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Mathematics : J EE Advance Paper – 1 : 20131 Mathematics : JEE Advance Paper – 1 : 2013 SECTION – 1: (Only one option correct Type) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 41. Perpendiculars are drawn from points on the line 3 z 1 1 y 2 2 x = ÷ + = + to the plane x + y + z = 3. The feet of perpendiculars lie on the line (a) 13 2 z 8 1 y 5 x ÷ ÷ = ÷ = (b) 5 2 z 3 1 y 2 x ÷ ÷ = ÷ = (c) 7 2 z 3 1 y 4 x ÷ ÷ = ÷ = (d) 5 2 z 7 1 y 2 x ÷ = ÷ ÷ = 41. (d) Any point B on line is (2ì – 2, –ì – 1, 3ì) Point B lies on the plane for some ì ¬ (2ì – 2) + (–ì – 1) + 3ì = 3 ¬ ì = 3/2 ¬ B ÷ (1, – 5/2, 9/2) The foot of the perpendicular form point (– 2, – 1, 0) on the plane is the point A (0, 1, 2) ¬ D.R. of AB = | . | \ | ÷ 2 5 , 2 7 , 1 ÷ (2, – 7, 5). Hence, feet of perpendicular lies on the line 5 2 z 7 1 y 2 x ÷ = ÷ ÷ = The concept used in the question is finding projection of line on the plan. The following questions in the book explain how to find the projection of given line in given plane, which is same is finding locus of feet of perpendicular from points on the given line on the given plane. Q. no. 12 : Concept Application exercise 3 : Vector and 3-D Geometry : Chapter 3 : 3-D Geometry : page no. 3.50 12. Find the direction ratios orthogonal projection of line 3 2 z 2 1 y 1 1 x ÷ = ÷ + = ÷ in the plane x – y+ 2z – 3 = 0 Also find the direction ratios of image of the line in the plane. Sol. Any point on the line t 3 2 z 2 1 y 1 1 x = ÷ = ÷ + = ÷ is (t + 1, – 2t – 1, 3t + 2), which lies on the given plane if t + 1 + 2t + 1 + 6t + 4 – 3 = 0 ¬ t = – 1/3 ¬ point of intersection of line and plane is P(2/3, – 1/3, 1) Also if foot of perpendicular from (1, – 1, 2) on the plane is Q(x, y, z) then 2 1 4 1 1 ) 3 4 1 1 ( 2 2 z 1 1 y 1 1 x ÷ = + + ÷ + + ÷ = ÷ = ÷ + = ÷ ¬ Q(x, y, z) is Q(1/2, –1/2, 1) Hence direction ratios of PQ are 1 1 , 2 1 3 1 , 2 1 3 2 ÷ + ÷ ÷ or 0 , 6 1 , 6 1 If the image of the point A(1, – 1, 2) in the plane is R, then Q is mid point of AR point R is (0, 0, 0) Hence direction ratio of PR or image of the line in the plane is <2/3, – 1/3, 1> Mathematics : J EE Advance Paper – 1 : 2013 2 Q. no. 75 : Single option correct type : Vector and 3-D Geometry : Chapter 3 : 3-D Geometry : page no. 3.67 75 the projection of the line 3 1 z 2 y 1 1 x ÷ = = ÷ + on the plane x ÷ 2y + z = 6 is the line of intersection of this plane with the plane (a) 2x + y + 2 = 0 (b) 3x + y ÷z = 2 (c) 2x ÷3y + 8z = 3 (d) none of these Sol. Equation of plane through (÷1, 0, 1) is a (x + 1) + b(y ÷ 0) + c(z ÷ 1) = 0 ….(i) Which is parallel to given line and perpendicular to given plane ÷ a + 2b + 3c = 0 .…(ii) and a ÷ 2b + c = 0 …..(iii) From Equation (ii) and (iii), c = 0, a = 2b. From Equation (i), 2b(x + 1) + by = 0 ¬ 2x + y + 2 = 0 Hence the correct answer is (a) 42. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than 2 2 , then (a) a + b – c > 0 (b) a – b + c < 0 (c) a – b + c > 0 (d) a + b – c < 0 42. (a) Solving given lines for their point of intersection we get point of intersection as , c c a b a b ÷ ÷ | | | + + \ . Its distance from (1, 1) is 2 2 1 1 2 2 c c a b a b | | | | + + + < | | + + \ . \ . ¬ (a + b + c) 2 < 4(a + b) 2 ¬ (a + b + c) 2 – (2a + 2b) 2 < 0 ¬ (c – a – b)(c + a + b) < 0 Since a > b > c > 0, (c – a – b) < 0 or a + b – c > 0 Here simple concepts of straight lines are used, like finding point of intersection and then comparing its distance, solving inequalities etc. One can easily attempt after going through the straight line chapter in coordinate geometry book 43. The area enclosed by the curves y = sin x + cos x and y = | cos x – sin x | over the interval | | 0, / 2 t is (a) ) 1 2 ( 4 ÷ (b) ) 1 2 ( 2 2 ÷ (c) ) 1 2 ( 2 + (d) ) 1 2 ( 2 2 + 43. (b) Since sin x and cos x > 0 for x e | | 0, / 2 t The graph of y = sin x + cos x always lies above the graph of y = |cos x – sin x| Also cos x > sin x for x e | | 0, / 4 t and sin x > cos x for x e | | / 4, / 2 t t ¬ Area = / 4 0 ((sin cos ) x x t + } – (cos x – sin x)) dx + / 2 / 4 ((sin cos ) x x t t + } – (sin x – cos x)) dx = 4 – 2 2 Mathematics : J EE Advance Paper – 1 : 2013 3 Here simple concept of finding the area bounded by curves is used. For function y = |cos x – sin x| we have to remove modulus sign depending upon the values of sin x and cos x in the interval ( ¸ ( ¸ t 2 , 0 . This is regular feature of the modulus function. One can easily evaluate the function if has gone through the book. 44. Four persons independently solve a certain problem correctly with probabilities 8 1 , 4 1 , 4 3 , 2 1 . Then the probability that the problem is solved correctly by at least