Bridge

10th11th About this BRIDGE Every year, hundreds of thousands of students like you cross the threshold of class 10th, and go to class 11th with the dream of preparing for engineering entrance examinations and making it to a prestigious college like IIT. For many, that dream remains just that, because they find the transition from class 10th to class 11th too difficult to cope up with. And that is because this transition is too abrupt, and catches even the best of students by surprise. This document called BRIDGE is to help you overcome the said transition with the least possible effort. In just about 30 pages, we give you a flavor of what topics you'll encounter over the coming two years, and what is the ideal perspective from which you should view these topics. This is not a full exposition, so you should not expect detailed discussions. This is just meant to an indicator, a guide to point you in the right direction and get you properly started on your subsequent journey. We hope you find the BRIDGE useful. LOCUS: Institute for IIT-JEE All IITian Faculty Contents: Part - A NUMBERS (a) Fundamental Facts About Number Systems (b) Significance of Arithmetic Operations (c) Exponentials and Logarithms Part - B ALGEBRA (a) Quadratics (b) Some Basic Series Part - C CALCULUS (a) What are Functions? (b) Case Study: Motion in one Dimension (c) Differentiation: Slopes (d) Integration: Areas Part - D GEOMETRY (a) Cartesian Coordinates (b) Vectors Part - E FAQS ABOUT IIT-JEE PART - A NUMBERS (a) FUNDAMENTAL FACTS ABOUT NUMBER SYSTEMS Numbers are the fundamental building blocks of mathematics, and yet we have seen so many students with basic misconceptions about numbers. Part of the reason for this is that sufficient time is never devoted to introducing successive levels of number systems properly. We’ll try to remedy this on this bridge - and we hope that after reading these pages, you’ll spend more time and effort to learn more about numbers. The best source for doing that is the internet, and with a simple search, you can find some good links where you can get started. We are going to talk about numbers from the perspective of sets. You are aware that the most basic number set is the set of natural numbers ¥ , consisting of the numbers 1, 2, 3, ... onwards. In ¥ , we have a smallest element, namely 1, but no largest element, since no matter how large a number you choose from ¥ , you can always find larger numbers. When the human race was evolving in its earlier stages, the set ¥ was their mathematical friend to count things: 1 tree, 2 cows, 21 stones, etc. But it was discovered later on that the set ¥ is insufficient in many ways. For example, we’ll consider the concept of solvability of equations. Consider the equation x + 2 = 3 . This equation has a solution in ¥ , which is x = 1, so we will say that this equation is solvable in ¥ . But equations like x + 2 = 2, and x + 2 = 1 , are not solvable in ¥ , since no numbers in ¥ satisfy these equations. We therefore arrive at higher extensions of ¥ . Whole numbers W Integers ¢ : 0, 1, 2, 3, .. onwards to infinity ∞ : − ∞,...., − 3, − 2, − 1, 0, 1, 2, 3, ....∞ Are these extensions purely mathematical considerations, or do they have practical value? Yes, they do. One of the most important values of ¢ is that it allows us to mark a number line in both directions, with our starting point as 0. ... -3 -2 -1 0 1 2 3 ... Therefore, a number like –5 makes sense: it is at a distance of 5 units to the left of 0. Note that there is neither a smallest number nor a largest number in ¢ , since it extends to both sides infinitely. Going from ¥ to ¢ has increased the solvability of equations: there is now a larger class of equations that we can solve. But it is easy enough to think of equations which are not solvable in ¢ . For example, 2 x = 3 , whose solution 3 is x = does not lie in ¢ . Physically also, we see that ¢ is insufficient, in the sense that there is nothing in ¢ 2 between two successive integers. How do you tell a person the distance she needs to walk to reach the mid-point of 1 and 2 ? Thus, there are physical lengths that ¢ cannot describe. , where p and q are integers, and q q is non-zero (you are already aware of the reason why q must be non-zero). Note that every integer m is also a m rational, since it can be written as . Using Q, the solvability of equations increases further, and physically too, we 1 can measure lengths which were not possible to measure in ¢ . Sometimes, students become confused about the 355 meaning we should attach to such numbers. For example, what is the physical significance of a number like ? 113 355 You can think of it as dividing a length of 355 units into 113 equal parts; then is the length of one such part. 113 Well, are we done? Is Q enough to solve all polynomial equations? Can every possible physical length in space be represented by a number in Q? Or do there exist numbers outside Q? ww w.locuseducation.org We therefore arrive at Q, the set of rationals. A rational is any number of the form p 1 but we have obtained a p common factor of 2.In the 6th century BC.org 2 . there exists no such number. we must have c = a + b c a c =a +b 2 2 2 b The problem occurs when we apply this theorem to a right triangle with sides 1 and 1 : d 1 d 2= 1 + 1 = 2 1 Now. We had initially assumed common factors between p and q. where m is some integer. But there soon arose a problem.e. Assume p = 2m. which is a contradiction. people believed that there do not exist any numbers outside Q. the q common factor could be canceled out. so it is even. That is. The q Brotherhood was shocked even more. Using a similar logic as earlier. You may remember this from an earlier class. and hence q. Now. p2 should be even. and that every natural phenonenon could be expressed in terms of these number-hence the name “rational”. because if they had. this discovery meant a lowering of the prestige of integers. that in musical instruments. Substituting this in p = 2q gives q = 2 m .. 2 2 ww w. This observation led him to believe that the harmony of the world was closely related to that of the rational numbers. We assume that there exists a p rational number equal to 2 . You see. Therefore. it is possible to extend Q further. which represents a valid physical length. one of the members of the Brotherhood wondered: what (rational) number d is such that d 2 = 2 ? That is. b. Up to that point. But before we get to that. for example. Assume q = 2n. this number. the unfortunate discoverer of this result was drowned! It is important to appreciate the magnitude of this discovery.locuseducation. because according to their beliefs. He thought that it must be possible to represent any length using an integer or a rational number constructed from integers. which correspond to real geometrical lengths. must have been rational. 2 ? q p Great was his shock and dismay when he discovered that there exists no rational number which equals 2 . So our original assumption that there exists a number equal to 2 q is incorrect. the Pythagorean Brotherhood had discovered the famous theorem concerning right-angled triangles. Pythagoras had found many natural phenomena that were described by rational numbers. c as shown. But it is not! For the Brotherhood. a great Greek mathematician named Pythagoras founded a Brotherhood of mathematicians and philosophers. We also assume that p and q have no common factors. let us quickly look at how it can be shown that 2 is not a rational number. But this result meant that there are numbers. what p number equals the square root of 2. i. should be even. 2 2 2 which says that in a right angled triangle with sides a. which means that p should 2 2 be even. and according to legend. p 2 2 = 2 ⇒ p = 2q q The right hand side is 2 times an integer. Now we see the contradiction. The easiest method is one of contradiction. yet do not have any rational interpretation. He had also realized. strings whose lengths are related by rational numbers have harmonious pitches. we see that q2. Thus. W and ¢ . and that solution is d = 2. If you had a camera. is at 2 .. there are still an infinite number of rationals between a and b.. because as we said. -3 2 2 -1 3 3 0 4 4 1 .This means that in Q. ∞ . there is no sense in asking the question “What number comes after 2”. Alright. no matter how close they are. Then consider . the mid-point of a and point of a and b. the mida+b 2 a+ a+b 2 = 3a + b . no matter how close to each other. which lies between a and b. In simple words. One such gap.. Because no matter how close a number you find to 2. Such numbers are called irrational..locuseducation.. you can still find an infinite number of rationals between a and b . the rationals are spaced very close to each other. has a solution in ¡ .org 3 . ¥ : W : ¢ : 1 0 – ∞ . To put it crudely. there exist numbers which are not rational. for example. and this is something really important. there