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Managing Editor Editor : : Mahabir Singh Anil Ahlawat (BE, MBA) cONteNts Maths Musing Problem Set - 147 rial 8 Mock Test Paper JEE Main - 2015 10 Mock Test Paper BITSAT - 2015 20 10 Challenging Problems 28 Mock Test Paper JEE Advanced - 2015 31 Math Archives 48 Mock Test Paper JEE Main - 2015 50 You Asked, We Answered 58 Mock Test Paper ISI - 2015 59 CBSE Board 2015 Sample Paper 71 Maths Musing - Solutions 82 Practice Paper JEE (Main & Advanced) & Other PETs 83 Olympiad Corner 88 Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited. In modern studies of diffraction of crystals, group theory is very important. Now, quantum mechanics embraces everything from the determination of energy levels, intensities and widths of spectral lines. If atomic and molecular physicists study X - ray diffraction, nuclear physicists study spectroscopy and diffraction of g - rays. When mathematics is so interesting and exciting, why do many students and even professors try to avoid mathematics? It is a purely psychological problem and some fear of punishment. The solution is novelty in packaging which should be so attractive that persons rush to study maths in schools. This is being done in many innovative modern schools. However, to make it accessible to everybody, the solution is simple – to publish popular books in science, mathematics, languages and so on. The key for making education a success is to publish popular books in a simple language so that all can understand these books on self studies on every topic. It is a success for European languages. Why not science and mathematics for every topic written in a popular style? Anil Ahlawat Editor subscribe online at www.mtg.in individual subscription Rates 1 yr. 2 yrs. 3 yrs. Mathematics Today 300 500 675 Chemistry Today 300 500 675 Physics For You 300 500 675 Biology Today 300 500 675 combined subscription Rates 1 yr. 2 yrs. 3 yrs. PCM 800 1200 1700 PCB 800 1200 1700 PCMB 900 1500 2100 Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44 Institutional Area, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent. MatheMatics tODaY | MARCH ’15 7 M aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced. Prof. Dr. Ramanaiah Gundala, Former Dean of Science and Humanities, Anna University, Chennai Set 147 jee main 2 2 1. In triangle ABC, if D = a – (b – c) , then sin A = 4 8 15 15 (a) (b) (c) (d) 11 17 22 17 x 2. If y( x ) = , then the sum of the digits of 1 − x4 5 d y (0) is dx 5 (a) 3 (b) 4 (c) 5 (d) 0 3. The planes x + y = 1, y –z = 2, z + x = 3 form a triangular prism with cross sectional area 4 8 (b) (c) 4 3 (d) 2 3 (a) 3 3 4. The area of the triangle, whose vertices are the 3 2 roots of the equation x + ix + 2i = 0, is (a) 2 (b) 3 π 5. (a) ∫x 0 2 π 3 2 (c) 5 (d) 2 π π (c) 6 3 jee advanced (b) − (d) − 7 2 π 6 10 C C C C C 6. If Cr = and 0 − 1 + 2 − 3 + ... + 10 r 10 11 12 13 20 1 = , then n is divisible by n (a) 11 (b) 13 (c) 17 (d) 19 comprehension Let S = {1, 2, 3, 4, ..., 25} and T = {x, y} ⊂ S. 2 2 2 8. The probability that x –y is divisible by 7 is 73 7 1 71 (a) (b) (c) (d) 300 30 3 300 integer match 8 8 8 8 9. Let 88 − 78 + 68 − 58 + 48 1 2 3 4 8 8 8 − 38 + 28 − 18 = n. The sum of the 5 6 7 digits of n is matching list 10. The sequence of positive integers a1, a2, a3, ... is such that a1, a2, a3 are in G.P., a2, a3, a4 are in A.P., a3, a4, a5 are in G.P., a4, a5, a6 are in A.P. etc. Let a1 = 1 and a5 + a6 = 198. Then match the following: sin 6 x dx = 2 2 7. The probability that x – y is divisible by 5 is 1 2 1 1 (a) (b) (c) (d) 4 5 5 3 Column-I P. Sum of the digits of a8 is Column-II 1. 5 Q. Sum of the digits of a9 is 2. 9 R. Sum of the digits of a10 is 3. 15 S. Sum of the digits of a11 is 4. 19 (a) (b) (c) (d) P 3 2 1 4 Q 2 1 4 3 R 1 4 3 2 S 4 3 2 1 See Solution set of Maths Musing 146 on page no. 82 Prof. Ramanaiah is the author of MTG JEE(Main & Advanced) Mathematics series 8 MatheMatics tODaY | March ’15 Let f (x) be differentiable on the interval (0. –7) (d) (–6. Find the probability that I. 7) (c) (6. The locus of the centre of circle which touches (y – 1)2 + x2 = 1 externally and also touches x-axis. Three balls are drawn at random. then the locus 2+|z| of z is (a) |z| = 5 (b) |z| < 5 (c) |z| > 5 (d) None of these 3. A bag contains 8 red and 5 white balls. then z x y (a) (b) (c) (d) A=B=C ABC = 1 A+B+C=0 none of these . All the three balls are white II. y). 2 2 p (c) 0. be a function defined by 2x f ( x ) = tan −1 . ∑ (r 2 − r + 3)x r −1 = r =1 (a) 3 + 2x(1 – x)–2 (c) 2 3( x + 1) − 4 x (1 − x )3 (d) none of these (1 − x )3 2. 7) (b) (–6. y ≥ 0} ∪ {(0. 2 2 p p (b) − . Then f(x) is 2 (d) px tan 2a 1 4x2 + 3x 3 1 x = (b) e2/p (d) none of these 6. B = z c − a . All the three balls are red III. One ball is red and two balls are white 10 MatheMatics tODaY | March ’15 III 5 29 40 143 143 143 5 28 40 (b) 143 143 143 7 28 40 (c) 143 143 143 (a) 3x + 2 (b) II (d) None of these 7. y ∈ R 9. 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 at (a) (6. x →a a (a) 2/p (c) e–2/p 2 t f ( x ) − x f (t ) = 1 for t−x t→x such that f (1) = 1.Exam on 4th April I ∞ 1. 2 8. ∞) 1 2x 2 + 3x 3 1 2 (c) − + 2 x x (b) − (a) x lim 2 − 5. y < 0} (b) x2 = y (c) y = 4x2 (d) {y2 = 4x} ∪ (0. C = and A = . The tangent at (1. 1) → B. y). a b c y z x If = 1 z x y 1 1 1 y b−c x a −b . is (a) {x2 = 4y. 4. and lim each x > 0. –7) | z |2 − | z | +1 If log 3 < 2 . 2 p (d) 0. then f is both one-one and 1 − x 2 onto when B is in the interval p p (a) − . Let f : (–1. . If 1 + sin2 q cos2 q 4 sin 4q sin2 q 1 + cos2 q 4 sin 4q 2 sin q then q is equal to 7 p 11p . then x = 2 5 ±1 (b) ± 2 16. Given a = 2 i + j − 2 k. 1 + 4 sin 4q 5p 7 p . 7. y. b. For 0 ≤ x ≤ sin2 x ∫0 (d) none of these p . unit sq .10. If ar be the coefficient of xr in the expansion of (1 – x)2011. then sum of which of the following pair vanishes ? (a) a777. The expression 3p 3 sin 4 − a + sin 4 (3p + a) 2 p −2 sin6 + a + sin6 (5p − a) is equal to 2 (a) 0 (b) 1 (c) 3 (d) sin 4a + cos 6a 13. then p : q is (a) 6 : 5 (b) 3 : 2 (c) 4 : 3 (d) 5 : 3 14. For each real x : –1 < x < 1. b = 2 i + j. z) satisfies the equation 1 1 1 + + = K . unit sq . The area bounded by the curves y = (x –1)2. a1357 (d) All of these 21. 5. | c | = a ⋅ c. Centroid (x. 9 without repeating the digits. p | c − a | = 4cos and angle between c & (a × b ) 4 is p/6. unit (c) (d) 4 5 ^ ^ ^ ^ ^ 15. a1234 (b) a1111. 24 24 p 7p . The value of K is 2 2 x y z2 1 1 (a) 9 (b) 3 (c) (d) 3 9 20. (a) 24 24 11p p . 1 y = (x + 1)2 and y = is 4 1 2 sq . (c) 24 24 2 cos q (b) = 0. 5-digit numbers are to be formed using 2. If p be the number of such numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000. A variable plane A. 3. 3 3 3 3 12. c are the sides of a DABC such that x2 – 2(a + b + c)x + 3l(ab + bc + ca) = 0 has real roots. If a. (d) l ∈ . a900 (c) a654. If tan −1 x + sin −1 x = (a) ± 5 −1 2 (c) ± 5 +1 2 17. unit (a) (b) 3 3 1 1 sq . the value of 2 sin −1( t )dt + ∫ cos2 x 0 cos−1( t )dt is equal to p p (b) 0 (c) 1 (d) 4 4 18. The value of (a) p (c) 4p p /2 sin2 x ∫− p /2 1 + 2x dx is p 2 p (d) 4 (b) . then matrix (1 − x )−1 1 + xy − x 1 (a) A(z) = A(x) – A(y) (b) A(z) = A(x) A(y) (c) A(z) = A(x) [A(y)]–1 (d) A(z) = A(x) + A(y) (a) − x y z + + = 1 at a unit a b c distance from origin cuts the coordinate axes at 19. Let A(x) be the 1 −x x+y and z = . (d) 24 24 11. B and C. then 5 4 (b) l > (a) l < 3 3 4 5 1 5 (c) l ∈ . then | (a × b ) × c | =is?equal to 12 MatheMatics tODaY | March ’15 (b) 1 (d) none of these (a) 5 (c) 3/2 p . e. where p ≠ q. 0) ∪ (1. x2 +(2x + 5)2 + 16x + 12 (2x + 5) + c = 0 ⇒ 5x2 + 60x + 85 + c = 0 must have equal roots. ∞) (–1. 2) ∪ (2. the value of q such that the sum of intercepts on axes made by this tangent is minimum. p/2)).22. 2. Tangent is drawn to ellipse p p p p (b) (c) (d) 6 3 8 4 29.M. ∞) (1 − x )S = 3 + 2 x + 4 x 2 + 6 x 3 + . then locus of P(x. n. for i = 1. to ∞ (− ) x ⋅ S = 3x + 5x 2 + 9 x 3 + . 0) ∪ (0.. The locus of the orthocentre of the triangle formed by the lines (1 + p)x –py + p(1 + p) = 0. If x1. of t and t′ = (d) None of these 2h g 2h g S = 3 + 5x + 9 x 2 + 15x 3 + .M. to ∞ (b) [0. ∞) (d) (0. 2) (1. where the function x f (x ) = is differentiable. A particle is projected vertically upward and reaches at a height of h after time t seconds. 2) ∪ (2. (d) : The tangent to the parabola x2 = y – 6 at (1. of t and t′ = (c) A. is 1+|x| \ (a) (– ∞. to ∞ = 3 + 2 x(1 + 2 x + 3x 2 + . ∞) 27. x2. 7) is y = 2x + 5 which is also a tangent to the given circle. is 28. The complex numbers z = x + iy which satisfy z − 5i = 1 .M... (1 + q)x –qy + q(1 + q) = 0 and y = 0. Then. If 2 + eiq represent a/an (a) Ellipse if l = 1. y) will 24. If a variable x takes values xi such that a ≤ xi ≤ b. 2) (–1.. x3 and x4 are the roots of the equation x4 –x3 sin 2b + x2 cos 2b –x cos b –sin b = 0. then 1 (a) A.. Let the roots be a.. i. of t and t′ = g (b) G.. a MatheMatics tODaY | March ’15 13 . 