one of them is (a) 256 235 (b) 256 21 (c) 256 3 (d) 256 253 44. (a) P (at least one of them solves correctly) = 1 – P (none of them solves correctly) = 1 – 256 235 8 7 4 3 4 1 2 1 = | . | \ | × × × The concept used in the question is that of independent events and how find the probability when “atleast one” is involved. The following question in based on the same concept. Q. no. 6 : Single option correct type : Algebra : Chapter 9 : Probability : page no. 9.22 6. A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is 4 1 and 3 1 , 2 1 . Probability that the problem is solved is (a) 3/4 (b) 1/2 (c) 2/3 (d) 1/3 Sol. P(a) = ( ) ( ) 4 1 C P and 3 1 B P , 2 1 = = ( ) ( ) 3 2 3 1 1 2 1 2 1 1 = ÷ = = ÷ = B P , A P and ( ) C P = 1 ÷ 4 3 4 1 = Reqd. Prob. = 1 ÷ P( ) A ( ) ( ) C P B P = 4 3 3 2 2 1 1 × × ÷ = 4 3 4 1 1 = ÷ 45. Let complex numbers o and o 1 lie on circles (x – x0) 2 + (y – y0) 2 = r 2 and (x – x0) 2 + (y – y0) 2 = 4r 2 , respectively. If z0 = x0 + iy0 satisfies the equation 2|z0| 2 = r 2 + 2, then |o| = (a) 1/ 2 (b) 1/2 (c) 1/ 7 (d) 1/3 45. (c) Given circles are (x – x0) 2 + (y – y0) 2 = r 2 and (x – x0) 2 + (y – y0) 2 = 4r 2 or |z – z0| = r …….(1) and |z – z0| = 2r …….(2) Now o and o 1 lies on circle (1) and (2) respectively, ¬ 0 z r o ÷ = and 0 1 2 z r o ÷ = ¬ 0 z r o ÷ = and 0 1 2 | | z r o o ÷ = ¬ 2 2 0 z r o ÷ = and 2 2 2 0 1 4 | | z r o o ÷ = Mathematics : J EE Advance Paper – 1 : 2013 4 Subtracting, we get 2 2 2 2 2 0 0 1 4 | | z z r r o o o ÷ ÷ ÷ = ÷ ¬ 2 2 2 2 2 2 0 0 0 0 0 0 1 ( ) 4 | | z z z z z z r r o o o o o o o + ÷ ÷ ÷ + ÷ ÷ = ÷ ¬ 2 2 2 2 2 2 2 0 0 1 4 | | z z r r o o o + ÷ ÷ = ÷ ¬ 2 2 2 2 2 0 (1 )(1 ) 4 | | z r r o o ÷ ÷ = ÷ Given 2|z0| 2 = r 2 + 2 ¬ 2 2 2 2 2 2 (1 ) 1 4 | | 2 r r r o o | | + ÷ ÷ = ÷ | \ . ¬ 2 2 2 2 2 (1 ) 4 | | 2 r r r o o | | ÷ ÷ = ÷ | \ . ¬ 2 2 1 8| | 2 o o ÷ = ÷ ¬ 2 1 7 o = ¬ 1 7 o = The concept used in the question is that of points satisfying the given equation and then the expansion |z1 + z2| 2 = | z1 | 2 + | z2 | 2 + z1 2 z + z2 1 z . This is given in the theory as properties of Modulus and the example 2.44 is using this expansion. Example 2.44 : Algebra : Chapter 2 : Complex Numbers : page no. 2.14 Example 2.44 Let 2 1 2 1 z z 2 z 2 z ÷ ÷ = 1 and |z2| = 1, where z1 and z2 are complex numbers, show that |z1| = 2. Sol. 2 1 2 1 z z 2 z 2 z ÷ ÷ = 1 ¬ 2 2 1 2 2 1 2 2 z z z z ÷ = ÷ ¬ ) z z )( z z ( ) z z )( z z ( 2 1 2 1 2 1 2 1 2 2 2 2 ÷ ÷ = ÷ ÷ ¬ ) z z )( z z ( ) z z )( z z ( 2 1 2 1 2 1 2 1 2 2 2 2 ÷ ÷ = ÷ ÷ ¬ 2 2 1 1 2 1 2 1 2 2 2 1 2 1 1 1 2 2 4 4 2 2 z z z z z z z z z z z z z z z z + ÷ ÷ = + ÷ ÷ ¬ 2 2 2 1 2 2 2 1 4 4 z z z z + = + ¬ 0 4 z 4 z z z 2 2 2 2 2 1 2 1 = ÷ + ÷ ¬ 0 1 4 1 2 2 2 2 2 1 = ÷ + ÷ ) z ( ) z ( z ¬ 0 4 1 2 1 2 2 = ÷ ÷ ) z )( z ( ¬ |z1| = 2 (as |z2| = 1) 46. The number of points in (– ·, ·), for which x 2 – x sin x – cos x = 0, is (a) 6 (b) 4 (c) 2 (d) 0 46. (c) Let f(x) = x 2 – x sin x – cos x ¬ f'(x) = 2x – x cos x f'(x) = 0 ¬ x(2 – cos x) = 0 ¬ x = 0 (2 – cos x > 0 for all real x) Also for x = 0 is point of minima. f(0) = – 1 < 0 and · ÷ x l i m f(x) ÷ ·, · ÷ ÷ x l i m f(x) ÷ ·. Hence it meets x –axis at two points, hence two solutions. Mathematics : J EE Advance Paper – 1 : 2013 5 The concept used in the question is how monotonocity and extremum of functions helps to identify the number of roots of the typical equations. Q. no. 25, single option correct type : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.24 25. Number of solutions of the equation x 3 + 2x 2 + 5x + 2cosx = 0 in [0, 2t] (a) one (b) two (c) three (d) zero Sol. : Let f(x) = x 3 + 2x 2 + 5x + 2 cos x ¬ f'(x) = 3x 2 + 4x + 5 ÷ 2 sinx Now least value of 3x 2 + 4x + 5 is ÷ a 4 D = ÷ ) 3 .( 4 ) 5 )( 3 ( 4 ) 4 ( 2 ÷ = 3 11 and Greatest value of 2 sin x = 2 ¬ 3x 2 + 4x + 5 > 2 sin x ¬ x e R ¬ f'(x) = 3x 2 + 4x + 5 ÷ 2 sin x > 0 ¬ x e R ¬ f(x) is strictly increasing function also f(0) = 2 and f(2t) > 0. Thus for the given interval f(x) never becomes zero. Hence number of roots are zero. Hence correct answer is (d) Matrix match 1 : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.33 Column I : Number of roots f(x) + p = 0 Column II : Values of p (a) two negative real roots (p) p > 120 (b) two real roots of opposite sign (q) ÷8 s p s ÷5 (c) four real roots (r) 3 < p s < 120 (d) no real roots (s) p < – 8 or – 5 < p < 3 1. (a) ÷ (r); (b) ÷ (s) ; (c) ÷ (q); (d) ÷ (p) f'(x) = 4x 3 ÷ 28x + 24 = 4(x 3 ÷ 7x + 6) = 4(x 3 ÷ x 2 + x 2 ÷ x ÷6x + 6) = 4(x ÷1)(x 2 + x ÷6) = 4(x ÷1)(x +3)(x ÷ 2) Now nature of roots of f(x) + p = 0 can be obtained by Shifting the graph of y = f(x) by p units upward or Or downward depending on p is positive or negative ÷3 (÷3, ÷120) (2, 5) O 2 1 (1, 8) Graph of y = f(x) Mathematics : J EE Advance Paper – 1 : 2013 6 Q. no. 17 : Subjective type : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.22 17. Discuss the number of roots of the equation e(k ÷ x log x) = 1, for different values of k Sol. e(k ÷ x log x) = 1 …...