are still gaps. consider . again between a and b. which had no solution in Q. so read carefully and with attention. so much for Q. The rationals and the irrationals 2 form the real number set. as the Pythagoreans discovered. and so on ad which still lies between a and b. Then consider 2 8 infinitum. But look at the picture when we extend ¢ to Q .. you would find that the line still looks filled up... 0 What ! The entire line seems to have filled up. denoted by ¡ . Our equation. What we want to do now is discuss the geometrical picture of the set Q and ¡ . and try to visualize. ∞ 2 3 . the number line is discrete. ∞ 1 -2 It is important to see that up-to now. But is Q continuous or are there gaps even in Q? Well. there is a minimum jump that you have to make. b a z oo m a b a+b To see an example of how there are infinite rationals between (any) two rationals a and b. 2 2 4 3a + b a+ 4 = 7 a + b . A better way of saying this is that if you have two rationals a and b. this implies that it is meaningful to ask questions like “What number comes after 2?” From one number to the next. d = 2 . We have already understood the geometrical (number line) pictures of the sets ¥ . and you could zoom onto the rational line between two points a and b. 0 2 Q ww w. there will still be infinitely more numbers between 2 and that number. . Do their exist numbers with physical significance.org 4 . We see that rational numbers have decimal representations that are either terminating. is still ∞. for example. Also understand that no matter how closely spaced two numbers are. Comment: It is not an exaggeration to say that the concepts contained in this discussion are some of the most important ones you’ll be studying in college. without any gaps or holes. which do not lie in ¡ ? The answer is yes. all these gaps fill up. For now.e. by decimals such as 3. or non-terminating but = 0. As you’ll learn later. no matter how large. conventionally there is no sense in talking about expressions like ∞ − ∞ and ∞ (b) SIGNIFICANCE OF ARITHMETIC OPERATIONS We need to first understand a basic difference between rationals and irrationals. for example. Having extended our number system to the set ¡ . close your eyes a few minutes.14159 or by rational repeating. More on this when you reach class 12th. these quantities make sense in a manner different from conventional mathematical operations. But we’ll not be studying the complex set C on the bridge. if you had a powerful camera with which you could analyze the Q-line. That is. that ∞ + a. Think of the chips of wood as rational numbers and the water as irrational numbers! What you need to really understand from all this is that once we extend the set Q and include all the irrationals to form the set ¡ .4142135623. Also.. there are gaps at all the irrational numbers. you’ll still find infinitely more numbers between these two numbers. which can sometimes seem confusing. irrational numbers have infinitely long decimal expansions that never form a repeating pattern. even though there are infinitely many rationals and infinitely many irrationals. 2 7 = 0 ⋅ 285714 . proper understanding upto ¡ is sufficient. Irrationals : 2 = 1. where a is any . π can never be written down exactly.. Sounds confusing? Think of it this way. and such numbers are called complex numbers. 2 Rationals : 5 = 0. i. Visualize a large lake with small chips of wood floating here and there. Before we close this section.4 . is not a number in the strict mathematical sense. 3. 3 In contrast. Infinity... In fact. = 0. is based on the fact that the real line is continuous So. represented by the symbol ∞.1415926535. This means that we can talk about the exact value of a rational number. All you can say about ∞ is that it is greater than any number. ∞ when you understand the concept of limits. Thus. and the number line becomes continuous. number. but only approximately. This means. that in their decimal representations.There are similar gaps at every length which cannot be represented using a rational number. However. is very large. The set of these gaps. the next question that arises is whether any further extension is possible. before reading further. 1 representations such as 22 7 or 355 113 . and visualize the continuity of the real line in your mind. Any very small volume of the lake selected at random will very likely contain no chips at all and just water. for example. the number of irrationals is infinitely more than the number of rationals.. let us briefly talk about the concept of infinity. ww w.3333.. For example. The whole concept of limits and calculus. π = 3. You can pick up any point on the number line. no matter where it is.locuseducation. the set of irrationls. and you’ll be able to find a corresponding number in the real set. you would notice that there are a lot many gaps of this form. ÷ .. The way you’ll do it is first ask. However.. but what about quantities 3⋅ 4 5⋅6 = 34 56 = 17 28 = { Length of one part when a length of 17 units is divided into 28 equal parts } However. suppose you were told to calculate 2 × π .. ww w. ??? Well. × 3 ⋅ 1415926535 . This is a consequence of the infinite nonterminating non-repeating decimal expansions of irrational numbers. the example we proposed should not give you the idea that the product of two irrationals must be irrational. we can never calculate it exactly as already mentioned. what sense do we attach to quantities like 2 × π . Try answering these questions: (i) Can the product of two rationals be irrational? (ii) Can the product of a rational and an irrational be (a) rational (b) irrational ? (iii) What if we talk about addition instead of multiplication? As in the case of multiplication.org 5 . × .. but you can choose to what accuracy you need to calculate the product: Low accuracy Medium accuracy High accuracy Very high accuracy : : : : 2 × π ≈ 1 ⋅ 4 × 3 ⋅ 1 = 4 ⋅ 34 2 × π ≈ 1 ⋅ 414 × 3 ⋅ 141 = 4 ⋅ 441374 2 × π ≈ 1 ⋅ 41421 × 3 ⋅ 14159 = 4 ⋅ 4428679939 2 × π ≈ 1 ⋅ 4142135 × 3 ⋅ 1415926 = 4 ⋅ 4428826664201 What we’ve done is successively better approximations of the product. It is easy enough to understand the geometrical significance of addition and subtraction in terms of lengths: A d d ition : a + b = a + b S u b tra ctio n : a – b What about multiplication and division? Quantities like 3 × 4 and like 3 ⋅ 4 × 5 ⋅ 6. This concludes our discussion on the four basic arithmetic operations : + . the product of two irrational numbers? We just saw that 2 and π have infinite decimal expansions. But no matter how accurately we calculate the product. or 3 ⋅ 4 ÷ 5 ⋅ 6 ? Easy enough: 3⋅ 4 × 5⋅6 = 34 10 × 56 10 = 1904 100 = 19 ⋅ 04 10 13 = a–b make perfect sense. −. what is the accuracy to which I need the product? You can never exactly calculate the product. A counter example is 2 × 2 = 2 .locuseducation.Now we talk about arithmetic operations. you can only approximate it. we can understand division too. so what do we mean by their product: 2 × π = 1 ⋅ 4142135623 . org 6 .. But as in the earlier case.. at the outset itself. ax is defined whether x is positive or negative or zero. of the form generally have a > 0.01 + 0. as we’ve seen. we cannot calculate it exactly. In general.. If a is a fixed constant and we let x be a variable real number.. to any term of the form a interpretation: . We clearly understand the meaning of these terms: −3 3 2 = 2×2×2 .. the following question arises: what if x is irrational? What since do we attach to a then? For example .. has a well defined meaning. = 1+ 4 10 + 1 100 + 4 1000 + 2 10000 + 1 100000 + ..4 + 0.. (Why should that be?) q . We can calculate 2 2 as accurately as we wish by taking as many terms after the decimal point into account. rational or irrational. Each of the terms in the product. 2 2 =2 1 ( 1+ 4 1 4 2 1 + + + + + .. so it is imperative that you take time in understanding these operations completely.00001 + . but you have to be clear about how to interpret 2 = 1. we can attach the following a p/q = a ( ) = (q 1/ q p th root of a ) p p So. 10 100 1000 10000 100000 4 10 1 100 4 1000 2 ) 1 100000 = 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅ .. the same relation becomes log a b = x x ww w. This discussion should make it clear that for a > 0.  1  = −4   81  a = c is equivalent to saying that log a c = b Now..41421. Additionally.. In terms of log.. 3 p/q is a real number which when cubed will give 3.. First of all. and therefore so does 2 2 . we see that we can attach sense to any term of the form a x where x is rational. Example: 2 = 8 is the same as saying that log 2 8 = 3 3 3 = −4 1 81 is the same as saying that log 3  b In general. This is how 2 can be expressed as an infinite series of rational terms.. 