is equal to A) 28. P2. then ÐA is a b c bc ca A) /2 38. R. 2 If the area and an angle of of a triangle are given. D) 2/3 In any ABC b2 sin2C +C2 sin2B is equal to A) 42. then the third side is A) 12 2 3 B) 12 2 3 C) 2 3 2 D) 2 3 2 If A + B + C = .s In a triangle ABC the value of 41. base angle a < p/4. cosAcosB 2 4 Given an isosceles triangle with equal side of length b. then A) R = 1 b cos ec 2 B) D = 2b2 sin2a b cos b sin 2 C) r = 21 cos 37. 2 sin cos 2CosA CosB 2CosC a b . If the angles are in A. 3 1 3 sin A sin B . There exists a triangle ABC satisfying A) tanA + tanB + tanC = 0 B) sin A sin B sin C 2 3 7 2 sin A cos A 3 C) (a+b)2 = c2+ab and D) sinA + sinB = 36. then the side opposite to the given angles is minimum when the triangle is isosceles with the length of the equal sides equal to A) 40. a tanA + a tan B = (a+b) tan then 2 224 . n = z then tan nA + tan nB +tan nC is equal to A) 0 C) tan nA tan nB tan nC 43. C) /6 2 sin 39. 3 2 B) 1 D) none of these AB If in a ABC.IIT. D) OI = B) /4 2 sin B) A) abc 2 C) sin 2 D) None bc ca a b r1 r2 r3 r B) s C) 2s D) 3s B) 2 C) 3 D) 4 In a triangle the length of the two larger sides are 24 and 22 respectively. I the centres of the circumcircle and incircle respectively.P. If in a triangle ABC.MATHS 35. r the radii and 0. C) –1 In a ABC if a2 sin (B-C) + b2 sin (C – A) + c2 Sin (A – B) = 0 then triangle is A) a 2 b 2 c 2 54. 1 If length of the sides of a DABC are 3.6 56. then AH is equal to D) 5+ 6 . 3 1 If in a triangle ABC a = 5. C) A = 2B B) a 2 b2 c2 2 C) 2 a 2 b2 c2 D) a 2 b2 c2 If the angles of a triangle are 30° and 45° and the included side is ( 3 1) cm then the area of the triangle is A) 47. A 75 then a + c 2 is equal to A) 0 45. b t 2 1c c = 2t then value of r r1 r2 r3 is A) 1 49. then which of the following is true A) right angled 53. D) 4 In a ABC if median AD is perpendicular to the side AB. 3 1 C) 1 B) 0 A) 2tanA+tanB = 0 51. D) B = 2A Let in a ABC such that A 45 . then the length of the third side be A) 5. if the angles are in A. D) 4 If in a ABC the median AD is perpendicular to the side AB then A) a 2 c 2 3b 2 52. C) In a if a = t 2 1. the lengths of the two larger sides are 10 and 9 respectively. D) ½ 1 In a triangle ABC 2ac sin ( A B C ) is equal to 2 A) 1 : 3 : 5 55..PROPERTIES OF TRIANGLE A) A = B B) A = -B 44. 1 3 1 B) 6 3 /2 225 B) 5 /2 C) ½ D) ¼ B) tanA+2tanB = 0 C) tanA–2tanB = 0 D) 2tanA–tanB = 0 B) a 2 b 2 3c 2 C) a 2 c 2 2 b 2 D) a 2 b 2 2 c 2 B) obtuse angled C) isosceles D) equilateral B) c 2 a 2 b 2 C) b 2 c 2 a 2 D) c 2 a 2 b 2 The sides of a triangle are in the ratio 1 : 3 : 2 then the angles of the triangle are in the ratio B) 2 : 3 : 4 C) 3 : 2 : 1 D) 1 : 2 : 3 In a triangle.4 and 5 then distance between its incentre and circumcentre is A) 50. B) b The value of C) 2b D) –b 1 1 1 1 2 2 2 is 2 r1 r2 r3 r A) 0 46. b = 4 and cos (A-B) = 31/32 then the third side ‘c’ is equal to A) 4 48.P. B) B) 3 3 C) 5 If H is the orthocentre of the triangle ABC. 2abc D) cosA B) 4/5 C) 3/5 D) – 3/5 If P1. ABC . 1 1 3 2 D) . a straight line is drawn from the midpoint of one of equal sides to the opposite angle.. C and D the area of the triangle then P1 2 P 2 2 P 3 2 is equal to A) 61. 2 C) cosA cos B cos C is equal to P1 P2 P3 A) 1/r 63. a 2 b2 c2 If P1. BC and AC of a triangle are 8cm. B) the altitudes are in H. B) 2R sinA B) 1 : 3 C) 3 : 5 D) None The cosine of the obtuse angle formed by the medians drawn from the vertices of the acute angles of an isosceles right angled triangle is A) – 4/5 60. sinC are in A. P2. In a 1 3 B) . P3 are respectively the perpendiculars from the vertices of a triangle to the opposite sides.P. C) altitudes are in G. then 62.1 1 2 3 3 C) .D and F are the points on BC such that ADB 2ACB and AEB 3 ACB .P.1 65. P3 are the altitudes of a triangle ABC from the vertices A.P. C) a cotA If in a triangle ABC. right angled at B. Then the ratio of the tangents of the two parts in two which it is divided by the line is A) 3 : 2 59. sinB. B. the tangents of the angles are roots of the equation A) qx 2 px 3 1 q x p 0 B) px 3 qx 2 1 p x q 0 C) 1 q x 3 px 2 qx p 0 D) qx 3 px 2 1 q x p 0 If length of the sides AB. P2. 15cm. 58. D) None of these In an isosceles right – angled triangle. B) 120 2 cm 23 B) 60 2 cm 23 C) 30 2 cm 23 D) 30 23 In a right angle ABC. then A) altitudes are in A. B / 3 and C / 4 Let D divide BC internally in the ratio 1 : 3 then sin BAD equals sin CAD 226 . 17 cm respectively.MATHS A) 2R cosA 57.P. then length of the angular bisector of ABC is A) 64. then ratio of DE and CD will lie in the interval 1 2 A) .IIT. sinA. abc 4 a 2 b2 c2 2 D) abc 2 B) 1/R C) 1/ D) r/ If P is the product of the sines of angles of a triangle and q the product of their cosines. PROPERTIES OF TRIANGLE A) 1/ 6 B) 1/3 66. C) 3 If in a triangle ABC.P. then the triangle is A) isosceles 227 3 3 2 If the angles A. then a2 . D) None of these If cos2A + cos2B +cos2C = 1. given b > c. b and c opposite these angles are in G.P. 2 /3 In a triangle ABC. D) In a triangle ABC. 71. c2 are in A) G. the length of the bisector of angle A is A) 68. b=4 and cos(A-B) = 31/32 then the third side c is equal to A) 2 69.P. AD is the altitude from A. B and C of a triangle ABC are the sides a. D) Let A 0 A1A 2A 3 A 4 A 5 be a regular hexagon inscribed in a circle of unit radius. B) 3 3 2bc sin A 2 B) 2bc cos bc A 2 bc abc C) 2R b C cosec A/2 D) 4 A cos ec bc 2 B) 3 C) 6 D) 12 B) 1130 C) 930 abc then B = b c2 2 D) None B) A. C) H. Then the product of the lengths of the line segments A 0 A1 . C) 1/ 3 B) right angled C) obtuse angled D) acute angled . a=5. A0A 2 and A0A 4 is A) ¾ 67. C = 230 and AD= A) 130 70. b2 .P. D) B + C = A In a DABC. A = 150. Then A) R = 3. cosA cosB = = sinA sinB 2 4 a c AC In a DABC. Then A) tanA tan B < 1 C) tanA + tanB + tanC < 0 9. If in a triangle ABC. B) (b-c)2 = a2 .ac Given an isosceles triangle with equal sides of length b. Then 2 ( a c2 ac A) B = /3 B) B = C C) A.B MULTIPLE ANSWER TYPE QUESTIONS 1. C) (c-a)2 = b2 . where a. c are the sides of a triangle. C are in A. then C) (a+b)2 = c2 + ab and 8.bc B) tanA tanB > 1 D) tanA + tanB + tanC > 0 If the sines of the angles A and B of a triangle ABC satisfy the equation c2x2-c(a+b)x+ab= 0.c2x + ab = 0. I the centres of the circumcircle and incircle. D) OI = 2 sin cos / 2 C) 5 3 1 cm A) tanA = a/b D) tan A + tan B = c2/ab 6. r the radii and O.MATHS SECTION . b cos3 / 2 C) r = 21 cos In DABC. 1 b cosec a 2 B) 15cm If ABC : a = 5. A = B) tan-1 (9/40) D) 2tan-1 (1/9) E) none of these B) tanB = b/a C) cos C = 0 2 2 2 E) sin A + sin B + sin C = 2 These exists a triangle ABC satisfying D) sinA + sinB = B) sin A sin B sin C 2 2 7 2 (sinA + cosA ) = 3 3 +1 3 .ab 2. b.IIT. B. b = 10 2 cm the value of ‘a’ for which these will be a unique triangle meeting theese requirement is A) can not be evaluated C) tan-1 (1/40) 5. 2 cos 2 . B = 60 then A) (a-b)2 = c2 .P. D) a2+b2+c2 = 2b2 + ac b sin 2 B) D = 2b2 sin 2a A) 10 2 cm 4. respectively. tanB are the roots of the quadratic abx2 . base angle < /4. + B for the value of angle C 2 A) tanA + tanB + tanC = 0 7. then the triangle 228 . tan C < 0. b = 4. D) 5 3 1 cm If tanA. which of the following is true ? A) cos A cos B cos C a b c B) C) sin A sin B sin C 3 a b c 2R D) cos A cos B cos C a 2 b 2 c 2 a b c 2abc sin 2A a 2 sin 2B b 2 sin 2C c2 If H is the orthocentre of triangle ABC. A < /2 15.angled C) is obtuse angled 10. 2 : These exists a triangle ABC satisfying the conditions A) b sinA = a. 2 : 2: 3 1 B) 2 : 2 : 3 +1 C) 4 2 :1 : 3 1 2 D) 2 2 4 C) 2cos2 4 D) 2 sin2 If l is the medium from the vertex A to the side BC of a DABC. B) a2 + b2 . then A) a/b = 4/5 B) a/b = 5/4 If two sides of a triangle are 12 and third side is A) 2 2 .PROPERTIES OF TRIANGLE A) is acute . 12. between sina and cosa. A < /2 C) b sinA > a. A < /2.ab < c2 3 < 0) then C) a2+b2 > c2 D) none of these For a triangle ABC. 3 x2-4x + In a ABC tanA and tanB satisfy the inequation A) a2+b2 + ab > c2 11. 1 If sinb is the G. r1 = 2r2 = 3r3.6 229 B) 2 cos2 A) 4l2 = 2b2+2c2-a2 C) 4l2 = a2+4bc cosA 19. then the C) 6 2 D) none of these . 3 1 D) B) b sinA > a. B) is right-angled D) satisfies sinA + cosA = (a+b)/c B) 2R sinA C) a cotA D) ratio A) 14.P. then cos2b is equal to 4 18. then the sides opposite these angles are in the A) 2R cosA 13.the angle opposite to the shorter side is 450.. cos () are in H. 2 If cos ().M. then AH is equal to 2abc cos A If the angles of a triangle are in the ratio 2 : 3 : 7. then cosq sec /2 is equal to A) -1 16. then B) 4l2 = 2b2+2bc cosA D) 4l2 = (2s-a)2-4bc sin2A/2 If in a ABC. b > a B) .2 A) 2 sin2 17. A > /2 D) b sinA < a. cos. C) 1 : B) 2 6 C) a/c = 3/5 D) a/c = 5/3 8 . 28. D A 27. 7.B D C B A A B C B 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 B A A C AB CD A B A. 14.C D A B D A. 2.A SINGLE ANSWER TYPE QUESTIONS 1. 26. C A B. 18.C A. 5. 10. 22. 11. 25. 20. 23 24.IIT. 6. 8. 13. 4. 3. 17. A A C B A A A A B B B 16. 15. 12. 30.MATHS SECTION . C B 29.B D A C A B B 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 A B B B B B C B D A A B C C 61 62 63 64 65 66 67 68 69 70 71 B A A D A C B C B A B A 230 . 19. 9. 21. 15. 3. 4. 14. 13. 2. 19. 17. 12.B MULTIPLE ANSWER TYPE QUESTIONS 1. 11.PROPERTIES OF TRIANGLE SECTION . AC AB BC BC CD 231 5. 9. 8. 18. CD ACD AD BD 16. 6. ABC CD AC AC BD AB DE BC AC ACD AD BC . 7. 10. IIT.MATHS 7 CO-ORDINATE GEOMETRY 232 . co-ordinates were introduced to help geometry. y) are called co-ordinates of point ‘P’. CARTESIAN CO-ORDINATES Let XOX and YOY be two fixed straight lines at right angles. Historically. From any point ‘P’ a line is drawn parallel to OY. In co-ordinate geometry all the properties of geometrical figures are studied with the help of algebraic equations. Here we will only discuss rectangular co-ordinate system in detail.t. triangular system etc. And so well did they do this job. The abscissa OM and the ordinate MP together written as (x. The directed line OM = x and MP = y.r. which determine the position of point ‘P’. co-ordinate geometry) : A point whose abscissa and ordinate both are integers. When the axes of co-ordinates XOX and YOY are not at right angles. rectangular. polar. 233 . The word ‘geometry’ today generally means coordinate geometry.CO-ORDINATE GEOMENTRY The co-ordination of algebra and geometry is called co-ordinate geometry. Remarks : Y nd st II quadrant I quadrant X O th IV quadrant IIIrd quadrant x y Lattice Point (w. For this purpose we require a co-ordinate system. Here OM is abscissa and MP is ordinate of the point ‘P’.g. this system of representation is called rectangular (or orthogonal) co-ordinate system. that the very identity of geometry was changed. There are various types of coordinate systems present in two dimension e. Since XOX YOY. XOX is called axis of x and YOY is called axis of y and O is named as origin. Students should note that the object of coordinate geometry is to use some known facts about a curve in order to obtain its equation and then deduce other properties of the curve from the equation so obtained. they are said to be oblique axes. Here (x. y) is an ordered pair of real numbers x and y. oblique. y2 ) is given by Q(x2. y1) and (x2.r. then sometimes. y2) y P (x1.r. ABC. y2). is to be determined. y) divides the line joining A(x1. The coordinates of centroid are given by 234 . then (i) Internal division: x = y= (ii) 2. in which a given line segment is divided. B (x2. my 2 ny1 mn External division: x = y= mx 2 nx1 mn mx 2 nx1 mn my 2 ny1 mn The coordinates of the mid-point of the line-segment joining (x1. If P(x.MATHS Distance between two points : The distance between two points P(x1. to the angle bisector on which they lie.t. Centroid : The point of concurrency of the medians of a triangle is called the centroid of the triangle. OA.y1) PQ = y2 x1 x 2 2 y1 y 2 2 y1 O x x1 x2 SECTION FORMULA 1. 2 2 Remarks: If the ratio.t. y3).r. we take the ratio : 1 and apply the formula for internal division. where A (x1. y1 ) & B(x2 . y2 ) in the ratio m : n. Incentre and Excentre of a triangle are harmonic conjugate of each other w. BC = a. y1) and Q(x2. y1).IIT. it is an internal division otherwise it is an external division. y2) are x1 x 2 y 1 y 2 . Centres connected with a Triangle : (w.t. CA = b & AB = c). If the value of turns out to be positive. Remarks: Points P and Q are said to be harmonic conjugate of each other w. C (x3. The centroid of a triangle divides each median in the ratio 2 : 1. for convenience (instead of taking the ratio m : n). i. . The triangle formed by joining the feet of altitudes in a is called the orthic triangle.CO-ORDINATE GEOMENTRY x1 x 2 x 3 y1 y 2 y 3 . Remarks : 1. The coordinates of the incentre are given by ax 1 bx 2 cx 3 ay1 by 2 cy 3 . Centroid G and Orthocentre H of a ABC are collinear. G 3 3 Orthocentre : The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle. Incentre : The point of concurrency of the internal bisectors of the angles of a triangle is called the incentre of the triangle. Circumcentre O. and similarly for excentres (I2 & I3) opposite to I1 º a bc a bc ax1 bx 2 cx 3 ay1 by 2 cy 3 . B and C are given by I2 º abc a bc A c B b L C I1 BL c AI bc .e. Here DEF is the orthic triangle of ABC. also 1 LC b I1L a ax1 bx 2 cx 3 ay1 by 2 cy 3 . G Divides OH in the ratio 1 : 2. OG : GH = 1 : 2 235 . . I3 abc a bc Circumcentre : The point of concurrency of the perpendicular bisectors of the sides of a triangle is called circumcentre of the triangle. I abc abc Excentre : Co-ordinate of excentre opposit e to ÐA is given by ax1 bx 2 cx 3 ay1 by 2 cy3 . . y1) to P. B. the point traces out a path. y1)... (x2. B(x2. y1).. If we plot the points A(x1.(1) . incentre and circumcentre lie on the same line and in an equilateral triangle all these four points coincide. We sometimes include some unknown quantities known as parameters. is 1 [x (y – y )+ x2 (y3 – y1) + x3 (y1 – y2)] 2 1 2 3 = x1 y1 1 1 x2 2 x 3 y2 1 y3 1 . y3). C of a triangle ABC. 236 .