0) ∪ (1. sin q) (where q ∈ (0. l = 0 (c) Circle if l ≠ m ≠ 1 (d) None of these (c) 25. is f (x ) = 2 4−x (a) (b) (c) (d) (1. ∞) (c) (–∞. It further takes t′ seconds to reach the ground. is (a) a hyperbola (b) a parabola (c) an ellipse (d) a straight line 23. The set of all points. lie on the equation z + 5i (a) x-axis (b) straight line y = 5 (c) a circle passing through the origin (d) None of these (a) 30. Let the greatest height attained be H... m = 2 (b) Pair of straight lines if m = 1. (c) : Let S = ∞ ∑ (r 2 − r + 3)x r −1 r =1 26. then tan–1 x1 + tan–1 x2 + tan–1 x3 + tan–1 x4 is equal to p −b (a) b (b) 2 (c) p – b (d) – b SolutionS 1.. ∞) x2 + y2 = 1 27 at (3 3 cos q.. .. Domain of definition of the function 3 + log10 ( x 3 − x ). then (b) a ≤ var(x) ≤ b (a) a2 ≤ var(x) ≤ b2 a2 ≤ var( x ) (d) (b – a)2 ≥ var(x) 4 3 = lx + imy . to ∞) = 3 + 2 x(1 − x )−2 = \ S= 3(1 − x )2 + 2 x 3( x 2 + 1) − 4 x (1 − x )2 = 3( x 2 + 1) − 4 x (1 − x )2 (1 − x )3 2. 60 ⇒ a = −6 5 \ a+a=− \ x = –6 and y = 2x + 5 = –7 | z |2 − | z | +1 3. 2 2 Hence. t 2 f ( x ) − x 2 f (t ) =1 t−x t→x ⇒ x2f ′(x) –2xf (x) + 1 = 0 4. 8. k) touching (y –1)2 + x2 = 1 and x-axis shown as y 2 2 1 x + 3 3x (0. 2x f ( x ) = tan −1 = 2 tan −1 x ( x 2 < 1) 1 − x 2 p p So.. function is one-one onto. k) x y′ distance between O and A is always 1 + |k|. 2 2 Given that. we get 1 f ( x ) = cx 2 + 3x Also f (1) = 1 ⇒ c = Hence f ( x ) = 2 3 5 C3 6. 4 4 p p ⇒ 2 tan −1 x ∈ − . P(One ball is red and two balls are white) = 8 C1 × 5C2 8 × 10 40 = = 13 13 12 11 × × 143 C3 3×2 p p 7. (a) : Given. x ∈ (–1. (a) : Let the locus of centre of circle be (h. P(All the three balls are red) 13 3 C3 8 ! × 10 ! 8×7×6 28 = = = 5 ! × 13 ! 13 × 12 × 11 143 III. i. P(All three balls are white) = 13 C3 5 ! × 10 ! 5×4×3 5 = = = 2 ! × 13 ! 13 × 12 × 11 143 8 C = II.e. (b) : log <2 3 2+|z| | z |2 − | z | +1 ⇒ < ( 3 )2 2+ | z | ⇒ ⇒ ⇒ ⇒ \ |z|2 – |z| + 1 < 3(2 + |z|) |z|2 –4|z| – 5 < 0 (|z| + 1) (|z| –5) < 0 ⇒ –1 < |z| < 5 |z| < 5 as |z| > 0 Locus of z is |z| < 5. (b) : Q Given limit is in 1∞ form x \ lim 2 − x→ a a = x px lim 2 − −1 tan 2a a e x →a =e = px tan 2a 1 − a lim px p x →a − cosec2 2a 2a x′ =e x 1− a lim px x →a cot 2a [L-Hospital’s Rule] 2 2 px sin 2a e x →a p = e2/ p lim 14 MatheMatics tODaY | March ’15 O |k| A B (h. y < 0 . (a) : Since. 1) ⇒ tan −1 x ∈ − . lim ⇒ x 2 f ′( x ) − 2 xf ( x ) + 2 2 (x ) 1 x 4 =0 d f (x ) 1 ⇒ =− 2 dx x x4 On integrating both sides. y ≥ 0 where. (h − 0)2 + (k − 1)2 = 1 + | k |. 1) C 1 5. f ( x ) ∈ − . ⇒ h2 + k2 – 2k + 1 = 1 + k2 + 2|k| ⇒ h2 = 2|k| + 2k ⇒ x2 = 2|y| + 2y y. (b) : I. | y | = − y. MatheMatics tODaY | March ’15 15 . (a) : The curves y = (x – 1)2 . b c a a a z x z z x z x ⋅ = = ⋅ = ⋅ x x y y x y y x ⇒ y c−a 1 x a −b ⇒ y z = x a a −b 1 c−a z = x ⇒C = B Similarly. (d) : p = The number of such numbers that exceeds 20000 =5! = 120 q = The number of those that lie between 30000 and 90000 = 5! – 4! – 4! = 120 – 24 –24 = 72 \ p 120 5 = = q 72 3 14. y < 0} 9. and R − . 24 24 2 11. (b) : 3 sin 4 − a + sin 4 (3p + a) 2 p −2 sin6 + a + sin6 (5p − a) 2 = 3(cos4a + sin4a) –2(cos6a + sin6a) = 3(1 – 2sin2a cos2a) –2(1 – 3 sin2a cos2a) = 3 – 6 sin2a cos2a –2 + 6 sin2a cos2a = 1 13. y) : x2 = 4y. therefore D ≥ 0. 3 8 3 8 3 . cos A = b2 c2 .. y ≥ 0} ∪ {(0. (a) : We have..(i) b2 + c 2 − a2 <1 2bc ⇒ + < 2bc 2 Similarly. it can be proved that A = B \ A = B = C. 10. y = (x + 1)2 and y = 1/4 are shown as y y = (x + 1)2 y = (x – 1)2 2 sin q cos q 1 + 4 sin 4q ⇒ sin2 q + cos2 q + 1 + 4 sin 4q = 0 7 p 11p 1 ⇒ sin 4q = − ⇒ q = . ⇒ ⇒ ⇒ 1/4 R (a) : Since. we get 4 3l – 2 < 2 ⇒ l < 3 16 MatheMatics tODaY | March ’15 Q –1 –1/2 O 1 y = 1/4 x where point of intersection are 1 1 ( x − 1)2 = ⇒ x= 4 2 and ( x + 1)2 = 1 1 ⇒ x=− 4 2 1 1 1 1 \ Q . (a) : Applying R1 → R1 – R3 and R2 → R2 – R3 we get.. roots are real.(ii) ( x − 1)3 1 = 2 − x 4 3 0 1 1 1 1 = 2 − − − − − 0 = sq . 1 0 −1 0 −1 1 2 =0 3p 12. unit . 4(a + b + c)2 –12l (ab + bc + ca) ≥ 0 (a + b + c)2 ≥ 3l (ab + bc + ca) a2 + b2 + c2 ≥ (ab + bc + ca) (3l – 2) ⇒ 3l − 2 ≤ a2 + b2 + c 2 ab + bc + ca Also.. 2 4 2 4 –a2 a2 + b2 + c 2 P \ Required area = 2 ∫ 1/ 2 0 2 1 ( x − 1) − 4 dx 1/ 2 . c + a2 – b2 < 2ca and a2 + b2 – c2 < 2ab ⇒ a2 + b2 + c2 < 2(ab + bc + ca) ⇒ ab + bc + ca <2 \ From (i) and (ii). y < 0 ⇒ x2 = 4y when y ≥ 0 and x2 = 0 when y < 0 \ {(x.\ x2 = 2y + 2y. y ≥ 0 and x2 = –2y + 2y. y). y= . B(0. 0. y . (d) : tr + 1 = 2011C (–x)r r = (–1)r · \ According to problem. a777 + a1234 = (–1)777 2011C777 + =– 2011C 77 7 + 2011C 2011C r 2011C r · xr (–1)1234 ·2011C1234 777 =0 MatheMatics tODaY | March ’15 17 . 0.. (b) : We have. \ x2 = 5 −1 2 17. ar = (–1)r · Also. A(x) A(y) 1 −x (1 − y )−1 = (1 − x )−1 − x 1 ^ k ^ ^ Now. dt = 2 sin z cos zdz and dt = – 2 cos u sin udu \ x= = x x ∫0 2z sin z cos zdz + ∫p /2 − 2u cos u sin udu x 2z dz − ∫ u sin 2udu ∫0 zI sin p /2 II x 1 − y − y 1 1 + xy −( x + y ) = [(1 + xy ) − ( x + y )]−1 − ( x + y ) 1 + xy 16. (d) : We have.S. if x < 0 then L.(i) \ = 1 or 2 2 a b c2 1 1 1 + + a2 b2 c 2 Let P be the centroid of triangle 19. c) and its distance from origin = 1 1 1 1 1 + + = 1 . of given equation becomes –ve and given equation is not satisfied. (a) : Since. z ) = .. (d) : Put t = sin2 z in 1st integral and t = cos2u in 2nd integral.H. tan −1 x = x cos 2 x sin 2 x p p − −x + − + 0 = 4 4 2 4 18.p 15.z= 3 3 3 \ From (i) and (ii).. b. C(0. 0). a+0+0 0+b +0 0+0+c \ P(x .. . 0). a × b = 2 1 −2 = 2 i − 4 j \ I= cos 2 x sin 2 x = −x + − {0 + 0} 2 4 x cos 2z sin 2z −u cos 2u sin 2u = − z + − + 2 4 0 2 4 p / 2 x + y 1 − 1 + xy −1 x+y = 1 − 1 + xy − x + y 1 1 + xy = A(z) x y z + + = 1 cuts the coordinate a b c axis at A(a. we get 1 1 1 1 1 1 + + =9=K + + =1 ⇒ 2 2 2 2 2 9x 9y 9z x y z2 \ K=9 20.(ii) ⇒ x= . 3 3 3 a b c . (a) : | c − a |= 4 cos = 2 2 ⇒ | c − a |2 = 8 4 ⇒ (c − a ) ⋅ (c − a ) = 8 ⇒ | c |2 + | a |2 − 2(a ⋅ c ) = 8 [ | a | = 3] ⇒ | c |2 − 2 | c | = 8 − 9 ⇒ |c |=1 ^ i ^ j 2 1 0 ⇒ | a × b | = 20 p \ | (a × b ) × c | =| a × b || c | sin 6 1 = 20 (1) = 5 2 p − sin −1 x = cos−1 x 2 1 ⇒ sec −1 1 + x 2 = sec −1 x 1 ⇒ 1 + x2 = ⇒ x4 + x2 − 1 = 0 2 x −1 ± 1 − 4 ⋅ 1 ⋅ (−1) −1 ± 5 = 2 2 5 −1 2 2 x </ 0 \ x = 2 5 −1 \ x=± 2 But. 1/ 4 3 + log10 ( x 3 − x ) 25. (d) : Since. \ Equation of altitude CM passing through C and perpendicular to AB is x = pq . 0) Line (ii) −q ( x + p) 1+ q 18 MatheMatics tODaY | March ’15 ⇒ 1= (3 − 2 lx )2 + (−2my )2 ( lx )2 + (my )2 ⇒ l2 x2 + m2y2 = 9 – 12lx + 4l2x2 + 4m2y2 ⇒ l2 x2 + m2y2 – 4lx + 3 = 0 . a1111 + a900 = (–1)1111 · =– 2011C 1111 + 2011C 2011C 1111 + (–1)900 · 2011C 1111 = 0 900 Similarly. 2) ∪ (2.(1) This is the locus of P(x. ⇒ 2 + cos q + i sin q = (1 − cos 2 x ) dx sin 2 x ⇒ 2I = x − 2 0 y= 23. ∞) And 4 –x2 ≠ 0 ⇒ x≠±2 Region is (–∞. (a) : 2 + eiq = 3 lx + imy 3 lx + imy (3 − 2 lx ) − 2imy ⇒ cos q + i sin q = lx + imy On taking modulus to both sides.. S. k) is x + y = 0.Also. a654 + a1357 = (–1)654 2011C654 + (–1)1357 2011C1357 = 2011C 21. 0) ∪ (1... x3 –x > 0 ⇒ x(x – 1)(x + 1) > 0 + + 0 1 Region is (–1. ( x − 2)2 12 + y2 = 1...(iv) (1 + q) Let orthocentre of triangle be H(h. On solving (iii) and (iv). we get x = pq. (d) : Let I = ⇒ I= p /2 ∫− p /2 ⇒ 2I = = p /2 2011C sin2 x 2 x sin2 x 1 + 2x ∫− p /2 sin ∫0 – 654 =0 ∫− p /2 1 + 2x dx p /2 p /2 654 2 ∫0 xdx = 2 2 sin xdx p /2 = p p 2 ⇒ I= 4 22. we get x = pq and y= –pq ⇒ h = pq and k = –pq \ h+k=0 \ Locus of H(h.(iii) 1 + q Q Slope of line (ii) is q \ Slope of altitude BN (as shown in figure) is −q . (1 + p)(1 + q)}. 1+ q y e Lin A (i) N C H(h. 0) ∪ (1. (d) : (1 + p)x – py + p(1 + p) = 0 . f ( x ) = 2 4−x For domain of f(x). m = 2. ∞). y) If l = 1.(ii) On solving (i) and (ii). y = (1 + p) (1 + q) \ Coordinates of C are {pq.(i) (1 + q)x – qy + q(1 + q) = 0 . 2) ∪ (2.. (d) : Since. k) M O Equation of BN is y − 0 = 24. then (1) becomes x2 + 4y2 – 4x + 3 = 0 ⇒ ( x − 2)2 + 4 y 2 = 1 ⇒ which is an ellipse. k) which is the point of intersection of (iii) and (iv). < Range ⇒ s ≤ (b – a) ⇒ s2 ≤ (b – a)2 dx p /2 −q ( x + p) ... ⇒ B x(p. \ Common region is (–1..D.. –2) ∪ (–2. ∞) . (i) 1 g (t + t ′) u = 2g 2g 2 t + t′ = 2 For maxima/minima f ′(q) = 0 1 ⇒ sin3 q = cos3 q 3/ 2 3 p 1 ⇒ tan q = . 2H g g (t + t ′) gtt ′ 1 1 h = ut − gt 2 = t − gt 2 = [From (i)] 2 2 2 2 2h g p . Sx1x2x3 = cos b and x1x2x3x4 = –sin b + y sin q = 1. 5) x O x (0. (b) : We have. 1 Thus.. f ′′(0) > 0. H t + t′ u \ = g 2 3 3 sin3 q − cos3 q \ Hence. sum of intercepts (2 sin b − 1) cos b −1 = tan −1 = tan (cot b) sin b (2 sin b − 1) 3 3 1 = + = f (q) (say) cos q sin q p p = tan −1 tan − b = − b 2 2 3 3 nn MatheMatics tODaY | March ’15 19 . \ tt ′ = p .x≥0 (1 + x )2 ⇒ f ′( x ) = 1 .e. i. q = 6 3 h t g (t + t ′) 2 2 sin2 q cos2 q 29. Sx1x2 = cos 2b. 27. then it is a perpendicular bisector of z1 and z2) . (b) : ⇒ ⇒ tt ′ = 2h g sin 2 b − cos b = tan −1 1 − cos 2 b − sin b x2 y2 + = 1. (b) : Given. (a) : Given. tangent is minimum at q = t 2u g f ′(q) = and at q = Hence. tangent is drawn at (3 3 cos q. –5) is x-axis. 27 1 \ Equation of tangent is x cos q \ tan–1 x1 + tan–1 x2 + tan–1 x3 + tan–1 x4 Sx1 − Sx1 x2 x3 = tan −1 1 − Sx1x2 + x1 x2 x3 x 4 28. 6 z − 5i =1 z + 5i ⇒ |z – 5i| = |z + 5i| (if |z –z 1| = |z –z 2|.x<0 (1 − x )2 1 \ RHD at x = 0 ⇒ lim x →0 and LHD at x = 0 ⇒ lim x →0 2 (1 + x ) 1 (1 − x )2 =1 Since total time = t + t′ = =1 ⇒ u= H= \ { } 2 = g (t + t ′)2 8 (0. x 1 + x . –5) y \ Perpendicular bisector of (0. 6 y Also. (b) : Given. Sx1 = sin 2b.x<0 1 − x 1 . x ≥ 0 x f x ( ) = = 26. 1+|x| x . 