2 = 1 2 3 = 1 2 1/ 2 × 1 2 × 1 2 1/ 3 Going further. it is helpful to think of the exp and log functions an ‘inverses’ of each other..... = 1 + 0. we can define the general exponential term as 10000 b = a . where x is any real a is called the base. we need to understand the exponentiation operation more carefully.004 + 0. we Now. how will we evaluate 2 this quantity: 2 x ? Is this quantity defined? The answer is yes.locuseducation. and x is the exponent.0002 + 0.(c) EXPONENTIALS AND LOGARITHMS It has been our experience that students are not very clear about these operations. 2 is a real number which when squared will give 2.. terms like 2 1/ 2 and 3 1/ 3 also make sense. Now.. B (a) QUADRATICS ALGEBRA Consider a variable x. x =1 ⇒ y=3 . the value of the ‘output’. PART . Exp x –4 –3 –2 –1 2 x Log x 0 . y will vary: x=0 ⇒ y=2 . say y.. We do this by calculating exp and log for a fixed base. A linear expression in x is of the form ax + b. i. We will pictorially try to understand this expression by drawing its graph. The linear expression becomes y = x + 2. we construct two perpendicular axes. ax + b varies. where a and b are real constants. and as x varies.locuseducation.0 6 2 5 0 . n o m atter w h at th e in p u t x .It is very important to get a numerical feel of the variations in the exp and log terms. x= 2 ⇒ y= To pictorially represent this variation.5 2 0 1 1 2 2 4 3 8 4 16 G en eral B eh avior x – . T h e lo g o p eratio n is d efin ed fo r p o sitiv e in p u ts o n ly. lo g 2 x 8 8 T h e o u tp u t 2 is alw ay s p o sitiv e. 1 = 0 . ‘linear’ suggests ‘line’ .0 6 2 5 16 1 = 0 .e.org 7 . lo g 2 x . As x varies. let a = 1 and b = 2. 2x 0 x x . How do you draw a graph? Well.2 5 4 1 = 0 . The important thing to ask here is: why is this expression called linear? Well. sin ce th e lo g o f a n eg ativ e n u m b er m ak es n o sense.and this is what we find. The general convention is to denote the output by another letter. the horizontal one representing the input x.5 1 2 4 8 16 x x log 2 x –4 –3 –2 –1 0 1 2 3 4 8 2 + 2 etc. and the vertical one representing the output y: y x 8 ww w. and particular values of the variable exponent x.1 2 5 8 1 = 0 . So y = ax + b is the output. 2 8 8 x G en eral B eh avior – 0. x is your ‘input’ variable.2 5 0 . For example.1 2 5 0 . say a = 2. In the case of y = x + 2. we obtain 0. Returning to quadratics: Consider a variable x – a quadratic in x is any expression in which the degree (highest power) of x is 2: 2 y = x (the most basic quadratic). and plotted as follows: y 3 (1 . since these two x-values are the roots of the expression. ww w. There will be (infinitely) many points which satisfy this expression. we mean those input values for which the output value is 0.locuseducation. in y = x + 2.org 8 . try factoring the quadratic expression y = x 2 + x + 1 . we say that x = 1 and x = 2 are ‘roots’ of this quadratic expression. However. y = ( x − 1) ( x − 2 ) = x − 3 x + 2 2 Conversely. if we substitute x = 1 or x = 2 in this quadratic expression.3 ) ( 2 . Lets take a look at the graph of y = x 2 − 3 x + 2 : y x 2 – 3 x + 2 = (x – 1 ) ( x – 2 ) 2 x 1 2 Note the curvature in the graph. Since its factors are (x – 1) and (x – 2). For example.2 ) (-2 . For example. y = ax + bx + c (general quadratic expression) 2 Note that if we multiply two linear expressions. any expression of the form y = ax + b gives us a straight line.Now. the general equation of a straight line. This point will be written as (1. We plot all such point on the ‘x–y plane’. Also note that the graph cuts the x-axis at x = 1 and x = 2. y = 3 is one particular point satisfying this expression. for the subsequent discussion. and so the y-value is 0 at these x-values. By the roots. y = 2 x 2 + 3 x + 4 . we plot the input-output pairs as points on this figure.3 ) x 1 Note that this is just one point satisfying y = x + 2. 2 + 2 ) (0 . x = 1. 2 Now. Thus. we say that the quadratic expression y = x 2 − 3 x + 2 can be ‘factored’ into two linear expressions (x – 1) and (x – 2). and join them. This special curve is actually called a parabola. and that’s how we obtain the graph. 0 ) x In general. we get a quadratic expression. factoring a quadratic expression may not always be possible (until the time you use complex numbers). we get a straight line: y=x+2 S o m e m o re po in ts h av e y b ee n p lotted as e x am p le s on y = x + 2 (1 . Many of you will be familiar with the equation y = mx + c. and it has a lot of interesting properties. 3). we use the expression y = x − 3 x + 2 . y = x 2 + 1. As x decreases and 2 approaches −∞ . say y = ( x − a )( x − b ) .locuseducation.Consider another example: y = x + 6 x + 5 = ( x + 5) ( x + 1) 2 y –5 x –1 Once again. To help you in this. if a quadratic expression can be written as a product of two linear factors.org -1 / 2 –2 x 9 . and the roots this time are at x = −5 and x = −1. (i) y = x 2 − 1 = ( x + 1)( x − 1) y (ii) y = x 2 = ( x − 0 )( x − 0 ) y T h e g rap h to u ch e s the x -ax is sin ce th e ro o ts are the sam e . we present a few more examples of quadratic expressions and their corresponding graphs. We write this as follows: > 0 if x < a or x > b y = (x − a) (x − b) : < 0 if a < x < b These are extremely important observations. y again approaches +∞ . observe this graph carefully and note some very important facts: (a) The output y has a least value at the point marked x. x = 1   1 2  ( x + 2) 2 1 1 x ww w. (d) The y-value is negative when the input is between a and b. 0 x –1 –1 +1 x (iii) y = x 2 − 2 x + 1 = ( x − 1) 2 (iv) y = 2 x 2 + 5 x + 2 = ( 2 x + 1)( x + 2 ) = 2  x + y y O n ce ag a in . y also increases and approaches +∞ . but positive when x < a or x > b . In general. Make sure you understand them carefully. th e gra ph ju st to u ch e s th e x -a xis sin ce th e tw o ro ots are x = 1. (c) The graph is symmetrical about the point marked x. This is because of the x term in the quadratic expression – the square of a negative real is positive. (b) As x increases and approaches +∞ . the graph will be a parabola which cuts the x-axis at the two roots x = a and x = b : y a b x Now. the graph is a parabola. Observe each one of them in detail. but no greatest value. So. the complex set C. you may have come across the quadratic formula: for a quadratic expression y = ax 2 + bx + c . The roots exist in the next higher extension of ¡ . In this case. we see that the graph of a quadratic expression may intersect the x-axis at two distinct points (two distinct roots) or it may touch the x-axis at one single point (two identical roots). the graph of y = x 2 + x + 1 does not cut the (real) x-axis. where D = b 2 − 4 ac (D is called discriminant) 2a How do we arrive at this expression? Well.org 10 . the roots of y = x 2 + x + 1 are non-real. We make this distinction graphically. Lets apply this to the expression y = x + x + 1: 2 x= −1 ± (1) − 4 × 1 × 1 2 2 = −1 ± 2 −3 The term under the square root is negative. In a lower class. we rearrange the quadratic in the following manner: 2 2 2 2  x2 + b x + c  = a  x2 + b x +  b  + c −  b   b b − 4 ac y = a        = a x +  −   a a a   2a  a  2a   4a  2a   x= −b ± D When we equate this to 0 and solve for x. there is no x for which 2 y = x + x + 1 becomes 0. Therefore. the constant a has a very special role to play. we get the required quadratic formula. but not in ¡ . since the roots are non-real. −3 is not a real number. Now. Does this mean that the roots do not exist? No! The roots do exist. we need to mention one very important fact. we asked you to try and factor y = x + x + 1 into (real) linear factors. because for no real value of x can y become 0. In the quadratic expression y = ax 2 + bx + c . a while back. Let us discuss this from an algebraic point of view. we want to draw your attention to the case when a < 0. y will approach − ∞ . a>0 a<0 U p w a rd s p ara bo la D o w n w ard s p arab o la ww w. try to see that as x approaches + ∞ or − ∞ .locuseducation. In other words. However. the roots are given by . Specifically. Before summarizing. You’ll have been unable to do so. Let us look at the graph of this expression and try to understand why you could not factorize it: 2 y 1 x No roots exist for this expression! The graph never intersects the x-axis. and not + ∞ as earlier. d istin ct ro ots a .org 11 . + n = 6 2 2 n ( n + 1) 3 3 3 3 1 + 2 + 3 + . but you could still try to prove them if you wish to. we summarize all the possible cases of a quadratic expression y = ax 2 + bx + c . id en tic al roo ts a ii) y > 0 if x ≠ a iii) y = 0 if x = a (2) D =0 a i) R ea l id e n tica l roots a ii) y < 0 if x ≠ a iii) y = 0 if x = a a E x a m p le : y = x 2 – 2 x + 1 E x a m p le : y = – x 2 + 2 x – 1 (3) D <0 i) N o re al ro o t ii) y is alw ay s p o sitive (3) D <0 i) N o re al ro o t ii) y is alw ay s n eg ativ e E x a m p le : y = x 2 + x + 1 In th is ca se . ww w. and hope that you will take time to observe each one in detail and construct your own examples. (A) Basic Number Series You just need to be familiar with the results of the following three summation series... ro o ts are 'im a gin ary ' E x a m p le : y = – x 2 – x – 1 In th is ca se . where you’ll study them in a lot of detail. ro o ts are 'im a gin ary ' () b SOME BASIC SERIES In class 11th. + n = 4 Since we’re still on the bridge. b ii) y > 0 if x < a or x > b iii) y < 0 if a < x < b a 2 (1) a D>0 b i) R ea l.Now. n ( n + 1) 1 + 2 + 3 + . knowing the justification of these results is not necessary.... d istin ct ro ots a .. there is a whole chapter dedicated to series and progressions. E x p re ssion : y = a x 2 + b x + c a>0 R o o ts : x = –b± D 2a a<0 D (d iscrim ina n t) = b 2 – 4 a c (1) D >0 i) R ea l..locuseducation. + n = 2 n ( n + 1)(2 n + 1) 2 2 2 2 1 + 2 + 3 + ... But it is still important to be familiar with sum basic series summations as soon as possible. We are going to discuss three classes of series.. b ii) y > 0 if a < x < b iii) y < 0 if x < a o r x > b b E x a m p le : y = – x 2 + 3 x – 2 E x a m p le : y = x – 3 x + 2 (2) D =0 i) R ea l. .