(2) While using formula (1) or (2). Procedure for finding the equation of the locus of a point : (i) If we are finding the equation of the locus of a point P. while finding the area of triangle ABC. orthocenter. y2) and C(x3. y1). y3) respectively be the coordinates of the vertices A... Equation of locus The equation to a locus is the relation which exists between the coordinates of any point on the path. y3) has not been taken into account.+ (xn – 1yn – yn – 1xn) ) + (xny1 – ynx1)| 2 LOCUS When a point moves in a plane under certain geometrical conditions.. This path of the moving point is called its locus.. assign coordinates (h. So. and which holds for no other point except those lying on the path. order of the points (x1. then area of polygon is given by 1 |(x1y2 – y1x2) + (x2y3 – y2x3) + .. then the area of the triangle as obtained by using formula (1) or (2) will be positive or negative as the point A.. we take modulus.MATHS 2. y2) and (x3. Then the area of triangle ABC.. . (ii) Express the given conditions in terms of the known quantities to facilitate calculations. (x2... Remarks : In case of polygon with vertices (x1. B.IIT. In an isosceles triangle centroid. C are in anti-clockwise or clockwise directions. y1). y2) and (x3.. In other words equation to a curve (or locus) is merely the equation connecting the x and the y coordinates of every point on the curve.. (x2.... yn) in order. (xn... y2).. Area of a triangle : Let (x1. k) or (x1.. (iii) Position of a given point relative to a given line : The fig. 2 1 Remark : (i) If 900 . b. (ii) If = 0°. m = tan`.axis. eliminate ‘t’ to obtain the relation in h and k and simplify this relation. If an ordinate is dropped from P to meet the line L at N. denoted by ‘m’ is tan. then the x coordinate of N will be x1. (iv)Replace h by x. The resulting equation would be the equation of the locus of P. then the locus of P is 8 x + 4y + 4 = 0 3 2x + 3y + 3 = 0. in the eliminant. Y = y .y= 1 3 1 3 from which X = 4 4 x. STRAIGHT LINE Any equation of first degree of the form ax + by + c = 0. m1 = m2 and vice versa.m2 = -1 and vice versa. 0º < 180º. where a. and k by y. 3 3 Substitute these values. x1 x2 are any two points. y2 ). y1) lying above a given line. c are constants always represents a straight line (at least one out of a and b is non zero) Slope If a straight line makes an angle ‘’in anticlockwise direction with the positive direction of x-axis. Shows a point P(x1. Putting x = x1 in the equation ax + by + c = 0 gives y coordinate of 237 N=– (ax1 c) b . m1. (a) If lines are parallel. k and known quantities. then the slope of the line. y1 ) and B(x2. Let m1 and m2 be slopes of two given lines. ¹90º. i. (b) if lines are perpendicular. x= 0 3X 0 3Y . m = 0 and the line is parallel to x-axis. then slope of the line passing through y 2 y1 A and B is given by m = x x . If h and k coordinates of the moving point are obtained in terms of a third variable ‘t’ called the parameter.CO-ORDINATE GEOMENTRY (iii)Eliminate the parameter. If A(x1. m does not exist and line is parallel to y .e. So that the eliminant contains only h. e. it would mean that P lies below the line ax + by + c = 0. then we have y1 > – (ax1 c) b i. length of y intercept = |b| (0. b Remark : If (ax1 + by1 + c) and (ax2 + by2 + c) have same signs. L ( x 1 . it would mean that P lies above the line ax + by + c = 0..e. y1 + i. y2) both lie on the same side of the line ax + by + c = 0. y1) and (x2.(1) b (ax1 c) >0 b Hence. it implies that (x1. length of x intercept = |a| x y intercept = b.. Intercept of a straight line on the axis : If a line AB cuts the x-axis and y-axis at A and B respectively and O be the origin then OA and OB with proper sign are called the intercepts of the line AB on x and y axes respectively. b) x y 1 a b x intercept = a.IIT..e. then they lie on the opposite sides of the line. y1 ) < 0. and if L ( x 1 . 0) (a. if P(x1. (ax1 by1 c) >0 b i. y1 ) > 0 .. y1) satisfies equation (1). 0) 238 ..MATHS If P(x1. where m = slope of the line = tanq c = y intercept 2 Intercept form : x/a + y/b = 1 y (0. y1) lies above the line. If the quantities ax1 + by1 + c and ax2 + by2 + c have opposite signs. Standard equations of straight lines : 1 Slope-intercept form : y = mx + c.. (Direction of increasing ordinate is called the positive direction of the line). the angle of slope being q. the slope of the line = m 4. y1) and (b) The direction of the straight line i.e. Y) Shown in the figure AP is regarded as a positive vector and AQ as a negative vector. y1). The positive direction of the line is in the sense BAP.4 Slope point form : Equation : y – y1 = m(x – x1). 0 2 and p is always positive.. From the general definitions of cosq and sinq we have cosq = x x1 y y1 . Here 2 1 y 2 y1 m= x x . where (x1. 2 1 Parametric equations of a straight line : In figure given below let BAP be a straight line through a given point A (x1.CO-ORDINATE GEOMENTRY 3 Normal form : x cos + y sin = p. (0. is the angle which the perpendicular to the line makes with the axis of x and p is the length of the perpendicular from the origin to the line. where (a) One point on the straight line is (x1. as indicated by the arrows. y1) and (x2. x x1 y y1 r cos sin 239 .0) X p+ ve L Y 4. where . y – y1 = AP sinq.y) and Q (X.4. y2) are the two given points. For the points P (x.4.5 Two point form : y 2 y1 Equation : y – y1 = x x (x – x1). sinq = AP AP or x – x1 = AP cosq. any line of the form L1 + L2 = 0 passes through a fixed point which is the point of intersection of the lines L1 = 0 and L2 = 0.MATHS or r (4 + 21 ) + 71 = 0 5 or r = – 355 . The distance between two parallel lines : The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is | c1 c 2 | a 2 b2 . Ax + By + C = 0 has the general form ax + by + c + (Ax + By + C) = 0 In which can have any real value . Rather it represents a family of straight lines. y1) on the line ax + by + c = 0 is x x1 y y1 ax by c = = – 1 2 12 . In other words if a linear expression L contains an unknown coefficient. If L1 = 0 and L2 = 0 are parallel lines. is parameter which can be evaluated specifically if some further condition is imposed. Hence the general equation of the family of lines through the point of intersection of two given lines is L + L = 0 where L = 0 and L = 0 are the two given lines. and is a parameter. y1) about the line ax + by + c = 0 is x x1 y y1 ax by c = = – 2 1 2 12 a b a b and the foot of perpendicular from a point (x1. then the line L = 0 can not be a fixed line. 240 . a 2 b2 c The length of the perpendicular from origin on ax + by + c = 0 is 2 a b2 . y1) on ax + by + c = 0 is ax1 by1 c . Reflection of a point about a line : The image of a point (x1. a b a b FAMILY OF LINES: (Equation of any straight line through the point of intersecton of two given straight lines). they will meet at infinity. here. Length of the perpendicular from a point on a line : The length of the perpendicular from P(x1. Conversely. Remarks : 1. The equation of any straight line passing through the intersection of the two lines ax + by + c = 0. The distance between A and P is thus 355/41 units. 41 the vector AP being in the negative direction of the line.IIT. CONCURRENCY OF STRAIGHT LINES : The condition for three lines a1x + b1y + c1 = 0. given below. m1 = tan 1 . a2x + b2y + c2 = 0. 2 1 m 2 m1 or. (iii) The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines... if the value is negative. In numerical examples the value of the right-hand side of (3) may be positive or negative . From (2). where k is a parameter. c3 (ii) There exist three constants . the angle is acute . from it we can calculate from the given-or deducible-values of the gradients of the two given lines... tan 2 tan 1 = 1 tan tan .. The angles of slope are 1 and 2... The family of lines perpendicular to a given line ax + by + c = 0 is given by bx .... tan = tan(2 – 1). LA. the corresponding gradients are given by Let.. the angle is obtuse.ay + k = 0. n (not all zero the same time) such thatL1 + mL2 + nL3 = 0.. if the value is positive.. m2 = tanq2 = 2 – 1 . 3...(3) This is the formula required . 241 .. where L1 = 0. m..(2) Thus.. THE ANGLE BETWEEN TWO STRAIGHT LINES : In fig.. L2 = 0 and L3 = 0 are the three given straight lines. and in the same sense as. tan = 1 m m 1 2 ...CO-ORDINATE GEOMENTRY 2. GAH and LAM are two straight lines meeting the Y K M H A 2 1 O G L X x-axis at G and L and intersecting at A. where k is a parameter. by means of (1).(1) . a3x + b3y + c3 = 0 to be concurrent is - (i) a1 b1 c1 a2 a3 b2 b3 c2 = 0.. The family of lines parallel to a given line ax + by + c = 0 is given by ax + by + k = 0. f is the angle through which GA has to be rotated about G in the counter-clockwise direction to be parallel to. e. a1 a2 + b1b2 0 and not parallel i.MATHS It is a convention to tell acute angle for the angle between the two lines. since m1 and m2 are unequal. which is the condition that the two lines should be perpendicular. then find tan = 1 m m 1 2 242 . which is otherwise obvious from (1). a1 b2 a2b1 then one of these equations is the equation of the bisector of the acute angle between two given lines and the other that of the obtuse angle between two given lines. m2 = m1. 1 2 Remarks : 1. Remarks: Whether both given lines are perpendicular or not but the angular bisectors of these lines will always be mutually perpendicular. The bisectors of the acute and the obtuse angles Take one of the lines and let its slope be m1 and take one of the bisectors and let its slope be m2. cot = m m and it 2 1 follows that. then 1 + m1m2 = 0 or m1m2 = – 1. where is the acute angle. The equations of the two bisectors of the angles between the lines a1x + b1y + c1 = 0 and a 1x b1 y c1 a2x + b2y + c2 = 0 are a12 b12 a 2 x b2 y c2 =± a 22 b 22 If the two given lines are not perpendicular i. If the lines are parallel then = 90º so that cot = 0 . by (2). from (3). If m1 m 2 be the acute angle between them. BISECTORS OF THE ANGLES BETWEEN TWO GIVEN LINES : Angle bisector is the locus of a point which moves in such a way so that its distance from two intersecting lines remains same. 1 m1m 2 2. If the lines are parallel then 2 = 1 and.e. thus. = 0 that tan = 0. For this purpose tan m 2 m1 = 1 m m .IIT. from (3). CO-ORDINATE GEOMENTRY If tan > 1 then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle. The equation of the bisector of the angle which contains a given point : The equation of the bisector of the angle between the two lines containing the point (a. The equation of reflected ray : Let L1 a1x + b1y + c1 = 0 be the incident ray in the line mirror L2 a2x + b2y + c2 = 0. k) be a point on L2. Let L3 be the reflected ray from the line L2.) is a 1x b1 y c1 2 1 2 1 a b = a 2 x b2 y c2 2 2 a b 2 2 or a 1x b1 y c1 2 1 2 1 a b =– a 2 x b2 y c2 a 22 b 22 according as a1 + b1 + c1 and a2 + b2 + c2 are of the same signs or of opposite signs. y1) and equally inclined with the lines a1x + b1y + c1= 0 and a2x + b2y + c2 = 0 are those which are parallel to the bisectors between these two lines and passing through the point P. then origin will lie inthe obtuse angle. Since L3 passes through A. (ii) Equation of straight lines passing through P(x1. If two lines are a1 x + b1y + c1 = 0 and a2x + b2y + c2 = 0. 243 . Clearly L2 will be one of the bisectors of the angles between L1 and L3. If 0 < tan < 1 then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angles. Then. For example the equation of the bisector of the angle containing the origin is given by a 1x b1 y c1 2 1 2 1 a b a 2 x b2 y c2 =+ a 22 b 22 for same sign of c1 and c2 (for opposite sign take –ve sign in place of +ve sign) Remarks: (i) If c1c2 (a1a2 + b1b2) < 0. then a 1x b1 y c1 a12 b12 C N a 2x b2 y c2 A a 22 b 22 P(x. y) M B will represent the equation of the bisector of the acute or obtuse angle between the lines according as a1a2 + b1b2 is negative or positive. then the origin will lie in the acute angle and if c1c2 (a1a2 + b1b2) > 0. so L3 L1 + L2 = 0 Let (h. k) lies on L2. a2h + b2k + c2 = 0 a12 + a22 2 + 2a1a2 + b12 + b222 + 2b1b2 = a12 + b12 = 0 or = 2(a1a 2 b1b 2 ) a 22 b 22 But = 0 given L3 = L1. then y2 – (m1 + m2)xy + m1m2x2 = 0 and y2 + 2h a xy + x2 = 0 are identical. ROTATION OF THE AXES (To change the direction of the axes of co-ordinates. y).IIT. If is the angle between the b b 244 . then x x cos y sin y x sin y cos x cos sin y = sin cos or. a 22 b 22 Remarks : Some times the reflected ray L3 is also called the mirror image of L1 in L2. with respect to which the coordinates of P are (x. both systems of co-ordinates being rectangular. are y = m1x and y = m2x. OY be given rectangular axes with respect to which the coordinates of a point P are (x. x y (in matrix form) PAIR OF STRAIGHT LINES The general equation of degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight a h g lines if h b f 0 g f c abc + 2fgh – af 2 – bg2 – ch2 = 0 and h2 ³ ab. The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin If lines through the origin whose joint equation is ax2 + 2hxy + by2 = 0. OV as a new pair of coordinate axes. y). We take OU. OV are the two perpendicular lines obtained by rotating OX. Suppose that OU. without changing the origin.MATHS | a 1h b1k c1 | 2 1 2 1 a b = | a 1h b1k c1 (a 2 h b 2 k c 2 ) | (a 1 a 2 ) 2 (b1 b 2 ) 2 .) Let OX. Since (h. Hence L3 L1 – 2(a 1a 2 b1b 2 ) L2 = 0. OY respectively through an angle a in the counter-clockwise sense. CO-ORDINATE GEOMENTRY m1 m 2 2 4m1m 2 two lines. 2 2 x my x my ax + 2hxy + by + (2gx + 2fy) c 0 is the equation of the lines n n 2 OA and OB 245 2 .e. the line not passing through origin) cuts the curve ax + 2hxy + by2 + 2gx + 2fy + c = 0 at two points A and B. then tan = ± 1 m1m 2 2 h 2 ab ab The lines are perpendicular if a + b = 0 and coincident if h2 = ab. ((n 0) i. Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Line and a Curve A B O If the lines x + my + n = 0. then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line i.e. and Solution : Since . Show that the sum of the ratios in which C and D divide AB is zero if ab’ + a’b = 2hh’. are real roots of the equation x 3 3px 2 3qx 1 0 .0).0) and D( .0 ) are such that .0).