5) and (0.. f(x) is differentiable for all x.. sin q) to 30. P. b. –1) (b) (–1. Assuming that straight line work as the plane mirror for a point.. 2. The values of 'a' for which (a2 – 1)x2 + 2(a – 1)x + 2 is positive for any x. 5. e ab2 are in A.P. x ≠ 1.. If the scalar product of the vector i + j + k with the unit vector in the direction of the resultant of the vectors 2i + 4 j − 5k and λi + 2 j + 3k is unity. lim x →0 (a) –1 2 sin x − sin 2 x x3 (b) 1 is equal to (c) 2 (d) –2 11. For any vector a. then is equal to (2a − b)2 (a) b (b) a (c) e (d) d 3. 2) in the line x – 3y + 4 = 0. –7) ∪ − . c are in H. ∞) 7. 7) ∪ − . find the image of the point (1. 4. if the repetition of the digits is not allowed? (a) 90 (b) 50 (c) 40 (d) 100 8. How many numbers lying between 100 and 1000 can be formed with the digits 0. –2) −6 −7 6 7 (c) . (d) 1/2 Solution of (2x + 1)(x – 3) (x + 7) < 0 is 1 (a) (– ∞. and c. are (a) a > 1 (b) a ≤ 1 (c) a > –3 (d) a < 1 −3 3π 5. 3. then ( 1.P. d. 3 2 1 (b) (– ∞. c. prove that | a × i |2 + | a × j |2 + | a × k |2 is equal to (a) 2 | a |2 (b) | a |2 (c) 3 | a |2 (d) 4 | a |2 12. (a) (–2. 1.Exam on 14th to 29th May x fofo. If f (x ) = x −1 19 times equal to 19 (a) x x −1 x (b) x − 1 (c) 19 x x −1 (d) x 2. (d) . If cosq = and p < q < . d are in G. –7) ∪ . b.. then l equals (a) 5 (b) 2 (c) 1 (d) –1 . – 7) ∪ (3. if (a) |z – 1| = 2 (b) arg (z – 1) = 2a (c) arg (z – 1) = a (d) |z| = 1 4.of )(x ) is .. then the value of 5 2 cosecθ + cot θ is sec θ − tan θ (a) 1/6 (b) 1/7 (c) 1/5 6.. The imaginary part of (z – 1)(cosa – isina) + (z – 1)–1 (cosa + isina) is zero. 3 2 20 MatheMatics tODaY | MARCH ’15 1 (c) (– ∞. If a. (a) (b) (c) (d) y2 x2 + = 1 represents a/an 14 − a 9 − a ellipse if a > 9 hyperbola if 9 < a < 14 hyperbola if a > 14 ellipse if 9 < a < 14 The equation 10.. 3 2 (d) (– ∞. 5 5 5 5 9. MatheMatics tODaY | MARCH ’15 21 . The first 12 letters of English alphabet are written in a row at random. (iii) (c) (i). The point on the curve y = (x – 2)2 at which the tangent is parallel to the chord joining the points (2. Let f : (2. If x. z are positive real numbers and x + y + z = 1. The determinant − sin θ − x x cos θ 1 19. x1 + x2 ≤ 5. (a) 80 at (3. 0 ≤ x2 ≤3. Find the equation of the plane. (iii) (d) (i). Then. 1) 25. The number of solutions of the equation x 1 + sin x sin2 = 0 in [–p. 4 9 16 then prove that the minimum value of + + is x y z (a) 80 (b) 81 (c) 85 (d) 82 In DABC. so that the probability of having atleast one head is more than 80% ? (a) Atleast 3 (b) Atleast 5 (c) Atmost 3 (d) Atmost 5 28. which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0. 3) → (0. y. 3 (a) (i). then (I + A)3 – 7A is equal to (a) A (b) I – A (c) I (d) 3A 22 MatheMatics tODaY | MARCH ’15 x y (b) e = x + C (d) ey = x + C 26. then which one 23. (a) 51x – 15y + 50z + 173 = 0 (b) 51x + 15y – 50z + 173 = 0 (c) 15x + 15y + 50z + 173 = 0 (d) none of these 27. (ii) (b) (ii). If f (x ) = if x > 0 4 x + 1. 3) (b) (3. 4 −π (iii) The principal value of tan–1 (− 3 ) is . (iii) (i) The principal value of cos–1 A2 21. p] is 2 (c) two (d) three (a) zero (b) one 18.13. if 8R2 = a2 + b2 + c2. 3) (c) 30 at (3. 3) . is (a) (1. The probability of the birth dates of all 6 persons to fall in only two different months is 341 341 341 541 (a) (b) (c) (d) 6 5 4 12 12 12 126 15. 0) (d) (0. 0) (d) 95 at (2. of the following is correct ? (a) f(x) is continuous at x = 0 for any value of l (b) f(x) is discontinuous at x = 0 for any value of l (c) f(x) is discontinuous at x = 1 for any value of l (d) None of the above 24. f –1(x) equals to (a) x – 2 (b) x + 1 (c) x – 1 (d) x + 2 20. is (a) isosceles (c) equilateral (b) right angled (d) scalene 17. if x ≤ 0 . 4). Maximize Z = 10x1 + 25x2. subject to 0 ≤ x1 ≤ 3. 1) be defined by f(x) = x – [x]. Solution of the differential equation x y x y ye dx = (xe + y 2 )dy ( y ≠ 0) is x y ]dx is equal to (a) e = y + C (c) ex + y + C 0 (a) 2 − 2 (b) 2 + 2 (c) (d) − 2 − 3 + 5 2 −1 x sin θ cos θ 1 is 22. 1) (c) (1. 2 ∫ [x 2 (a) (b) (c) (d) independent of q only independent of x only independent of both q and x none of the above λ(x 2 − 2 x ). How many times must a man toss a fair coin. then the triangle 16. 2 6 π –1 (ii) The principal value of cosec (2) is . 0) and (4. If A is a square matrix such that = A. 2) (b) 75 at (0. The probability that there are exactly four letters in between A and B is 5 7 1 1 (a) (b) (c) (d) 66 66 11 22 14. (ii). Which of the following is/are true? 3 π is . 2 2 dy dy + 1+ . If three of these numbers are –1. g(x) = and h(x) = logex. 3 (d) 2. dx dx (c) 1. (a) ±(2i − 4 j + 4k ) (b) ±(i − 2 j + 2k ) (c) ±(i − j + k ) (d) ±(i + j + k ) 43. For all n ∈ N.29. b)S(c. then x is equal to (a) {2. Evaluate tan x ∫ sin x ⋅ cos x dx (a) 3 tan x + c (b) tan x + c tan x +c 2 39. is x →α (a) (α − β)2 2 2 (a) 1 (b) (a – b)2 2 2 2 31. Number of ways in which 3 boys and 3 girls (all are of different heights) can be arranged in a line so that boys as well as girls among themselves are in decreasing order of their heights (from left to right). 1 and 2. Using cofactors of elements of third row. 2} (d) {1. then the equation of the ellipse is x2 y2 x2 y2 1 + = + =1 (a) (b) 25 10 50 100 (c) x2 y2 + =1 5 5 (d) (c) 0 (b) –1 (d) 2 x2 a (α − β) α (a − b) (d) 2 2 30. –3} (c) {–3. If F(x) = (hogof)(x). If the sum is even. b = 4 i − 2 j + 3 k and ^ ^ ^ c = i − 2 j + k . Two integers are selected at random from integers 1 to 11. 2} MatheMatics tODaY | MARCH ’15 23 . 3 40. 1 4 2 3 (a) (b) (c) (d) 5 5 5 5 41. then find a vector of magnitude 6 units which is parallel to the vector 2a − b + 3c . then F″(x) is equal to (a) a cosec3x (b) 2cotx2 – 4x2cosec2x2 2 (c) 2x cot x (d) –2 cosec2x 38. then the other two numbers are (a) – 5 and 3 (b) – 4 and 2 (c) – 3 and 1 (d) – 2 and 0 (c) 36. 41n – 14n is a multiple of (a) 26 (b) 27 (c) 25 (d) none of these 34. If the trace of the matrix 0 2 5 x − 1 3 2 4 1 x −2 A= 1 −2 x −3 −1 2 2 0 4 x − 6 is 0. if defined y = x (a) 1. 2 then x2 + xy + y2 is equal to 3 1 1 (a) 0 (b) (c) (d) 2 8 2 37. The relation S defined on the set N × N by (a. where (x − α)2 a and b are the roots of the equation ax2 + bx + c = 0. If the latus rectum of an ellipse with axis along x-axis and centre at origin is 10. then find the probability that both the numbers selected are odd. is (a) 6! (b) 3! × 3! × 2! (c) 10 (d) 20 π 35. 1 z x+y x2 y2 + =1 100 50 33. 3} (b) {–2. then x = (a) 102 (b) 103 (c) 10 (d) 104 32. If a = i + j + k. The mean of five numbers is 0 and their variance is 2. 1 x y+z evaluate ∆ = 1 y z + x . If cos −1 x + cos −1 y = and tan–1 x – tan–1 y = 0. If the third term in the expansion of 5 1 log10 x x + x is 1000. The value of lim 1 − cos(ax 2 + bx + c) . distance between foci = length of minor axis. Let f(x) = sinx. Find the order and degree of the differential (c) 2 tan x + c (d) equation. 2 (b) 2. d) ⇒ a + d = b + c is an (a) equivalence relation (b) reflexive but not symmetric (c) only transitive (d) only symmetric ^ ^ ^ ^ ^ ^ 42. . We first have to count the permutations of 6 digits taken 3 at a time. d.. 6 ! 5! So.(iv) On putting the values of c and d from (iv) and (v) in (ii). This number is 5P2.(ii) . (a) : (a2 – 1)x2 + 2(a – 1)x + 2 is positive for all x. then the maximum volume of the box is (a) c3 6 3 (b) c2 c3 6 3 sOlutiOns (c) (d) c3 3 3 x x 1.(i) . these permutations will include those also where 0 is at the 100th place.(i) . (c) : Let Q(h.19 times)(x) = x −1 2ac 2. (c) : Q a.. \ b = a+c b.... we have 1 r sin(θ − α) − sin(θ − α) = 0 r ⇒ r2 – 1 = 0 ⇒ r = 1 ⇒ |z – 1| = 1 or. (2x + 1) (x – 3) (x + 7) < 0 1 when x < – 7 or − < x < 3 2 1 \ x ∈ (– ∞. . If f(x) defined by the following is continuous at x = 0. \ d = 2 From (i). (a) : (2x + 1)(x – 3)(x + 7) = 0 1 ⇒ x = – . 2 2 2 2 2 2 2 45. 3 . But. 2 7..P. (d) : Every number between 100 and 1000 is a 3-digit number. (c) : Let z – 1 = r(cosq + isinq) = reiq 1 \ Given expression = reiθ ⋅ e −iα + iθ ⋅ eiα re Clearly. d are in G. 1 (b) (c) 3 . we fix 0 at the 100th place and rearrange the remaining 5 digits taking 2 at a time. d = + e (Using (iii) and (iv)) .. we get 2 1 = rei(θ− α) + e −i(θ− α) r Since. (a) : ( fof )x = f = x −1 = x x − 1 x x − − 1 1 x ⇒ (fofof)x = f(fof)x = f(x) = x −1 x \ (fofof.44. This number would be 6P3. \ c2 = bd c+e and c.(iii) . – 7) ∪ − .P. b.(v) 2 2a − b a 2 b2 3.. the required number = 6P3 − 5P2 = − 3! 3! = (4 × 5 × 6) – (4 × 5) = 100 8. ab + bc = 2ac ab ⇒ c= 2a − b b ab = + e a b 2 2 − (2a − b) ⇒ e= ab2 (2a − b)2 24 MatheMatics tODaY | MARCH ’15 4. sin(q – a) = 0 ⇒ q – a = 0 ⇒ q = a ⇒ arg(z – 1) = a 5. 1 (d) − . if a2 – 1 > 0 and 4(a – 1)2 – 8(a2 – 1) < 0 ⇒ a2 – 1 > 0 and (a – 1)(a + 1) > 0 ⇒ a < –1 or a >1 6. k) be the image of the point P(1. –7 2 . if x > 0 3 2 bx 1 1 1 1 (a) − 3 . An open box with a square base is to be made out of a given quantity of metal sheet of area c2. 2) in the line x – 3y + 4 = 0 . if x = 0 2 x + bx − x .P.. then the values of and c are sin(a + 1)x + sin x ... if x < 0 x f (x ) = c ... e are in A. c are in H. c. (a) 1 ab Also.. To get the number of such numbers. 