(i) ¬ k ÷ e 1 = x ln x Then equation (1) has solution where graphs of y = x ln x and y = k ÷ e 1 intersect Now consider the function f(x)= x logex f'(x) = 1 + logex, f'(x) = 0 ¬ x = 1/e, f'' (x) = 1/x ¬ f'' (1/e) = e > 0 ¬ x = 1/e is point of minima. Also 0 x lim x 1 x 1 lim x 1 x log lim x log x lim 0 x 2 0 x e 0 x e 0 x = ÷ = ÷ = = ÷ ÷ ÷ ÷ Hence the graph of f(x) = x loge x is as follows : f(1/e) = –1/e Hence equation k ÷ e 1 = x logex has two distinct roots if ÷ e 1 < k ÷ e 1 < 0 ¬ 0 < k < e 1 Equation has no roots if k ÷ e 1 < ÷ e 1 ¬ k < 0 Equation has one root if k ÷ e 1 = ÷ e 1 or k ÷ e 1 > 0 ¬ k = 0 or k k > e 1 47. Let f : ( ¸ ( ¸ 1 , 2 1 ÷ R (the set of all real numbers) be a positive, non–constant and differentiable function such that f'(x) < 2f(x) and f(1/2) = 1. Then the value of } 1 2 / 1 dx ) x ( f lies in the interval (a) (2e – 1, 2e) (b) (e – 1, 2e – 1) (c) | . | \ | ÷ ÷ 1 e , 2 1 e (d) | . | \ | ÷ 2 1 e , 0 47. (d) Given f'(x) – 2f(x) < 0 ¬ f'(x)e – 2x – 2 e – 2x f(x) < 0 ¬ 2 ( ( ) ) 0 x d f x e dx ÷ < Thus g(x) = f(x) e – 2x is decreasing function. Mathematics : J EE Advance Paper – 1 : 2013 7 Also f(1/2) = 1, g(x) < g(1/2) ¬ f(x) e – 2x < f(1/2) e – 1 ¬ f(x) < e 2x – 1 ¬ 0 < } } ÷ < 1 2 / 1 1 2 / 1 1 x 2 dx e dx ) x ( f < } ÷ < 1 2 / 1 2 1 e dx ) x ( f The concept used in the question is multiplying the given inequality by integrating factor and having singe function derivative, then from the given data discussing its monotonocity and then analyzing the inequality. This concept is discussed with much emphasis in the book, citing that it is one of the categories of the problems. Also concept of inequality in definite integration is used. The topic “Inequality is discussed in the book” Q. no. 10 : Subjective type : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.22 10. If f is a real function such that f(x) > 0, f'(x) is continuous for all real x and ax f'(x) > 2 ) x ( f ÷ 2af(x), (a.x = 2), show that x ) 1 ( f ) x ( f > , x > 1 Sol. Thought : we have to prove that x ) 1 ( f ) x ( f > , x > 1 or ) 1 ( f . 1 ) x ( f x > This suggest that we have to consider function which involves ) x ( f x Now given ax f'(x) > 2 ) x ( f ÷ 2af(x), dividing both sides by ) x ( f 2 we have 0 1 ) x ( f a ) x ( f 2 ) x ( ' f ax > ÷ + ¬ ( ) 0 x ) x ( f ax dx d > ÷ ¬ Hence x ) x ( f ax ÷ is increasing functions ¬ x > 1 then f(x) > f(1) ¬ x ) x ( f ax ÷ > 1 ) 1 ( f a ÷ ¬ ) x ( f ax > ) 1 ( f a + x ÷1 > ) 1 ( f a (as x > 1) ¬ x ) 1 ( f ) x ( f > Q. no. 18 : Single option correct type : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.23 18. If |(x) is a polynomial function and |'(x) > |(x), ¬ x > 1 and |(1) = 0, then (a) |(x) > 0 ¬ x > 1 (b) |(x) < 0 ¬ x > 1 (c) |(x) = 0 ¬ x > 1 (d) None of these Sol. Given |'(x) ÷ |(x) > 0 ¬ x > 1 ¬ e ÷x {|'(x) ÷ |(x)} > 0 ¬ x > 1 ¬ 0 ) x ( e dx d x > | ÷ ¬ x > 1 e ÷x |(x) is an increasing function ¬ x > 1 Since |(x) is a polynomial ¬ e ÷x |(x) > e ÷1 |(1) ¬ e ÷x |(x) > 0 [ |(1) = 0] Hence the correct answer is (a) Mathematics : J EE Advance Paper – 1 : 2013 8 Q. no. 9 -10 : Comprehension type : Calculus : Chapter 6 : Monotonocity and Maxmima-Minima of Functions : page no. 6.31 Paragraph 4 : Question Nos. 9 to 10 If |(x) is a differentiable real valued function satisfying |'(x) + 2|(x) s 1, then it can be adjusted as e 2x |'(x) + 2e 2x |(x) s e 2x or 0 2 2 2 s | | . | \ | ÷ | x x e ) x ( e dx d or 0 2 1 2 s | . | \ | ÷ | ) x ( e dx d x Here e 2x is called integrating factor which helps in creating single differential coefficient like above. 9. If P(1) = 0 and ) x ( P dx ) x ( dP > for all x > 1 then (a) P(x) > 0 ¬ x > 1 (b) P(x) is constant function (c) P(x) < 0 ¬ x >1 (d) None of these 10. If H(x0) = 0 for some x = x0 and ) x ( cxH ) x ( H dx d 2 > for all x > x0, where c > 0 then (a) H(x) = 0 has root for x > x0 (b) H(x) = 0 has no roots for x > x0 (c) H(x) is constant function (d) None of these Paragraph 4 : Question Nos. 9 to 10 9.(a) ) x ( P dx ) x ( dP > ¬ 0 ) x ( P dx ) x ( dP > ÷ ¬ ( ) x e ) x ( P dx d ÷ > 0 ¬ P(x) .e ÷x is an increasing function. ¬ P(x).e ÷x > P(1).e ÷1 ¬ x > 1 ¬ P(x).e ÷x > 0 ¬ x > 1 ¬ P(x) > 0 ¬ x > 1. 10.