(2) Let C divides AB in the ratio : 1 246 . .MATHS WORKED OUT ILLUSTRATIONS ILLUSTRATION : 01 If .y) the centroid of ABC. are the roots of equation ax 2 2 hx b 0 and . + + = 3q. C (. Let G(x.. B (. ..q) ILLUSTRATION : 02 The four points A (. = 1 1 1 1 Let A = . B .IIT. and C . . are the roots of a ' x 2 2h ' x b' 0 then 2h ' b' and a' a' ……. Solution : Since . are those of equation a' x 2 2 h' x b' = 0. . are the roots of x 2 2hx b 0 + = 2h b and a a ……. then X= 3p p 3 3 1 1 1 3q And y q 3 3 Hence co-ordinates of the centroid of ABC are (p. are the roots of the equation x 3 3px 2 3qx 1 0 + + = 3p. (1) and . Find the centroid of the triangle 1 1 1 whose vertices are . 11) 247 1 . -7 points are (3. Then 1 and let D divides AB in the ratio : 1 then . Let P (t. 1. then y = 4-t.CO-ORDINATE GEOMENTRY 1. Solution : Let x = t. 1 but given + = 0 0 2 2 0 2h 2h ' 2h 2b ' 0z a' a a ' a or ab’ + a’b = 2hh’ ILLUSTRATION : 03 Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0. 4 – t) be an arbitrary point on the line x +y = 4 Distance of P from 4x + 3y – 10 = 0 is unity 4t 3 4 t 10 | 42 32 |t + 2| = 5 t+2 ±5 or t = -2 ± 5 t = 3.1) & (-7. 2) and B is (2.(2) Solving (1) and (2). 28) ILLUSTRATION : 05 Find the co-ordinates of the orthocentre of the triangle formed by the lines y = 0.MATHS ILLUSTRATION : 04 Find a point P on the line 3x + 2y + 10 = 0 such that |PA .PB 2 PA PB AB 2 PA PB AB | PA PB | 2 2 Maximum value of |PA . (1+t)x–ty+t (1+t)=0 and (1+u)x–uy+u (1+u) = 0 respectively.4) Solution : Let P be x1 . and show that for all values of t and u.e.PB Since cos 1 2 2 PA PB AB 2 2 PA.PB| is 2 2 when = 0 i. the orthocentre lies on the line x +y = 0. P lies on the line AB as well as on the given line. we get P (-22. y1 And APB 2 3 PA PB AB Then cos = 2 2 PA. (1+t)x – ty + t (1+t) = 0 and (1 +u) x – uy + u (1+u) = 0 t 0 .k) be the orthocentre then Slope of OB x slope of AC = -1 248 . Solution : Let equations of BC.PB| is maximum where A is (4. equation of AB is y2 42 x 4 24 y – 2 = -x + 4 x+y=6 and given line 3x +2y + 10 = 0 …… (1) …….IIT. CA and AB are y=0. Let (h.PB 2 2 1 2 PA PB AB 2 PA. we get K (t -u) = -h (t – u) h = k = 0 locus of orthocentre is x + y = 0 Putting h = -k in (1) Then we get k = -t u h = tu then orthocentre is (t. (1) and slope of OC x slope of AB = -1 k0 h t k 1 u 1 u u h t k ku uh ut 1 u ……(2) Subtracting (2) from (1) .CO-ORDINATE GEOMENTRY k 0 1 t 1 hu t k (1+t) = -th – tu k=- th tu k tk th tu 1 t ……. Show that the area of 33 11 2 c the triangle formed by the liens y = m1x. y m 2 x and y c is 4 Solution Since m 1 and m 2 are the roots of the equation x2 32x 3 1 0 then m1 m 2 3 2 249 m1 m 2 = 3 4 4 = 11 3 1 0 m1 m 2 2 4m1m 2 34 34 .u – tu) ILLUSTRATION : 06 If m 1 and m 2 are the roots of the equation x 2 32x 3 1 0 . 1) is 7x – 4y + = 0 250 . Find the equations of other three sides Solution : Since (-3.1) does not lie on AD Co-ordinate of C is (1.1).1) lies on 4x + 7y + 5 = 0 And (1.1) is 7x – 4y + = 0 . (c/m1.1) and (1.IIT. Hence the required area of triangle = 0 1 c 2 m1 c m2 0 1 c 1 c 1 = 1 2 1 1 c 2 m1 m 2 = 1 2 m 2 m1 c 2 m 1m 2 = 1 2 c 2 = 1 2 c . Two of its vertices are (-3.c). equation of AD is 7x – 4y + 25 = 0 (1.0).21 – 4 + = 0 = 25 Therefore. 2 11 3 1 33 11 2 c } 4 3 1 11 3 1 3 1 ILLUSTRATION : 07 One side of a rectangle lies on the line 4x + 7y + 5 = 0.c) and (c/m2.1) Equation of BC which is parallel to AD and Passing through (1.MATHS and co-ordinates of the vertices of the given triangle are (0. which is to 4x +7y + 5 = 0 And passing through (-3.1) does not lie on 4x + 7y + 5 = 0 Equation of AD. 2k + sin ] Points B. AD are in H.C. Solution : Given lines are 7x – y – 16 = 0 …… (1) 5x – y – 8 = 0 …… (2) x – 5y +8 = 0 …… (3) Let the equation of line passing through A (k+1. r2 . AD = r3 ) B [(k+1) + cos . AB. r3 ) (if AB = r1 . r2 .D respectively. AC = r2 . 2k + sin ] C [(k+1) + cos . (2) and (3) respectively Then r1 r2 = r3 251 91 k 7 cos sin 31 k 5 cos sin 91 k 5 sin cos .CO-ORDINATE GEOMENTRY 7x 1 – 4 x 1 + = 0 = -3 equation of AD is 7x – 4y –3 = 0 Equation of DC which is parallel to AB passing through (1.1) is 4x + 7y + = 0 4 x 1 + 7 + 1 + = 0 = -11 Therefore equation of DC is 4x + 7y –11 = 0 ILLUSTRATION : 08 A line through the variable point A (k+1. Prove that AC. C. D satisfying (1). 5x – y – 8 = 0 at B. 2k + sin ] D [(k+1) + cos . 2k) meets the lines 7x +y – 16 = 0. 2k) making an angle with the + v e direction of x – axis be x k 1 y 2k r1 .P. r3 cos sin (if AB = r1 . 3. 252 . where r = 1.P. ILLUSTRATION : 09 The three sides of a triangle are L r x cos r y sin r p r 0 .MATHS 1 1 5 cos sin 5 sin cos r2 r3 31 k 91 k = 15 cos 3 sin 5 sin cos 91 k = 14 cos 2 sin 91 k 2 = r 1 Hence r2 .IIT.2. Show that the orthocentre is given by L1 cos 2 3 L 3 cos 3 1 L 3 cos 1 2 Solution : The given lines are L1 x cos 1 y sin 1 p1 0 L 2 x cos 2 y sin 2 p 2 0 L 3 x cos 3 y sin 3 p 3 0 Now equation of AD is …. r3 are in H. (1) L 2 L 3 0 (xcos 2 y sin 2 p 2 ) + (x cos 3 y sin 3 p 3 ) 0 (xcos 2 cos 3 ) + (x sin 2 sin 3 )p 2 p 3 0 cos cos 2 3 Slope of AD = sin sin 2 3 cos 1 and slope of BC = sin 1 Since AD BC Slope of BC x slope of AD = -1 cos 2 cos 3 cos 1 1 sin 2 sin 3 sin 1 cos 1 cos 2 cos 3 cos 1 sin 1 sin 2 sin 3 sin 1 cos 1 2 cos 3 1 0 cos1 2 cos3 1 Now from (1). r1 . 0) for any position of U in the plane. ) is a variable point on the plane.y 6 6 6 6 .. AU and BU meet the y – axis at C and D respectively and AD meets OU at V. the equation of AD is x 6 y 1 3 6 and the equation of OU is x = y Solving (1) and (2). we can obtain equation of altitude BE as L 3 cos1 2 L1 cos 2 3 . we get L1 cos 2 3 L 2 cos 3 1 L 3 cos 1 2 ILLUSTRATION : 10 A (3.. we get x 6 6 .0) are two fixed points and U (.0) and B (6. 3 3 Now.. Solution : The equation of BU is Y-= 0 x 6 So that the coordinates of D are 0.……. Prove that CV passes through (2.(3) From (2) and (3). 6 6 Similarly the coordinates of C are 0. …. Hence coordinates of V are 6 6 Then the equation of CV is 6 3 3 y 6 3 x 0 6 3 0 6 253 ..CO-ORDINATE GEOMENTRY L2 cos 1 2 L3 0 cos 3 1 L 2 cos 3 1 L 3 cos 1 2 .(2) Similarly. k) then h = r3 cos . A point S is chosen on the variable line such that mn m n . ILLUSTRATION : 11 A variable line is drawn through O to cut two fixed straight lines L 1 and L 2 in R and S. P r3 cos . r2 sin .OS r2 and OP r3 Then co-ordinates of R. Show that the locus P is a OP OR OS straight line passing through the point of intersection L 1 and L 2 . R lies on and lies on L 1 and S lies on L 2 r1 sin c and ar2 cos br2 sin 1 r1 c 1 and r2 sin a cos b sin From the given condition mn m n r3 r1 r2 From the given condition mn m n r3 r1 r2 mn m n r3 r1 r2 mn m sin n a cos b sin = r3 c …… (1) Let co-ordinates of P be (h. S and P are R r1 cos .MATHS y 3 9 x 3 63 y= 3 x 1 3 2 Which pass through the point (2.IIT. Sr2 cos . r1 sin . mn mr3 sin n ar3 cos br3 sin c mn mk n ah bk c locus of P is 254 . k r3 sin from (1).0) for all values of (. r3 cos . Solution : Let the equation of the variable line through ‘O’ be x y and let cos sin OR r1 . ). CO-ORDINATE GEOMENTRY n ax by my m n c y n ax by 1 m 1 0 c ax by 1 m y c 0 nc L1 m L2 0 nc m where nc L1 L 2 0 Locus of P is point of intersection of L 1 and L 2 255 . 3y sin = 6 A) 5 B) independent of q C) cos2q D) 7sin2q 256 .2) 5 . D) hyperbola The equation of a line which passes through (acos3. t - 1 2 9t ) 2 where u.4) or (13.0) C) (2. 2sin) internally in the ratio 2 : 3 for all is a A) straight line 8. C) a parabola The equation of the line passing through the intersection of the line x – 3y + 1 = 0 and 2x + 5y – 9 = 0 and 0 at distance 5 from the origin is A) 2x – y = 5 6. C) an ellipse If A and B are fixed point then the locus of a point which moves in such a way that the angle APB is a right angle is A) a circle 4.-2) D) (1. D) hyperbola One vertex of an equilateral triangle with centroid at the origin and one side as x+y–2=0 is A) (-1.0 to the straight line 2x cos . B) an ellipse B) x + 2y = 5 D) x + 2y = 1 C) (-2.0). g are constants. a sin3) and perpendicular to the line xsec + ycosec = a A) xcos . u sina.13) D) (1.0 5 . B) a parabola B) (1.1) C) (2. B) (7.ysin =2a cos2 C) xsin + ycos = 2a cos2 5.0) and (2 cos. The locus of the moving point is A) a circle 3. C) 2x + y = 5 B) (2.t.-2) respectively and P is a point such that PA = PB and area of triangle PAB = 10 sq units then co-ordinates of P are A) (7.2) 10.2) or (1. (0.1) The position of a moving point in the x – y plane at time is given by (ucosa.1) B) circle C) pair of st lines D) parabola The equations of the lines through (-1. If A and B are two points having co-ordinates (3.7) or (4.MATHS SECTION A SINGLE ANSWER TYPE QUESTION 1.1) The product of the perpendiculars from the points D) (2.4) and (5. -1) 7.2) or (2. a.1) to a variable line is zero.2) A) x2 – xy + x – y = 0 C) xy +x + y = 0 9.-1) and making an angle 45° with the line x + y = 0 are given by B) xy + x – y – y2 = 0 D) xy + x +y +1 = 0 Let the algebraic sum of the perpendicular distances from the points (2.2) 2.IIT. then the line passing through a fixed point whose co-ordinates are A) (1. B) xsin + ycos = 2a sin2 D) xsin -y cos =2a sin2 The locus of a point P which divides the line joining (1.2) and (1. x + 3y – 4 = 0 form a triangle which is A) isosceles 21. D) (-1. D) –1 The equation of the line through the point of intersection of the lines x – 3y + 1 = 0 and A) 3x + 2y – 7 = 0. D) 24 If co-ordinates of the vertices B and C of a ABC are (0. -1) and parallel PS is A) 2x – 9y – 7 = 0 C) 2x + 9y – 11 = 0 257 5 are The number of integer values of m for which the x –coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is A) 2 23.-1) 12. 3x – 5y + 2 = 0 D) 2x – y + 5 = 0 B) equilateral C) right angled D) none of these B) 3 : 4 C) 2 : 1 D) 4 : 3 B) 0 C) 4 D) 3 Let PS be the median of the triangle with vertices P(2.CO-ORDINATE GEOMENTRY 11.1) and (-1. C) (1.2) Q(6.3) respectively then value of cosB + cosC is A) 1 16.3). D) 1 C) ½ 2x + 5y – 9 = 0 and whose distance from the origin is 19.2) and (0.1) and (-1. D) 10 units A straight line through the origin 0 meets the parallel lines 4x + 2y = 9 and 2x + y +6 = 0 at points P and Q respectively then the point O divides the segment PQ in the ratio A) 1 : 2 22.1) B) 2x – 9y – 11 = 0 D) 2x + 9y + 7 = 0 The orthocentre of the triangle formed by the lines xy = 0 and x +y = 1 is .0) and (5. C) 2 : 1 externally 2 2 If the points at 1 2at 1 at 2 at 2 and (a.1) B) .0) are collinear then value of t 1t 2 is A) –1 15. The equation of the line passing through (1. 5x – 7y + 12 = 0 C) 2x + y – 5 = 0 20.54 B) –2 C) 2 B) 0 B) 2 B) 5 units C) 1 C) triangle not possible B) 2x + y – 7 = 0. B) (-1. D) 1 : 2 externally Let equation of the BC of a ABC is 2x + 3y –1 = 0 co-ordinates of the incentre and circum centre of ABC are (2.-1) (3. C) a and b both true Number of lines passing through (3.0) respectively and co-ordinates of incentre of ABC is (3.4) and whose difference of the intercept is 2 A) 4 17. Circum center of the triangle whose vertices are (2. 3x + y – 4 = 0. D) ½ The straight line x + y = 0.-1) Ratio in which the join of (2.4) then length of the side AC is A) 25 units 18.-1) R (7. B) 1 : 1 externally Distance between the lines 5x + 12y – 1 = 0 and 10x + 24y + k = 0 is 2 then the value of K is A) 50 14.2) is divided by the line x + 3y + 5 = 0 A) 1 : 1 internally 13.3) is A) (1. 0) and B(2. C) (0. B) 5 B) 1/2 C) 2 D) 3 The vertices of a triangle are A(-1. B (5. the same line has intercepts p and q.IIT. The locus of the middle point of AB is B) a circle C) a parabola D) an ellipse The range of values of the ordinate of a point moving on the line x=1. when the axes are rotated through a given angle.-7) . then A) a straight line 31. B) 0 1 1 1 1 1 1 1 1 A) a 2 b 2 p 2 q 2 B) a 2 b 2 p 2 q 2 C) a 2 p 2 b 2 q 2 D) a 2 p 2 b 2 q 2 Two points A and B move on the x–axis and the y–axis respectively such that the distance between the two points is always the same. Area of the triangle formed by the lines y2 – 9xy +18x2 = 0 and y = 9 is A) 27/4 29.1).1) 32. then the line ax + by + c = 0 A) has a fixed direction B) always passes through a fixed point C) form a triangle with the axes whose area is constant D) always cuts intercepts on the axes such that their sum is zero 28.1) and C(1. 24. 30. C) 4 B) x + 7y – 2 = 0 C) x + 6y + 2 = 0 D) x – 7y – 2 = 0 If a. 3 3 3 1 A) 2 .MATHS 1 1 2 2 1 1 3 3 A) . 2 3 3 3 1 . keeping the origin fixed.4). 1 1 4 4 D) . and always remaining in the interior of the triangle formed by the lines y=x. is A)(0. b are in G. c. B) . 2 2 B) 1 3 1 3 . D) 27 Line L has intercepts a and b on the co–ordinate axes.4) D) None of these Let A=(1. C) 9/3 B)(0.1) C)(0.0) The area of the triangle formed by joining the origin to the points of intersection of the line 5x 2y 3 5 and circle x2 + y2 = 10 is A) 6 25. The line AB turns about A through an angle /6 in the clockwise sense. The equation of the bisector of ABC is A) x – 7y + 2 = 0 27. the x–axis and x+y=4. The B’ has the coordinates.P. and the new position of B is B’. 2 2 C) D) None of these 258 . D) 3 The area bounded by the curves x + 2|y| = 1 and x = 0 A) 1/3 26. 1 D) tan If the lines x . The perpendicular bisector of AB meets the line through (0.0) respectively. The bisector of the acute angle formed between the lines 4x–3y+7=0 and 3x–4y+14=0 has the equation A) x+y–7=0 34. D) =4 C) (0. D) None of these 2 2 1 t m tan C) t 2 m2 B) /2 A) =2 37.2). b) If P(1+t/ 2 . C) 3x+y–11=0 B) 91 sq units C) 48 sq units D) 100 sq units The line PQ whose equation is x – y = 2 cuts the x – axis at P and Q is (4.–1) B) =–3 3 x+y=4 5 2 5 1 2 2 1 5 2 2 B) . 2 The coordinates of two consecutive vertices A and B of a regular hexagon ABCDEF are (1. The equation of the line PQ in the new posi- . include an angle A) /3 36. D) x=2y–12=0 The equations of the three sides of a triangle are x=2.b) be an end of a diagonal of a square and the other diagonal has the equation x–y=a then another vertex of the square can be A) 39. 3x + y . B) x2 = 9y2 D) None of these D) None of these C) x2 – 9y2 = 0 D) y2 – 4x2 = 0 The point P (1.0) respectively. 2 5 2 D) 5 . then the value of expression 4p2 +P’2 B) 3a2 C) 2a2 D) 4a2 The line 3x + 2y = 24 meets y – axis at A and x – axis at B. D) (a+b.1) is translated parallel to y = 2x in the first quadrant through a unit distance.4 = 0 and lx + 4y + l2 = 0 are concurrent.