3. imaginary part of given expression is zero... \ A and B can take the following places. (a) : Let a = a1 i + a2 j + a3 k. the number of total outcomes is 126..... (b) : Since the birth date of any person can fall in anyone of the 12 months.. (b) : lim 2 sin x − sin 2 x x →0 = lim x3 (2 x )3 (2 x )5 x3 x5 + − ... Therefore required 14 7 probability is = 11 × 12 66 14. | a × i |2 + | a × j |2 + | a × k |2 0 P(1.(ii) h +1 k +2 −3 + 4 = 0 or h – 3k = –3 . − 0 + . Since we want 4 letters in between A and B. 2 2 6 7 Solving (ii) and (iii). (c) : A a n d B c a n b e a r r a n g e d i n 1 2 P 2 = 11 × 12 ways.(iii) Also... we get h = and k = .. (b) : Since x + y + z = 1.. = − + = = 1 3! 3! 6 6 6 11.. 2 x − 3! 5! 3! 5! x3 x →0 = lim 8 x 3 32 x 5 −2 x 3 2 x 5 + − .... y = .. Then. k) x y+ –3 4= Therefore... Let E : Event that the birth dates of all 6 fall in two different months. 3! 5! 5! 3! x →0 x 3 2 2 8 6 8 + 0 − . we have =− i j a × i = a1 a2 1 0 k a3 = a3 j − a2 k 0 13. (b) : Given.Y Q(h.. 2) X = (a22 + a32 ) + (a12 + a32 ) + (a12 + a22 ) = 2(a12 + a22 + a32 ) = 2 | a |2 X O Y Hence. (c) . | a × i |2 = a22 + a32 Similarly. we have 4 9 16 4 9 16 + + = (x + y + z ) + + x y z x y z 4 z 16 x 9z 16 y 4 y 9 x = (4 + 9 + 16) + + + + + + x z y z x y ≥ 29 + 2 64 + 2 144 + 2 36 = 81 2 3 4 Equality holds if and only if x = . slope of line PQ −1 = Slope of line x − 3 y + 4 = 0 so that k − 2 −1 or 3h + k = 5 = h −1 1 3 12. z = 9 9 9 MatheMatics tODaY | MARCH ’15 25 . 14 − a 9 − a The equation will represent an ellipse if 14 – a > 0 and 9 – a > 0 ⇒ a < 14 and a < 9 ⇒ a < 9 a hyperbola if 14 – a > 0 and 9 – a < 0 ⇒ – a > –14 and a > 9 ⇒ 9 < a < 14 10. − + . Place for A 1 2 3 4 5 6 7 Place for B 6 7 8 9 10 11 12 A and B can be interchanged. − 2 x − + − . 5 5 2 2 y x + =1 9.. | a × j |2 = a12 + a32 and | a × k |2 = a12 + a22 Hence.. P (E ) = 12 C2 (26 − 2) 12 6 = 66 × 62 12 6 = 11 × 31 12 5 = 341 125 15.. (b) : 8R2 = 4R2(sin2 A + sin2B + sin2C) 1 − cos 2 A 1 − cos 2 B 1 − cos 2C ⇒ 2= + + 2 2 2 ⇒ –1 = cos2A + cos2B + cos2C = –1 – 4cosAcosBcosC Hence. 100 0 n 80 8 1 1 \ 1 − P ( X = 0) > ⇒ 1 − nC0 > 100 2 2 10 1 4 1 1 ⇒ 1− n > ⇒ . X is the number of heads. p] 2 1 0 0 18. According to question. (b) : Since (x + 2y + 3z – 4) + l(2x + y – z + 5) = 0 ⇒ x(1 + 2l) + y(2 + l) + z(3 – l) – 4 + 5l = 0 . where.(i) This is perpendicular to the plane 5x + 3y + 6z + 8 = 0 \ 5(1 + 2l) + 3(2 + l) + 6(3 – l) = 0 29 ⇒ 7l + 29 = 0 ⇒ λ = − 7 On putting l = –29/7 in (i). cosA cosB cosC = 0 So one of the angles of the triangle is a right angle.(i) = 0 ⇒ x + y = −2 5 .. (i) < ⇒ 2n > 5 n 5 5 2 2 Inequality (i) is satisfied for n ≥ 3. (a) : Let ∆ = − sin θ − x cos θ 1 x ⇒ ∆=x −x 1 − sin θ 1 − sin θ − x − sin θ + cos θ 1 x cos θ x cos θ 1 26 MatheMatics tODaY | MARCH ’15 = x(–x2 – 1) – sinq (–x sinq – cosq) + cosq (–sinq + x cosq) = –x3 – x + x sin2q + sinq cosq – sinq cosq + x cos2q = –x3 23.. 80 P(X ≥ 1) > . (d) 29. −1 + 1 + 2 + x + y . (a) : Since. (d) : The given function is f : (2. (b) : Slope of chord joining (2. 1). (b) 24. 1) defined by f(x) = x –[x] Let y ∈ (0. (a) 26. coin must be tossed 3 or more times. (d) : Let the other two numbers be x and y. (c) 21... the required point is (3. (a) : Let man tosses the coin n times. (c) : Here. Hence. (d) 30. 0) and (4. we have the equation of the required plane as 145 58 29 29 x 1 − + y 2 − + z 3 + − 4 − =0 7 7 7 7 ⇒ 51x + 15y – 50z + 173 = 0 27.16.(i) = = =2 x2 − x1 4 − 2 2 Equation of given curve is y = (x – 2)2 dy ⇒ = 2(x − 2) dx Now.. 1 + sin x sin2 = 0 2 1 − cos x ∴ 1 + sin x = 0 2 ⇒ 2 + sinx – sinx cosx = 0 ⇒ sin 2x – 2 sinx = 4 which is not possible for any x in [–p. from (i). 25.. 28. (d) : ∫ [x 2 ]dx = ∫ [x 2 ]dx + + 3 2 ∫ [x 2 ]dx 0 ∫ [x 2 ]dx + 2 1 = ∫ 0 dx + 0 2 3 1 2 2 ∫ [x 2 ]dx 3 2 ∫ 1dx + ∫ dx + ∫ 3 dx 3 = 0 + 1( 2 − 1) + 2( 3 − 2 ) + 3(2 − 3 ) = 5− 3 − 2 19. x 17. 4) is y2 − y1 4 − 0 4 . A2 = A \ (I + A)3 – 7A = I3 + A3 + 3I2 A + 3A2I – 7A = I + A3 + 3A + 3A2 – 7A = I + A·A – A = I + A – A = I (Q A2 = A) x sin θ cos θ 1 22. we get 2(x – 2) = 2 ⇒ x = 3 ⇒ y = (3 – 2)2 = 1 Hence.. 1) be such that y = x – 2 {Q 2 < x < 3 ⇒ [x] = 2} ⇒ x = y + 2 ⇒ f –1(x) = x + 2 20. 3) → (0.. 41k – 14k = 27l . (a) : Given. F = Event that the sum of chosen number is even in integers from 1 to 11. (a) : y = x + 1 + dx dx 2 ⇒ dy dy y−x = 1+ dx dx 2 2 dy dy (x 2 − 1) − 2 xy + y 2 − 1 = 0 dx dx (squaring on both sides) dy which represents a quadratic polynomial in . dx \ Order of the differential equation = 1 and degree of the differential equation = 2 40 (d) : Let E = Event of selecting both odd numbers. Therefore. (c) dy dy 39. (d) : Since order of boys and girls are to be maintained in any of the different arrangements. i. Hence. 6×5 6 C2 6×5 3 2 ×1 ∴ P (E ) = = = = 11 × 11 10 × 11 10 11 C2 × 2 1 and P(F) = P(both numbers are even) + P(both numbers are odd) 6×5 5× 4 6 5 2 ×1 2 × 1 C2 C2 5 = + = + = 11 11 C2 C2 11 × 10 11 × 10 11 × 2 × 1 2 1 ⇒ 6 × 5 2 × 1 3 = and P (E ∩ F ) = 11 = 11 10 × 11 C2 2 × 1 E P (E ∩ F ) 3 = \ Required probability = P = F P (F ) 5 41. 3!3! ⇒ 2t – 3 = 35. (i) For n = k + 1. (x – y) = x + y – 2xy = 4 – 0 = 4 ⇒ x – y = ±2 . (c) 38. y = 0 (from (i) and (iv) 31.e. the statement is true for all natural numbers n..(iv) ⇒ x = 0. (a) nn MatheMatics tODaY | MARCH ’15 27 . (a) : Let the vector be r = λ(2a − b + 3c ) ^ ^ ^ ^ ^ ^ ^ ^ ⇒ r = λ(2 i + 2 j + 2 k − 4 i + 2 j − 3 k + 3 ^i − 6 j + 3 k) ^ ^ ^ ⇒ r = λ( i − 2 j + 2 k ) ^ ^ ^ ∴ |r | = ± λ | i − 2 j + 2k | 6 C2 ⇒ ± λ 1+ 4 + 4 = 6 ⇒ ± λ ⋅3 = 6 ⇒ λ = ± 2 ^ ^ ^ ^ ^ ^ Therefore.. P(1) = 411 – 141 = 27 = 1 × 27 which is a multiple of 27 \ P(1) is true.(iii) 2 2 2 Now.. (c) 37. 41k+1 – 14k+1 = 41k41 – 14k14 = (27l + 14k) 41 – 14k14 [using (i) ] = (27l × 41) + (14k × 27) = 27(41l + 14k) which is a multiple of 27... neglected) \ log10x = 2 ⇒ x = 102 = 100 32.. Let P(k) be true i. (a) ⇒ x = −3. from the principle of mathematical induction.. here 5 even and 6 odd integers. (a) 42. s2 = 2 (−1 − 0)2 + (1 − 0)2 + (2 − 0)2 + (x − 0)2 + ( y − 0)2 =2 ⇒ 5 ⇒ 1 + 1 + 4 + x2 + y2 = 10 ⇒ x2 + y2 = 4 .e. (d) 33. 34. y = –2 or x = –2. T3 = 1000 \ 1 C2 x 5 3 (x log10 x ) = 1000 2 2 log x ⇒ 10 x– 3 × x 10 = 1000 ⇒ (2 log10 x − 3) = log x 102 ⇒ (2log10x – 3) = 2 log10 x 2 . (d) 36.. (b) : Let P(n) be the statement given by P(n) : 41n – 14n is a multiple of 27. r = ± 2 ( i − 2 j + 2 k) = ± (2 i − 4 j + 4 k) 43. For n = 1. (c) : Trace of matrix is defined as n ∑ aii = 2x 2 + 2x − 12 = 0 i =1 44. 2 45.Also.(ii) ⇒ (x + y)2 – 2xy = 4 ⇒ xy = 0 .. P(k+1) is true when P(k) is true. 6! \ Required number = = 20. where t = log10x t ⇒ (2t + 1) (t – 2) = 0 ⇒ t = 2 (t ≠ – 1/2.. Find the value of x by evaluating the given series 1 1× 3 1× 3 × 5 1+ + + + . Inside a square ABCD points P and Q are positioned so that DP || QB and DP = PQ = QB. r19 r29 r39 r49 r59 4... i Thus n = 1 2. ∞ = x . we find z 1 z Rearranging. Of all configurations that satisfy these requirements. we get 1 1 1 iz 2 (z − 1) = z + 1 + + + + . Sarvapriya Vihar... n lim ∑ r3 − 8 n→∞ r =3 r 3 +8 iz 2 (z − 1) = y=x O 9. 5. what is the minimum possible value of ∠ADP. 50-C. sOlutiOns 1.. Multiplying both sides of the equation by z. Given that 3.... r2. Notice that each term is of the following form: n ∏ (2k − 1) = 7. 2 3 z z z z4 and z = n ± −i . Find the sum of all (distinct) complex values of c for which the polynomial fc(x) = x4 – (c2 – 7c + 11)x2 + (18 – 21c + 8c2 – c3) has strictly less than four distinct complex zeroes. Let P be a point (other than the origin) lying on the parabola y = x2.. Find the value of 1 1 1 1 1 + + + + . 9).2 3 4 5 iz 2 = 1 + + + + + . Find the value of 2nπ −1 ∫0 max{sin x .... 2. Kalu Sarai. The minimum possible value for the area bounded by the line PQ and the parabola is 1/ x 1 10.. Find the value of lim ∫ (by + a(1 − y ))x dy x →0 0 where b > a. New Delhi-16 28 MatheMatics tODaY | MARCH ’15 n = 1 2n 10n n . r3. z z2 and subtracting the original equation from this one.. one anywhere on the y-axis and one anywhere on the line y = x is y 6.. we get 3 4 iz 3 = z + 2 + + + . Draw the graph of k =1 n ∏ (5k ) n k =1 n ∏ (2k − 1) ⋅ ∏ (2k ) k =1 = k =1 n n (n !)5 ∏ (2k ) = (2n)! n (n !)5 (n !)2 k =1 n Here = nCr r Whizdom Educare. The minimum possible perimeter of a triangle with one vertex at (3. 2 z z z3 Using the formula for an infinite geometric series... If r1. we get x = 1− z2 z −1 1 iz 2 (z − 1)2 = z 2 ⇒ (z − 1)2 = ⇒ z = 1 ± −i . x ∈Q 5 5 × 10 5 × 10 × 15 1. (in degrees)? 8.. Find the value of n. the same. r5 are the complex roots of the equation x5 – 3x4 – 1 = 0.. r4.. The normal line to the parabola at P will intersect the parabola at another point Q.sin (sin x )}dx. 1 2n n n=0 10 n Now since the result is the square root of a rational number, let’s find s2 . Using the Cauchy Product (with 1/10 as the independent variable), we get the following formula. ∞ 1 n 2k 2(n − k ) s2 = ∑ ∑ n n=0 10 k =0 k n − k ∞ Hence we need to find s = ∑ Now it can be shown that for all whole numbers n we have n 2k 2(n − k ) n ∑ =4 k =0 k n − k Hence, we have ∞ n ∞ 2 1 5 = 4n = ∑ = = n 2 3 n=0 10 n=0 5 1− 5 Thus, x = 5/3 3. From the given polynomial, we have s2 = ∑ 1 ∑ ri = 3 and −1 ∑ ri =0 The sum of the reciprocals of the roots come from the fact that the polynomial with reciprocal roots has its coefficients reversed. x5 – 3x4 – 1 = 0 ⇒ x4(x – 3) = 1 and this happens for two distinct values of c which sum to 5. Thus the sum of the possible values of c is 2 + 3 + 5 = 10. 