(B) Arithmetic progressions (AP) An AP is a series in which successive terms have the same difference.. + ( a + ( n − 1) d ) .. 8. a + 2d.. On the right... and thus S= (C) 12  1  14 + 11 × −  = 51 2  2 Geometric progression (GP) A GP is a series in which successive terms have the same ratio.. + 1 2 1 2 There are 12 terms in this AP. We make use of this fact by writing S in reverse order. Let the first term be a. our series sum S can be written as ... a = 1 and d = 3. 4. 2 S = n × ( 2 a + ( n − 1) d ) ⇒ S= n 2 ( 2 a + ( n − 1) d ) Lets apply this formula on two examples..... 2 2.... + ar + ar 2 n −1 Subtracting the two. . Note that the nth term will be a + ( n − 1) d .. (1) S = 1 + 4 + 7 + 10 + .. and the (common) difference between successive terms be d. ( r − 1) S = ar − a ⇒ n S= a ( r − 1) n r −1 ww w. So.. etc Suppose that we want to sum n terms of an AP. 4. S 2 = 1. Also. Examples: 3 5 S1 = 1.. 2. 5. . so that 100 S= ( 2 + 99 × 3 ) = 14950 2 (2) S = 7+6 1 2 +6+5 1 2 + . On the left. 2... the first and the last term. 2 2 S 3 = 11.. S = ( a + ( n − 1) d ) + ( a + ( n − 2) d ) + . . whose first term is a and the (common) ratio is r.. The terms in the GP will be a.org 12 .. 7. the term ( a + ( n − 1) d ) is generated n times.. 2 4 8 S 3 = 1....(1) How do we find S? Note that terms at the same distance from the start and the end (for example. + 298 It should be easy enough to figure out that this AP has 100 terms. etc) have the same sum 2a + (n – 1)d. Note that d can be positive as well as negative (or even 0). a + d. 10.... Thus. 2 etc. Once again our problem is to find the sum S of n terms of a GP.. . ar .. 7.. 9.... 2... Therefore.. Examples: 1 1 1 S1 = 1. Also... we get 2S... the nth term being ar n −1 . . + ar 2 n −1 n rS = ar + ar + .. S 2 = 1. a = 7 and d = − ... the second and the second-last term. ar . ⇒ S = a + ar + ar + .locuseducation.(2) Now..... + ( a + d ) + ( a ) S = ( a ) + ( a + d ) + ( a + 2 d ) + .. we sum (1) and (2). .. 2 . The terms in the AP will thus be a.. y ( x . So here’s the picture you have to keep in mind. i. Thus. + 1 2048 1× 2 − 1 12 ( 2 −1 ) = 4095 In this case.e.Let us apply this extremely useful result on two examples: (1) S = 1 + 2 + 4 + 8 + . A function is generally denoted by a letter. another number is assigned through a specific rule. y = x2 tells us that there is an input variable ‘x’. the input. But since space is short in this introductory document. while the output f(x) or y. P is a function of I: 2 P = g (I ) = I R Both of these examples describe a rule whereby. which takes in a input value and gives out an output value according to a specific rule. In a graph. DO NOT think of it as multiplying f with x! Here are some examples: (a) The area A of a circle depends on its radius r.. to produce a value ‘y’. and helps us to visualize the behaviour of the function pictorially.C CALCULUS (a) WHAT ARE FUNCTIONS? We are not exaggerating when we say that if you understand the concepts and applications of functions properly. 2 The most common method for visualizing a function is its graph. f ( x )) f (x) f (1 ) 0 1 2 f (2 ) x x 13 ww w. and also much in physics.locuseducation. say f. The graph of a function is like a ‘life history’ of the function.... or a black box. f(x) is read as ‘f of x’. to a given number... S = (2) S = 1+ 1 2 + 1 4 + . i. r = 1 2 1×   . A function is a rule which tells us how one variable is related to another. we’ll have to make do with a brief overview.org . For example. x f f (x ) = y The symbol ‘f(x)’ should be thought of as an operator ‘f ’ acting on a variable ‘x’. A is a function of r: A = f (r ) = π r (b) The power P dissipated in a resistor depends on the current input I.e. so that S =   1 12   − 1  2   = 4095 2048  1 − 1   2  PART . is along the horizontal axis. say x. + 2048 There are 12 terms in this GP.. you will already have accomplished a lot in covering pre-college mathematics. is along the vertical axis. and the output variable ‘y’ is the square of ‘x’. with a = 1 and r = 2. x In p ut F u n ction O u tpu t y You can think of a function as an operator. the car is moving to the right.Since the y-coordinate of any point (x. which tells us how the velocity of the car is changing. totally smooth. which means that if the car is moving with a certain speed. which means that the velocity of the car is decreasing very fast. Suddenly. So. Here’s another example of a graph. x. So. which means that the velocity is increasing very fast. Are these variables sufficient to describe the motion of the car? Well. if the accelerator of the car is pressed full on. Suppose that the car moves with this acceleration for 20 seconds. the car has attained a certain velocity v. even the acceleration of the car keeps changing. a. so that the car starts moving with a constant acceleration a0.. If v = 0.org 14 . What all variables will we need to describe the motion of the car? Well.7 etc For now. y) on the graph is y = f(x). v. and the car starts decelerating at the same rate a0 at which it was initially ww w. the acceleration of the car is 0. We have the velocity v of the car. because the accelerator is not pushed to the same level all the time. x=0 t=0 Now. there’s the time variable t. for our purposes. we can read the value of f(x) from the graph as being the height of the graph above the point x. Finally. y 1 0 1 x From this graph f (1) = 3 f (2 ) = 4 f (5 ) ≈ −0. this much background on functions is sufficient to follow the subsequent discussion. so that it moves with that same velocity v for another 20 seconds. the acceleration is highly negative. You are driving the car. Now. we also need to take into account the acceleration a of the car. these four variables are sufficient. After these 20 seconds. For example. and if it is negative. and the driver pushes the brakes very hard. the car is moving to the left. we need to consider four variables: t. Now you remove your foot from the accelerator. and the car keeps on moving with that same speed indefinitely. you push the brakes. because there is no change in velocity. the car is accelerating very fast. i. and we’ll assume the convention that if v is positive. an obstacle appears on the road. the car is stationary. you can have even more variables. which can initially be assumed to be x = 0. However.locuseducation. Then we have the position x of the car. we want you to do a thought experiment. there is nothing to slow down the car. which is initially t = 0. that is. the car is decelerating very fast. (b) CASE STUDY: MOTION IN ONE DIMENSION We will now study the (practically simple) case of a car moving on a straight road. Now. we see that to accurately describe the car’s motion. So we could have another variable j telling us how the acceleration changes (we do have such a variable and it is called jerk). you are at the position x = 0 at t = 0 . and the road is ‘frictionless’.e. For example. Obviously. you push the accelerator to a certain level. Note that during these 20 seconds. and you remove your foot from the accelerator. t graph from 0 to 20 seconds. What we want to do now is to plot the variation of the motion variables x. b u t at a n o n u nifo rm rate. at t = 20 seconds.org 15 .locuseducation. this should be easy to understand. At t = 60 seconds. Since the acceleration is uniform. th e v e lo city g o e s from 2 0 a 0 to 0 t 20 x 800a0 6 0 0 a0 40 60 C a r co v ers a d ista nc e o f 2 00 a 0 at n o n-u nifo rm d ec rea sin g ra te in th e se 2 0 se co n d s C a r co v ers a d ista nc e o f 4 00 a 0 'u nifo rm ly ' in th ese 2 0 se c o nd s 2 0 0a 0 C a r co v ers a d ista nc e o f 2 00 a 0 in the first 2 0 se c o nd s. the velocity is 20 a0 which is the area under the a . S o a fte r 2 0 se co n d s. If we denote the area under a graph from 0 to t by the symbol ∫ . so a fte r 2 0 se co n d s. we want to draw your attention to the following facts. the velocity is 0. For example. At t = 60 seconds therefore. and emphasize that you try to understand each and every part of these graphs carefully and in detail. and