–a) B) x+ 3 y+4=0 A) y2 = 9x2 43.y sin = a cos2 respectively.4) and (6.1 1 5 B) 1 1 . 41. Then area of D ABC is A) 182 sq units 259 C) x+ 3 y=4 Let O be the origin and A. t m 2lm 2 If (a.1 5 5 1 C) 5 . C) =4 B) (a. B)x–y+3=0 B) (2. C) .4) The diagonals of the parallelogram whose sides are lx+my+n=0. The co-ordinates of the new position of P are 2 5 . 1 5 If P and P’ be the perpendiculars from the origin upon straight lines xsec + ysec = a and xcos .-1) parallel to x – axis at C. mx+ly+n=0.6 = 0.a) 38.0) A) . C) (0. The equation of the diagonal CE is 3 2 42. y+1=0 and x+2y=4. The co–ordinates of the circumcentre of the triangle are A) (4. B be the two points having co-ordinates (0.0) 35. The line PQ is rotated about P through 45° in the anticlockwise direction. If a point P moves in such a way that the area of the OPA is always twice the area of POB then P lies on A) a2 45.2y .2+t/) be any point on a line then the range of values of t for which the point P lies between the parallel lines x+2y=1 and 2x+4y=15 is A) 1 44. lx+my+ n =0. then A) (a–b.CO-ORDINATE GEOMENTRY 33.0) and (2. 16 16 11 11 .0) and R (3. then the intercepts made by the line L on the new axes are respectively A) 1 and 50. C) The equation of the lines through the point (2.1).2) and equation of the perpendicular bisectors of AB and AC are 3x + 4y – 1 = 0 and 4x + 3y – 5 = 0 then the equation of median AD is 47.2 2 3 B) 2 3 x y 1 54.3) and making an intercept of length 2 units between the lines y +2x =3 and y + 2x = 5 are A) x + 3 = 0. 4x – 3y = 6 D) none of these Let P (-1. If the equation of the line QR is 2x +y = 3 then the equation representing the pair of lines PQ and PR is A) 3x 2 3y 2 8xy 20 x 10y 25 0 B) 3x 2 3y 2 8xy 20x 10 y 25 0 C) 3x 2 3y 2 8xy 10x 15y 20 0 D) 3x 2 3y 2 8xy 10x 15y 20 0 In an isosceles triangle ABC. 3x +4y = 18 52. 8 8 D) A ray of light traveling along the line x + y = 1 is incident on the x – axis and after refraction is incident on the x-axis and after refraction it enters the other side of the x-axis by turning /6 away from the x-axis. 31 31 . 3 3 ) be three pts then the equation of the bisector of the angle PQR is A) 53. If co-ordinates of its orthocentre and circum centre are (1.IIT.0) Q (0. 16 16 C) Let L be the lines 2x + y = 2. 16 16 A) 49.MATHS tion is A) y = 2 46. The equation of the line along which the refracted ray travels is A) x + 2 3 y 1 51. B) 2 2 2 .1) and (2. 2 and 1 D) 1 and 2 3 C) y 2 3 x 2 D) x y 4 B) y – 2 = 0. If the axes are rotated by 45° without transforming the origin.6) respectively and co- 21 21 ordinates of its orthocentre is . then radius of the circum circle of ABC is A) 3 48.0) respectively. 15 15 . B) 3x x y0 2 B) x 3y 0 C) 3x y 0 D) x 3 y0 2 Let PQR be a right angled isosceles triangle.0) and (0. B) y = 2 C) x = 2 D) x = -2 I f the co-ordinates of the vertex A of a ABC is (1. 3x + 4y = 12 C) x – 2 = 0. then co-ordinates of its circum center is 4 4 13 13 . the coordinates of the points B and C on the base BC are respec260 . B) 10 C) 2 2 D) 2 Let co-ordinates of the vertices A and B of a triangle ABC are (6. A) 11x – 10 y + 9 = 0 B) 10 x – 11y + 12 = 0 C) 3x + 4y –11 = 0 D) 3x + 4y + 12 = 0 Let the equation of the side BC of ABC is x + y + 2 = 0. right angled at P(2. 1). If p denote the perpendicular distance of a side of this square from the origin. 3y+x–3=0 A) (1. one of which has the equation 3x + 4y = 2.3). a square is constructed away from the origin. 3) and its orthocentre is (–6.y) be a point such that (x-1) (x-3) + (y-2) (y-4) = 0. C) y = 1 X then the equation of the line AC is 2 5 . If the equation of the line AB is y = A) 2y = x + 3 55.6) C) (1. Then its third vertex is B) (–1. D) 3x + 4y = 2 B) x – y = 1 C) x = 0 5 2 D) x =1 B) c2 – a2 = m2 C) 2c2(1+m2) = a2 D) c2 + a2 = m2 C) y–3x+9=0.4) and let C (x.6) then the equation of BC is . If ar D (ABC) =1 then the maximum number of position of C in the x – y plane is A) 2 56. side AB has the equation 2x+3y=29 and the side AC has the equation x+2y=16.1) and (1. The co–ordinates of the point A is 3 .2) B (3. with this portion as one of its side. C) 8 If a ray traveling along the line x = 1 gets reflected from the line x + y = 1 then the equation of the line along which the reflected ray travels is A) y = 0 58.2). B) 3x + 4y = If one of the diagonal of a square is along the line x=2y and one of its vertices is (3.6) 62. 3y–x=3=0 D) y–3x+3=0.0 D) None of these In a ABC. D) y = x –1 There are two parallel lines.–6) D) None of these A ray of light coming from the point (1. 3y+x–3=0 A) 63 62 2 A) y– 3x+9=–.0) then its sides through this vertex are given by the equation A) 61.CO-ORDINATE GEOMENTRY tively (2. If the mid point of BC is (5. 3y+x+9=0 On the portion of the straight line x+y=2 which is intercepted between the axes. D) 5 The pair of straight lines joining the origin to the common points of x2 +y2 = a2 and y = mx +c are perpendicular to each other if A) 2c2 =a2 (1+m2) 59.–2) and (–2.0 13 B) C) 7. If the lines cut an intercept of length 5 on the line x +y = 1 then the equation of the other line is A) 3x 4 y 57. B) y = 2x B) 4 6 2 2 60.0 5 261 C) 3x + 4y = 7 B) y+3x+9=0. then the maximum value of p is 2 B) 2 2 C) 3 2 D) 4 2 Two vertices of a triangle are (3.2) is reflected at a point A on the x–axis and then passes through the point (5. 1 x 1 2 Let A = (1. 3) Two sides of an isosceles triangle are given by the equation 7x – y + 3 = 0 and x+y–3=0.y + 1 = 0 70. Then length of PQ is 262 .MATHS D) None of these B) x = -3y C) y = 2x + 1 D) 2x + 6y +1 = 0 The ratio in which the line 3x + 4y+2 = 0 divides the distance between 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is A) 7 : 3 72. B) 3 : 7 B) 3x – 4y +25 = 0 C) 4x + 3y – 25 = 0 D) 4x – 3y +25 =0 Point Q is symmetric to P(4.2) the diagonal BD has the equation B) x + y – 1 = 0 C) x + y + 1 = 0 D) x + y – 7 = 0 If each of the points (x. -1) and (5. 3x + y – 7 = 0 C) x – 3y – 31 = 0 . 88 3 B) C) 1 B) x 2 1 y 2 4 p 69.-3) then the point P(x1y1) lies on the line A) x = 3y 71. 4 7 The locus of the midpoint of the position intercepted between the axes by the line x cos + y sin = P where P is a constant is A) x2 + y2 = 4p2 66. 3x + y + 7 = 0 In a rhombus ABCD the diagonals AC and BD intersect at the point (3.-1) with respect to the bisector of the first quadrant.3) by the line x – y = 0 is A) x . D) None of these 3 2 2 C) x y p 2 2 A) 2(x2 + y2) = a2 + b2 C) x2 + y2 = a2 + b2 67. C) 2x–y=17 ABC is an equilateral triangle such that the vertices B and C lie on two parallel lines at a distance 6. 4) and (-2.A) 2x+y=7 64.-3) C) (-1. 4 The locus of the point of intersection of lines x cos + y sin = a and xsin . B) x+y=1 IIT.y) lies on the line joining the points (2. If its third side passes through the point (1. 3x + y – 31 = 0 B) x – 3y – 31 = 0 . 1 D) The image of the point (-1.-4) then the equation of the line is A) 3x – 4y = 25 73. If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is A) 8 65. C) 2 :3 D) 1 : 2 If the foot of the perpendicular form the origin to the straight line is at the point (3. If the point A is (1.1) D) (3.4). -1) 68. x 2 1 y 2 2 p2 B) x2 – y2 = a2 – b2 D) x2 + y2 = a2 – b2 B) (1.-10) then its equation are A) x – 3y – 7 = 0 .ycos = b is ( is a variable) A) (3. 3x + y + 7 = 0 D) x + 3y – 31 = 0 . 1 x at an angle 45° is C) y = ½ C) 2 2 2 2 m n D) y =1 D) 4 B) m n 1 C) m n 1 D) m n If the vertices P. 2 . y = nx and y = nx+1 equals A) 85.0).CO-ORDINATE GEOMENTRY A) 5 2 B) 5 74.1) 7 7 2 2 B) .1) 75. which of the following points of the triangle PQR is always irrational A) Centroid 86.3) and making intercepts of equal .5) (3. 2 2 C) Locus of all such points which is equidistant from (1. D) Area of the parallelogram formed by the lines y = mx. 5 In center of the triangle whose vertices are (6. 3 are A) concurrent 79. B) B) y = ¼ B) 2 mn 263 2 .-1) 9 9 2 2 C) (1.2) then area of triangle is A) 5 sq units 77. D) Orthocentre of the triangle whose vertices are (1. y = mx +1. D) 1. B) (1.1 If the points a 1b1 a 2 b 2 a 3b 3 are collinear then the lines aix biy 1 0 i = 1.2) 82.1) (3. C) (0.2) and x – axis is A) x2 – 2x – 4y + 5 = 0 C) x2 – 2x + 4y + 5 = 0 78. 11 11 .-1) (2.0) is A) 10 80.R of a triangle PQR are rational points.-1) (2.6) and (7. (0.7) is A) .3) (3.|x| +1 is A) 1 84. B) 6 sq units D) 4 sq units B) x2 + 2x – 4y + 5 = 0 D) x2 + 2x + 4y – 5 = 0 B) identical B) C) parallel D) none 1 10 2 C) 5 2 D) 5 a2 2 D) 3a 2 Area of the quadrilateral formed by |x| + |y| = a is A) a 2 81.1 The line which is parallel to x-axis and crosses the curve y = A) x = ¼ 83.Q. 76. 2. D) (3.0) is A) (-1. C) 11/2 sq units Distance between orthocentre and circumcentre of the triangle whose vertices are (3.1) (0. B) Incentre C) circumcentre D) orthocentre The equation of the straight line passing through the point (-2.2) The area a bounded by the curves y = |x| -1 and y = .5) If the Co-ordinates of the midpoints of the sides of a triangle are (1. C) 2 5 B) 2a 2 C) If one vertex of an equilateral triangle of side 2 is the origin and another vertex lies on the line x = x 3 y then the third vertex can be A) (0. 1) C) (1. D) x +y –2 =0 The slope of one of the lines represented by ax2 + 2hxy +by2 = 0 be n times the other than n 1 n A) ab 2h 89. where a. a/2.P C) b.2).2) and the axes enclose an area . If ar(DABC) = 1 then maximum number of positions of C in the x–y plane is.2) meets the x and y–axis at A and B respectively. C) a – b = 2h If bx +cy=a. c are in G.0) and (0. 2a. Then locus of middle point of all such line segment is a A) parabola 90.2) D) None of these B) 2x 1 y 1 1 C) x 1 2 y 1 1 D) 2x 1 2 y 1 1 B) ay–bx+2b=0 C) ax+by+2b=0 D) None of these If a ray traveling the line x=1 gets reflected the line x+y=1 then the equation of the line along which the reflected ray travels is A) y=0 95.1) 92. –2a.a. The intercepts on the axes made by the line are given by the two roots of A) x 2 2 | | x | | 0 B) x 2 | | x 2 | | 0 C) x 2 | | x 2 | | 0 D) None of these Let A=(1.y) be a point such that (x–1)(x–3)+(y–2)(y–4)=0.P B) (2.c are in G.c are the same sign. 2 96. be a line such that the area enclosed by the line and the axes of reference is 1/8 unit2 then A) b. 2 n 2h B) ab 1 n A line passing through P(4.b.1). D) (a . P 91. The line L will always pass through.c are in G. 2 1 2h C) ab 1 n The straight line y=x–2 rotates about a point where it cuts x–axis and becomes perpendicular on the straight on the line ax+by+c=0 then its equation is A)ax+by+2a=0 94.2) from the line is equal to zero.D) b.IIT. If O is the origin. B) x – y = 5 B) x–y=1 C) x=0 D) None of these A line passing through the point (2.(2. then locus of the center of the circumcircle DOAB is A) x 1 y 1 2 93. A) 2 B) 4 C) 8 D) None of these 264 .b)2 = 4h2 L is a variable line such that the algebraic sum of the distances of the points (1. C) x – y + 5 = 0 If one of the lines of the pair ax2 + 2hxy + by2 = 0 bisects the angle between positive direction of the axes. B=(3.P. c are in A. D) none of these B) ellipse C) hyperbola D) circle B) b. A) (1.4) and let C=(x. B) a + b = -2h Let AB be a line segment of length 4 with the point A on the line y=2x and B on the line y=x.MATHS length on the axes is A) 2x + y + 1= 0 87. then a. b. h satisfy the relation A) a + b = 2h 88. If one vertex of an equilateral triangle is at (2. If the extremities of the base of an isosceles triangle are the points (2a. then the length of each side is A) 3 2 B) 2 3 C) 2 3 D) 3 2 105.2) are vertices of a ABC then as varies the locus of its centroid is A) x2 + y2 – 2x . A variable line meet the axes at P and Q so that BP is always perpendicular to AQ. Locus of the centres of the circles touching the line 3x – 4y + 1 = 0 and 12x + 5y – 1 = 0 are A) 21x+77y –18 = 0 265 B) 99x–27y+8=0 C) (A) and (b) both D) none of these .1) is rotated through A in anticlockwise direction through an angle 15°. then the equation of the line L is A) x + 5y + 5 = 0 B) x + 5y ± 2 0 C) x + 5y ± 5 0 D) x 5y 5 2 0 102.0) and (0. The equations of the lines on which the perpendicular form the origin make 30° angle with x – axis and which form a triangle of area 50/ 3 with axes are A) x + 3 y ± 10 = 0 B) 3x y 10 0 C) x 3y 10 0 D) 3x y 10 0 103. cos) C(1. Let co-ordinates of the two fixed points A and B are (a. B) (5.4y + 1 = 0 B) 3 (x2+ y2) – 2x – 4y + 1 = 0 C) x2 + y2 – 2x – 4y + 3 = 0 D) x2 + y2 +2x + 4y – 3 = 0 100. Then locus of the point of intersection of BP and AQ is B) x 2 y 2 ax by 0 D) x 2 y 2 ax by 0 A) ax + by + a + b = 0 C) y 2 4a x b 107.2) C) (2. If a line joining points A(2.CO-ORDINATE GEOMENTRY 97.–4) B) |a| = 1 C) |a| < 1 D) |a| < 1/2 If A (cos.0) and (0.2) 106. sin) B (sin. The limiting position of the point of intersection of the lines 3x+4y=1 and 91+c)x+3c2y=2 as c tends to 1 is A) (–5.4) 98.1) then co-ordinates of a point which is equidistant from the vertices of the triangle is A) (2.a) and the equation of one of the sides is x = 2a then the area of the triangle is A) 5 sq units B) 5/2 sq units C) 25/2 sq units D) 2 sq units 104.b) respectively.2) B) (3. If a straight line L perpendicular to the line 5x – y = 1 such that the axes of the D formed by the line L and the co-ordinate axes is 5.0) and B(3. If co-ordinates of orthocentre and centroid of a triangle are (4.3) D) (1.-1) and (2.a) falls between the lines x + y = 2 then A) |a| = 2 99.–5) D) None of these If the point (a. then the equation of the line in the new position is A) 3x y 2 3 B) 3x y 2 3 C) x 3y 2 3 D) 3x y 3 101. C) (4.-1) and the base is x + y – 2 = 0. The equation of perpendicular dropped from C to the internal bisector of angle A is A) y – 5 = 0 B) x – 5 = 0 C) 2x + 3y – 7 = 0 D) x + 5 = 0 112. If a pair of lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 is such that each pair bisects the angle between the other pair then 266 .).1) B) [0.5) respectively. The range of values of the ordinate of a point moving on the line x = 1 always remaining in the interior of the triangle formed by the lines y = x and the x –axis. sin) and Q (cos ( . The equation of the line parallel to x – 4y = 0 that divides the quadrilateral in two equal areas is A) x – 4y + 5 = 0 B) x – 4y –5 = 0 C) 4y = x + 1 D) 4y + 1 = 0 116. x +y =4 is A) (0. Equation of the line passing through (1.4] D) (0.MATHS 108.0) the centroid of the triangle is 1 3 A) 2 . P(3. Let 0 < < /2 be a fixed angle.1] C) [0. In the ABC the co-ordinates of B are (0.1) and give an intercept between the lines 5x+12y+7=0 and 5x + 12y – 32 = 0 of length 3 unit is A) 12x – 5y – 7 = 0 B) 12x+5y+7 = 0 C) 12x + 5y – 7 = 0 D) 12y + 5x – 7 = 0 109.