5. Let A be the vertex (3, 9), B be the vertex on the y-axis and C be the vertex on the line y = x. Also let D (–3, 9) be the reflection of A in the y-axis and E(9, 3) be the reflection of A in the line y = x. Then AB = BD and AC = CE, and thus the perimeter of DABC is equal to DB + BC + CE. But the shortest distance between two points is a straight line, so DB + BC + CE ≥ DE = (9 − (−3))2 + (3 − 9)2 = 180 = 6 5 This minimum can be obtained by then choosing B and C as the points of intersection of the line DE with the y-axis and the line y = x, respectively. This 15 gives us the points B 0, and C(5, 5). This will 2 yield a perimeter for DABC of 6 5. 6. 33 − 8 43 − 8 n3 − 8 lim ....... n→∞ 33 + 8 43 + 8 n3 + 8 3 − 2 32 + 4 + 2(3) 4 − 2 42 + 4 + 2(4) n−2 n = lim . . ...... . n+2 n (x − 3) n→∞ 3 + 2 32 + 4 − 2(3) 4 + 2 42 + 4 − 2(4) − 9 4 –1 x = (x – 3) ⇒ x = x 2 4 − 2 42 + 4 + 2(4) n − 2 n2 + 4 + 2(n) 2 2= lim 3− 2 . 3 + 4 + 2(3) . ...... . (r − 3) r − 6ri +9 −9 = ∑ i n→∞ ∑ ri = ∑ i 3+ 2 32 + 4 − 2(3) 4 + 2 42 + 4 − 2(4) n + 2 n2 + 4 − 2(n) ri r i n − 2 3 − 2 4 − 2 5 − 2 −1 = lim . . ......... = ∑ (ri − 6 + 9ri ) = ∑ ri − ∑ 6 + 9∑ ri−1 n + 2 n→∞ 3 + 2 4 + 2 5 + 2 = 3 –5(6) + 0 = –27. 32 + 4 + 2(3) 42 + 4 + 2(4) n2 + 4 + 2(n) . ......... 4. The polynomial fc(x) will fail to have four 2 2 n2 + 4 − 2(n) 3 + 4 − 2(3) 4 + 4 − 2(4) distinct complex zeroes when the quadratic polynomial 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6.... 19 ⋅ 28 ⋅ 39 ⋅ 52 ⋅ 63.... = lim gc(x) = x2 – (c2 – 7c + 11)x + (18 – 21c + 8c2 – c3) n→∞ 5 ⋅ 6 ⋅ 7 ⋅ 8.... 7 ⋅ 12 ⋅ 19 ⋅ 28 ⋅ 39 ⋅ 42 ⋅ 63.... either has repeated roots or has 0 as a root. 2 2 Case1: One of the roots of gc(x) is zero precisely = lim (n + 5)(n + 2n + 4)(1 ⋅ 2 ⋅ 3 ⋅ 4) 1 = 2 7 ⋅12 7 (n − 1)n(n + 1)(n + 2) n→∞ when 3 2 2 c – 8c + 21c – 18 = (c – 2)(c – 3) = 0 7. Without loss of generality let the corners of the and so precisely when c = 2, 3 square be Case 2 : The roots of gc(x) are repeated when its A(0, 2), B(2, 2), C(2, 0) and D(0, 0). Now let point discriminant is zero, so that P have coordinates (a, b); then by symmetry the (c2 – 7c + 11)2 – 4(18 – 21c + 8c2 – c3) = 0 coordinates of point Q must be (2 – a, 2 – b). Then c4 – 10c3 + 39c2 – 70c + 49 = 0 since DP = PQ, we have that (c2 – 5c + 7)2 = 0 a2 + b2 = (2 – 2a)2 + (2 – 2b)2 2 MatheMatics tODaY | MARCH ’15 29 ⇒ 3a2 + 3b2 – 8a – 8b + 8 = 0 2 2 4 4 8 ⇒ a − + b − = 3 3 9 This means that P lies on a circle centered at O 4 4 2 2. 3 , 3 with radius r = 3 Therefore, ∠ADP will be minimized when DP is tangent to this circle. Now by symmetry OD makes an angle of 45° with 4 the x-axis and has length 2. 3 Thus ∠PDO = 45° – ∠ADP, and so sin(∠PDO) = sin(45° – ∠ADO) 2 2 OP 3 1 = = = OD 4 2 2 3 This implies that 45° – ∠ADP = sin–1(1/2) = 30°, and so the minimum value for ∠ADP is 45° –30° = 15° y 2 1 1 P2 = − − x0 , − − x0 2 x0 2 x0 Hence the area bounded by P1P2 and the parabola is x0 2 1 x0 + − x − x 2 dx ∫ 1 2 − 2 x + x0 0 (using the equation of normal) Evaluating it we get A = 4 1 x0 + 3 4 x0 3 By applying AM-GM inequality we get 1 x0 + 3 4 x0 1 1 ≥ ⇒ x0 + ≥1 2 4 4 x0 4 Finally Amin = 3 10. L = lim (∫01 (by + a(1 − y ))x dy )1/ x I x →0 = ∫01 (by + a(1 − y ))x dy Let by + a(1 – y) = t, (b – a)dy = dt 8. x O The integral can be divided as π = n[∫0π/2 xdx + ∫π/2 (π − x )dx + ∫π2 π sin xdx] π2 π2 1 π2 = n + − × π − − 2 2 2 4 8 π2 − 8 Solving we get, I = n × 4 9. We take a point P1 = (x0, x02) on the parabola. Then slope of tangent is = 2x0 −1 Hence slope of normal is 2 x0 So equation of normal is (x – x0) = –2x0(y – x02) Solving it with the parabola we get 1 x = x0 , − − x0 2x 0 So the other point is 30 MatheMatics tODaY | MARCH ’15 I=∫ tx t x +1 dt = +C (b − a) (x + 1)(b − a) (by + a(1 − y ))x +1 I= computed at y = 1 and (x + 1)(b − a) y=0 1/ x b x +1 − a x +1 b x +1 − a x +1 , L = lim I= (x + 1)(b − a) x →0 ( x + 1)(b − a) (b x +1 − a x +1 ) ln (x + 1)(b − a) ln(L) = lim x x →0 As x approaches 0, denominator and numerator approaches 0. Hence, we can use L-hospital’s rule b x +1 ln(b)a x +1 ln(a) 1 ln(L) = lim − (x + 1)(b − a) x +1 x →0 1 bb 1 bb (b −a) 1 ln(L) = ln − 1 ⇒ L = e aa b − a aa nn *ALOK KUMAR, B.Tech, IIT Kanpur sectiOn-i Single CorreCt AnSwer type this section contains multiple choice questions. each question has 4 choices (a), (b), (c) and (d), out of which only one is correct. 1. The sum of all positive integral values of a, a ∈ [1, 500] for which the equation [x]3 + x – a = 0 has solution is ([.] denote G.I.F) (a) 462 (b) 512 (c) 784 (d) 812 2. Let P(x) be a polynomial with degree 2009 and leading co-efficient unity such that P(0) = 2008, P(1) = 2007, P(2) = 2006, …, P(2008) = 0 and the value of P(2009) = (n!) – a where n and a are natural number, then value of (n + a) is _____ (a) 2010 (b) 2009 (c) 2011 (d) 2008 3. If circumcentre of an equilateral triangle inscribed in x2 y2 = 1 with vertices having a 2 b2 eccentric angles a, b, g respectively is (x1, y1), then ∑ cos a ⋅ cos b + ∑ sin a ⋅ sin b is 9 x12 + 9 y12 3 (a) + − 2 2 2 2a 2b (c) x12 y12 5 + − 2 2 9 9a 9b x12 y12 5 + − (b) 2a2 2b2 2 (d) x12 y2 1 + 1 − a 2 b2 2 4. Let f : [0, ∞) → R be a continuous and strictly increasing function such that x f 3 (x ) = ∫ t ⋅ f 2 (t )dt , ∀ x > 0 . The area enclosed by 0 y = f (x), the x-axis and the ordinate at x = 3, is (a) 1 (b) 3/2 (c) 2 (d) 3 5. p(x ) ; x ≠2 Consider the function f (x ) = x − 2 7 ; x =2 where P(x) is a polynomial such that P′′′(x) is identically equal to 0 and p(3) = 9. If f(x) is continuous at x = 2, then p(x) is (a) 2x2 + x + 6 (b) 2x2 – x – 6 2 (c) x + 3 (d) x2 – x + 7 6. A particle starts to travel from a point P on the curve C1 : |z – 3 – 4i| = 5, where |z| is maximum. From P, the particle moves through an angle tan–1(3/4) in anticlock wise direction on |z – 3 – 4i| = 5 and reaches at point Q. From Q, it comes down parallel to imaginary axis by 2 units and reaches at point R. Complex number corresponding to point R in the Argand plane is (a) (3 + 5i) (b) (3 + 7i) (c) (3 + 8i) (d) (3 + 9i) 7. 6 If 15 = 2P1 3P2 5P3 7 P4 11P513P6 , then ∑ Pr is r=1 (a) 24 (b) 23 (c) 22 (d) 21 1 1 f (2 x 2 − 1) + f (1 − x 2 ) ∀ x ∈ R, 4 2 where f ′′(x) > 0 ∀ x ∈ R, g(x) is necessarily increasing in the interval 2 2 (a) − , 3 3 8. Let g (x ) = 2 2 − 3 , 0 ∪ 3 , ∞ (c) (–1, 1) (d) None of these (b) *Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants. MatheMatics tODaY | march ’15 31 9. Let A = {x1, x2, x3, ...., x7}, B = {y1, y2, y3}. The total number of functions f : A → B that are onto and there are exactly three elements x in A such that f (x) = y2, is equal to 14·7C (a) 2 (c) 7·7C2 (b) (d) 14·7C3 7·7C3 6 ∫ ( x + 12 x − 36 + x − 12 x − 36 )dx is equal to 3 (b) 4 3 (c) 12 3 (d) 2 3 11. The number of the positive integer pairs (x, y) 1 1 1 such that + = where x < y is x y 2007 (a) 5 (b) 6 (c) 7 (d) 8 12. Let f (x) = x2 + lx + mcosx, l being an integer and m a real number. The number of ordered pairs (l, m) for which the equations f (x) = 0 and f (f (x)) = 0 have the same (non empty) set of real roots is (a) 4 (b) 6 (c) 8 (d) infinite 13. a > 0 (a ≠ 1), b > 0 (b ≠ 1) such that x x a(loga b) = b(logb a) , then x = (a) 1 (b) –1 (c) 1/2 (d) 2 3 2 + x , −3 ≤ x < −1 14. If the range of f (x ) = 2/3 −1 ≤ x ≤ 2 x , 3 is [0, n ] where n ∈ N, then n = (a) 1 (b) 2 (c) 4 (d) 6 15. A variable straight line with slope m(m ≠ 0) intersects the hyperbola xy = 1 at two distinct points. Then the locus of the point which divides the line segment between these two points in the ratio 1 : 2 is (a) an ellipse (b) a hyperbola (c) a circle (d) a parabola 2 16. If ∫ (2 − 3 sin x ) sec x dx = 2 f (x ) g (x ) + c and f(x) is non constant function, then (a) f 2(x) + g2(x) = 1 (b) f 2(x) – g2(x) = 1 (c) f(x) g(x) = 1 (d) f (x) = g(x) 32 MatheMatics tODaY | march ’15 23571113 is (a) 2385 (b) 2835 18. 10. The value of (a) 6 3 17. The number of ordered pairs of positive integers (a, b) such that L.C.M of a and b is (c) 3825 (d) 8325 (−1)K −1 10 .( CK ) = K K =1 10 ∑ 1 1 1 1 (a) 1 + + + + ...... + 2 3 4 11 1 1 1 1 (b) 1 + + + + ...... + 2 3 4 10 1 1 1 1 (c) 1 + + + + ...... + 2 3 4 9 1 1 1 1 (d) 1 + + + + ...... + 2 3 4 12 19. The numbers a, b are chosen from the set {1, 2, 3, 4, ....., 10} such that a ≤ b with replacement. The probability that a divides b is 5 29 (a) (b) 11 55 27 (c) (d) None of these 55 20. The area of the loop of the curve y2 = x4(x + 2) is [in square units] 64 2 32 2 (a) (b) 105 105 128 2 256 2 (c) (d) 105 105 ∞ 21. The sum ∑ k k =1 (3 (a) 1 (b) 2 6k = − 2k )(3k +1 − 2k +1 ) (c) 3 (d) 4 22. The coefficient of x6 in the expansion of 2 x x2 x3 x 4 x5 + + + is 1 + + 1 2 3 4 5 4 31 2 (a) (b) (c) 15 360 15 23. (a) p 2 x(1 + sin x ) −p 1 + cos2 x ∫ p2 4 (b) p2 (d) 2 45 (d) p 2 dx = (c) 0 A hyperbola does not have mutually perpendicular tangents. then integer function) and an+1 = (n + 1) lim (an ) is n→∞ (a) 1 (b) e2 (c) ln2 (d) ln3 x 26. f (x). Then which of the following statement(s) is/are correct? (a) g(1) + h(1) = 70 (b) g(1) – h(1) = 28 (c) g(1) + h(1) = 60 (d) g(1) – h(1) = –28 p→∞ 34. g(x) and h(x) are all continuous function at x = 1. If f (x ) = ∫ dt 1 2+t (a) (c) 1 f (2) < 3 1 f (2) = 3 4 f (2) > 1 3 (d) f (2) > 1 (Multiple CorreCt AnSwer type) this section contains multiple correct answer(s) type questions. (b).. 