(b) Given that ) x ( cxH 2 ) x ( H dx d > ¬ 0 ) x ( cxH 2 ) x ( H dx d > ÷ ¬ 0 ) e ) x ( H ( dx d 2 cx > ÷ ¬ H(x) 2 cx e ÷ is an increasing function. But H(x0) = 0 and 2 cx e ÷ is always positive. ¬ H(x) > 0 for all x > x0 ¬ H(x) can’t be zero for any x > x0. Theory : Calculus : Chapter 8 : Definite Integration : page no. 8.19 INEQUALITIES Property I : If at every point x of an interval [a, b] the inequalities g(x) s f (x) s h (x) are fulfilled then } b a g(x) dx s } b a f(x) dx s } b a h (x) dx, a < b Proof : It is clear from the figure. Area of curvilinear trapezoid aAFb s Area of curvilinear trapezoid ABEb s Area of curvilinear trapezoid aCDb i.e., } b a g(x) dx s } b a f(x) dx s } b a h (x) dx Mathematics : J EE Advance Paper – 1 : 2013 9 48. Let k ˆ 2 j ˆ i ˆ 3 PR ÷ + = and k ˆ 4 j ˆ 3 i ˆ SQ ÷ ÷ = determine diagonals of a parallelogram PQRS and k ˆ 3 j ˆ 2 i ˆ PT + + = be another vector. Then the volume of the parallelepiped determined by the vectors PQ , PT and PS is (a) 5 (b) 20 (c) 10 (d) 30 48. (c) Area of base (PQRS) = 2 1 4 3 1 2 1 3 k ˆ j ˆ i ˆ 2 1 | SQ PR | ÷ ÷ ÷ = × = 3 5 | k ˆ j ˆ i ˆ | 5 | k ˆ 10 j ˆ 10 i ˆ 10 | 2 1 = + ÷ = ÷ + ÷ Height = projection of PT on 3 2 3 3 2 1 k ˆ j ˆ i ˆ = + ÷ = + ÷ . Volume = (5 3 ) | | . | \ | 3 2 = 10 cm. unit The concept used in the question is finding the volume of the parallelepiped when the diagonals of the base and the edge not containing the diagonal is given The following questions is on the same concept. Q. no. 58 : Single option correct type : Vector & 3-D Geometry : Chapter 2 : Different product of vectors and their geometrical applications : page no. 2.57 58. If the two diagonals of one of its faces are k ˆ 6 i ˆ 6 + and k ˆ 2 j ˆ 4 + and of the edges not containing the given diagonal is k ˆ 8 j ˆ 4 c ÷ = ÷ then the volume of parallelepiped is (a) 60 (b) 80 (c) 100 (d) 120 Sol. Let k ˆ 8 j ˆ 4 c , k ˆ 2 j ˆ 4 b , k ˆ 6 i ˆ 6 a ÷ = + = + = ÷ ÷ ÷ then ÷ ÷ × b a = ÷ k ˆ 24 j ˆ 12 i ˆ 24 + ÷ = 12( ) k ˆ 2 j ˆ i ˆ 2 + ÷ ÷ Area of base of the parallelepiped = | b a | 2 1 ÷ ÷ × = ) 3 12 ( 2 1 × = 18 Height of the parallelepiped = length of projection of ÷ c on ÷ ÷ × b a = | b a | | b a . c | ÷ ÷ ÷ ÷ ÷ × × = ( ) 36 | 16 4 12 | ÷ ÷ = 3 20 Volume of the parallelepiped = 18 × 3 20 = 120 Hence the correct answer is (d) Mathematics : J EE Advance Paper – 1 : 2013 10 49. The value of cot | | | . | \ | | | | . | \ | + ¿ ¿ = = ÷ n 1 k 23 1 n 1 k 2 1 cot is (a) 25 23 (b) 23 25 (c) 24 23 (d) 23 24 49. (b) cot | | | . | \ | + + ¿ = ÷ 23 1 n 2 1 ) 1 n n ( cot = cot | | | . | \ | | | . | \ | + + ÷ + ¿ = ÷ ) 1 n ( n 1 n 1 n tan 23 1 n 1 = cot 23 1 1 1 (tan ( 1) tan ) n n n ÷ ÷ = | | + ÷ | \ . ¿ = cot 1 1 (tan 24 tan 1) ÷ ÷ ÷ ¬ cot 23 25 25 23 tan 1 = | | . | \ | | . | \ | ÷ The concept used in the question is finding the series involving tan – 1 x function in which formula 1 1 1 tan tan tan 1 x y x y xy ÷ ÷ ÷ ÷ ÷ = + is used. Writing the given general term as a difference of two tan – 1 x function and then writing the terms in such a manner the terms gets cancelled. Following question illustrate the same method. Q. no. 4, 5 : Subjective Type : Trigonometry : Chapter 4 : Inverse Trigonometric Functions : page no. 4.36 4. Find the sum cot ÷1 2 + cot ÷1 8 + cot ÷1 18 + ….. to infinity. Sol. Let tn denotes the nth term of the series, then tn = cot ÷1 2n 2 = 2 1 n 2 1 tan ÷ = ) 1 n 4 ( 1 ) 1 n 2 ( ) 1 n 2 ( tan 2 1 ÷ + ÷ ÷ + ÷ = tan ÷1 (2n + 1) ÷ tan ÷1 (2n ÷ 1) Putting n = 1, 2, 3, …… etc in (1), we get t1 = tan ÷1 3 ÷ tan ÷1 1 t2 = tan ÷1 5 ÷ tan ÷1 3 t3 = tan ÷1 7 ÷ tan ÷1 5 … … … … … … tn = tan ÷1 (2n + 1) ÷ tan ÷1 (2n ÷ 1) Adding Sn = tan ÷1 (2n + 1) ÷ tan ÷1 1 as n ÷ ·, tan ÷1 (2n ÷ 1) ÷ t/2 Hence the required sum = 4 t 5. Find the sum to the n term of the series cosec ÷1 170 ec cos 50 ec cos 10 1 1 ÷ ÷ + + + ……….. + cosec ÷1 ( ) ( ) 2 n 2 n 1 n 2 2 + + + Mathematics : J EE Advance Paper – 1 : 2013 11 Sol. Let u = cosec ÷1 ( )( ) 2 n 2 n 1 n 2 2 + + + ¬ cosec 2 u = (n 2 + 1) (n 2 + 2n + 2) = (n 2 + 1) 2 + 2n (n 2 + 1) + n 2 + 1 = (n 2 + n + 1) 2 + 1 ¬ cot 2 u = (n 2 + n + 1) 2 ¬ tan u = ( ) ( )n 1 n 1 n 1 n 1 n n 1 2 + + ÷ + = + + ¬ u = tan ÷1 ( ) ( ) ( ¸ ( ¸ + + ÷ + n 1 n 1 n 1 n = tan ÷1 (n + 1) ÷ tan ÷1 n Thus, sum n terms of the given series = (tan ÷1 2 ÷ tan ÷1 1) + (tan ÷1 3 ÷ tan ÷1 2) + (tan ÷1 4 ÷ tan ÷1 3) + ……. + (tan ÷1 (n + 1) ÷ tan ÷1 n) = tan ÷1 (n + 1) ÷ t/4 The following question in books has same series which is asked in the exam. Q. no. 9 : Integer type : Trigonometry : Chapter 4 : Inverse Trigonometric Functions : page no. 4.51 9. If n is number of terms of the series cot –1 3, cot –1 7, cot –1 13, cot –1 21, ............................, whose sum is 2 1 cos –1 | . | \ | 145 24 , then the value of n – 5 is 9.