-2) (5. 2 5 1 B) 3 .y) are three points such that the angle PRQ is a right angle and the area of PQR = 7 then the number of such points R is A) 0 B) 1 C) 2 D) 4 111. The angle between a pair of tangent s drawn from point P to the circle x 2 y 2 4x 6 y 9 sin 2 cos 2 0 is 2. 3 4 3 1 C) 3 . If P (cos. Equations of the straight lines.1) (4. 3 D) none 115. inclined at 30° to the axis of x such that the length of its (each of their) lines segments between the co-ordinates axes is 10 units is A) x 3y 5 3 0 C) x 3y 5 3 0 B) x 3y 5 3 0 D) x 3y 5 3 0 110. The equations of locus of point ‘p’ is A) x 2 y 2 4x 6 y 4 0 B) x 2 y 2 4x 6 y 9 0 C) x 2 y 2 4x 6 y 4 0 D) x 2 y 2 4x 6 y 9 0 114.2) 117.0) AB = 2 ABC = /3 and the middle point BC has the co-ordinates (2.) then Q is obtained from P by A) clockwise rotation around origin through an angle B) anticlockwise rotation around origin through an angle C) reflection in the line through origin with slope tan D) reflection in the line through origin with slope tan /2 113. sin ( .5) and R(x. The four sides of a quadrilateral are given by the equation xy (x-2) (y-3) = 0.IIT.1) Q(6. The vertices of a triangle ABC are (1. The point R on the x–axis such that PR+RQ is the minimum is 5 3 A) .0). o C) (3. Then the final co–ordinates of the point are A) (4.a) B)(a.4) 121. In the ABC.2) B) (3.0 1 3 B) . If () be an end of a diagonal of a square and the other diagonal has the equation x–y= then another vertex of the square can be A) (a–b.2). ABC= D) (7/2. 2 B) .0).1) undergoes the following three transformations successively (I) Reflection about the line y=x 267 . Let P=(1.4) C) (1. There are two parallel lines. AB=2.1)undergoes the following two successive transformations : A) reflection about the line y = x B) rotation through a distance 2 unit along the positive x–axis. 3 3 1 3 C) D) None of these. 122.CO-ORDINATE GEOMENTRY A) pq = -1 1 1 C) p q 0 B) pq =1 1 1 D) p q 0 118. 3 4 3 1 .0) C) (0.a) D) None of these 125.1) and Q=(3. 2 / 2 ) be any point on a line then the range of values of t for which the point P lies between the parallel lines x+2y=1 and 2x=4y=15 is A) 4 2 5 2 3 6 B) 0 5/ 2 6 C) 4 2 0 3 D) None of these 120. The point (–4. 7/2) and the middle point of BC has 3 the coordinates (2. the coordinates of B are (0. If the lines cut an intercept of length 5 on the line x+y=1 then the equation of the other line is A) 3x 4 y 6 2 2 B) 3x 4 y 62 2 C)3x+4y=7 D) None of these 123. The point (4. The centroid of the triangle is 1 3 5 A) 2 .0) D)None of these 124. one of which has the equation 3x+4y=2. The four straight lines given by the equations 12x 2 + 7xy – 12y2 = 0 and 12x2 + 7xy – 12y2 – x + 7y –1 = 0 lie along the sides of a A) square 119 B) parallelogram C) rectangle D) rhombus If P (1+ / 2 . IIT- MATHS (II) Transformation through a distance 2 units along the positive direction of x–axis (III) Rotation through an angle p/4 about the origin in the anticlockwise direction. The final position of the point is given by the co–ordinates 1 A) 2 , 7 2 B) 2,7 2 C) 1 , 2 7 2 D) 2 ,7 2 126. P is a point on either of two lines y– 3 x 2 at a distance of 5 units from their point of intersection. The co–ordinates of the foot of perpendicular from P on the bisector of the angle between them are 1 1 4 5 3 or 0, 4 5 3 ,( depending on which line the point P is taken) 2 2 1 45 3 2 A) 0, B) 0, C) 0, 1 45 3 2 5 5 3 D) 2 , 2 127. The equation of a line through the point(1,2) whose distance from the point(3,1) has the greatest possible value is A) y=x B) y=2x C) y=–2x D) y=–x 128. The point P(2,1) is shifted by 3 2 parallel to the line x+y=1, in the direction of increasing ordinate, to reach Q. The image of Q by the line x+y=1 is. A)(5,–2) B)(–1,–2) C)(5,4) D)(–1,4) 268 CO-ORDINATE GEOMENTRY SECTION -B MULTIPLE ANSWER TYPE QUESTIONS 1. The points (2, 3) (0, 2) (4, 5) and (0,t) are concyclic of the value of t is A) 1 2. 3. B) 1 C) 17 The point of intersection of the lines D) 3 x y x y =1 and 1 lies on a b b a A) x-y = 0 B) (x+y) (a+b) = 2ab C) (lx + my) (a + b) = (1 + m) ab D) (lx–my) (a+b) = (1-m) = ab The equations (b-c)x + (c-a) y + a-b= 0 (b3-c3) x + (c3-a3)y + a3-b3 = 0 will represent the same line if A) b =c 4. 6. 269 C) x2-y2 = 2(ax+by) B) bx = ay B) = -3 C) = 4 B) x + y - 3 = 0 B) D) P can be (a, b) D) = -4 C) x - 3y - 5 = 0 D) x - 3y + 5 = 0 3 a / 2, a / 2 C) (0, -a) D) 3 a / 2, a / 2 If the lines ax + by + c = 0, bx + cy =a = 0 and cx + ay + b = 0 are concurrent (a+b+c 0) then A) a3 + b3 + c3 - 3abc = 0 C) a = b = c 11. D) (-1/4, 11/4) If one vertex of an equilateral triangle of side a lies at the origin and the other lies on the line x 3y =0, the co-ordinates of the third vertex are A) (0, a) 10. C) (7/2, 13/2) Equation of a straight line passing through the point of intersection of x-y+1 = 0 and 3x+y-5 = 0 are perpendicular to one of them is A) x + y + 3 = 0 9. B) (3/4, -3/2) If the lines x - 2y - 6 = 0, 3x + y -4 = 0 and x + 4y + 2 = 0 are concurrent, then A) = 2 8. D) a + b + c = 0 The points (k, 2-2k), (-k+1, 2k) and (-4-k, 6-2k) are collinear for 1 A) any value of k B) k=1 C) k = D) no value of k 2 If the point P(x, y) be equidistant from the points A(a+b, a-b) and B(a-b, a+b) then A) ax = by 7. C) a = b The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x+3. The co-ordinates of the third vertex can be A) (-3/2, 3/2) 5. B) c = a B) a = b D) a2 + b2 + c2 - bc - ca - ab = 0 If the co-ordinates of the vertices of a triangle are rational numbers then which of the following points of the triangle will always have rational co-ordinates A) centroid 12. B) incentre Let S1, S2 .... be squares such that four each n ³ 1, length of a side of Sn equals the length of a diagonal of Sn+1. If the length of a side of S1 is 10cmm, then for which of the following values of n is the area of Sn less than 1 sq. cm ? A) 7 13. B) 8 C) 9 1 unit2 then 8 A) b, a, c, are in G.P C) b, 15. A) bisector of the angle including origin B) bisector of acetic angle C) bisector of obtuse angle D) none of these Two roads are represented by the equations y - x = 6 and x + y = 8. An inspection bunglow has to be so constructed that it is at a distance of 100 from each of the roads. Possible location of the bunglow is given by B) (1-100 2 , 7) C) (1, 7 + 100 2 ) D) (1, 7 - 100 2 ) Angles made with the x-axis by two lines drawn through the point (1, 2) cutting the line x + y = A) 5 and 12 12 6 /3 from the point (1, 2) are B) 7 11 and 12 12 C) 3 and 8 8 D) none of these If (a, b) be an end of a diagonal of a square and the other diagonal has the equation x-y = a then another vertex of the square can be A) (a–b, a) 18. D) b, -2a, c are in G.P. Consider the straight lines x + 2y + 4 = 0 and 4x + 2y - 1 = 0. The line for 6x + 6y + 7 = 0 is 4 at a distance 17. B) b, 2a, c are in G.P. a , c are in A.P.z 2 A) (100 2 + 1, 7) 16. D) 10 If bx + cy = a, where a, b, c are the same sign, be a line such that the area enclosed by the lime and the axes of reference is 14. C) circumcentre IIT- MATHS D) orthocentre B) (a, 0) C) (0, –a) D) (a+b, b) The points (p+1, 1), (2p+1, 3) and (2p+2, 2p) are collinear if A) p = -1 B) p = 1/2 C) p = 2 D) y = - 1 2 270 CO-ORDINATE GEOMENTRY SECTION - A SINGLE ANSWER TYPE QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. B B A A C C B D B B C 16. 17. 18. 19. 20. 21. 22. 23 24. 25. 26. 12. 13. B C 27. 28. 14. 15. C A 29. 30. A C C A B A D C B B A C A B B 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 A A B A B A,B B,D C A A,B B B A A C 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 A B D B A C C B D A A A B A 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 B C B C A C D D B A C A 271 C A IIT- MATHS 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 B A A B B A C B D B C B B B,D 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 A B B A C B A C B 106 107 108 109 B 121 A C A A 122 123 124 B B B A D 110 111 112 113 114 115 116 C B D B A B A B A B B D 117 118 119 120 B B B D 125 126 127 128 D 272 CO-ORDINATE GEOMENTRY SECTION - B MULTIPLE ANSWER TYPE QUESTIONS 1. 2. 4. 5. 6. 10. 11. 12. 13. 14. 15. AC AB AB AC CD CD BC BD AD BD AB AB CD CD AC D BC BD D AB AB CD 16. 17. 18. AB BD CD 273 3. 7. 8. 9. z 2 are two points. Z ) in new axes. . y 2 . with X. z1 Q x 2 . cos . y4 . . y 3 . y 2 . y1 . y1 . cos is called a triple of the direction cosines of the given line. Q x 2 .C’s) of a line are components of a unit vector parallel to the given to the given line and two such triplets of D. 4 . the new axes are parallel to original axes and passing through the point x1 . y3 .C’s of OP are x1 y1 z1 . . z 2 in space is given by 2 2 PQ x 2 x1 y 2 y1 z 2 z1 1 2 2 ___ If P x1 . 2 2 2 2 x2 y2 z2 x1 y1 z1 x1 y12 z12 1 1 1 274 . z4 are the vertices of a xi yi zi tetrahedron. z1 ] then x X+ x1 . y1 . x1 x 2 y1 y 2 z1 z 2 ___ . Y. cos180 . z3 and S x4 . y1 .C’s exist for any line namely cos . z1 then D.IIT. z1 through parallel shifting of axes [i. the triple cos . y1 . Note that direction cosines (D. z1 . the centroid (or the center of gravity) of DPQR is G. y 2 . 4 PARALLEL SHIFTING OF AXES If the origin is changed to x1 . If a line makes angles . cos . cos180 . Y. y Y+ y1 . z1 and Q x 2 . Z axes respectively. z = Z+ z1 where P x. y. (i) If cos . we say R divides PQ internally and when the ratio is negative. z changes to (X. its center of gravity (or) centroid is given by 4 .MATHS THREE DIMENSIONAL GEOMETRY Distance bet ween any two point s P x1 . R lies between P and Q on PQ and if the ratio of negative. cos180 2 (ii) 2 2 If ‘O’ is the origin and P = x1 . y1 . z 2 R x3 . the points R which divides PQ in ratio l : m is given mx1 lx 2 my1 ly 2 mz1 lz 2 . Then (iii)If P x1. by R lm lm lm ___ If the ratio l : m in positive. . ___ we say R divides PQ externally. we get cos cos cos 1. cos . cos . z2 R x3 . y2 . (i) Mid point of PQ = 2 2 2 (ii) If P x1. y1 . cos is a t riple of direct ion cosines of a line. R lies on ___ ___ extension of PQ when l : m is positive.e. z1 Q x2 . z 3 are vertices of a triangle.. y. y. z Projection of the line segment joining P x1 .C’s of a line is called a triple of direction ratios (D. If a. y1 . n is x 2 x1 l y 2 y1 m z 2 z1 n PLANE The line joining two points on a plane totally lie on the same plane. m1 . m2 .C’s of a line. then lk . the angle between them is given by l m cos l1l 2 m1m2 n1 n2 (or) sin 1 l 2 m1 2 2 The lines are perpendicular if and only if l1l 2 m1m2 n1n 2 0 l1 m1 n1 The lines are parallel if and only if l m n 2 2 2 If two rays have D.e.0 .axis = x 2 x1 (ii) on Y . .CO-ORDINATE GEOMENTRY (iii) Any triple of numbers proportional to the D. in YOZ plane is 0.axis = y 2 y1 (iii) on Z . then (iv) a b c . z and in XOZ plane is x. y1 . n2 respectively. If l . b1 . y 2 . i. b2 . m.R’s a1 . z 2 (i) on X . z1 and Q x 2 . y. m. nk .C’s l .0. (i) 275 The general equation of a plane is ax by cz d 0 .R’s) of the line. c 2 respectively.R’s of PQ is given by x2 x1 . z 2 are two points. mk .C’s l1 .C’s of the given line. y 2 . the angle between them is given by (vi) cos a1a 2 b1b2 c1c 2 a12 b12 c12 a 22 b22 c 22 (or) sin a b 1 2 2 a 2 b1 a12 b12 c12 a 22 b22 c 22 The lines are perpendicular a1a 2 b1b2 c1c 2 0 a1 b1 c1 The lines are parallel a b c 2 2 2 PROJECTION Projection of a point P x. z in XOY plane is x. z2 z1 (v) If two rays have D. k R -{0} is a triple of D. n1 and l 2 .R’s of the given line. n is a triple of D.R’s of a line. c is a triple of D. a triple of D. y 2 y1 . b. If P x1 . c1 a 2 . 2 2 2 a2 b2 c2 a2 b2 c2 a b c are triple of D.axis = z 2 z1 (iv) on any line with the D. z1 and Q x 2 . y1 . z1 is A x x1 B y y1 C z z1 0 (vi) The equation of the plane passing through three non collinear points A x1 . n and is of length ‘p’ then the equation of the plane is lx my nz p (Normal form).0 0. y1 .0.C’s l . (iii) If the perpendicular from (0. 0.MATHS If a plane cuts the X. B x 2 .0) is (iv) .0 and the perpendicular distance of P to Y-axis is (iii) x2 z 2 foot of the perpendicular from P to Z -axis is 0. y y1 . 0) to a plane has D.(ii) IIT. b.R’s of the line PQ and 276 . Y. Y. z is any point in the space. the perpendicular distance from (0.R’ss a. c is given by a x x1 b y y1 c z z1 0 . y 2 . Z . b. y 3. z1 . its equation is given by x y z 1 (Intercept form) a. y1 . 0. (i) foot of the perpendicular from P to X -axis is x. z 2 we get that x x1 . 0. The equation of the plane through x1. c and the D. z1 .0. Also the equation of the plane parallel to Ax By Cz D 0 through x1 . m. From the above we observe that the D.R’s of any normal to the plane ax by cz d 0 are proport ional to a. .0 . z1 and Q x 2 . c are called the X. Hence for a given plane |d| ax by cz d 0 .intercepts of the plane respeca b c tively. z 2 z1 are two triples of D. c respectively. z is any point on the line passing through P x1 . whose normal has D. y 2 y1 . y 2 .0. y.0 and the perpendicular distance of P to X-axis is (ii) foot of the perpendicular from P to Y -axis is 0.C’s of any line normal to this plane are a b c . z 2 x x1 x x1 and C x3. b. Z axes at a. y1 .0. y. z and the perpendicular distance of P to Z-axis is (iv) y2 z2 x2 y2 If R x. z 3 is 2 x3 x1 y y1 y 2 y1 y3 y1 z z1 z 2 z1 0 z3 z1 STRAIGHT LINE If P x. y . z z1 and x2 x1 . b. 2 2 2 a2 b2 c2 a2 b2 c2 a b c (v) a2 b2 c2 . CO-ORDINATE GEOMENTRY they must be proportional (i. y y1 t y 2 y1 .) x x1 y y1 z z1 t (say) x 2 x1 y 2 y1 z 2 z1 Hence we get x x1 t x 2 x1 . Therefore any point on the line can be taken in the form for some t R. z z1 t z 2 z1 .e. 277 . t = -1 278 . 3.7.2t) 0 i.2t If it is required foot of the perpendicular of (1. Hence using 1st leation i.C’s of two liens satisfy the relations 3l m 5n 0 and 6mn 2nl 5lm 0 .7) and (9.e. 5) (d) (3. 