52. Line AP cut BC at M. such that y ≤ 16 − x y p tan −1 ≤ is x 3 16 (a) p 3 sectiOn-ii 31. then (b) 30. If a. unit then area of DABC is (a) 20 sq. If a . b. (b) G. 0 such that 2 sin a sin a (sin a + sin b) + = 0 . (sin a + sin b) = −1 sin b sin b and l = lim n→∞ 27.units (d) 10 sq. b . If a1 is the greatest value of f (x). b. then (b) l = 2 (d) l = 1 33. DAPB are equal. Area of the region in which point P(x. (c) and (d). Considering the following figure. e are in H. Let A = {12. out of which one or More is/are correct.24.units 2 and 8p (b) + 8 3 3 (c) (4 3 − p) (d) ( 3 − p) 28. x ≠ 1 and 7 x p + 3x + 1 f (1) = 7. . d.P.P..units (c) 30 sq. and c. (x > 0) lies. If 9 elements are selected from set A to make a 3 × 3 matrix. P is any point inside a triangle such that area of DBPC. then a. Which of the following quantities would vary as a varies ? (a) Eccentricity (b) Ordinate of the vertex (c) Ordinates of the foci (d) Length of the latus rectum + MatheMatics tODaY | march ’15 33 ... each question has 4 choices (a).units (b) 25 sq. then its eccentricity can be (a) 7/6 (b) 7/5 (c) 3/2 (d) 2 . e are in (a) A. then det(A) will be divisible by (a) 9 (b) 36 (c) 8 (d) 64 1 + (2 sin a)2n (2 sin b)2n p 6 p (c) a = − 3 (a) a = − . DAPC.] denotes greatest 2 + [sin x] (−1)n+2 + an . a is equal to (a) (c − 2b ) −1 (c − 2b ) (b) 3 1 (c) (c − 2b ) 3 (d) None of these 29. (c) H. y). c. p 32. d are in G.P.P. Consider the ellipse x2 y2 = 1 . where tan2 a sec2 a a ∈ (0. c are the position vectors of points on the circumference of a circle having centre at origin.}. c.P. (d) none of these 25. area of DPMC is 5 sq. If a. p/2).P. c are in A. If f (x ) = lim x p g (x ) + h(x ) + 7 . b ∈ − . In DABC. 32. where 1 f (x ) = (where [. DC = 8 cm. 5] R ∩ R′ defines a function | x − 3| . Based upon each paragraph. then (a) the triangle is necessarily acute angled D A 4 (b) tan = 2 7 (c) perimeter of the triangle ABC is 42 cm (d) Area of DABC is 84 cm2 40. 1) (c) (5. 46 to 48 Define a function f : N → N as follows : f(1) = 1. f(Pn) = Pn – 1(P – 1). c are in G. if p is prime and n ∈ N. multiple choice questions have to be answered. Tr = . y ≥ x 2 . then which of the following is/are always true ? (a) f (x ) = 0. y) : x. If log 2 (log1/2 (log 2 (x ))) = log 3 (log1/3 (log 3 ( y ))) = log 5 (log1/5 (log 5 (z ))) = 0 for positive x. 1] (d) f (1) = k 42.. x2 + y2 ≤ 25} and 4 R ′ = (x . then 9 (a) (b) (c) (d) domain of R ∩ R′ = [–3. each question has 4 choices (a). f : [0. Let R = {(x. 46. 1 1 such that P (E ∩ F ) = . (n ∈ N ) . y ∈ R.. The number of isosceles triangles with integer sides if no side exceeds 2008 is (a) (1004)2 if equal sides do not exceed 1004 (b) 2(1004)2 if equal sides exceed 1004 (c) 3(1004)2 if equal sides have any length ≤ 2008 (d) (2008)2 if equal sides have any length ≤ 2008 45. (b). 2011) (b) (0. then 2 3 n 38. f(8n + 4) where n ∈ N is equal to (a) f(4n + 2) (b) f(2n + 1) (c) 2f(2n + 1) (d) 4f(2n + 1) ..35. 0) 1 1 1 44. f(mn) = f(m) f(n) if m and n are relatively prime natural numbers.. then z = Let z = c − ib ib ia ia (a) (b) (c) (d) 0 c b c 41. Suppose three real numbers a. a + ib .. 1] (c) f (x ) ≠ 0. 3] range of R ∩ R′ = [0. x ≥1 2 36. then which of the following is/are NOT true? (a) z < x < y (b) x < y < z (c) y < z < x (d) z < y < x 1 43. If Sn = 1 + + + . If BD = 6 cm.. A circle of radius 4 cm is inscribed in DABC which touches the side BC at D. ∀x ∈[0. The function f (x ) = x 3x 13 is 4 − 2 + 4 . 1] → R is a differentiable function such that f(0) = 0 and |f ′(x)| ≤ k|f (x)| for all x ∈ [0. y and z. P (E c ∩ F c ) = and 6 3 (P (E ) − P (F ))(1 − P (F )) > 0 . If E and F are two independent events. y ) : x . x < 1 (a) (b) (c) (d) continuous at x = 1 differentiable at x =1 continuous at x = 3 differentiable at x = 3 (k > 0). then 1 1 (a) P (E ) = (b) P (E ) = 4 2 2 1 (c) P (F ) = (d) P (F ) = 3 3 39. 4] range of R ∩ R′ = [0. If lim x tan −1 x + m 4 x →∞ pair(s) (l. + . Paragraph for Question Nos. then ordered 37. 34 MatheMatics tODaY | march ’15 S1 + S2 + . m) can be (a) (2000. ∀x ∈[0. (c) and (d). out of which only one is correct.P. + Sn – 1 is equal to (a) nSn – n (b) nSn – 1 (c) (n – 1)Sn – 1 – n (d) nSn – 1 – n + 1 sectiOn-iii CoMprehenSion type this section contains paragraphs. ∀ x ∈ R (b) f (x ) = 0. b. y ∈ R. 1]. 3) (d) (1. then (here r ∈ N) r r + 1 + (r + 1) r (a) Tr > Tr +1 (b) Tr < Tr + 1 (c) 99 ∑ Tr = r=1 9 10 (d) n ∑ Tr < 1 r =1 x +l p − = 1 . ∫ 0 (1 + x )m+n dx (a) g(n. n) (c) g(m – 1. then the value of n is (a) 3 (b) 4 (c) 5 (d) 6 Paragraph for Question Nos. Tr(A3) = (a) 149 (c) 128 (b) 101 (d) 133 51. For any such matrix A. It is known that. Six or fewer steps is 1 (a) 16 3 (c) 64 (b) (d) 1 46 15 44 1 32 5 (d) 64 (b) Paragraph for Question Nos. n) = g(n. ∞ x m−1 dx = (1 + x )m+n (a) g(m. then a non-trivial solution X = (x y z)T such that AX = lX(l ∈ R) yields 3 values of l say l1. all the four being equally likely. n + 1) (m + 1)n+1 m 1 x m−1 + x n−1 57. m) (c) g(m –1. 58 to 60 200 (b) 3 128 1 (d) 256 If (1 + x + x 2 )100 = ∑ ar x r r =0 58. n) (d) g(m. Answer the 1 1 2 following questions for matrix A = 2 2 1 2 1 2 49. l’s are called eigen values and corresponding X’s are called eigen vectors. ∫ x log e dx = x 0 f (n) f (n + 1) (a) (b) n (m + 1)n+1 (m + 1) f (n + 1) (c) (d) g (m + 1. 0 It is known that f(n). Tr(A) = l1 + l2 + l3. for any 3 × 3 matrix. Exactly 4 steps is 5 (a) 128 1 (c) 128 53.47. ∫ 0 1 (b) g(m – 1. right. Which of the following is true? (a) a28 = a72 (b) a56 = a144 (c) a200 = a300 (d) none of these MatheMatics tODaY | march ’15 35 . n – 1) 55. n) = ∫ x m−1 (1 − x )n−1 dx . n – 1) n 1 56. 52 to 54 Starting at (0. n > 0. l3. l2. Each step is left. Which of the following is false? (a) ∃ a non-trivial solution X such that AX = (2 + 7 ) X (b) ∃ a non-trivial solution X such that AX = X (c) ∃ a non-trivial solution X such that A−1 X = (2 − 7 ) X (d) The total number of non-trivial solutions X such that AX = lX is 3. n be two positive real numbers and define ∞ f (n)= ∫ x n−1e − x dx and 0 1 g (m. up or down. The number of natural numbers ‘n’ such that f(n) is odd is (a) 1 (b) 2 (c) 3 (d) 4 48. If f(7n) = 2058 where n ∈ N. n – 1) (b) g(m – 1. Paragraph for Question Nos. Then. 49 to 51 If A is 3 × 3 matrix. n – 1) Paragraph for Questions Nos. 2) in 52. detA = l1 l2 l3. each of length 1 unit. 55 to 57 Let m. Exactly 6 steps is 6 (a) 44 6 (c) 46 54. Tr(A–1) = (a) 1/3 (b) 1/2 (c) –1/2 (d) –1/3 50. The probability that object reaches (2. n + 1) (d) g(m + 1. m) for m. for n > 0 is finite and g(m. 0). an object moves in xy-plane via a sequence of steps. Statements in Column i are labeled as A. B. 2 tan–1(sin x – cosx) is strictly increasing. 2 f (x) f (–x) = 9 for all real x. q .59. D – s.. p) (D) p Column II 2 2 ∫ x(sin (sin x ) + cos (cos x ))dx (p) 0 p2 / 4 ∫ (2 sin x + x cos x )dx (q) 0 p p/ 4 ∫ ln( 1 + sin 2 x )dx ln 2 − p/4 p 8 x 3 cos 4 ∫ 0 x sin2 x p2 − 3px + 3x 2 dx p2 16 p2 2 (r) p2 4 (s) p2 8 (t) p2 32 63. q. r and s. is (B) Interval in which (q) − p p 2 . p 2 one of the points of local maximum of cosx + sin2 x lies. then m is 4 (D) Number of term in expansion (s) of (1 + 31/3)6 which are free from radical sign is 5 (t) 1 . is (C) Interval containing the (r) p 3p value of the integral 4 . if the correct matches are A – p. C – p. 4 3 2 ∫ [x − 5x + 6]dx . is (D) Interval containing the value (s) 3 dx of ∫ if −3 3 + f ( x ) p p 3 . Match the definite integrals in Column–I with their values in Column–II. the answers to these questions have to be appropriately bubbled as illustrate din the following examples. each question contains statements given in two columns which have to be matched. where k is k (n !)2 (B) Number of interior point (q) when diagonals of a convex polygon of n side intersect if no three diagonal pass through the same interior point is nCl . r . Match the following: Column I Column II (A) Number of ways to select (p) 3 n objects from 3n objects of which n are identical and rest are dif ferent is 1 (kn)! k 2k −1 + . Match the following:Column I Column II (A) Interval in which at least (p) 0.. (b) 64a35 + 148a36 (d) none of these (A) (B) sectiOn iV (C) MAtrix-MAtCh type this section contains questions. 37a37 is equal to (a) 64a36 + 165a35 (c) 56a32 + 168a22 62. s . is (t) 36 MatheMatics tODaY | march ’15 (–p. then l is 2 (C) Five digit number of different (r) digit can be made in which digit are in descending order is 10Cm . a0 + a1 + a2 + . C and D whereas statements in Column ii are labeled as p. B – q. where 2 [x] is the greatest integer function. + a99 is equal to 399 − a99 2 100 3 − a100 (c) 2 (a) (b) 3101 − a99 2 Column I (d) none of these 60. then the correctly bubbled matrix will look like the following: 61. then tanx + cotx can a be written as c where a. 3. 2) and latus rectum is of 12 units is (x – 1)2 + (y – 2)2 = a2 . Two lines zi − zi + 2 = 0 and z (1 + i) + z (1 − i) + 2 = 0 intersect at a point P. 70. Find [|x|] (where [. R then 2 is R1 (C) Thre e p oints D. 1 2 r2 r3 r2. 