(6) Tn = | | . | \ | + + ÷ + ÷ 1 )· 1 n ( 1 1 1 n tan 1 = tan –1 (n + 1) – tan –1 (n) Hence S n = tan –1 (n + 1) – tan –1 1 = | | . | \ | + + ÷ + ÷ 1 )· 1 n ( 1 1 1 n tan 1 = | . | \ | + ÷ 2 n n tan 1 = | . | \ | ÷ 145 24 cos 2 1 1 ¬ 2 | . | \ | + ÷ 2 n n tan 1 = | . | \ | ÷ 145 24 cos 1 ¬ | . | \ | + + + ÷ 2 n 2 n ) 1 n ( 2 cos 2 1 = | . | \ | ÷ 145 24 cos 1 ¬ | . | \ | + + + 2 n 2 n ) 1 n ( 2 2 = | . | \ | 145 24 ¬ 12(n + 1) 2 – 145 (n + 1) + 12 = 0 ¬ ( )( ) 1 ) 1 n ( 12 12 ) 1 n ( ÷ + ÷ + = 0 ¬ n + 1 = 12 ¬ n = 11 Q. no. 56, 57, 58, 59 : Single option correct type : Trigonometry : Chapter 4 : Inverse Trigonometric Functions : page no. 4.41 – 4.42 56. | | . | \ | + ÷ ÷ E ÷ = ) 1 r ( r 1 r r si n 1 n 1 r is equal to (a) tan ÷1 ) n ( ÷ 4 t (b) tan ÷1 4 ) 1 n ( t ÷ + (c) tan ÷1 ) n ( (d) tan ÷1 ) 1 n ( + 56.(c) | | . | \ | ÷ + ÷ ÷ = | | . | \ | + ÷ ÷ ÷ ÷ ) 1 r ( r 1 1 r r tan ) 1 r ( r 1 r r si n 1 1 ¬ ¿ = ÷ | | . | \ | + ÷ ÷ n 1 r 1 ) 1 r ( r 1 r r si n = ¿ = ÷ ÷ ÷ ÷ n 1 r 1 1 ) 1 r tan r (tan = tan ÷1 n 57. | | . | \ | + E ÷ ÷ ÷ = 1 r 2 1 r 1 n 1 r 2 1 2 tan is equal to (a) tan ÷1 (2 n ) (b) tan ÷1 (2 n ) ÷ 4 t (c) tan ÷1 (2 n+1 ) (d) tan ÷1 (2 n+1 ) ÷ 4 t Mathematics : J EE Advance Paper – 1 : 2013 12 57.(b) ¿ = ÷ ÷ ÷ | | . | \ | + n 1 r 1 r 2 1 r 1 2 1 2 tan = | | . | \ | + ÷ ÷ = ÷ ¿ 1 r r 1 r n 1 r 1 2 . 2 1 2 tan = ¿ = ÷ ÷ ÷ | | . | \ | + ÷ n 1 r 1 r r 1 r r 1 2 . 2 1 2 2 tan = ¿ = ÷ ÷ ÷ ÷ n 1 r 1 r 1 r 1 )] 2 ( tan ) 2 ( [tan = tan ÷1 (2 n ) ÷ tan ÷1 (1) = tan ÷1 (2 n ) ÷ 4 t 58. ¿ = ÷ | . | \ | + + n 1 m 2 4 1 2 m m m 2 tan is equal to (a) | | . | \ | + + + ÷ 2 n n n n tan 2 2 1 (b) | | . | \ | + ÷ ÷ ÷ 2 n n n n tan 2 2 1 (c) | | . | \ | + + + ÷ n n 2 n n tan 2 2 1 (d) None of these 58.(a) We have ¿ = ÷ | | . | \ | + + n 1 m 2 4 1 2 m m m 2 tan = ( ) ( ) ¿ = ÷ | | . | \ | + ÷ + + + n 1 m 2 2 1 1 m m 1 m m 1 m 2 tan = ( ) ( ) ( )( ) ¿ = ÷ | | . | \ | + ÷ + + + + ÷ ÷ + + n 1 m 2 2 2 2 1 1 m m 1 m m 1 1 m m 1 m m tan = ¿ = ÷ n 1 m 1 [tan (m 2 + m + 1) ÷ tan ÷1 (m 2 ÷ m + 1)] = (tan ÷1 3 ÷ tan ÷1 1) + (tan -1 7 ÷ tan ÷1 3) + (tan ÷1 13 ÷ tan ÷1 7) + …… + [tan ÷1 (n 2 + n + 1) ÷ tan ÷1 (n 2 ÷ n + 1)] = tan -1 (n 2 + n + 1) ÷ tan ÷1 1 = tan ÷1 | | . | \ | + + + n n 2 n n 2 2 59. Value of | . | \ | + + E ÷ · = 2 1 0 r r r 1 1 tan is equal to (a) t/2 (b) 3t/4 (c) t/4 (d) None of these 59.(a) | | . | \ | + + ÷ + = | | . | \ | + + ÷ ÷ ) 1 r ( r 1 r 1 r tan r r 1 1 tan 1 2 1 = ) r ( tan ) 1 r ( tan 1 1 ÷ ÷ ÷ + ¬ ¿ = ÷ ÷ ÷ + n 0 r 1 1 )] r ( tan ) 1 r ( [tan = tan ÷1 (n + 1) ÷ tan ÷1 (0) = tan ÷1 (n + 1) ¬ 2 ) ( tan r r 1 1 tan 1 2 0 r 1 t = · = | | . | \ | + + ÷ · = ÷ ¿ 50. A curve passes through the point | . | \ | t 6 , 1 . Let the slope of the curve at each point (x, y) be | . | \ | + x y sec x y , x > 0. Then the equation of the curve is (a) | . | \ | x y si n = log x + 2 1 (b) cosec | . | \ | x y = log x + 2 (c) sec | . | \ | x y 2 = log x + 2 (d) cos | . | \ | x y 2 = log x + 2 1 50. (a) Mathematics : J EE Advance Paper – 1 : 2013 13 . x y sec x y dx dy + = Let y = vx ¬ Given equation reduces to v sec dv = x dx ¬ } vdv cos = } x dx ¬ sin v = ln x + c ¬ sin | . | \ | x y = log x + c The curve passes through | . | \ | t 6 , 1 ¬ sin | . | \ | x y = log x + 2 1 . Similar type of question which uses homogenous differential equation Q. no. 26 : Single option correct type : Calculus : Chapter 10 : Differential Equations : page no. 10.17 26. The slope of the tangent at (x, y) to a curve passing through | . | \ | t 4 , 1 is given by x y ÷ | . | \ | x y cos 2 then the equation of the curve is (a) | | . | \ | | . | \ | = ÷ x e l og tan y 1 (b) | | . | \ | | . | \ | = ÷ e x l og tan x y 1 (c) | | . | \ | | . | \ | = ÷ x e l og tan x y 1 (d) None of these 26.(C) We have, | . | \ | ÷ = x y cos x y dx dy 2 Putting y = ux, so that dx d x dx dy u + u = , we get u + x u ÷ u = u 2 cos dx d ¬ x dx cos d 2 ÷ = u u ¬ sec 2 u du = ÷ dx x 1 On integration, we get tan u = ÷log x + log C ¬ C l og x l og x y tan + ÷ = | . | \ | This passes through (1, t/4), therefore 1 = log C So, 1 x l og x y tan + ÷ = | . | \ | ¬ e l og x l og x y tan + ÷ = | . | \ | ¬ y = x tan ÷1 | | . | \ | | . | \ | x e l og Section – 2 : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 51. A line l passing through the origin is perpendicular to the lines l1 : (3 + t) i ˆ + (– 1 + 2t) j ˆ + (4 + 2t) k ˆ , – · < t < ·, l2 : (3 + 2s) i ˆ +(3 + 2s) j ˆ + (2 + s) k ˆ , – ·< s < · Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is (are) (a) | . | \ | 3 5 , 3 7 , 3 7 (b) (– 1, – 1, 0) (c) (1, 1, 1) (d) | . | \ | 9 8 , 9 7 , 9 7 51. (b, d) Mathematics : J EE Advance Paper – 1 : 2013 14 The common perpendicular is along 1 2 2 2 2 1 k ˆ j ˆ i ˆ = k ˆ 2 j ˆ 3 i ˆ 2 ÷ + ÷ Let M ÷ (2ì, – 3ì, 2ì) So, 2 4 2 2 1 3 1 3 2 ÷ ì = + ì ÷ = ÷ ì ¬ ì = 1 So, M ÷ (2, – 3, 2) Let the required