4. the angle between them is 1 1 (a) Cos 6 1 1 (b) Cos 3 (c) 1 3 (d) Cos 4 2 Ans : (A) Solution : Eliminating m from the given relations we get l 2 3 ln 2n 2 0 i. 9) Ans : (B) Solution : Any point on the line joining the given points can be taken as 6 3t.e.3) to the line joining the points (6. l 2n l n 0 . 7 2t.MATHS SOLVED PROBLEM ILLUSTRATION : 01 The perpendicular distance of (3. 5) from the Z-axis is (a) 26 (b)5 (c) 17 (d) 10 Ans : (B) Solution: The perpendicular distance to Z -axis = x 2 y 2 9 16 5 ILLUSTRATION : 02 If D. 2 1 1 1 2 1 The angle between these lines is given by cos 2 2 1 4 11 1 4 1 1 6 ILLUSTRATION : 03 The foot of the perpendicular from (1. 9. 3) we get 3(5 3t) 2(5 2t) . 7 . l n l n or 2 1 1 1 l m n l m n or hold.IIT. 9) (b) (3. 2.2(4 .9.5) is (a) (5. 9.2.e. 9) (c) (3. 5. 4) are given points. 2. 3). 7.e.C’s of CD are 3 . 3 . x 3 y 6 z 8 0 ILLUSTRATION : 06 If (2. i. -1) is the foot of the perpendicular from (4. 5) and D (3. 279 .e. B(6. (1. x 1 3 2 y 10 3 z 16 0 0 2 1 0 2 1 i. 2. 1) to a plane. 2). then the projection of AB on CD is (a) 1/3 (b) 4/3 (c) 25/3 (d) 5/3 Ans : (D) Solution : D. -2.2. -1) or (-2. 2) is (a) x 3 y 6 z 8 0 (b) x 3 y 6 z 14 0 (c) x 3 y 6 z 4 0 (d) x 3 y 6 z 20 0 Ans : (A) Solution : The required plane equation is x 1 y 1 z 1 x 1 y 1 z 1 3 3 1 3 3 1 = 0. 3. -1. 1) and (1. 1.R’s of CD are (2. 0. 2. 1) 2 2 1 D. 8) C(1. then equation of the plane is (a) 2 x y 2 z 3 0 (b) 2 x y 2 z 9 0 (c) 2 x y 2 z 5 0 (d) 2 x y 2 z 1 0 Ans : (D) Solution : The line joining the given points is normal to the plane.CO-ORDINATE GEOMENTRY ILLUSTRATION : 04 If A (1. -2. 3 2 2 1 2 2 1 5 ___ Projection of AB on CD = 6 1 7 2 8 3 = 5 3 3 3 3 3 ILLUSTRATION : 05 The equation of the plane passing through the three points are (-2. 4.e. . 0) (b) (5. 0) 1 (b) 0. -15. 0) intersect at (a) (5.MATHS D. -1) and B(1. 21 2t 22 t 34 4t 14 0 (i. -1. The equation of the plane is 2 x y 2 z 4 3 2 ILLUSTRATION : 07 The equation of the plane parallel to x y 2 z 3 0 through (1.) 4t 2t 12t 2 4 12 14 0 (i. . 3.1.2. 12) Ans : (B) Solution Any point on the given line is 1 2t . 4. -4) 280 . 2. .IIT. -1) lies in the plane. i.0 3 1 5 (c) 0.) x 3 y 3z 1 0 1 This cuts Y-axis at 0. 4 4t This lies in the given plane.) 14t 28 0 t = 2 Intersection point is (5. -6) (d) (10.e. -4) (c) (-3. 5) on Y-axis is (a) (0.R’s of normal are (2. 4) is (a) x y 2 z 11 0 (b) x y 2 z 6 0 (c) x y 2 z 11 0 (d) x y 2 z 8 0 Ans : (C) Solution : Plane parallel to the given plane can be taken in the form x y 2 z k 0 .e. -2. 2. 4. 2 t . . k 1 2 8 11 ILLUSTRATION : 08 The point which is equidistant from A(3.e. 4) and (3..0 3 3 Ans : (C) Solution : ___ The plane that perpendicularly bisects AB is 2 x 2 6 y 1 6 z 2 0 (i.0 3 ILLUSTRATION : 09 The plane 2 x 2 y 3z 14 0 and the line joining (1. This pass through (1. 2) and (2. -1. 3.0 (d) 0. 2. 4). -1. -1) (2. 5) are 1. 4.4. -1. 5. -3. -6 The required plane is x 5 y 6 z 2 15 6 19 281 (d) x 5 y 6 z 11 0 . 5) is (a) x 5 y 6 z 19 0 Ans : (A) (b) x 5 y 6 z 7 0 (c) x 5 y 6 z 23 Solution: DR’s of line joining (3. -1) and (2.CO-ORDINATE GEOMENTRY ILLUSTRATION : 10 The equation of the plane through (2. 1) which is perpendicular to line joining the points (3. -1. b) 3 b) (3.-13.0.-5) d) (3. (6.0.3.4.4. 2 2 2 c) .1) b) -4 c) ± 4 d) 0 a b c 2 2 2 b) . b) 7/6 The co-ordinates of a point equidistant from the four points O(0.5) and C(5.6) is a) 7/3 5. –4) is a) (18/5.0) B(0.0) (4. d) (0. 4) with respect to B (3.4) (1. . .0. The projection of the line PQ joining the points P(3.MATHS SECTION .0.6) and C(5.b. c) (1/2. 1/3.c) is a) . 3.3) with the xoz plane a) (1. 2).4.-4.5) c) (-3.-16. 0.3) on x-axis is a) 1 2. 4) 4. b) (11.1) to the line joining B(1. 2.5) The angle between the points passing through the points (8. a b c . 5) then its centroid is a) (3. 9 7 3 .6. 9 3 3 .11) are collinear then a = a) 4 10.6) (3.IIT.4) is a) (-3. 5. a b c 2 2 2 c) 21 d) 7/9 b) 5 c) 5 2 d) 2 5 b) 7 : 5 c) 9 : 11 d) 11: 9 b) 4x+y-3z-26=0 c) 2x-4y+3z+23=0 d) 3x+5y-2z+12=0 b) (-1.0.6.4) is 282 . –17. –5.-4.5.1/4) If the projections of a line segment on the axes are 3. 3 7 7 2 2 2 b) . .-2.0) and C(0. 2 2 2 d) The foot of the perpendicular drawn from the point A (1.5 then the length of the line segment is a) 12 6.-7) and (-3.7) on the line whose D.4.2) (-9.1.2.A SINGLE ANSWER TYPE QUESTIONS 1.1) B(2. C (6. c) –2 The harmonic conjugate of A (2.-7) d) (7. . d) –1 If the orthocentre and circumcentre of a triangle are (–3.7) 9.2. The points A(1.–2) is a) 3 : 5 7.5) Q(4.-7) c) (1.0.2.2) The projection of the join of the two points (2. 6) The ratio in which the plane 2x–3y+5z–2=0 divides the line segment joining (1.-3. 2 2 4 c) The point of intersection of the line through (-2.-5) and perpendicular to the planes x+2y+2z-8=0 and 3x+3y+2z+5=0 is a) 2x-3y+2z+15=0 8. 11.2.a.0) A(a. . 2 2 2 d) . 3.1. –2.4/5) 3.4.r’s are (6. The equation of the plane passing through (2. 2).4.3. a b c .2.0.3) (2.5) 12. 0) (1. –3) is a) y2+z2–6x–8z+25=0 c) x2 + y2 + z2 – 25 = 0 16.r’s of the two lines are a) 30o 18. l2 – m2 + n2 = 0 then the angle between the lines is a) 2x+3y-3z = 5 19. 14. m1.4) is 4 is a) (1. 4 2 63 1 63 1 a) Cos 63 x y z =3 a b c a b c b) x y z 3 c) ax+by+cz = 3 d) x2 + y2 + z2 = a2+b2+c2 A variable plane is a at a constant distance 3p from the origin and meets the axes in A.3. 17 2 The equation of the locus of the point which moves in such a way that the sum of its distances from (2. (l2.-1.m.3.n of two lines are connected by the relations l+m+n = 0. (2. then the d.c) cuts the coordinate axes at A.2.CO-ORDINATE GEOMENTRY 2 13. 1 d) Cos 63 1 c) Cos If (l1. B and C.1. 1 b) Cos a) l1 l 2 m1 m 2 n 1 n 2 . 2 2 2 b) l1 + l2.B. Then the locus of the centroid of the triangle is a) 22. .2.-2.c’s l. 2 2 2 The distance between orthocentre and circumcentre of the triangle formed by (1.2) d) (1.2) c) (1. 4.n of two lines are connected by the relations 2l+2m-n = 0 and mn+nl+mn =0 then the d.c’s l. n1 + n2 c) l1 l 2 m 1 m 2 n 1 n 2 .-2).-2) (2. m1 + m2 .-1.4) and (-2. The locus of the centroid of the triangle ABC is . m2. n2) are d.5)(4.m.c’s of two lines inclined at an angle 1200.2.3)(3.1.-2) b) 45o c) 60o b) 2x-3y+3z+5= 0 283 d) –2x-3y-3z+5 = 0 b) 2x + 4y – 3z – 39 = 0 d) 2x + 4y – 3z = 0 The equation of the plane lying mid way between 2x + 3y – 6z + 1 = 0 and 2x+3y-6z+7 = 0 is b) 2x-3y+6z – 4 =0 d) 2x+3y+6z-8 = 0 A plane passing through the fixed point (a. n1). 0. l1 l 2 m 1 m 2 n 1 n 2 . c) 12x+6y-5z = 1 If (2. 3 3 3 d) b) 1 2 c) 66 b) x2+y2–6y+ 25 = 0 d) x2 + y2 – 6y – 8z + 25 = 0 b) (1.-1.2) (2. –3) is foot of perpendicular drawn from origin to plane.1.-1) is a) 2x + 4y – 3z – 29 = 0 c) 2x + 4y + 3z + 29 = 0 20.2) (2.2. then the equation if the plane is a) 2x+3y-6z+4 = 0 c) 2x+3y-6z+8 = 0 21.C.c’s of the line bisecting the angle between them are a) 0 15. d) 90o Equation of the plane passing through the points (1.-2. 66 If the d.-2) 17.b. .1) and (-2. . d) If the d.2. 5.5) b) (-1/2. l+m+n = 0 is a) /6 b) /4 c) /3 d)/2 The volume of the tetrahedron included between the plane 3x+4y-5z-60=0 and the coordinate planes is a) 60 33.10).b. 5. c) 13 The image of (1.5.5. c) 3. b) (1.-5) d) (1/2. 3. 4) in the plane 2x – y +z + 3 = 0 is a) (3. -2) c)(1. 5.2) and containing the line joining the points (1.16. -8.4.-10) d) (1. –2) c) ax by cz a 2 b 2 c 2 32.1.-1. 0) b) (3. 2) b) 450 c) 600 x 1 y 1 z 10 is 2 3 8k a) (3.-1.4) and (1.19).0) B(1. -4. -2.6) and (3.4. b) 4.8) If the foot of the perpendicular from the origin to a plane is (a.7. 30.0.MATHS 1 1 1 1 1 1 1 3 1 1 1 9 1 1 1 16 a) x 2 y 2 z 2 p 2 b) x 2 y 2 z 2 p 2 c) x 2 y 2 z 2 p 2 d) x 2 y 2 z 2 p 2 23. the angle which it makes with the z -axis is a) 300 29.1.-1. 0) c) (–3.4. -2.4 The extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to reference planes are (-2.c) the equation of the plane is x y z 3 a b c b) ax by cz 3 d) ax by cz a b c The angle between the lines whose direction cosines are given by the equations l2+m2–n2 =0. 0) d) (3.5 A point lying on the line joining the points (-3.5) has its x-coordinate as 1 then the point is a) (1.-5. 0) 27.-1.3 b) 600 c) 720 d) none of these The plane passing through the point (-2. 2. c) (1.5) 26. The end points of a diagonal of a rectangular parallelopiped with faces parallel to the coordinate planes are (2. b) 2 13 The circumcentre of the triangle formed by the points A(1.5. –5.1) and (1.-3.2) makes intercepts on the coordinates axes the sum of the whose lengths is a) 3 b) 4 c) 6 d) 12 284 . The length of the base diagonal is a) 13 25.-5) The reflection of the point P(1. c) (-1/2.-2.1) is a) (1/2.1.2. d) 3. d) None A line makes an angle of 600 with each of x and y-axis.0) in the line a) 31.-4) d) None of these d)(2. 2) 28. 2) b)(5.3 24. 2.1) and C(0.-1.5) and (5.IIT.1.3. The lengths of its edges are a) 5. the equation of the plane is a) x ry r 2 z 3r 2 39. then cos 2 cos 2 cos 2 cos 2 is equal to a) 1/3 42. The locus of the point is a) x 2 y 2 z 2 1 37. 2. y = 1 . 0.-1) B(2. . b)k =1 or -1 The ratio in which the yz plane divides the line joining the points (-2. 2 and 2.18.54 35.4. -1 is a) x 2 y 2z 0 41.2. b) -27.-3) is a) 4 5 43. r2). 285 b)3 c)6 d)None of these b) xyz 8k 3 c) x y z 6k d) x 3 y 3 z 3 64k 3 A plane meets the co-ordinate axes in A.C such that the centriod of the triangle is the point (1.-5. -1) and (3. -2. c) 4/3 The area of the triangle with vertices A(3. c) -27. d) 8/3 A variable plane makes with coordinate planes a tetrahedron of constant volume 64k3.6). . r. The lines b) x 2 y 2 z 2 2 c) x y z 1 d) x y z 2 b) 3:2 c) 4:5 d) -7 :8 b) r 2 x ry z 3r 2 c) x ry r 2 z 3 d) r 2 x ry z 3 b) 19/2 c)-22/3 d) 26/3 Equation of the plane passing through the origin and containing the lines whose direction ratios are 1. A line segment has length 63 and direction ratios are (3.-18. -2. z= 1 is 10 units. 3.-54 A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = 1 . C such that the centroid of the triangle ABC is the point (a. B. The locus of the centroid of the tetrahedron is a) xyz 6k 3 44. d)27.-54 x 2 y 3 z 4 k 1 y 4 z 5 and are coplanar if 1 1 k k 2 1 a) k = 0 or -1 36. with four diagonals of a cube. c).7) and (3.B. d) k = 3 or -3 Algebraic sum of the intercepts made by the plane x+3y-4z+6 =0 on the axes is a) -13/2 40. 2) and parallel to the line .54 b) 2x 3y z 0 c) x 5 y 3z 0 d) 4x 5y 7 z 0 If a line makes angles .CO-ORDINATE GEOMENTRY 34. Then the equation of the plane is a) 45. b. the components of the line vector are a) 27. -18.1) and C(3. b) 2/3 x y z 3 a b c b) a b c 3 x y z c) ax + by + cz = 3 d) none of these The equation of the plane through the points (1.18.8) is a) 2 :3 38. If the line makes an obtuse angle with x-axis. c) k = 0 or -3 A plane meets the coordinates axes in A.4.4. 19/3) b) (5/3. 2. 4) c) (1. (2. (2. 4) to the line joining B(0. 3. 19/3) c) (5/3. 4) d) (1. 4 2 1 1 1 1 d) none of these b) 1 c) -1 d) 0 The coordinates of the foot of the perpendicular from the point A (1. 4 The coplanar points A.2. 5.1) and Q () are two points in the space such that PQ = d) none 27 . 3.3. D are (2-x. 8. 2/3. 2-y. C. 1 1 1 1 . 4 2 b) plane c) sphere P (1. the value of can be 286 . d) none of these If a. -3.clockwise direction it is found that a vector a has the components 2 3 . 2. . b) . 2. B(-1. 0. 2).c’s of a line which makes equal angles with the axes is 1 1 1 . 4 51. 4.1) and D(x. 6) is a) 3/ 74 47. 1. 7) and (5. 4) 49. The components of a in the OXYZ coordinate system are a) 5. 3) and C(2. 3 2 and 4.1.5) are coplanar is a) 2 54. -1. b) 3/2 65 c) 65 / 3 c) 3 d) 1 The centre of the circle given by r ( i 2 j 2k ) = 15 and r ( j 2k ) 4 is b) (-1. b) 5. 1 The value of x for which the points A (1. 19/3) d) none The locus of a point which moves so that the difference of the squares of its distance from two given points is constant is a a) straight line 55. -5.3). -2/3. 2). 2/3.MATHS x 1 y 1 z 1 is 1 2 3 a) 4x+y+2z = 6 46. -1. g are angles which a line makes with the axis then the value of sin2a + sin2b + sin2g is equal to a) 2 48. 3. b. 1. -4) The d. 3 3 3 a) 1.IIT. 0).0. -1. -1) is a) (-5/3. -3. then a) x + y + z = 1 52. B. c) -1. 1) respectively.2. (3. c) 4x-y-2z = 6 The area of the triangle whose vertices are (0.4) C(1. d) none of these b) 4 a) (1. 2-z) and (1. d) -1. b) 4x-y+2z = 6 d) None of these When a right handed rectangular cartesian system OXYZ is rotated about the Z-axis through an angle p/4 in the counter . c) 2 2 2 50. 1 b) x y z = 1 c) 1 x 1 y 1 z =1 53. 0) and (0.b. R. b) x2 + y2 + z2 + x + y + z = 0 d) x2 + y2 + z2 . The equation of the plane in its new position is x-4y+6z=k. n = 3 65. c) 2 b) 73 a) parallel lines 63. 3.C. 5. 0. S are four coplanar points on the sides AB. c) 16/ 3 AP BQ CR DS equals. -1) The distance of the point A (-2. c) and meet the axes at A. CD. 0) (0. where k is a) 106 62. a tetrahedron of constant volume 64K3. d) 5/ 3 The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x + 3y + 10z = 25. The lines b) 14 / 3 b) -1 The lines d) -2 c) -69 d) 28 X 1 Y 1 Z 2 X 1 Y Z 1 . 1. d) none P. Then the locus of the centroid of tetrahedron is the surface. 0.x . PB QC RD SA a) 1 61.B.CO-ORDINATE GEOMENTRY a) -4 b) 2 56. c) m = -1. BC. a) xyz = 6k2 64.z . d) none of these The equation of a sphere which passes through (1.z = 0 c) x2 + y2 + z2 + x + y + z = 0 57.C and parallel to the co-ordinate plane is a) 287 b) m = 1. 1) from the line PQ through P (-3. 2) which make equal angles with the axes is a) 2/ 3 59. n = 3 A variable plane passes through the fixed point (a.y . 1) and whose centre lies on the curve 4 x y = 1 is a) x2 + y2 + z2 .2 = 0 x 1 y 1 z 1 x 4 y 0 z 1 and 3 1 0 2 0 3 a) do not intersect 58.y . c) -2 b) xy + yz + zx = 6k2 The plane cont aining t he t wo lines c) x2 + y2 + z2 = 8k2 d) none X 3 Y 2 Z 1 X 2 Y 3 Z 1 and is 1 4 5 1 4 5 11x+my+nz = 28 where a) m = -1. n = -3 a b c 2 x y z b) a b c 1 x y z c) a b c 1 x y z a b c d) x y z 2 . n = -3 d) m = 1. DA of a skew quadrilateral. b) intersect at (4. Q. The product 60. The length of the line is a) 7 b) 17 c) 21 d) 25 A variable plane makes with the coordinate planes. 