the (algebraic) area under the a – t graph from 0 to 60 seconds is also zero. a C a r is a cc e le ra ting a0 C a r is m o v in g w ith u n ifo rm sp ee d 40 20 -a 0 v 60 t C a r is d ec e le ra ting C a r's v eloc ity is in cre asin g c on tin u ou sly a t a ra te o f a 0 p e r se co n d . we can express this fact as 0 t v (t ) = Z ∫ ( a − t g rap h ) 0 t v elo city as a fun c tio n o f tim e ww w. you have come to a stop. it will also take 20 seconds at the deceleration rate a0 to go from v to 0.accelerating. How much time will it take for the car to stop? Since the car took 20 seconds at the acceleration rate a0 to go from 0 to v. v and a with time. sin ce th e ve loc ity is inc rea sin g co n tin uo u sly t 20 40 60 Once you’ve analyzed these graphs to your satisfaction. (a) The velocity of the car at any time instant t is the area under the acceleration-time graph from 0 to t. a t th e ra te o f a 0 p e r se c on d . We do this below. th e v elo city is 20 a 0 C a r's v eloc ity is co n stan t a t 2 0a 0 2 0a 0 C a r's v eloc ity is de c re asin g c o ntin u o usly. Y Ve ry stee p X N o t so ste ep Returning to our graphs. the determination of the (varying) slope of the x – t graph may not yet be obvious to you. which is the 2 value of x at t = 20 seconds. For example. but you can still verify this assertion for the middle 20 seconds.locuseducation. But the important thing to note is the relation of x(t) in terms of the area under the v – t graph: increasing uniformly in the first 20 seconds.org 16 . You do know how to calculate the slope m of a straight line segment XY. → → a (t )  v(t )  x(t ) 0 0 ∫ t ∫ t Now. you can intuitively try to understand that the tangent at any point on the curve is a measure of the steepness of the curve at that point. the area under the v – t graph (in the form of a right triangle) is × 20 × 20 a0 = 200 a0 . For example. How did this happen? Well. the average velocity in this period is 0 + 20 a0 x (t ) = ∫ (v − t graph ) 0 t Thus.(b) The position of the car at any time t is the area under the velocity-time graph from 0 to t. For example: t = 25 seconds ⇒ v = 20 a0 = Slope of x − t graph at (t = 25 s ) ww w. we make two further observations. we observe that the acceleration at any time instant t is equal to the slope of the velocitytime graph at that instant. one way to think of it is this: since the velocity is = 10 a0 . For example. For the first 20 seconds and the last 20 seconds.t graph at any time instant t gives us the value of v(t) at that instant. but for now. at t = 20 1 seconds. t = 15 seconds t = 25 seconds t = 45 seconds ⇒ ⇒ ⇒ a = a0 = Slope of v − t graph at t = 15 s a = 0 = Slope of v − t graph at t = 25 s a = − a0 = Slope of v − t graph at t = 45 s Do we have a similar relation between v(t) and the x . and so 2 the distance covered is 10 a0 × 20 = 200 a0 .t graph? Yes. we do! The slope of the x . Y a b a θ m = tan θ = X b What about the slope of a curved segment XY? The precise way to understand the slope of a curve will become clear in the next section. Since all the three graphs have time t as the variable on the horizontal axis. dt (v − t time graph ) dt dt (It will become clear later why we’re using this symbol. For now. Thus. so calculating slopes and areas are easy. we have Thus. What if the v – t graph is something like this: v ( t) t In this case. In this sense: Calculation of graph slopes d     dt  Calculation of AND areas under graphs ARE t  ∫  0  Inverse Operations Now. Combining with our previous result. how do you calculate t (v − t graph ) dt It is in answer to such questions that the subject of calculus originated. Let us denote the action of calculating the slope of a graph at the time instant t by the symbol . v (t ) = d ( x − t time graph ) and a (t ) = d  →  → a (t ) ← v (t ) ← x(t )   d d 0 0 ∫ t ∫ t dt dt Thus. ww w. we considered a very simple case of a car accelerating. and hence has two branches: x (t ) = a (t ) = 0 ∫ (v − t graph ) and d Problem Calculating Slopes Symbol d dt t Name of branch Differential Calculus Integral Calculus Calculating Areas ∫ 0 We now turn our attention to these two problems separately. just stick with it).org 17 . going from x → v → a involves calculation of slopes of graphs. (c) DIFFERENTIATION : SLOPES We want you to make this association permanent is your mind: differentiation of a curve means calculating the slope (steepness) of the curve at a given point. In this case.locuseducation. calculus is about two questions. the slopes are being calculated ‘versus d time’. while going from a → v → x involves the calculation of areas under graphs. the variations in velocity consist of straight lines. with one end point at A. y = 1 ) is th e p oin t at w h ich w e w ish to c alc ulate the slo pe. y = t 2 ) is a p oin t o n the c u rv e lyin g to the rig h t of A .e . a s sh o w n .Once again. te lls us th e insta n ta n eo u s d irec tio n of m ov e m e nt a t X . t2 1 A 1 t B C 2 x ww w.. i. What we do is consider a straight line segment on the curve y = x2. B Slope of AB is constant and equals BC AC C The problem is in finding the slopes of ‘curvy’ segments: B A θ m = tan θ = What is the slope or steepness of the curve at X ? A As we mentioned earlier. which lies on the curve to the right side of A: y 4 A ( x = 1 . We consider the function y = x2. T h e slo p e o f th e ta n g en t a t X .org 18 . what is your ‘instantaneous’ direction of movement at the point X? That is what the slope of the tangent tells us: B X X A θ T h e d irec tio n o f the c ar is ch a n gin g a t ev ery in stan t. the slope of the tangent at X should be taken as the slope of the curve at X. at some given point. Suppose that x varies from 0 to 2. and the other end-point at B. y will therefore vary from 0 to 4. We intend to calculate the slope of this curve at x = 1: y 4 1 A θ x 1 2 The solution is obtained as follows: we know how to calculate the slope of a straight line segment. tan θ. B ( x = t .locuseducation. The problem now boils down to finding the slope of the tangent to the graph of a function y = f(x). we note that it is easy to calculate slopes of straight lines. Think of it this way: if you are driving a car from A to B. Let us take a concrete example to illustrate how this problem was first solved more than 300 years ago. 1) to B(t. Secant AB → Tangent at A In other words. Average slope of curve from A to B → Instantaneous slope of curve at A We just calculated the average slope of y = x2 from A(1.e .org 19 . try to visualize this variation in your mind. The brilliant idea that solved this problem was as follows: to calculate the instantaneous slope at A from the overall slope between A and B. which means that As B → A. As B → A. without any consideration of how the curve varies between A and B. t2) as t + 1. i. x 2 ) x2 A y C x x ww w. a t A ( x . Exactly at the point A. y 4 B 1 A 1 2 x As the diagram may not to be too clear. As B comes closer and closer to A. then. we calculated the slope of y = x2 when x = 1. the slope must therefore be 2. y y2 B W e n e ed to ca lcu la te th e slo p e o f th e c urv e a t an y g ive n v a lu e o f x . the average slope of the curve between A and B. If we denote this movement by B → A. we have evaluated the ‘overall behaviour’ of the curve between A and B. the secant AB ‘almost’ becomes the tangent at A. Now. goes closer and closer to 2. As B moves towards A. As B → A. which is t + 1. t + 1 → 2 This is the answer we were looking for! By making B move closer and closer to A. t → 1. What we actually wanted to do was calculate the slope of the curve exactly at the point A. When B comes very close to A.. Note that as B → A. What is the slope of the curve at any given value of x? We repeat the process of movement above once again. In other words.What is the slope of the segment AB? We know how to do this: 2 BC t −1 = = t +1 m AB = AC t −1 Easy enough! But what we have calculated above is the ‘average’ slope of the curve from the point A to the point B. we make B move along the curve towards A. the average slope between A and B comes closer and closer to the instantaneous slope at A.locuseducation. the secant AB ‘comes closer and closer’ to the tangent at A. This also verifies our earlier calculation: the slope at x = 1 is 2. We differentiated y = x at x = 1 2 and and We obtained a slope of 2 We obtained a slope of 2 x d We differentiated y = x at any given x 2 In the previous section. Once again. Now. for which we want to evaluate the slope at a given point A. so that m → 2 x .locuseducation. and make B move towards A: y y = f (x ) f( x + h ) B ∆y T h e x -co o rd in a te o f B is x + h . As B A. This process that we carried out is what is known as differentiation. so (x ) 2 at x =1 2 =2 = 2x d (x ) Now. ∆y = f ( x + h ) − f ( x ) . from this figure. Therefore. we see that ∆x = ( x + h ) − x = h. the average slope of the curve y = f(x) between the point A and B is m AB = BC AC = ∆y ∆x = f ( x + h) − f ( x) h ww w.org 20 . y → x. we take another point B to the right of A. We consider a general curve y = f(x). the instantaneous slope of y = x2 at a given value of x is 2x. Therefore. h 0 f (x ) A ∆x x C x+ h x Any change in a given variable is represented using ∆. let us quickly try to understand why the symbol for differentiation is dx .The average slope of the curve from A to B is m AB = What happen when B BC AC = y −x 2 2 y−x AB = y+x → A? Well. we mentioned that the symbol of differentiation is d dx d dx dx . . 