point be P. Given that PM = 17 ¬ (3 + 2s – 2) 2 + (3 + 2s + 3) 2 + (2 + s – 2) 2 = 17 ¬ 9s 2 + 28s + 20 = 0 ¬ s = – 2, – 9 10 . So, P ÷ (– 1, –1, 0) or | . | \ | 9 8 , 9 7 , 9 7 Example 3.19 : Vector and 3-D Geometry : Chapter 3 : 3-D Geometry : page no. 3.14 Example 3.19 Find the equation of line which passes through point A(1, 0, ÷1) and perpendicular to the straight lines ) k 3 j 7 i 2 ( k j i 2 r . . . . . . ÷ ÷ + ì + + ÷ = and ) k 5 j 2 i 2 ( k 3 j i 3 r . . . . . . ÷ + ÷ ì + + ÷ = Sol. Since line to be determined is perpendicular toe given two straight lines it is directed towards vector × ÷ + . . . ) k 3 j 7 i 2 ( ) k 5 j 2 i 2 ( . . . + ÷ = 5 2 2 3 7 2 k j i ÷ ÷ . . . = . . . + ÷ k 18 j 16 i 41 Hence equation of lien passing through point A(1, 0, ÷1) and along vector . . . ÷ + ÷ k 18 j 16 i 41 is 18 1 z 16 y 41 1 x + = ÷ = ÷ 52. Let f(x) = x sin tx, x > 0. Then for all natural numbers n, f’(x) vanishes at (a) a unique point in the interval | . | \ | + 2 1 n , n (b) a unique point in the interval | . | \ | + + 1 n , 2 1 n (c) a unique point in the interval (n, n + 1) (d) two points in the interval (n, n + 1) 52. (b, c) We have f'(x) = sin tx + tx cos tx = 0 ¬ tan tx = – tx ¬ tx e | . | \ | t + t + ) 1 n ( , 2 1 n 2 ¬ x e | . | \ | + + 1 n , 2 1 n or (n, n + 1) The concept used in the question is that of elementary differentiation and then solving the equation using graphs. There are many questions in the function and other chapters where we have to solve the equation using graphs particularly when two different type of functions are there in one equation. Mathematics : J EE Advance Paper – 1 : 2013 15 Q. no. 101 : Single option correct type : Calculus : Chapter 1 : Functions : page no. 1.46 101. The number of solutions of tan x ÷ mx = 0, m > 1 in | . | \ | t t ÷ 2 , 2 is (a) 1 (b) 2 (c) 3 (d) m 101.(c) In | . | \ | t ÷ 0 , 2 the graph of y = tan x lies below the line y = x which is the tangent at x = 0 and in | . | \ | t 2 , 0 it lies above the lies y = x. For m > 1, the line y = mx lies below y = x in (– t/2, 0) and above y = x in (0, t/2). As |tan x| mono-tonically increases from O at x = 0 to · for | x| = 2 t , the graphs of y = tan x and y = mx, m > 1, meet at three points including x = 0 in | . | \ | t t ÷ 2 , 2 independent of m. Q. no. 9 : Matrix match type : Calculus : Chapter 1 : Functions : page no. 1.53 9. Column I : Equation Column II : Number of roots (a) x 2 tan x = 1, x e [0, 2t] (p) 5 9. (p) y = tan x = 2 x 1 From graph it is clear that it will have two real roots Q. no. 4 : Concept Application exercise 3.8 : Trigonometry : Chapter 3 : Trigonometric Equations : page no. 3.19 4. Prove that the least positive value of x, satisfying tan x = x + 1 lies in the interval. | . | \ | t t 2 , 4 . Sol. Let; f(x) = tan x and g(x) = x + 1; which could be shown as : From the above figure tan x = x + 1 has infinitely many solutions but the least positive value of x e | . | \ | t t 2 , 4 . Mathematics : J EE Advance Paper – 1 : 2013 16 53. Let Sn = ¿ = + ÷ n 4 1 k 2 2 ) 1 k ( k k ) 1 ( . Then Sn can take value(s) (a) 1056 (b) 1088 (c) 1120 (d) 1332 53. (a, d) Sn = ( 1) ( 1) 4 2 2 2 2 2 2 1 0 ( 1) ((4 4) (4 3) (4 2) (4 1) ) k k n n k r k r r r r + ÷ = = ÷ = + + + ÷ + ÷ + ¿ ¿ = ¿ ÷ = + + + ) 1 n ( 0 r )) 4 r 8 ( 2 ) 6 r 8 ( 2 ( = ¿ ÷ = + ) 1 n ( 0 r ) 20 r 32 ( = 16 (n – 1)n + 20n = 4n (4n + 1) = ¹ ´ ¦ = = 9 n for 1332 8 n for 1056 The concept used in the question is series having terms with alternate sign. The question asked has two +ve and two –ve terms alternatively. The similar question is in the book which has series having +ve and –ve sign alternatively. Example 3.83 : Algebra : Chapter 3 : Progression and series : page no. 3.23 Sum to n terms the series 1 2 ÷ 2 2 + 3 2 ÷ 4 2 + 5 2 ÷ 6 2 + ….. Sol. Clearly, nth term of the given series is negative or positive according as n is even or odd respectively. Case I : n is even 1 2 ÷ 2 2 + 3 2 ÷ 4 2 + 5 2 ÷ 6 2 + ….. + (n ÷ 1) 2 ÷ n 2 = (1 2 ÷ 2 2 ) + (3 2 ÷ 4 2 ) + (5 2 ÷ 6 2 ) + ……. + ((n ÷ 1) 2 ÷ n 2 ) = (1 ÷ 2) (1 + 2) + (3 ÷ 4) (3 + 4) + (5 ÷ 6) (5 + 6) + …… + ((n ÷ 1) ÷ (n)) (n ÷ 1 + n) = ÷ (1 + 2 + 3 + 4+ ….. + (n ÷ 1) + n) = ÷ ( ) 2 1 n n + Case II : n is odd (1 2 ÷ 2 2 ) + (3 2 ÷ 4 2 ) + …… + {(n ÷ 2) 2 ÷ (n ÷ 1) 2 } + n 2 = (1 ÷ 2) (1 + 2) + (3 ÷ 4) (3 + 4) + ….. + ((n ÷ 2) ÷ (n ÷ 1)) ((n ÷ 2) + (n ÷ 1)) + n 2 = ÷ (1 + 2+ 3 + 4 + …… + (n ÷ 2) + (n ÷ 1)) + n 2 = ÷ ( ) ( ) 2 n 2 1 1 n 1 n + + ÷ ÷ = ( ) 2 1 n n + 54. For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct ? (a) N T MN is symmetric or skew symmetric, according as M is symmetric or skew symmetric (b) MN – NM is skew symmetric for all symmetric matrices M and N (c) MN is symmetric for all symmetric matrices M and N (d) (adj. M) (adj. N) = adj (NM) = adj (MN). 