12 and 8.B. 1. The locus of the point of intersection of the planes through A. 0. -1) c) intersect at (1. are 1 2 1 2 1 4 b) intersecting lines c) perpendicular lines d) none of these The projections of a line on the axes are 9.x . i j 2k 7 A mirror and a source of light are situated at the origin O and at a point on OX respectively. If the DR’s of the normal to the plane are 1. Q (5. r2) the equation of the plane is a) x + ry + r2z = 3r2 73. IIT. 1 2 2 . 26) d) none If P(3. -6) and R(9. -18. 2. P2. 3 3 3 c) x+y+z = 1 d) none b) r2x + ry + z = 3r2 c) x + ry + r2z = 3 d) r2x + ry + z = 3 c) 7 : 8 d) 1 : 1 A line segment has length 63 and direction ratios are (3. The direction cosines of the line so directed that the angle made by it with positive direction of X-axis is acute. -18. -54 d) 27. . -54 X 2 Y3 Z4 X 1 Y 4 Z 5 and are coplanar if 1 1 k k 2 1 b) k = +3 c) k = 1 d) k = 0 If P1. P3 denote the distances of the plane 2x . 3. are a) 67. 1 2 2 . c) r. . -7. 3 3 3 X 2 Y 1 Z 2 meets the plane x-2y+z = 20 is 3 4 12 b) (8. 2 2 1 .i j 2k 1 70. b) b) 3 : 2 externally c) 2 : 1 internally The equation of the plane perpendicular to the line d) 2 : 1 externally X 1 Y 2 Z 1 and passing through 1 1 2 the point (2. 8) is a) 2 : 3 74. . the components of the line vector are The lines a) k = -1 76. 18. If the line makes an obtuse angle with X-axis. -1) and (2. -1. 26) c) (7.66. c) The ratio in which the plane 2x . b) r. -8. 3 3 3 The point in which the line a) (7. 18. 8. 4.3y+4z-6 = 0 respectively then 288 . 4x-6y+8z+3 = 0 and 2x . -3. -5. .3y + 4z + 2 = 0 from the planes 2x-3y+4z+6 = 0.i j 2 k 1 b) -27. -54 c) -27. -10) are collinear then R divides PQ in the ratio a) 3 : 2 internally 69. -4) . 8. 3 3 3 d) 2 2 1 . 26) 68. 3 3 3 X 1 Y 3 Z 2 and the point (0. -7) is 3 2 1 b) x+y+z = 2 b) 4 : 5 a) 27. . 54 75. . .MATHS A line passes through the points (6. . 1). C such that the centroid of the triangle is the point (1. 3 3 3 c) 2 2 1 .1 = 0 divide the line joinig (-2. r. 6). 4. 7. d) none A plane meets the coordinate axes in A. -2. 3 3 3 b) d) 1 2 2 . 2 2 1 . 1 2 2 . 3 3 3 The equation of plane containing the line a) x+y+z = 0 72. A ray of light from the source strikes the mirror and is reflected. 7) and (3. 1) is a) r. B. 1 then DC’s of the reflected ray are a) 71. 7. b) y-axis b) 4 planes r r r b) x .P3 = 0 b) P3 = 16P2 77. then the DC’s of the internal bisector of the angle between these lines are a) 1 2 m1 m 2 n 1 n 2 . . c) z-axis If (x. n2 are DC’s of the two lines inclined to each other at an angle q. 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2 b) 1 2 m1 m 2 n 1 n 2 . 4) in the plane 2x .2y . . 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2 c) 1 2 m1 m 2 n 1 n 2 .plane c) a) sin2 + cos2 + sin2 = 1 c) cos2+cos2+cos2 = 1 81. 2) 82. n2 are DC’s of the two lines inclined to each other at an angle q then the DC’s of the external bisector of the angle between the lines are a) 1 2 m1 m 2 n 1 n 2 . c) 5 planes If be the angles which a line makes with the coordinate axes. 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2 d) 1 2 m1 m 2 n 1 n 2 . 2a Graph of the equation y2 + z2 = 0 in three dimensional space is a) x-axis 78. 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2 If 1 . then other end is If one end of a diameter of sphere x2 + y2+z2 . d) yz. r r r d) none of these b) sin2 + sin2 + sin2 = 1 d) cos2+cos2+sin2=1 b) (-3. d) 6 planes The image of the point P(1. m2. then a) (3. c) P1 + 2P2 + 3P3 = b) 6 c) 8 d) none If 1 . 86. 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2 1 3 3 . 2 cos θ / 2 2 cos θ / 2 2 cos θ / 2 c) 1 2 m1 m 2 n 1 n 2 . z = 0 if c = 3 2 1 b) ± 1/3 c) ± 5 d) none The number of spheres of radius r touching the coordinate axis is a) 4 84. z) be the coordinates of a point P and OP = r then the direction cosines of OP are a) rx. . . 3. n1 and 2 . m1.y + z + 3 = 0 is a) ± 1 83. m1. y. ry. . 2) c) (3. 5. . 5. n1 and 2 .2x .CO-ORDINATE GEOMENTRY a) P1 + 8P2 . . m2. . (C) Tetrahedron is bounded by a) 3 planes 79. z The line 85. 2) d) (3. rz 80. (B) . .2z + 2 = 0 is 1 2 2 2 . . -5. y . 5. 289 x y z . 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2 b) 1 2 m1 m 2 n 1 n 2 . d) (A) . -2) X 2 Y 1 Z 1 intersects the curve xy = c2. 2 sin θ / 2 2 sin θ / 2 2 sin θ / 2 d) 1 2 m1 m 2 n 1 n 2 . . c is a) 91. 6.8 14 +1. C(1. are the vertices of a . 0) are a) 9. 0). -4) 95.z = 5 1 2 2 . 2) and C(-9. -3) and perpendicular to the plane 2x+3y+z + 5 = 0 is a) 89. 0. 1 1 .13. The equation of a plane is = 0. 1. 8 16 16 19 57 17 .IIT.4 c) (-7. a) rhombus b) square c) a regular tetrahedron d) none If P is the length of perpendicular from the origin on to the plane where intercepts on the axes are a. 2 2 2 a) x 1 y 2 z 3 1 1 1 c) x 1 y 2 z 3 2 3 1 b) x 1 y 2 z 3 1 1 1 a b c 2 abc b) | a x b b x c c x a | |axbbxccxa | d) none A (3. The 290 . 2. -3) are the vertices of a triangle ABC. 3. -12-1. 0. c) 1 b) 1/4 The coordinates of a point on the line c) 4/3 d) 2/3 x 1 y 1 =z at a distance 4 14 from the point 2 3 (1. then coordinates of D are 19 57 17 .MATHS a) 1 1 1 . . . . g. 0). then the value of cos2+cos2+ cos2 + cos2 is equal to a) 3/4 94. d) none The length of the perpendicular from the origin to the plane passing through three non-collinear points a . If the centroid of ABC meets BC at D. 8 16 16 b) 19 57 17 . b. -1. 88. B(5. c then d) none A line makes angle a. 2 2 2 1 Equation of a line passing through (-1. 11. d) none c) a b c 90. b. 0) and D 3 3 3 a) a 1 b 1 c 1 p 1 b) a 2 b 2 c 2 p 2 c) a + b + c = p 93. b. 12 14 -1. d) none The Cartesian equation of the plane r ( 1 )i ( 2 ) ( 3 2 2 )k is a) 2x + y = 5 92. d with the four diagonals of a cube. B(2. d) 2x . 8 16 16 c) b) 2x . 0). . . b) 1 1 3 3 . b) . 4 The position vectors of points A and B are and respectively. 2. 2 2 2 1 87. -4 d) 8+1.y = 5 c) 2x + z = 5 The points A (0. 291 b) are on the same sides of the plane d) none of the above The extremities of a diameter of a sphere lie on positive y and positive z-axis at distance 2 and 4 from the origin respectively. 1. then a) sphere passes through origin b) centre of the sphere is (0. 2) c) radius of the sphere is d) all the above 5 .CO-ORDINATE GEOMENTRY points A and B a) lie on the plane c) are on opposite sides of the plane 96. IIT. 4) b) 8 14 1. The coordinates of a point on the line x 1 y 1 = z at a distance 4 14 from the point (1. 2 and 2. . 4) and meets the axes of reference in A. 11. 1 2 1 c) 1. 1) d0 (4. 1. c) 5 The plane passing through the origin and containing the line whose direction cosines are proportional to 1. 12 14 1. a) (9. 1. 7) x 2 y3 z 4 x 1 y 4 z 5 and are coplanar if 1 1 k k 2 1 a) k = 0 5. 1. r. -1) c) (3. 1) If a plane passes through a fixed point (2. -2. 1 2 8. the coordinates of the centroid can be 1 a) 1. -4) d) 8 14 1.MATHS SECTION . 4 14 If a plane is at a distance 3/2 from the origin O. B and C. B and C. -1 passes through the point a) (1. 15) if r is equal to a) –3 3. C such that the centroid of the triangle ABC is the point (1. C parallel to the coordinate planes can be 292 . and meets the axes in A.B MULTIPLE ANSWER TYPE QUESTIONS 1. 3. -8. 2 d) (1.-1. B. c) 5 b) k = -1 c) k = 2 d) k = -3 An equation of the line passing through 3i – 5j + 7k and perpendicular to the plane 3x–4y+5z= 8 is a) x 3 y5 z 7 3 4 5 b) c) r = 3i – 5j + 7k + l (3i – 4j + 5k) x 3 y 4 z 5 3 5 7 d) r = 3i – 4j – 5k + m (3i + 5j + 7k) ( are parameter) 6. The plane passes through the point (4. -13. r2). 2 3 0) are 7. 1 b) . B. 12 14 1. the point of intersection of planes through A. -2. 4 14 c) (-7. 3. 3. 0. 1. The line whose vector equations are r = 2i – 3j + 7k + (2i + pj + 5k) and r = i + 2j + 3k + (3i – pj + pk) are perpendicular for all values of l and m if p = a) –1 2. d) 6 A plane meets the coordinate axes in A. 2) 4. b) 2 b) 2 The lines d) – 5 b) (2. -1. a) (0. 0. 19) d) (8. 9. 3). 0) If l. -9. Equation of a plane through the line a) 4y – 3z + 1 = 0 10. -4. 1) and (3. -4) x 1 y 2 z 3 and parallel to a coordinate axis is 2 3 4 b) 2x – z + 1 = 0 c) 3x – 2y + 1 = 0 d) 2x + 3y + 1 = 0 The line joining the points (2. -3. m. m. n are the direction cosines of the line of shortest distance between the lines x 3 y 15 z 9 x 1 y 1 z 9 and then 2 7 5 2 1 3 a) 3l – 25m + 9n = 0 c) l = m = n = 1/ 3 12. 0. 5) and B (11. 1) c) (-1. direction cosines of the normal to p are l. 4. 13) 11. and it contains the line joining the origin to the point (1.CO-ORDINATE GEOMENTRY a) (6. b) (4. 3. 16) b) (0. -1) d) (2. c) (1. b) x – y + z = 1 c) x + y + z = 1 d) x – y – z = 1 If a plane p passes through the point (1. 0. -5) cuts a coordinate plane at the point. -1) lies on the plane a) 2x + y + z = 6 13. The foot of the perpendicular from the origin to the join of A(-9. 12. n. 1). 1. 1. 12) 9. 2. then a) l + 2m + 3n = 0 293 b) 2l – 7m + 5n = 0 d) 2l + m – 3n = 0 b) l + m + n = 0 c) l + m – n = 0 d) l – m + 2n = 0 . 35) Equation of a line L. 3) and B (13. 4. -4 PASSAGE 2: a = 6i + 7j + 7k. b) 3i + 2j – 2k b) (5. 16 c) 3. Coordinates of the point P which divides the join of A and B in the ratio 2 : 3 internally are a) (33/5. 3).MATHS COMPREHENSION TYPE QUESTIONS PASSAGE 1: A (-2. perpendicular to the line AB is a) x 2 y 2 z3 15 5 10 b) x 2 y 2 z3 3 13 2 c) x 2 y 2 z3 3 13 2 d) x 2 y 2 z3 15 5 10 Direction ratios of the normal to the plane passing through the origin and the points A and B are a) 15. 2. A point P moves in the space such that 3PA = 2PB. 9) 3. 7) If A is the point with position vector a the Area of the PLA in sq. -2/5. Sum of the lengths of the intercepts made by the plane on the coordinate axes is 294 . 5. 11) 7. 0. 15) d) (17. b = 2i – j + 2k 8. -7) c) (5. -1. units is equal to a) 3 6 b) 7 17 / 2 c) 17 d) 7/2 PASSAGE 3: P(2. then the locus of P is a) x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0 c) x2 + y2 + z2 + 28x – 12y – 10 z + 247 = 0 2. 13) L is a line through A 1. (2i + 3j – 4k) = 7 9. -3. P(1. c) 3i + 5j + 9k d) 9i + 9j + 5k The image of the point P in the line r = a + b is a) (11. -5. 10 b) 11. 2 d) 7. b) x2 + y2 + z2 – 28x + 12y + 10z – 247 = 0 d) x2 + y2 + z2 . the foot of the perpendicular from P on the line r = a + lb is a) 6i + 7j + 7k 6. 13. -4). 7) c) (32/5. O being the origin is a) 2x – y + 2z + 7 = 0 b) 2x – y + 2z – 7 = 0 c) 2x + 3y – 4z + 7 = 0 d) 2x + 3y – 4z – 7 = 0 10. 0. 2. -12/5. 8. 17/5) d) (20.(2i + 3j – 4k) = -7 d) r . 16.IIT.3. The position vector of L. Vector equation of a plane passing through the point P perpendicular to the vector b is a) r. 2. (2i – j + 2k) = 7 c) r.(2i – j + 2k) = -7 b) r. 12. b = 3i + 2j – 2k. Cartesian equation of a plane passing through the point with position vector b and perpendicular to the vector OP .28x + 12y – 10 z + 247 = 0 b) (4. 13. 3. -1) ii) Point on the line common to the plane x + y + z + 3 = 0 b) (2. a) 3 12. 11. 4) d) (2. ax + by + cz + d = 0. 1. 0. 3.CO-ORDINATE GEOMENTRY a) 14 b) 91/12 c) 9/7 d) 5/7 PASSAGE 4: L: x 1 y 1 z 1 2 3 4 p1: x + 2y + 3z = 14. 2.and the coordinates of P are (). 3. L : 295 x 2 y3 z4 is a line them 3 4 5 i) Point on the line at a distance 10 2 from (2. 6) MATCHING TYPE QUESTIONS 1. -5) iii) x 5 y 2 z 2 c) direction cosines are 2/ 30 5/ 30 1/ 30 1 3 4 iv) x 1 y 1 z 1 2 5 1 2. 4) a) (-1. 4) iii) Point on the line at a distance c) (8. 0) c) (0.y + 3z = 27 If the line L meets the plane p1 in the point P. 1. -1. 6. a) (1. then the coordinates of the mid-point of PQ are 13. i) x 2 y 7 z 5 3 4 2 a) Perpendicular to the plane 3x + 4y + 2z = 1 ii) x 1 y 2 z 7 3 4 2 b) Passes through (2. 7. 1) b) (0. 14) 29 from the origin . 4. 0. p2: 2x . 1) If the line through P perpendicular to 1 meets the plane 2 in the point Q. b) 14 c) 28 d) 29 The line through P perpendicular to the plane p1 passes through the point a) (1. d) lies in the plane 7x – y – z = 35 x y z m n i) lines is perpendicular the plane a) if al + bm + cn = 0 ii) line is parallel to the plane ifb) if al + bm + cn = 0 and aa + bb + cg + d = 0 iii) line lies in the plane c) if a/l = b/m = c/n 3. 0) d) (0. 9) c) (2. then is equal to 11.3) b) (3. ax + by + cz + d = 0. 3. -1) ii) Point on the line common to the plane x + y + z + 3 = 0 b) (2. 3. -1. 4) iii) Point on the line at a distance c) (8. -5) iii) x 5 y 2 z 2 c) direction cosines are 2/ 30 5/ 30 1/ 30 1 3 4 iv) x 1 y 1 z 1 2 5 1 2. 14) 29 from the origin 296 . L : x 2 y3 z4 is a line them 3 4 5 i) Point on the line at a distance 10 2 from (2. 11.MATHS MATCHING TYPE QUESTIONS 1. 4) a) (-1. i) x 2 y 7 z 5 3 4 2 a) Perpendicular to the plane 3x + 4y + 2z = 1 ii) x 1 y 2 z 7 3 4 2 b) Passes through (2. 7.IIT. d) lies in the plane 7x – y – z = 35 x y z m n i) lines is perpendicular the plane a) if al + bm + cn = 0 ii) line is parallel to the plane ifb) if al + bm + cn = 0 and aa + bb + cg + d = 0 iii) line lies in the plane c) if a/l = b/m = c/n 3. 3 4 12 4. If d is the shortest distance between the lines r = 3i + 5j + 7k + l (i + 2j + k) and r = -i – j – k m (7i – 6j + k) then 125 d2 is equal to. -5. then 3d3 is equal to. If the point of intersection of the line r = (i + 2j + 3k) + l (2i + j + 2k) and the plane r. B. 11. 2.16). S are the points (1.(2i – 6j + 3k) + 5 = 0 is ai + bj + ck. then 10n2–20n+k=0. 11). 11).15. 11. then square of the 15 9 8 4 17 11 distance of P from the origin is.(2i – 6j + 3k) + 5 = 0 lies on the plane r. then the sum of the square s of the intercepts made by the plane on the coordinate axes is equal to. 53. 297 . The lines x 4 y 17 z 11 x 15 y 9 z 8 and intersect at the point P. The sum of the squares of the intercepts made by the plane on the coordinate axes is 2. (8. If the foot of the perpendicular from the origin on a plane is (11.a = 0. If s denotes the projection of PQ on RS then 29s2 + 29 is equal to. If d is the distance between the point (-1. 2. Square of the distance of the plane from the point (52. If the line x 1 y 1 z 1 intersect the curve 6x2+5y2=1. Q. If q is the angle between the line x 1 y 1 z 2 and the plane 2x+y–3z+4=0. 8. then 19 a + 17 is equal to. 3) and (3. (i + 75j + 60k) .-5. (1.where the 15 16 n value of k is. C such that the centroid of the DABC is the point (12. 10. z=0. If the position vector of the point of intersection of the line r=(i+2j+3k)+l (2i+j+2k) and the planer r. 7. then (50a + 70b + 75c)2 is equal to.CO-ORDINATE GEOMENTRY NUMERICAL SUBJECTIVE TYPE QUESTIONS 1. -10) and the point of intersection of the line x 2 y 1 z 2 with the plane x – y + z = 5. 