1 1 2 3 4 5 6 7 8 9 2 1 2 3 4 5 6 7 8 9 3 1 2 3 4 5 6 7 8 9 4 1 2 3 4 5 6 7 8 9 5 1 2 3 4 5 6 7 8 9 66. the appropriate bubbles below the respective question numbers in the orS have to be darkened. 100} and Y be a subset of X such that the sum of no two elements in Y is divisible by 7. Tangents are drawn at P and Q to meet at R. Let X = {1. 0) and (0. y ∈ R and f ′(0) = 0 is 9A. BE and CF are BD ⋅ CE ⋅ AF concurrent. There is a complex number a = x + iy at a distance of 2 units from the point P which lies on line z (1 + i) + z (1 − i) + 2 = 0 .64. value of A is 67. The minimum area bounded by the function y = f (x) and y = ax + 9 (a ∈ R) where f satisfies 1 5 the relation f (x + y ) = f (x ) + f ( y ) + y f (x ) ∀ x . 4 and 5 (say) are 2. 4. Let x be in radians with 0 < x < 65. (q) 2 If R1 and R2 are circumradius of DDEF and DABC respectively. F are (r) taken on side BC. 1. 2 If sin(2sinx) = cos(2cosx).. then DC ⋅ AE ⋅ FB is equal to (D) (s) 3 r1 r1 − − 1 1 is (where r . respectively then the correct darkening of bubbles will look like as given. 2. Let P. Match the following: sectiOn-V Column I Column II (A) Number of triangles to which (p) 3 an acute angle triangle ABC can act as a pedal triangle is (B) DEF is a pedal triangle of ABC. E.] represents greatest integer function). For example. Q be two points on the ellipse (s) 0 (t) 5 MatheMatics tODaY | march ’15 37 . . ranging from 0 to 9. 2. 3. c ∈ N. 2 and 3.. then PQ is equal to (D) A movable parabola touches the x and y-axes at (1.. 1) then radius of locus of focus of parabola is integer AnSwer type Column II (p) 2 (q) 4 (r) 3 a +b+c value of is 25 69. 2 2 ) on t he parab ola y2 = 8x cut the line y = 2 2 at Q. then a= (C) A line drawn through the focus F and parallel to tangent at P(1. CA and AB such that AD. then find m + n(m : n is in simplified form). If the chord PQ divides the joint of C and R in the ratio m : n (C being centre of ellipse). the answer to each of the questions is a single digit integer. Then the p −b 68. If the maximum possible number of element in Y is 40 + l. b. Match the following: Column I (A) Number of mutually perpendicular tangents that can be drawn from the curve y = ||1 – ex| – 2| to the parabola x2 = –4y (B) Locus of vertex of parabola whose focus is (1. if the correct answers to question numbers 1. then l is p . x2 y2 + =1 25 16 whose eccentric angles differ by a right angle. r3 are radii of excircles of DABC which is right angled at A) (t) 4 this section contains questions. b and c 4. If a = ei2p/7 and f (x ) = A0 + ∑ Ak x k and k =1 the value of f(x) + f(ax) + f(a2x) + . (2) . (2) and (3). (4) f ′(2x2 – 1) < f ′(1 – x2) ⇒ 2x2 – 1 < 1 – x2 2 2 ⇒ x ∈ − . y1) ∑ a cos a ∑ b sin a . A = ∫ x 2dx = 3 / 2 3 6 60 5... 1 = ∑ sin a ⇒ a b 2 2 9x 1 9 y1 ⇒ 2 + 2 − 3 = 2(∑ cos a cos b + ∑ sin a sin b) b a 38 MatheMatics tODaY | march ’15 2a 2 + 9 y2 3 − = Σ cos a cos b + Σ sin a sin b 2 2b 2 x x2 13 ⇒ f (x ) = . g(x) = x2 – 2. is 20 75. p(2) = 0 ⇒ 4a + 2b + c = 0 . (d) : a is integer then x must be integer.. (a) : 15! = 211 × 36 × 53 × 72 × 111 × 131 6 \ ∑ Pr = 11 + 6 + 3 + 2 + 1 + 1 = 24 r =1 8. bsing) Centroid = circumcentre = (x1. sOlutiOns 1.... x2. (x – 2008) Put x = 2009 ⇒ P(2009) + 1 = 2009! 3. 3 3 2 (4) ∩ (5) ⇒ x ∈ − .... (a) : P(x) – 2008 + x = x(x – 1)(x – 2)(x – 3). (6) . 1 72.. (5) .. 0 ∪ . then the value of g(x1)g(x2)g(x3)g(x4)g(x5) – 30g(x1x2x3x4x5).. p 73. (a) : A(acosa. x ∈ I 2 7 7 ⋅8 7 ⋅8 3 + = 812 ∑ ai = ∑ (x + x ) = 2 2 x =1 2. bsina). (b) : Since p′′′(x) = 0 Let p(x) = ax2 + bx + c. then find the value of 1/2 12 (sin−1 a + cos −1 b) .. The remainder when the 2003rd term of new sequence is divided by 2048.∞ 3 3 2 (1) ∩ (2) ⇒ x ∈ . x4 and x5.. A sequence is obtained by deleting all perfect squares from set of natural numbers. 7) 7. then find the value of k. we get a. (b) : f ′′(x) > 0 ⇒ f ′(x) is increasing function. To find where g is necessarily increasing : g is increasing ⇒ g′ > 0 1 1 ⇒ ⋅ f ′(2 x 2 − 1)(4 x ) + f ′(1 − x 2 )(−2 x ) > 0 2 4 ⇒ x{ f ′(2 x 2 − 1) − f ′(1 − x 2 )} > 0 Case 1 : For x > 0 . (b) : f ′(x ) = 0 and 9x 2 6.. B(acosb.. . (3) Solving (1).. ∪ . (1) p(3) = 9 ⇒ 9a + 3b + c = 9 .. (b) : |z – 3 – 4i| = 5 ⇒ (x – 3)2 + (y – 4)2 = 25 ⇒ R is (3. f ′ = p 2 3/2 2 ∫ f (x )dx = p + 1 . Let p(x) = x5 + x2 + 1 have roots x1.e.. is − 74. ∞ 3 . Find the value of 0 1 4 ∫ (1 − x 4 )6 dx ⇒ . (2) p′(2) = 7 ⇒ 4a + b = 7 . ∞ 3 3 .. [x] = x ⇒ a = x3 + x 1 ≤ a ≤ 500 ⇒ 1 ≤ x ≤ 7.. = 3 3 3x1 3y = ∑ cos a.. (1) f ′(2x2 – 1) > f ′(1 – x2) ⇒ 2x2 – 1 > 1 – x2 2 2 ⇒ x ∈ −∞. If f (x ) = a cos( px ) + b. 0 3 \ g is increasing in x ∈ (3) ∪ (6) 2 2 ⇒ x ∈ − . + f(a6x) is k(A0 + A7 x7 + A14 x14). (3) Case II : For x < 0 .. i. x3. bsinb).1 29 ∫ (1 − x 4 )7 dx 71.. C(acosg... a2. (c) : 1 1 1 ⇒ (x + y)2007 = xy + = x y 2007 ⇒ xy – 2007x – 2007y = 0 ⇒ (x – 2007)(y – 2007) = 20072 = 34 × 2232 The number of pairs is equal to the number of divisors of 20072 that is (4 + 1) × (2 + 1) = 15. t 1 Given m = − 2 1 1 or t1t2 = − t1t2 m Also by section formula. y3} f : A → B is onto.S consider 1 1 − (1 − x )n 1 1 In = ∫ dx ⇒ In+1 − In = ∫ (1 − x )n dx = x n +1 0 0 = 10C1 − 10 10 MatheMatics tODaY | march ’15 39 . x5. 3). (a) : Let a be a root of f (x) = 0. (1. (3. t1 .. which is always a 2 2 hyperbola as 25m − 4m2 = 9m > 0. (2. (c) : The given function has local maximum at x = –1. Since x < y. then there exist f (x ) = y2 Exactly 3 elements of x is associated with y2. 1). ⇒ f (0) = 0 ⇒ m = 0 Then we have. a2) can be (0. f (2) = 2 3 = 4 3 \ Range of f(x) = [0. of ways = 24 – 2C1(2 – 1)4 = 14 \ Total no. (b) : a. (b) : Let the points of intersection be 1 1 . total number of choices = 7 × 15 × 27 = 2835 18.H. + 10 ⇒ 10C 10x 10 (1 − x )10 − 1 = −[10 C1 − 10C2 x + . f (x) = x(x + l) and thus a = 0. b2) has 15 choices and (c1. max{a1. g (x ) = cos x 17. c1.C. we get (log a b)x logb a = (logb a)x \ (log a b)x = (logb a)x −1 1 \ 1− x = x ⇒ x = 2 14. t and t2 . 0) (one of the number is 3 and other number can be any where from 0 to 3 ) giving us 7 choices. of onto functions = 7C3 × 14 3 10. x6.M of a... Similarly (b1.. t 2 and eliminating them gives 2m 2 x 2 + 5mxy + 2y 2 = m.9.. (b) : A = {x1. f (f (x)) = x(x + l)(x2 + lx + l) We want l such that x2 + lx + l has real roots... This can be done in 7C3 ways. (3. Hence (a1. 2a2 5b2 11c2 where a1. (c) : Taking logb on both sides. –l. 3). c2 are non negative integers. x2... a2} = 3... so required number of pairs = 7 12.. x3. − 10C10 x 9 ]dx 0 10 C3 C10 C2 + − . b2} = 7 and max{c1. c2} = 13.. ∀m ≠ 0 4 4 16.. Remaining four elements of A is associated with 2 elements of B. Since L. − 10C10 x 9 ] x 1 1 − (1 − x )10 ⇒ ∫ 0 x 1 dx = ∫ [10 C1 −10 C2 x + .. b is 23571113. b1. c2) has 27 choices. (3. x4. 13. t2 + 2t1 = 3x 3y t1 + 2t2 = − m Solving for t 1 . max {b1. Hence. − 1 2 3 10 consider (1 – x)10 = 10C0 – 10C1x + 10C2x2 – . x7}. so we have f (a) = 0 and thus f (f (a)) = 0. 3). 4 ] 15. 3). (a) : ∫ 2 − 3 sin2 x cos x dx = ∫ = 2 ∫ (cos x ) cos x dx − ∫ 2 cos2 x − sin2 x sin2 x cos x cos x dx dx = 2sin x cos x + c ⇒ f (x ) = sin x . − 2 3 10 To find L. B = {y1. y2. (b) : Required value is 10 C10 C1 10 C2 10 C3 − + − . f(–1) = 1.. \ No. ⇒ 0≤l≤4 We can easily find that 0 ≤ l < 4.. 2). 2 /3 3 f (−3) = 2 − 3 . (a) : ∫ 0 3 ( (x + 3) + 2 3 ) x + (x + 3) − 2 3 x dx 3 ∫ (( x + 3 ) + ( 3 − x ))dx = ∫ 2 3 dx = 6 3 0 0 11. b are factors of the form 2a1 5b111c1 . b2. minimum at x = 0 and f(0) = 0. (3.. ⇒ c2 = bd . + 1 10 9 8 19.P.P. c... + 1 = 55 No.V.T. (a) : f ′(x ) = k dx = 4 ∫ dx 1 + cos2 x p sin x sin x \ 2I = 4 p ∫ dx = 8 p ∫ dx = 2 p2 2 2 0 1 + cos x 0 1 + cos x 0 1 + cos dx = p ( p − x )sin x x sin x 1 2 + x4 f (2) − f (1) By L..P. d. ⇒ a + c = 2b .P. b.. (1) b.. (c) : Coefficient of x6 is 11 1 1 1 1 1 1 11 + + + + 51 4 2 3 3 2 4 1 5 = 1 6 [ C1 +6 C2 +6 C3 +6 C4 +6 C5 ] 6 = 1 6 31 (2 − 2) = 6 360 40 3k +1 MatheMatics tODaY | march ’15 2 x 0 p /2 I = p2 24... 2) 2 −1 1 ⇒ f (2) = as f (1) = 0 2 + c4 Since 1 < c < 2 ⇒ 3 < 2 + c4 < 18 ⇒ f (2) < 1/3 27.t a ≤ b = 10 + 9 + 8 + . (b) : I = 4 ∫ 1 1 1 + I = + + I ≈ .. (b) : Required area is the area of shaded region (APOQ) = area of DOAQ + area of sector (OAP) 1 p(4 × 4) 8 p = +8 3 = ×4×4 3+ 3 2 6 . (d) : Area = 2 ∫ y dx = 2 ∫ x 2 x + 2 dx 2 = 4 ∫ (z 2 − 2)2 z 2dz (where x + 2 = z ) 0 2 z 7 4z 5 4z 3 256 2 = 4 − + = 5 3 105 7 0 21.. (2) 2ce .. (b) : = 6 (3 − 2 )(3k +1 − 2k +1 ) k k 3k 2k (3k − 2k )(3k +1 − 2k +1 ) ⇒ = 3k 3k − 2k − 3k +1 − 2k +1 ∞ 3k 3k +1 3 3n − lim ∑ k k − k +1 k +1 = k =1 3 − 2 3 − 2 3 − 2 n→∞ 3n − 2n =3–1=2 22. = ln 2 2 3 4 n→∞ 26.. ⇒ c+e (a + c)ce (1) × (3) ⇒ = bd = c 2 c+e \ (a + c)e = c(c + e) ⇒ ae = c2 ⇒ a. of ways of choosing a and b s.... (c) : a1 = 1 ⇒ a2 = 1 − 1 2 1 1 ⇒ a3 = 1 − + ……………… 2 3 1 1 1 lim (an ) = 1 − + − + . c are in A. (b) : a.. 10 9 10 9 8 \ 1 1 1 = + + + . (c) : Number of ways of choosing a and b s. f ′(c) = for some c ∈ (1..M. 25..