3 dy dx d x or ( ) ? First. What is average slope of this curve from x to x + h. Instantaneous slope of y = f ( x ) = dy dx =  ∆y    ∆x  = as ∆x → 0 f ( x + h) − f ( x) h as h → 0 Let us apply this differentiation process on the curve y = x . both ∆x and ∆y become smaller and smaller. ∆y ∆x ∆y → dy dx dy This is how you can think of it: ∆x represents the instantaneous slope of the curve.e .As B → A . dy represents the average slope of the curve over a large ‘interval’. ∆ x → 0 and ∆ y → 0 . but dx Summarizing. Thus.e. Now. since dx is so small. i . 1 1 x ww w. so that we obtain the instantaneous slope at x: dy dx = d x ( )= 3 dx (3 x 2 + 3 xh + h 2 ) as h → 0 = 3x 2 For example. the instantaneous slope of the curve. ∆x → 0 . i. y T h e slo p e o f y = x 3 a t x = 1 is 3 .org 21 . can be obtained from the average slope. we calculate the 3 dx ∆y ∆x = (x + h ) h 3 −x 3 = 3 x h + 3 xh + h 2 2 3 h = 3 x + 3 xh + h 2 2 Now. As B → A .locuseducation. one of the originators of calculus. we make ∆ x → 0. h → 0 . visualize the movement process carefully and try to see that as B → A . Therefore. ∆y ∆x . by making the interval over which you’re taking that average smaller and smaller. but the ratio ∆ y / ∆ x ‘approaches’ the instantaneous value of the slope at the point A. Leibnitz. the slope of y = x3 at x = 1 will be 3. In fact. dx . started using the notation dx and dy instead of ∆x and ∆y for small changes in x and y. y y= f(x) W e w ish to ca lcu la te th e sh ad e d are a A . do not be disheartened if you have not fully grasped the essence of this process in one reading. As we increase the number of rectangles n (or decrease the width of each rectangle). we get an approximation to the area A. Now. b u t y o u a re ex p ec te d to v isua liz e a la rge n u m be r o f re ctan g les. work out each minor step and detail on your own. Consider a curve y = f(x). You should try this yourself: draw a random curve. Read it again and again.org x 22 . and consider the area under it from x = a to x = b. First. if we sum the areas of all these rectangles. (d) INTEGRATION : AREAS Once again. and we want to calculate the area under this curve from x = a to x = b. we do know how to calculate the area of a rectangle: l b A A = lb The idea behind integration is to divide the area we wish to calculate into rectangles of very small width: y y = f (x) W e h av e d ivid ed th e are a A w e w ish to ca lc u la te in to rec ta n gles w ith v ery sm a ll w id th s. y = f(x ) a b x a b ww w. and there’s no reason why you shouldn’t understand it eventually. we want you to make the following association permanent in your mind: integration is the process of calculating areas under curves. This is expected: the process of differentiation is not exactly trivial and represented a major advance in mathematics at the time it was discovered.Comment: Many among the readers may not understand this discussion in one go. In th is figu re . is re pre se nted b y A= A a b x ∫ f (x) dx a b W e w ill d iscu ss m o re o n th is n o ta tio n so o n . like the one shown above? Well. So. only 7 rec ta n g le s are sh o w n . w h ich a s w e h av e a lre ad y m en tio n ed . approximate y that area using only 3 rectangles. So how do we go about calculating areas under arbitrary curves. this approximation to A becomes better and better.locuseducation. The height of the rectangle at x (shown shaded). For now. taught us. is f(x). is actually an elongated ‘S’. g (1) = 1 3 7 3 Step . we use the symbol . But how actually is this process carried out? How do we calculate the sum of areas of so many rectangles? Well. The area of the rectangle at x is therefore f(x)dx. In this case.2: Calculate g(x) for the lower and upper limits of integration. 23 . which as we’ve seen in the previous section. approximate this area using 8 rectangles: y f( x ) y = f (x ) x a x b In the second case. We need to sum the areas of all these rectangles between x = a and x = b. the lower limit is x = 1 and the upper limit is x = 2. so we make the width dx. To represent the sum. what do we mean by the notation ∫ f ( x)dx ? Well. Suppose that we want to calculate the area under y = x2 from x = 1 to x = 2.3: Take the difference. we get the exact value of A! So. which if you notice. the approximation to A becomes better and better. g (2) = 8 3 . This suggests that as n increases. you just need to be familiar with the result of that theorem. We wish to keep the width of this rectangle very small. because the number of rectangles n is larger. That is the area you wish to calculate! In this case. The area A thus becomes A= ∫ ∫ f ( x) dx : a b Summing the areas of all the rectangles of width dx This then is the process of integration.1: Calculate something called the anti-derivative of the function f(x). In this case. y 1 1 FTC does this in three steps Step .org . What they both independently discovered is something called the Fundamental Theorem of Calculus (FTC) which you’ll learn about in class 12th. As n → ∞ .locuseducation. is a very small quantity.Now. this is what Newton and Leibnitz. f ( x) = x 2 3 2 x has the anti-derivative > g ( x) = x 3 Step . both geniuses in their own respects. consider any point x between a and b as shown in the figure a b above. So. the approximation is better. A = g (2) − g (1) = ww w. you can still try to learn more about all this stuff on your own. In fact. 0 ) –3 –2 –1 –1 –2 (– 1. A Cartesian coordinate system specifies each point uniquely in a plane by a pair of numerical coordinates. So we'll recap the concept very briefly here. equivalently. may inspire you to read more on your own! A= ∫ 2 x dx = 3 (2) 4 − (1) 4 = 15 PART .locuseducation. the purpose of this document is only a brief introduction. 1 ) 1 (0 . The coordinates can also be defined as the positions of the perpendicular projections of the point onto the two axes.org 24 . you don’t feel lost! This said. so that when you start reading on them in more detail in the actual course.5) –3 1 2 2 x (2 . y 3 2 (– 3. many students who are really good in Physics and Math already know a lot about differentiation and integration even before starting class 11th. So.5. and the point where they meet is its origin. which are the signed distances from the point to two fixed perpendicular directed lines. (i) How is doing all this justified? (ii) Why the name anti-derivative? (iii) How do we know a function’s anti-derivative? (iv) What the heck? Well. The anti-derivative of f ( x ) = x is g ( x ) = 3 x 4 tween x = 1 and x = 2 will be 4 . expressed as signed distances from the origin. This. Each reference line is called a coordinate axis or just axis of the system. Its not at all difficult. – 2. as we said earlier. z) O x y X z Y ww w. by its perpendicular projection onto three mutually perpendicular lines). we hope.Here’s another example. 3 ) One can use the same principle to specify the position of any point in three-dimensional space by three Cartesian coordinates. if you are curious enough. its signed distances to three mutually perpendicular planes (or. just short hi's and hellos to the subject. y.D GEOMETRY (a) CARTESIAN COORDINATES Most of you are already familiar with Cartesian coordinates. the area A under the x3-curve be- 4 4 4 1 That’s it! But there may be a lot of question arising in your mind. Z ( x. a n d y o u are a sk ed to fin d o u t h o w fa r ap a rt th e y are . c 2 = (5 − 1) + (1 − ( −2 )) c = 2 2 (5 − 1) + (1 − ( −2 )) 2 2 = ( 4 ) + (3 ) 2 2 = 16 + 9 = 25 = 5 ww w. each point in the plane has an algebraic 'label' (x. usin g the se p o in ts a s tw o o f th e co rn ers: It's ea sy to fin d th e le n g th s of the h o riz o n ta l an d v e rtica l sid e s o f th e rig h t trian g le : ju st su btrac t the x -v alue s an d th e y -v alu es: Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle): c = a +b 2 2 2 So. 1 ) an d (1 . 5 ).The invention of Cartesian coordinates in the 17th century by René Descartes revolutionized mathematics by providing the first systematic link between geometry and algebra. Yo u c an d raw in th e lin e s th at fo rm a rig ht-an g led trian g le . Here's how we get from the one to the other: y y y 5–1 =4 x x 1 – (– 2 ) = 3 x S u p p o se y ou 're g iv en th e tw o p oin ts (– 2 . 