2.3) on the line x 5 y 2 z6 then 100 3 4 5 (PQ)2 is equal to 5. R. A plane passes through (1. 10. P. -2). 6. 7) respectively. then 64 cosec2 q 3 2 4 is equal to 3. 9. 57) is. 12. 2) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. If Q is the foot of the perpendicular fro the point P(4. A plane meets the coordinate axes in A. 5. 2. 14. 12. 6. 3. 20.IIT. 8. 7. 15. 22. A C A A C B A D A B B C B B C 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 C B D B C B D B A D C D A A C 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 B A A B D B C A B C A B A A 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 C B A C C A B B B D A B C D B D 298 . 30. 13. 17. 5. 11. 4. 10. 18. C B 27. 23 24. 19. 28. A A A A C C B B D B D 16. B A 29. 26. 9. 21.A SINGLE ANSWER TYPE QUESTIONS 1.MATHS SECTION . 25. C A.B A. 13. 3. 5. (i) (b). 4.CO-ORDINATE GEOMENTRY 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 D A B D C B C C B C A C A C 91 92 93 94 95 96 C B C C C D A SECTION .C A.B C C.C A. (ii) (a).D B.B MULTIPLE ANSWER TYPE QUESTIONS 1. 7. (iv) (c) 2. C D .B A. 13. 10.B A. (ii) (a). A.D C 5. (iii) (d). 6. 299 (i) (c). 6.C A. (iii) (b) 12. 9. (i) (c). A B C D C C B A C B B MATCHING TYPE QUESTIONS 1. 11. 2. 12. 4. A.D C D C COMPREHENSON TYPE QUESTIONS 1.B A.B A. 11. 8. (iii) (b) 3. 3. 7. 10. 9. (ii) (a). 8.C B. 2. 6591 7. 2563 4. 7225 12. 2630 300 .MATHS NUMERICAL SUBJECTIVE TYPE QUESTIONS 1. 1624 6.IIT. 5202 9. 8129 10. 1398 11. 3267 3. 5625 5. 2116 2. 1828 8. –). –) In case of three dimensional geometry we take three mutually perpendicular lines which divide the space in eight parts called octants. +). y1. –. 0) where a is the distance of foot of perpendicular from P to x–axis from the origin with suitable sign. +. –) and (+. Co–ordinates of a point P on x–axis will be of the form (a. y and z axes. +. Y O X Z CO-ORDINATES OF A POINT ON AXES 1. +) (–. y2. +). (–. 3. z2) Then PQ = 301 (x1 x 2 )2 (y1 y 2 ) 2 (z1 z 2 ) 2 . a is positive or negative according as it lies on positive or negative direction of x–axis. (+. Then the line segment LM is called the projection of the line segment AB on the plane. z1) and Q º (x2. The sign of co–ordinates of the points in the four parts are (+ . +. The position of a point in space can be determined with reference to three mutually perpendicular lines called x. In case of two dimensional geometry two mutually perpendicular lines are taken and they divide the plane (xy–plane) in four parts. 0). –). y) of the point with refersence to two mutually perpendicular lines called x and y axes are known. +. –) (–. (–. +). (–. 0. 0. (+. and L and M be the feet of normals from A and B to a given plane. The four parts are called the quadrants. –). Co–ordinates of a point on z–axis is of the form (0. But all points of space donot lie in a plane and so in order to locate a point in space two coordinate axes are insufficient. The sign of co–ordinates of the points in the 8 parts in which the space is divided are (+. (–.CO-ORDINATE GEOMENTRY THREE DIMENSIONAL GEOMENTRY We know that the position of a point in a plane can be determined if the co–ordinates (x. –. Co–ordinates of a point on y–axis is of the form (0. –. 4. +) (+. DISTANCE FORMULA Let P º (x1. 2. b. +). In order to locate a point in space we need three–co–ordinate axes. g). –. Projection of a Line segment on a Plane: Let AB be a line segment. MATHS We draw PL xy plane. R to yz–plane we can show that x= mx 2 nx1 mn Again by drawing perpendiculars from P. y2. 0) and M º (x2. SECTION FORMULAE I. Section Formula for Internal Division Let P º (x1. y2. y1. y1. y. y2. y1. R draw PL. R to zx plane we can show that 302 . Q. Then R º mn mn mn II. z1). Q. QM and RN perpendiculars to xy–plane Again draw PH RN and QK RN RH PR From similar DPHR and DQKR RK QR z z1 m PR m Þ nz – nz1 = mz – mz2 z z 2 n QR n (m – n) z = nz2 – nz1 z= nz 2 nz1 mn Similarly by drawing perpendiculars from P. z2) Let R divide the line segment PQ externally in the ratio m : n. Q º (x2. z) From P. Then L º (x1. 0) PL = z1. Section Formula for External Division Let P º (x1. . Q. then PR m RQ n Let R º (x. QM = z2 From DPHQ Let PQ = PH 2 QH 2 = 2 2 2 LM 2 QH 2 = {(x1 x 2 ) (y1 y 2 ) } (z 2 z1 ) { LM2 = (x1 – x2)2 + (y1 – y2)2] = (x1 x 2 )2 (y1 y 2 ) 2 (z1 z 2 ) 2 . z1) and Q º (x2. z2) Let R divide the line segment PQ internally in the ratio m : n mx 2 nx1 my 2 ny1 nz 2 nz1 .IIT. r1 r2 2. y2.r 2r1r2 1 2 1 2 1 2 cos = l1l2 + m1m2 + n1n2) … (1) or = cos–1 (l1l2 + m1m2 + n1n2) ANGLE BETWEEN TWO LINES IN TERMS OF DIRECTION RATIOS Let a1. Direction cosines of the two lines are given by l1 a1 a12 b12 c12 . Thus R º mn mn mn ANGLE BETWEEN TWO LINES Let AB and CD be two given lines having direction cosines l1 . n1 a12 b12 c12 b2 2 2 2 2 a b c 2 2 . m1.OP. z2) and OP = r1. n2 respectively and q be the angle between them. m1.CO-ORDINATE GEOMENTRY my 2 ny1 y= mn mx 2 nx1 ny 2 ny1 mz 2 nz1 . m 2 r . r12 r22 [(x1 x 2 )2 (y1 y 2 )2 (z1 z 2 )2 ] OP 2 OQ 2 PQ 2 cosq = = 2. m1 a2 and l2 = 2 2 2 2 a b c 2 2 b1 a12 b12 c12 . n1 and OP || AB Therefore d. z1). m1 r .OQ r12 r22 [(x12 y12 z12 ) (x 22 y 22 z 22 ) 2x1x 2 2y1y 2 2z1z 2 ] = 2r1r2 x1 x 2 y1 y 2 z1 z 2 r12 r22 (r12 r22 2x1x 2 2y1y 2 2z 1z 2 ) = = r . c2 be the direction ratios of two lines AB and D respectively. Let O be the origin. Q º (x2. n1. n 2 r 1 1 1 2 2 2 From PQO.r r . Through O we draw OP parallel to AB and OQ parallel to CD. b1. m2 c1 . Let P º (x1. then cos = l1 l2 + m1m2 + n1n2 303 c2 2 2 a b 22 c22 . n1 and l2. OQ = r2 Since direction cosines of AB are l1. m2.r r . c1 and a2.cs of OQ will be l2. n1 r and l2 = r . n2 If be the angle between the two lines. y1. m1. Similarly d. m2. .cs of OP are l1. b2. n2 x1 y1 z1 x2 y2 z2 Now l1 = r . m. z always represents a plane. z1) and B(x2.(1) a(mx2 + nx1) + b(my2 + ny1) + c(mz2 + nz1) + (m + n) d = 0 mx 2 nx1 my 2 ny1 mz 2 nz1 or. y2. n. y. (a1b2 – a2b1)2 + (b1c2 – b2c1)2 + (c1a2 – c2a1)2 = 0 or. = l(x2 – x1) + m(y2 – y1) + n (z2 – z1) The length of projection of the line segment joining (x1. Locus (1) will be a plane if the line joining any two points on the surface fully lies on the surface. z2) on the line having direction cosines l. c is non–zero. .(2) + n. y2.IIT. z2) on the line having direction cosines l. a1a2 + b1b2 + c1c2 = 0 sin = (l1m 2 l2 m1 )2 (m1m 2 m 2 n1 ) 2 (n1l2 n 2l2 ) 2 The two lines will be parallel iff = 0 or sin = 0 or. z1) and (x2. it is clear from mn mn mn (3) that P() lies on locus (1). a b c +d=0 mn mn mn … (3) mx 2 nx1 my 2 ny1 mz 2 nz1 Let P () where denotes . b. a1 b1 c1 sin tanq = = a 2 b2 c2 cos (a1b 2 a 2 b1 ) 2 (b1c 2 b 2c1 ) 2 (c1a 2 c 2a1 ) 2 a1a 2 b1b 2 c1c 2 PROJECTION OF THE LINE SEGMENT JOINING TWO POINTS ON A LINE Projection of the line segment joining (x1. = |l(x2 – x1) + m(y2 – y1) + n(z2 – z1)| GENERAL EQUATION OF A PLANE General equation of the first degree in x. then ax1 + by1 + cz1 + d = 0 … (1) and ax2 + by2 + cz2 + d = 0 … (2) m. Let A(x1. m. y and z be ax + by + cz + d = 0 … (1) where at least one of a. z2) be any two points on locus (1). z1) and (x2. (a2b2 – a2b1)2 = 0. (b1c2 – b2c1)2 = 0. Let the general equation of second degree in x. y1. But P is the point dividing the line segment joining A and B in the ratio 304 . y1. y2. y1. (c1a2 – c2a1)2 = 0 or.MATHS a1a 2 b1b 2 c1c 2 = 2 1 a b12 c12 a 22 b 22 c 22 The two lines will be perpendicular iff or. n. Note 1: Equation of the xy–plane Let the equation of the xy–plane be ax + by + cz + d = 0 … (1) Since O(0. 0) lies in the xy–plane d=0 Since (1. n be the direction cosines of the normal to it is lx + my + nz = p Equation of a plane parallel to a given plane: General equation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0. 1. Where k is a constant. therefore it also lies on xy–plane b=0 Hence from (1). y and z–axis respectively is x y z 1 a b c Equation of a plane in Normal Form Equation of the plane upon which the length of perpendicular from origin is p and l.CO-ORDINATE GEOMENTRY m : n and m and n are arbitrary. Angle between two planes: Angle between two planes is equal to the angle between their normals Angle q between two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by 305 . Thus locus (1) is a plane. Hence all points on the line joining A and B lies on the locus (1). 0. equation of the plane becomes cz = 0 or. 0) lies on x–axis. 0) lies on y–axis. z = 0 Thus equation of xy–plane is z = 0 Equation of a plane in Intercept Form Equation of the plane which cuts intercepts a. therefore it also lies on the xy–plane a=0 Again (0. m. c on x. b. 0. c1 are the direction ratios of the normal to the first plane and a2. Distance of a Point From a plane Length of perpendicular from point (.IIT. d2 > 0 (i) If a1a2 + b1b2 + c1c2 > 0. b2 c2 are the direction ratios of the normals to the second plane. then origin lies in the obtuse angle between two planes and the equation of bisector of the acute angle between two planes is 306 . ) to the plane ax + by + cz + d = 0 is given by P= a b c d a 2 b2 c2 Equation of the planes bisecting the angle between two planes Equation of the planes bisecting the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is a1x b1y c1z d1 = 2 1 2 1 2 1 a 2 x b2 y c2z d 2 a 22 b 22 c 22 a b c Bisector of the angle between two planes containing the origin Let the equation of the two planes be a1x + b1y ++ c1z + d1 = 0 … (1) and a2x + b2y + c2z + d2 = 0 … (2) where d1 and d2 are positive then equation of the bisector of the angle between planes (1) and (2) containing the origin is a1x b1y c1z d1 2 1 2 1 2 1 a 2 x b2 y c2z d 2 a b c a 22 b 22 c 22 Bisector of the acute and obtuse angle between two planes Let the equation of the two planes be a1x + b1y + c1z + d1 = 0 … (1) and a2x + b2y + c2z + d2 = 0 … (2) where d1. .MATHS a1a 2 b1b 2 c1c 2 cos = 2 1 a b12 c12 a 22 b 22 c 22 here a1. b1. the bisector of the acute angle makes with either of the planes an angle which is less then 45° and the bisector of the obtuse angle makes with either of them an angle which is greater than 45°.CO-ORDINATE GEOMENTRY a1x b1y c1z d1 a x b2 y c2z d 2 2 2 2 2 a1 b1 c1 a 22 b 22 c22 (ii) If a1a2 + b1b2 + c1c2 < 0. each bisecting plane bisects. This gives a test for determining which angle. then origin lies in the acute angle between two planes and the equation of bisector of the acute angle between two planes is a1x b1y c1z d1 2 1 2 1 2 1 a b c a 2 x b2 y c2z d 2 a 22 b 22 c 22 . 307 . Otherwise. acute or obtuse. which is the intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is a1x + b1y + c1z + d1 = 0 … (1) and a2x + b2y + c2z + d2 = 0 … (2) Equation (1) and (2) taken together is the equation of line of intersection of planes (1) and (2) in asymmetrical form. Let the given line PQ be 308 . One and only one line can be drawn through two given points. y2. Thus a straight line in space will be determined uniquely if (i) it passes through a fixed point and is parallel to a fixed line (ii) it passes through two fixed points. m. y1. therefore in general a line is obtained by the intersection of two planes. z2) is x x1 y y1 z z1 x1 x 2 y1 y 2 z1 z 2 The Equation of a Line (General Form) Intersection of two planes is a straight line. n is x x1 y y1 z z1 = l m n Equation of the line joining points (x1. z1) and having direction rations (or direction cosines) l. y1. Intersection of two non parallel planes is a unique line. z1) and (x2 . (iii) it is the intersection of two given non parallel planes. 2. one and only one line can be drawn through a given direction. Angle between a line and plane is the complement of the angle between the line and the normal to the plane. Angle between a line and a plane.IIT. Equation of the line passing through point (x1. Equation of a line as the intersection of two planes Equation of the line. 3.MATHS THE STRAIGHT LINES Introduction: We know that 1. x x2 y y2 z z2 l1 l2 m1 m2 n1 n2 0 To Find the shortest distance between two lines 309 x x1 y y1 z z1 l1 l2 m1 m2 n2 n2 =0 .CO-ORDINATE GEOMENTRY x x1 y yi z z1 … (1) l m n and the given plane be N Q 90° ax + by + cz + d = 0 … (2) P L Let q be the angle between the line and the plane. c and direction ratios of line PQ are l. then the angle between the line and normal PN to the plane is 90° –. b. n al bm cn Now cos(90° – ) a 2 b 2 c2 l 2 m 2 n 2 al bm cn or sin = a 2 b2 c2 l 2 m2 n 2 General Equation of the Plane containing a line Equation of any plane containing line x x1 y y1 z z1 is l m n A(x – x1) + B(y – y1) + C(z – z1) = 0 Where Al + Bm + Cn = 0 Equation of the Plane containing two given lines Equation of the plane containing lines x x1 y y1 z z1 x x 2 y y2 z z 2 and l1 m1 n1 l2 m2 n 2 is or. m. Direction ratios of PN are a. IIT. b. y1. z2) Let l. z1) & (x2. g) 310 . n be the direction ratios of the shortest distance LM. Image of a Point in a Plane To find the image of the point P(a. l m n m1n 2 m 2 n1 n1l2 n 2l1 l1m 2 l2 m1 = 1 (m1n 2 m 2 n1 )2 From the above equation l. y2. g) in the plane ax + by + cz + d = 0 Given plane is ax + by + cz + d = 0 … (1) P Q P (a. b. y2. Since LM AB and LM CD ll1 + mm1 + nn1 = 0 … (3) andll2 + mm2 + nn2 = 0 … (4) From (3) & (4). m & n can be found out and the shortest distance between the lines is projection of line joining the point (x1. z2) on the line with the direction cosines l. m & n. y1. m.MATHS x x1 y y1 z z1 x x 2 y y2 z z 2 and l1 m1 n1 l2 m2 n2 Given lines are x x1 y y1 z z1 … (1) l1 m1 n1 x x 2 y y2 z z 2 And CD : l m n 2 … (2) 2 2 AB : A(x1. z1) B 90° 90° C M D (x2. cr + g) Let Q (ar + a. . c equation of lien PQ is x y z = r (say) a b c Co–ordinates of any point on line PQ may be taken as (ar + a. therefore direction ratios of PQ are a.CO-ORDINATE GEOMENTRY Let Q(x1 .y1 . br + b. 2 2 2 Since L lies on plane (1) ar br cr a b c d = 0 2 2 2 (a2 + b2 + c2) r = –(aa + bb + cg + d) 2 Thus Q º (a + ar. b. z1) be the image of point P in the plane (1) Let PQ meet plane (1) at L. cr + g) Since L is the middle point of PQ ar br cr L . br + b. c Since PQ plane (1). Direction ratios of normal to plane (1) are a. g + cr) Where r = – 311 2(a b c d) a 2 b2 c2 r=– 2(a b c d) a 2 b2 c2 . b + br. b.