54. (c, d) (a) (N T MN) T = N T M T N = N T MN if M is symmetric and is – N T MN if M is skew symmetric (b) (MN – NM) T = NTMT = M T N T = NM = MN = –(MN – NM). So, (MN – NM) is skew symmetric (c) (MN) T = N T M T = NM ÷ MN if M and N are symmetric. So, MN is not symmetric (d) (adj. M) (adj. N) = adj(NM) = adj (MN). The question is based on simple properties of matrices mainly symmetric, skew-symmetric, adjoint etc. Some of the statements in the question are direct or indirect statements in the following questions in the book Mathematics : J EE Advance Paper – 1 : 2013 17 Q. no. 65 : Single option correct type : Algebra : Chapter 8 : Matrices : page no. 8.27 65. If A and B are symmetric matrices of the same order and X = AB + BA and Y = AB ÷ BA, then (XY) T is equal to (a) XY (b) YX (c) ÷YX (d) None of these 65.(c) Given that X = AB + BA ¬ X = X T and Y = AB ÷ BA ¬ Y = ÷Y T . Now (XY) T = Y T X T = ÷YX Q. no. 5 : Multiple option correct type : Algebra : Chapter 8 : Matrices : page no. 8.29 5. Let A and B are two non singular square matrices, A T and B T are the transpose matrices of A and B respectively, then which of the following is correct (a)B T AB is symmetric matrix iff A is symmetric (b)B T AB is symmetric matrix iff B is symmetric (c)B T AB is skew symmetric matrix for every matrix A (d)B T AB is skew symmetric matrix if A is skew symmetric 5.(A, D) (B T AB) T = B T A T (B T ) T = B T A T B = B T AB iff A is symmetric B T AB is symmetric iff A is symmetric Also (B T AB) T = B T A T B = B T ( – A) B = – (B T A T B) B T AB if A is skew symmetric if A is skew symmetric 55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (a) 24 (b) 32 (c) 45 (d) 60 55. (a, c) Let the sides of rectangle be 15k and 8k and side of square be x then (15k – 2x)(8k – 2x)x is volume. v = 2(2x 3 – 23kx 2 + 60k 2 x) 0 dx dv 5 x = = ¬ 6x 2 – 46kx + 60k 2 |x = 5 = 0 ¬ 6k 2 – 23k + 15 = 0 ¬ k = 3, k = 6 5 . Only k = 3 is permissible. So, the sides are 45 and 24. See the almost similar question in the book Q. no. 95 : Single option correct type : Calculus : Chapter 6 : Monotonocity and Maxma-Minima of Functions : page no. 6.27 95. A box is constructed from a rectangular metal sheet is 21 cm by 16 cm by cutting equal squares of sides x from the corners of the sheet and then turning up the projected portions. The value of x so that volume of the box is maximum is (a) 1 (b) 2 (c) 3 (d) 4 95.(c) The dimensions of the box after cutting equal squares of side x on the corner will be 21–2x, 16–2x and height x. V = x (21 – 2x) (16 – 2x) = x (336 – 74x + 4x 2 ) dx dV = 0 gives x = 3 for which 2 2 dx V d is – ive and hence max. SECTION – 3 : (Integer value correct Type) This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive). 56. Consider the set of eight vectors V = }} 1 , 1 { c , b , a ; k ˆ c j ˆ b i ˆ a { ÷ e + + . Three non-coplanar vectors can be chosen from V in 2 p ways. Then p is ________ Mathematics : J EE Advance Paper – 1 : 2013 18 56.(5) Let (1, 1, 1), (–1, 1, 1), (1, –1, 1), (–1, –1, 1) be vectors d , c , b , a rest of the vectors are d , c , b , a ÷ ÷ ÷ ÷ and let us find the number of ways of selecting co–planar vectors. Observe that out of any 3 coplanar vectors two will be collinear (anti parallel) Number of ways of selecting the anti parallel pair = 4 Number of ways of selecting the third vector = 6 Total = 24 Number of non co–planar selections = 8 C3 – 24 = 32 = 25 , p = 5 p = 5 Here the mixed concept of counting principal of permutation and combination and vectors. One can solve the problem easily if go through the concepts discussed in book at various location. 57. Of the three independent events E1, E2, and E3, the probability that only E1 occurs is o, only E2 occurs is | and only E3 occurs is ¸. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations (ì – 2|) p = o| and (| – 3¸) p = 2|¸. All the given probabilities are assumed to lie in the interval (0, 1). Then 3 1 E of occurrence of obabi l i ty Pr E of occurrence of obabi l i ty Pr 57.(6) Let P(E1) = x, P (E2) = y and P (E3) = z then (1 – x) (1 – y)(1 – z) = p x (1 – y) (1 – z) = o, (1 – x) y (1 – z) = |, (1 – x) (1 – y) (1 – z) = ¸ So o = ÷ p x x 1 x = p + o o , Similarly z = p + ¸ ¸ . so o + ¸ + = o + o ¸ + ¸ = + ¸ ¸ + o o = p 1 p 1 p p p p ) E ( P ) E ( P 3 1 also given | ÷ o o| = p = ¸ ÷ | |¸ 3 2 ¬ | = ¸ + o o¸ 4 5 Substituting back ¸ + o o¸ o = | | . | \ | | | . | \ | ¸ + o o¸ ÷ o 4 5 . p 4 5 2 ¬ op – 6p = 5o¸ ¬ | | . | \ | + ¸ 1 p = 6 | . | \ | + o 1 p ¬ 1 p 1 p + o + ¸ = 6. Here concept of independent events is used. If one go through this topic in the book, can easily solve the problem. 