3 2 1 y 2 2 x +y =4 2 –2 –3 –1 –1 –2 1 3 x –3 The important thing right now is not to worry about why we get this equation. Using the Cartesian coordinate system. but to appreciate the fact that it is possible to represent geometric shapes using algebraic equations. THE DISTANCE FORMULA The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. the circle of radius 2 may be described as the set of all points whose coordinates x and y satisfy the equation x2 + y2 = 22.org 25 . geometric shapes (such as curves) can be described by Cartesian equations: algebraic equations involving the coordinates of the points lying on the shape. y).locuseducation. For example. This is possible because in the Cartesian system. and then find the length of the hypotenuse. Given two points.28. Here’s another example. regardless. You will be studying Cartesian coordinate geometry in great detail in class 11th. which is one of the most basic results in the Cartesian system. Find the distance between the points (–2. For now.locuseducation. Since this format always works. it can be turned into a formula: Distance Formula: Given the two points ( x1 . you first go to Mumbai (M). (b) VECTORS Suppose that you are on a trip. The distance will be the same. D K M B What we wish to do is mathematically represent this trip. Whichever one you call "first" or "second" is up to you. Q. 4). you can always plot them. that is. draw the right triangle. that you have two points. then Bangalore (B). and finally Kolkata (K). Starting from Delhi (D).28 Then the distance is sqrt(53). The best way to do that is by using vectors. ww w.This format always holds true. rounded to two decimal places. y 2 ) the distance between these points is given by the formula: d = ( x2 − x1 ) + ( y2 − y1 ) 2 2 Don't let the subscripts scare you. y1 ) and ( x2 .org 26 . we just plug the coordinates into the Distance Formula: d = ( −4 − ( −2 )) + ( 4 − ( −3)) 2 2 = ( −4 + 2 ) + ( 4 + 3 ) 2 2 = ( −2 ) + (7 ) 2 2 = 4 + 49 = 53 ≈ 7. or about 7. Well. –3) and (–4. They only indicate that there is a "first" point and a "second" point. The length of the hypotenuse is the distance between the two points. you need to only know the distance formula. y) be any point in the plane. which means that we covered a distance BK = 1881 km. y ) y O x Q x 27 ww w. you ended up at a distance of DK = 1461 km. from your starting point D. that is..org . note how different vector addition is from normal addition. The net effect of the trip can therefore be represent by DK . In other words. resulted in the net uur effect D K . DM This quantity has just a magnitude. Finally. suppose that we have a Cartesian coordinate-axes with the origin O. It is very important that you think of this in terms of movements and net effects: Moving from D to M to B to K has a net effect of moving from D to K. This is a vector. Lets represent this movement by DM .In the first leg of your trip.locuseducation. in the direction B to K. D M . in the direction M to B. it is just a number equal to 1407 km. in the direction D to K. The net effect is that he moved from A to C. Here’s another example. this becomes uu r uu r uu r + = AB BC AC A r 4 + r 3 = = r 5 ↑ Direction A to B ↑ Direction B to C ↑ Direction A to C Once again. Let P(x. which tells us that we covered a distance MB = 998 km. This notation tells us that there is a magnitude (distance) as uuu r well as a direction involved. In vector notation. uuu r from D to M. So. We write this as DM + M B + BK = DK uur uu r uur uuu uur r uur uuu r uur uur uur Observe that this a totally new kind of addition. Now. But you did that in a particular direction. and is called vector addition. Consider the right triangle shown below C 3 5 B 4 A person starts from point A. Now. the three legs of the trip. So DM is different from DM in that sense: DM This quantity has a magnitude as well as a direction involved. M B and BK . let us ask and answer two simple questions: (a) What is the total distance you covered in the trip? This is simply DM + MB + BK = 3286 km (b) What is the net effect of this trip? The net effect is that you started from Delhi and ended up in Kolkata. We r want to analyze the vector OP in more detail. goes to B and then to C. in a sense . the third leg of the trip can be represented by BK . This is a scalar uuu r The next leg of the trip can be represented by M B . y P ( x. you covered a distance DM equal to 1407 km . .e. OP = OQ + QP uu r uur uu r uu r uur uu r ˆ The crucial step comes now: Let us denote a unit vector (a vector of length 1) along the x-axis by i y x 1 ˆ The magnitude of i is 1. 1 ) 1 O 3 y P (– 1 . i. –2 ) y OP = –2j O 2 P (0 . it would mean that the length of the vector doubles. j Thus. 2 ) O P = – i^ + 2 j^ 2 1 O x x O y O P = 2^ – 2^ i j x 2 P (2 .locuseducation.. go x units along the direction i and then y units along the direction ˆ j Here are some more examples y OP = 3i + j P (3 . if the unit vector along the positive y-axis is represented by ˆ .org 28 .e.e. –2 ) ^ x ww w. then Q P can be written as y ˆ i. y times j j the unit vector ˆ . and its direction is along the positive x-axis ^ i ˆ ˆ If we multiply i by some number. but it still points in the same direction: y 2^ i x ˆ ˆ This suggests that O Q in the figure above can be written as x i . y ). ˆ ˆ O P = O Q + Q P = xi + yj uur 2 uu r uu r uur uu r You should interpret this as follows: ˆ To go from O to P( x. say 2. what does the resulting quantity 2i mean? Well. x times the unit vector i . i... Similarly.Note that O P is the net effect of O Q and Q P . Let us resolve a vector as an example. which is of length uur uu r uu r components of O P are O Q and Q P . 6) The net effect of the motion is r ^ i i^ r r r u u r r u r r r r ˆ j ˆ ˆ ˆ a + b + c = OA + AB + BC = ( i + ˆ ) + ( i + 2 ˆ ) + ( i + 3 ˆ ) = 3i + 6 ˆ = OC j j j Comment: This brief discussion should have given you a basic idea about vectors. Finally.org 29 . 1). This may motivate you to study vectors in greater detail before formally starting your class 11th course. it reaches B(2.ˆ In general. j Thus. b and c in succession.locuseducation. This sum represents the ‘net effect’ on the particle after it undergoes the movements a . this sum represents the final position of the particle. 3). the two components of O P are O Q = 2 iˆ and Q P = 2 ˆ . the particle is at (1. Vectors will be studied in much more detail in both Physics and Math. Writing a vector as a sum of its components is called resolving the vector into its compoj nents. we want you to consider the following example. QP = 2 2 sin 45º = 2. uu r y P 2 2 45º The vector we want to resolve is O P . After b . The two uu r uur uu r Since OQ = 2 2 cos 45º = 2. O P is resolved as: uu r OP uu r = 2iˆ Horizontal component + 2ˆ j Vertical Component Finally. However. we must still mention that understanding vectors properly and being comfortable with their use is extremely important in doing good Physics. we see that any vector O P will have two parts or ‘components’. Thus. ww w. Suppose that three vectors are given to us: r r r ˆ a = iˆ + ˆ. uu r O Q i 2 2 and inclined at 45º to the horizontal. suppose a particle is initially at the origin. after the effect of c . which is sufficient for now. b = i + 2 ˆ. j j c = iˆ + 3 ˆ j r r r What do we mean by a + b + c ? Well. the particle ends up at C(3. and one along the ˆ -direction. one along the i -direction. j^ c B r r r C ^ 3j b A a ^ i ^ 2j ^ ^ j i O After the movement a . you wouldn’t have wasted one year. Dropping one year specially for the JEE is not advisable. The testing system at such institutes reflects the JEE pattern.locuseducation. though if you are sure you can crack the JEE in the second attempt.00. about than 3. you would automatically have prepared for the AIEEE and DCE. Is it necessary to join some coaching institute for JEE preparation? A8. if you fail to clear the JEE in the second attempt too. For any engineering aspirant.org 30 . Being in an institute has the additional advantage of providing you with a competitive environment. not way will ask you whether you did your degree from IIT or some other college. True. Q6. and this number grows every year. Q5.PART .E . does that mean that I am at any disadvantage? A5. But two years after graduation. Not at all. For AIEEE and DCE. Q3. No. is in the range of 10. This is because a) they have great faculty b) they have excellent infrastructure c) they have a good curriculum and d) they are autonomous. Q8. The tag of the “toughest exam” is because of the low selection rate. Is it true that the selection rate in the JEE is only around 3%? A3. Having an IIT degree will at the most help you get a job more easily (even that is not really true recession!). Yes. What you make of yourself is entirely up to your individual capabilities. IIT is one of the best possible destinations. which helps too. The number of seats.000 students worote the JEE – 09. and the other one next year.AQS ABOUT IIT-JEE Q1. ww w. making sure you are studying and practicing regularly. and if your fundamentals are strong. Q2. Two. It is however much better to enroll in some other college and simultaneously prepare for the JEE. How many attempts do I have at the JEE? A4. it may be done. If your school teachers are good. and keeping you on your toes. then you don’t need to join any institute. Are the IITs the best engineering institutes of our country? A2.000. and you are motivated enough to take the initiative to cover the JEE syllabus with their help. do I have to prepare separately? A7. This way. If you prepare for the JEE properly. The JEE is a test of fundamentals. If I cannot make it to IIT. Q7. Is the JEE very tough? I’ve heard that it is the toughest exam in the world. This question does not have any unambiguous answer. But this seldom happens – and most students find it mandatory to join a special course for the JEE. is it advisable to drop one year? A6. you won’t find it tough at all. A1. Q4. If I can’t clear the JEE with my 12th Boards. One immediately after your 12th Boards. on the other hand. I need to study more than 10 hours everyday? A14. How can I effectively manage both my school studies and JEE preparation? A10.org 31 . for others. do the same for your coaching classes. This way. which are extremely important for personality development. as we’ve already said. while others take a lot of time? This has to do with the two reasons mentioned. staying at home and solely studying and practicing JEE syllabus? A13. even 10 hours might seem insufficient. or doing your engineering from IIT? Secondly. This should be reason enough to convince you that there is no need for preparing for the JEE and your school tests/Boards separately. No! This is one of the worst things a JEE aspirant could do. Not true. Q10. you can plan much better. Skipping school has several disadvantages: a) you lose out on your social interactions. Strike a balance between how much importance you give to your school commitments and how much to cracking the JEE. What if you were already familiar with both these concepts. How should I decide which coaching institute to join? A9. For some. you are being taught Laws of Motion in your school. Is it true that to crack the JEE. This is one of the most frequently encountered problems of JEE aspirants. you won’t be able to do much with it unless you have a good personality b) you have to put in additional efforts when appearing for your Boards c) you become generally dull.Q9. 3 hours per day may be enough. Suppose that right now.locuseducation. you will automatically perform well on all your school tests. and in your coaching institute (if you’ve joined one) – and there shouldn’t be any problem. but Kinematics in your coaching. Have you noticed how some people learn new things very easily. It all depends on a) the individual’s capabilities in understanding and application of new concepts and b) whether he/she is following an effective preparation strategy. and iron out all the conflicts between the two. Judge any coaching institute on these four parameters: a) faculty members b) content – in the form of study material and assignments provided c) feedback system – how do they test a student d) past results. What should I do? A12. and I can’t give ample time to JEE preparation. You surely have to make some compromises. Q11. Be regular in school. if you are doing your JEE preparation properly. The sequence of syllabus coverage followed in my school is different from that followed in my coaching. The syllabus for the JEE is the same as your school syllabus. even if your knowledge was only limited to the basics? How easier would it have been? Take the initiative to find out from your school teachers the sequence of syllabus coverage they will follow. but more important is leading an interesting life! Q14. What is more important to you: scoring full marks on your Monday test. The best way of judging an institute is to talk to one of their former or current students – this should surely be done since it will immediately let you know all about the institute. Is it true that the best way of preparing for the JEE is skipping school. Q13. What should I do? A11. I have a lot of Monday tests and other commitments in my school. and you fail to find the time to practice both the topics. ww w. The best solution is to cover as much syllabus in advance as you can. even though you may be acquiring skills like solving JEE problems. The number of hours required per day is a subjective figure. Regularity/consistency is necessary. Getting into IIT may be an important goal. Q12. Even if you possess an IIT degree. and you still join a test series. Make sure you choose any such course only after properly browsing through some samples of the study material they have to offer. Can I rely on a correspondence course to provide me with all the syllabus coverage necessary for the JEE? A23. I can’t perform in an actual test. What should I do? A21. This is the only way available. Q24. What’s the ideal way of studying? A17. by joining some test series. Q21.org 32 . but only after you think you have read all the concepts sufficiently well. I’m able to understand all the concepts.locuseducation. There are so many books and study materials available.Q15. you will most likely perform poorly and that will do nothing but lower your morale. and in a planned manner d) don’t differentiate between preparing for Boards and JEE e) don’t succumb to peer pressure ww w. In fact. No. Q20. there has to be a balance – preparation should be your first priority. There are no guarantees. You need to build an examination temperament by appearing for as much tests as you can. Q18. What should I do? A19. Q16. Generally yes. How do I prevent this? A20. What’s the ideal amount of time I should put in everyday for studying? A15. See the question above. Remember to a) make a strategically suitable study plan in advance b) study in a relaxed and quiet environment c) not study for too long at a time d) avoid thinking of peers and how much they have done while you study e) clear all your doubts as soon as you can f) practice questions as much as you can. I go blank. 18 Q23. like sports for example. Everyday. But obviously. you should surely be involved in some other activity. The only solution is practice of questions from good sources. from the right sources. What are the books I should refer to for my preparation? A22. Best leave it to the experts! Q19. allocate some time to revising what you’ve already covered. Q22. If I learn something today. Refer to your teachers. How can I differentiate between them? A18. or your seniors who’ve cracked the JEE. However. Does preparing for the JEE mean that I cannot spend time in other activities? A16. See Q. but when it comes to applying them. while I can easily solve all the questions at home. but you can maximize your chances of getting selected if you stick to our approach: a) attend school regularly b) join a good JEE course c) study consistently. If you haven’t. since at your stage you are not really in a position to judge the quality of content you are reading. both at home (in test-like conditions) and in reality. Regular revision. I’ll forget it in a few days. Q17. What is the crux? How can I make sure I crack the JEE? A24. to counter monotony during preparation. joining a test-series is advised only once you’ve covered the entire syllabus. AIR 64 in JEE 2009 40% selections in JEE 2008 Here are 4 reasons why you should prefer LOCUS to any other institute for your IIT-JEE and AIEEE preparation. STUDY MATERIAL Detailed explanations Written by JEE top rankers Includes effective diagrams Aided by recorded video lectures .Why Join LOCUS ? PAST RESULTS 51% selections in JEE 2009 AIR 55. FACULTY All IIT graduates Highly experienced faculty Many JEE top rankers on the faculty panel A mentor faculty for every student FEEDBACK SYSTEM Tutorials in every class Regular phase tests All India Test series with detailed reports Periodic student-parentmentor meetings. 226012 Tel: 0522-3294460.JEE 2012 IIT . New Delhi . Road. Kanpur Road.2011 & 2012 (for XII Pass Students) th CONTACT Head Office & Delhi Centre 65 . Mob: 09450212488 Lucknow Centre (2) C .B IIT-JEE . Ompro Tower. Near IIT Flyover.org REGISTRATION OPEN REGISTRATION OPEN YOUR PATH TO IIT JEE All IITian Faculty IIT . Kalu Sarai. Hazratganj Lucknow . Tel: 011 .A. Pratap Bhawan. Nawal Kishore Road.: 01146080145 / 6 . Lucknow . Panipat. Ph.2012 (for XII Studying Students) th (for XI Studying Students) th One Year Classroom Program .COURSES One Year Classroom Program .2011 Two Year Classroom Program IIT-JEE .D. Ompro Tower.46080145/6 Lucknow Centre (1) 43 . Sector .226001.110016. Colony. Haryana.A IIT-JEE . L.2011 Correspondence Courses for IIT-JEE .T. G. Kalu Sarai.123. Tel: 0180-3058100-01 www. Mob: 09450212488 Panipat Centre DPS Panipat City.D. 77 Milestone.locuseducation.110016.A.JEE 2011 4th successive year of quality teaching & mentoring by IITians 65-A.C. Near IIT Flyover New Delhi . Tel: 0522-3294460.