58. The coefficients of three consecutive terms of (1 + x) n + 5 are in the ratio 5 : 10 : 14. Then n = _____________ 58.(6) let Tr – 1, Tr, Tr + 1 are three consecutive terms of (1 + x) n + 5 Tr – 1 = n + 5 Cr – 2 (x) r – 2 , Tr = n + 5 Cr – 1 x r – 1 , Tr + 1 = n + 5 Crx r , where, n + 5 Cr – 2 : n + 5 Cr – 1 : n + 5 Cr = 5 : 10 : 14. So 10 C 10 C 5 C r 5 n 1 r 5 n 2 r 5 n + ÷ + ÷ + = = Comparing first two results we have n – 3r = – 3 …….(1) Comparing last two results we have 5n – 12r = – 30 …….(2) From equation (1) and (2), n = 6 The concept used in the question is that of how two handle situation when three consecutive binomial coefficients are involved. This topic is discussed in detail in the book with supporting illustrations. Theory : Algebra : Chapter 6 : Binomial Theorem : page no. 6.8 RATIO OF CONSECUTIVE TERMS AND THEIR COEFFICIENTS : Ratio of coefficients of x r and x r + 1 are n Cr – 1 and n Cr respectively. Also we know that r 1 r n C C 1 r n r n + ÷ = ÷ . Similarly 1 r r n C C r n 1 r n + ÷ = + (replacing r by r + 1) Mathematics : J EE Advance Paper – 1 : 2013 19 1 r 1 r n C C r 1 n 1 r 1 n + + ÷ = + + + (replacing r by r + 1 and n by n + 1) etc. Example 6.32 : Algebra : Chapter 6 : Binomial Theorem : page no. 6.8 32. The coefficients of three successive terms in the expansion of (1 + x) n are 165, 330 and 462 respectively, then find the value of n Sol. Let the coefficient of three consecutive terms i.e. (r + 1) th , (r + 2) th , (r + 3) th in expansion of (1 + x) n are 165,330 and 462 respectively then, coefficient pf (r + 1) th term = n Cr = 165 Coefficient of (r + 2) th term = n Cr + 1 = 330 and Coefficient of (r + 3) th term = n Cr + 2 = 462 2 1 r r n C C r n 1 r n = + ÷ = + ¬ n ÷ r = 2(r + 1) ¬ ( ) 2 n 3 1 r ÷ = and 165 231 2 r 1 r n C C 1 r n 2 r n = + ÷ ÷ = + + ¬ 165(n ÷ r ÷ 1) = 231(r + 2) or 165n ÷ 627 = 396r ¬ ( ) 2 3 1 396 627 165 ÷ × × = ÷ n n ¬ 165n ÷ 627 = 132(n ÷ 2) or n = 11 59. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = ________ 59. (5) Clearly, 1 + 2 + 3 + … + n – 2 s 1224 s 3 + 4 + … n ¬ 2 ) 1 n ( ) 2 n ( ÷ ÷ s 1224 s 2 ) 2 n ( ÷ (3 + n) ¬ n 2 – 3n – 2446 s 0 and n 2 + n – 2454 > 0 ¬ 49 < n < 51 ¬ n = 50 2 ) 1 n ( n + – (2k + 1) = 1224 ¬ k = 25 ¬ k – 20 = 5 Exactly the similar questions is in the book in comprehension type. Q. no. 22-23 : Comprehension type : Algebra : Chapter 3 : Progression and series : page no. 3.38 Paragraph 8: Question Nos. 22 to 24 Two consecutive numbers from 1, 2, 3…… n are removed. Arithmetic mean of the remaining numbers is 4 105 22. The value of n lies in (a) [45, 55] (b) [52, 60] (c) [41, 49] (d) none of these 23. The removed numbers are (a) lies between 10 and 20 (b) greater than 10 (c) less than 15 (d) none of these 24. Sum of all numbers (a) exceeds 1600 (b) less than 1500 (c) lies between 1300 1500 (d) none of these Paragraph 8: Question Nos. 22 to 24 22. (a), 23. (c), 24. (b) Mathematics : J EE Advance Paper – 1 : 2013 20 Sol. Let m and (m + 1) be removed number from 1, 2, .., n then sum of remaining numbers. = 2 ) 1 n ( n + – (2m + 1) From given condition ) 2 n ( ) 1 m 2 ( 2 ) 1 n ( n 4 105 ÷ + ÷ + = ¬ 2n 2 – 103n – 8m + 206 = 0 since n and m are integers so n must be even let n = 2k we get, m = 4 ) k 1 ( 103 k 4 2 ÷ + Since m is an integer then (1 – r) must be divisible by 4. Let k = 1 + 4t, we get n = 8t + 2 and m = 16t 2 – 95 t + 1, Now 1 s m < n ¬ 1 s 16t 2 – 95 t + 1 < 8t + 2 Solving we get t = 6 ¬ n = 50 and m = 7. Hence removed no. are 7 & 8. Also sum of all numbers = 2 ) 1 50 ( 50 + = 1275 60. A vertical line passing through the point (h, 0) intersects the ellipse 3 y 4 x 2 2 + = 1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If A(h) = area of the triangle PQR, A1 = 1 h 2 / 1 mx s s A(h) and A2 = 1 h 2 / 1 min s s A(h), then 5 8 A1 – 8A2 = _____________________. 60.(9) 1 3 y 4 x 2 2 = + y = 2 h 4 2 3 ÷ at x = h let R (x1, 0) PQ is chord of contact, so 1 5 xx 1 = ¬ x = 1 x 4 which is equation of PQ, x = h so ! x 4 = h ¬ x1 = h 4 A(h) = area of APQR = 2 1 × PQ × RT = 2 h 4 2 3 2 2 1 ÷ × × (x1 – h) = 2 h 4 2 3 2 2 1 ÷ × × 9x1 – h) = h 2 3 (4 – h 2 ) 3/2 A’(h) = 2 2 h 2 ) h 2 4 ( 3 + ÷ 2 h 4 ÷ which is always decreasing so A1 = maximum of A(h) = 8 5 45 at h = 2 1 , A2 = minimum of A(h) = 2 9 at h = 1 so 2 9 . 8 8 5 45 5 8 8 5 8 2 1 ÷ × = A ÷ A = 45 – 36 = 9. The concept used in the question is finding area of triangle, chord of contact of ellipse and finding its extremum. Overall this is mixed concept question. All the concepts used in the question are there in the form of theory and examples in respective chapters.