\ In+1 = 1 1 − (1 − x )10 \ p 1 +I n +1 n I10 = ∫ x 0 23... e are in G. c. d are in G.t a divides b =10 + 5 + 3 + 2 + 2 + 5 = 27 27 \ Required probability = 55 0 0 −2 −2 20. (3) =d c. e are in H. Hence. b. we get lim =1 32.. no. d) : tan B 2 C 1 = .35. (a. of triangles = 1004 × 2008 = 2(1004)2 ⇒ \ Total no. (a. f (1) = lim 3x + 1 p→∞ 7 + P x g (1) ⇒ 7= ⇒ g (1) = 49 7 \ g(1) – h(1) = 28 ⇒ g(1) + h(1) = 70 34. b.. tan = ⇒ 2s = 3a = 3 × 14 = 42 2 2 s \ Perimeter = 42 \ D = r · s = 84 cm2 ⇒ tan MatheMatics tODaY | march ’15 41 . So for any a ∈ N. m) can be (5.b then the triangle forms only when 2a > b. b can take any value from 1 to 2008 But a has 1004 possibilities. d) : a2 = b2(1 – e2) ⇒ (sec2a)e2 = 1 ⇒ e = cosa Also. 3 f ′(1+) = f ′(1–) = –1 Not differentiable at x = 3 x →∞ 2 x 2 ( l − m) x 2 + 2 x(l + m) + (m2 + l2 ) =1 l−m = 1 ⇒ l − m = 2 \ (l. tan = 2 3 2 2 A 4 = 2 7 B C s−a tan . b) : x2 + y2 ≤ 25 and 9 9y ≥ 4x2 ⇒ x 2 ≤ y 4 y 9 Substituting x 2 = in 4 x2 + y2= 25. d) : Let the hyperbola be For non-existence of ^r tangents b2 x2 a 2 > − a2 y2 2 b ⇒ e> 2 −1 −1 ⇒ a = b = –30° and sin b = 2 2 h(1) + 7 33.a. (c) : P is centroid of DABC \ Area of DABC = 6 × 5 = 30 sq. b) : When x < 1. (a. 3] and [0.+ (2(1004) – 1) = (1004)2 And if 1005 ≤ a ≤ 2008. (a. b. of isosceles triangles = 3(1004)2 39. l = 2a2 b x+l p − tan −1 x + m 4 37. f (1) = 3 +1 h(1) + 7 ⇒ 7= ⇒ h(1) = 21 4 h(x ) 7 g (x ) + p + p x x When x > 1. c. 4] 36. c) : f (1 + h) = 2 = f(1 – h) f(3+) = f(3–) = 0 ⇒ Continuous at x = 1. b) : (sina + sinb)2 = 1 ⇒ sina + sinb = –1 ⇒ sin a = 9y + y 2 − 25 = 0 4 ⇒ 4y2 + 9y – 100 = 0 ⇒ (y – 4)(4y + 25) = 0 ⇒ y = 4 Domain and range of R ∩ R′ are [–3. (c. c) : If the sides are a. (a. we get ^ 28.. b can change from 1 to 2a – 1. (a. (c) : lim =1 1 x →∞ x Apply L' hospital rule and simplifying. where a ≤ 1004 ⇒ Number of triangles = 1 + 3 + 5 .. (a.units 31. 3) 2 38. (c) : (2n + – (2m + = 4(m + n + 1)(n – m) = multiple of 8 30. (c) : a = y ^c − x b c yc^ ^ b xb a 30° x 2 |a | = sec 30° ⇒ x = |a | 3 y 1 |a | = tan 30° ⇒ y = |a | 3 1 2 ⇒ a= c− b 3 3 1)2 1)2 29. b. c. ... we get ⇒ P (E ) = . l3 l1 + l2 + l3 = 5. |A – lI| = 0 2 1 2 ⇒ l3 – 5l2 + l + 3 = 0 . S2 : 1 + .P. b) : Let r be common ratio of G. (b) : f(7n) = 2058 7n – 1(7 – 1) = 2058 7n – 1 = 343 or n – 1 = 3 \ n = 4 (49-51) : 49. (c) and (d) are correct. 40. d) : Tr = = = r r + 1 − (r + 1) r 1 r − −r 2 − r 1 r ( r + 1) − (r + 1) r r 2 (r + 1) − (r + 1)2 r = (r + 1) r r r + 1 − r (r + 1) r (r + 1) r +1 99 1 1 1 1 1 9 ⇒ ∑ Tr = − + − . + − [1 + 1 + 1 + . d) : S1 : 1. 2 2 2 \ All angles are acute. But f(0) = 0 ⇒ both function g1 and g2 have a value zero at x = 0 42... (a. f(pn) = pn – 1(p – 1) f(mn) = f(m) · f(n) f(8n + 4) = f(4(2n + 1)) = f(4)·f(2n + 1) = f(22) · f(2n + 1) = 2 · f(2n + 1) 47. + 2 3 n −1 Adding (n − 2) (n − 3) n − (n − 1) + + . (a. ⇒ n – 1 = 0 ⇒ n = 1 \ f(1) = 1 = f(2) 48. (d) 50. (c). ⇒ f(pn) is odd ⇒ pn – 1(p – 1) is odd.. we have. (b) : (f ′(x))2 – k2(f (x))2 ≤ 0 ⇒ (f ′(x) – kf(x))(f ′(x) + kf(x)) ≤ 0 ⇒ (f(x)e–kx)(f(x)ekx) ≤ 0 ⇒ Exactly one of the functions g1(x) = f (x)e–kx or g2(x) = f(x)ekx is non decreasing. (b) 51. l2.. S3 : 1 + + 2 2 3 1 1 1 Sn – 1 : 1 + + + ... y = 31/3. l1l2 + l2l3 + l3l1 = 1. P (F ) = 2 3 ⇒ P ( E ) + P (F ) = 46... det(A) = –3 1 −1 Also. z = 51/5 31/3 > 51/3 as 35 > 53 Also... Q tan 43. c. (b. c... AX = lX ⇒ A X = X l 1 1 1 1 ∑ l1l2 ⇒ Tr ( A−1 ) = + + = =− l1 l2 l3 l1l2 l3 3 AX = lX ⇒ A3 X = l3 X ⇒ Tr ( A3 ) = l13 + l32 + l33 = (l1 + l2 + l3)(l12 + l22 + l32 – l1l2 – l2l3 – l3l1) + 3l1l2l3 = 5(25 – 3(1)) – 9 =101 . we get x = 21/2 . (b) : f(n) is odd. (n − 1) 2 3 1 1 1 = n 1 + + + .A B C .(1) whose roots are l1. (c) : f(1) = 1.. d) : Solving.. − =1− = 1 2 2 100 100 10 r =1 Hence (a)... (a. tan . (c) 1 1 2 For the matrix A = 2 2 1 . The only value p can take is p = 2 \ f(2n) is odd. 21/2 < 31/3 as 23 < 32 and 21/2 > 51/5 as 25 > 52 ⇒ y>x>z Hence (b). c) : P (E ∩ F ) = P (E ) ⋅ P (F ) = P (E c ∩ F c ) = (1 − P (E ))(1 − P (F )) = 1 3 1 6 …(i) 5 . l1l2l3 = –3 ⇒ Tr(A) = 5. (a. Q p is prime.... (ii) 6 As (P(E) – P(F))(1 – P(F)) > 0 ⇒ P (E ) > P (F ) 1 1 Solving (i) and (ii).... + 1] 2 3 n − 1 = (n − 1) + = nSn – 1 – (n – 1) = nSn – n 42 MatheMatics tODaY | march ’15 45. ⇒ 2n – 1(2 – 1) = 2n – 1 is odd. 1 1 1 44. tan all are less than 1. a 1 +i +i ib ia i or z = or =r = z= b c c b r −i r −i b 41. & (d) are NOT true. L or R. Hence probability for exactly six steps 2(60 − 12) 6 = = 46 44 6 6 3 Probability for six or fewer steps = 4 + 4 = 64 4 4 y 55. we get ar = a200 – r 59. (b) : ∑ ar = 1 + + x x2 r =0 x = ⇒ ⇒ 200 ∑ ar x r =0 200 −r 1 x 200 (x 2 + x + 1)100 2 = (x + x + 1)100 200 200 r r ∑ ar x = ∑ a200−r x r =0 r =0 Equating the coefficients of x200 – r. + a200 = 3100 But ar = a200 – r \ 2(a0 + . 1 2+ 7 2− 7 i. U \ Probability of reaching (2. we get 200 100(1 + 2 x )(1 + x + x 2 )99 = ∑ rar x r −1 r =0 MatheMatics tODaY | march ’15 43 . +a99) + a100 = 3100 1 100 ⇒ a0 + a1 + . + a99 = (3 − a100 ) 2 60.e. 2− 7 2+ 7 . 2) can be reached in 4 steps if the sequence of steps is some permutations of R. (a) : In g(m. D can be permuted in 2(60) ways of which 2 × 12 correspond to exactly 4 steps. L or U. (a) 54.. n). U. D in same order. U. U. the values of l yield non trivial solutions are 1. U. 1 . we get 1+ y 1 dx = dy and we get (1 + y )2 ∞ g(m. R. U. put x = 1/t ⇒ I2 = ∫ t n−1 1 1 + t ∞ ∞ =∫ 1 1 x m−1 ∞ x m−1 dx m +n x m−1 (1 + x )m+n −dt 2 t dx ∞ x m−1 dx + ∫ dx m +n m +n 0 (1 + x ) 1 (1 + x ) \ I=∫ =∫ 0 (1 + x )m+n dx = g (m. U. U. l3 = 2 − 7 . n) = ∫ 0 ∞ =∫ 0 y m−1 1 1 . (c) : Put x = 1 ⇒ a0 + a1 + a2 + . 2) in 6 or fewer steps. which by theory yield non-trivial solutions. putting x = . . 1.. (c) Since the net movement must be two steps right (R) and two steps up (U) there must be atleast 4 steps to reach (2. R. In particular for A–1. −3 −3 Hence (c) is false. dy (1 + y )m−1 (1 + y )n−1 (1 + y )2 x m−1 (1 + x )m+n dx 1 56.. (c) : Putting log e = t x ⇒ x = e −t n 1 Now. n) 100 r 200 1 1 1 58..Solving (1) gives l1 = 1. R. l2 = 2 + 7 . (b) 53. 2) in 4 steps 4! 2 2 6 3 ! ! = = = 4 4 128 4 4 A six step sequence moludes the steps R.. (a) : I = ∫ 1 =∫ x dx m +n 0 −1(1 + x ) m 1 m +n 0 (1 + x ) x n−1 dx + ∫ m +n 0 (1 + x ) = I1 + I2 1 1 In I2. (2. ∫ x m log 1 dx e x 0 0 ∞ ∞ 0 = ∫ e −mt t n (−e −t ) dt = ∫ t ne −(m+1)t dt = 1 ∞ f (n + 1) 0 (m + 1)n+1 n −t ∫ t e dt = n+1 (m + 1) 1 x m−1 + x n−1 57. R. (52-54) : 52. (a) : Differentiating. R.. U in same order as well as a pair of steps consisting of R. R. t) p (A) cosx + sin2x has a local maximum at x = 3 −p and x = in the interval (–p.. r. 31/3. + 2nCn = 22n−1 + 2 (n !)2 (B) n C 4 (each quadrilateral gives one point of intersection) (C) x4 > x3 > x2 > x1 > x0 10C (5 distinct digits selection) 5 (D) Terms is involving 30. p) 3 (B) tan –1(sinx – cosx) is strictly increasing in p 3p − 4 . (D) → (p) (A) Required number of selection 1 (2n)! = 2nC0 + 2nC1 + . (C) → (r). q. (D) → (p) (A) ABC is pedal triangle of acute angle l1l2l3 and of obtuse angled triangle ll1l2. (D) → (p. (A) → (q). (B) → (q). 2 I = p ∫ 2dx = 2 p ⋅ p / 2 = p ⇒ I = p2/2 0 (B) Let f (x ) = 2 sin x Then. t). I = p ∫ sin2 (cos x ) + cos2 (sin x ) dx 0 p /2 2 Adding. . r. (C) → (s). (A) → (q). s). q. 2 I = p ∫ (sin2 (sin x ) + cos2 (cos x ))dx p /2 0 ⇒ 2 I = 2 p ∫ (sin2 (sin x ) + cos2 (cos x ))dx ⇒ I = p ∫ (sin2 (sin x ) + cos2 (cos x ))dx 0 p /2 0 p /2 Also. ll2l3 and ll3l1. 32/3 → 3 64. (A) → (r). (C) → (s). 2 ∫ cos 4 x sin2 x dx = p ∫ cos 4 x sin2 x dx 2 0 0 = p⋅ (3 ⋅1) p p2 ⋅ = 6 ⋅ 4 ⋅ 2 2 32 63. (B) → (r). (D) → (t) 44 p/ 4 ∫ ln 1 + sin 2 x dx = ∫ ln |sin x + cos x | dx I=1 (A) I = ∫ x(sin2 (sin x ) + cos2 (cos x ))dx p2 p p2 × sin = 4 2 2 p/ 4 (C) 3 3I = ∫ p2 / 4 I = ∫ ( f (x ) + x f ′(x )) dx = x f (x ) 0 −p − p2 2 ln = 4 4 dx x sin2 x p2 − 3px + 3x 2 dx p Adding 2 I = p ∫ cos 4 x sin2 x dx 0 ⇒ p /2 p p /2 I = .. x f ′(x ) = x. cos x x = x cos x MatheMatics tODaY | march ’15 p /2 p /2 p /2 0 0 0 0 ∫ dt + ∫ ln sin t dt p p p = ln 2 − ln 2 = − ln 2 4 2 4 \ p p/ 4 p 2 ∫ ln 1 + sin x dx = ln 2 ln 2 − p/4 p x 3 cos 4 (D) I = ∫ 0 x sin2 x p2 − 3px + 3x 2 p ( p − x )3 cos 4 I= ∫ 0 p 0 p p /2 = ∫ ln| 2 sin t | dt = ∫ ln 2 dt + ∫ ln 2 sin t dt 62. (C) → (q.61. 4 2 5 1 1 2 (C) x − 5x + 6 = x − − ≥ − for all x 2 4 4 −1 ≤ x 2 − 5x + 6 < 0 4 3 ⇒ [x2 – 5x + 6] = –1 \ ∫2 [x 2 − 5x + 6] dx = −1 Q 2<x <3 ⇒ 3 3 3 3 dx = ∫ (D) 3I = ∫ −3 3 + f ( x ) −3 3 + f (− x ) 3 3 3 f (x ) f (x ) dx = ∫ dx −3 3 f ( x ) + 9 −3 3 + f ( x ) = ∫ ⇒ \ 3 3 ( f (x ) + 3) − 3 3 dx = ∫ 1dx − ∫ dx f (x ) + 3 −3 −3 −3 3 + f ( x ) p2 / 4 0 =2 × − p/ 4 p/ 4 − p/ 4 = ∫ ln | 2 sin(x + p / 4)| dx − p/ 4 p /2 0 = ln 2 ⇒ I = ∫ (p − x )(sin2 (sin x ) + cos2 (cos x ))dx 0 p Adding. (B) → (q). r. (A) → (p. (B) → (p. t) . MatheMatics tODaY | march ’15 45 . f4(sin x ) q) Q be (−5 sin q. 4 cos q) 70. 97}... 11. if AD. (7) : Let I = ∫ (1 − x 4 )7 ⋅1 dx Yi = 7m + i. 13. n(Y0) = 14 1 Y1 = {1.. b = 16. (8) : f ′(x ) = lim h→0 ⇒ lim h→0 f (x + h) − f (x + 0) h f (x ) + f (h) + h f (x ) − f (x ) − f (0) − 0 f (x ) h Y5 = {5.(B) R2 =2 R1 (C) In DABC.. n(Y1) = 15 = [x(1 − x 4 )7 ]10 + 7 × 4 ∫ x(1 − x 4 )6 x 3dx 0 Y2 = {2... m ∈ I 0 Y0 = {7. (5) : Let Y i be the subset of X such that 1 71. n(Y3) = 14 = −28 ∫ (1 − x 4 )7 dx + 28 ∫ (1 − x 4 )6 dx Y4 = {4... 95}.. (D) → (s) (A) Two tangent can be drawn because curve y = ||1 – ex| – 2| intersect the line y = 1 at two points (B) (x – 1)2 + (y – 2)2 = 32 ⇒ a = 3 (C) PQ = a + x = 2 + 1 = 3 (D) Locus of focus of parabola is 2x2 – 2x + 2y2 – 2y + 1 = 0 \ radius is zero 66. (1) : Solving the equation of the lines we get z = −z ⇒ z = i sin x + cos x = | a − i | = 2. p 68. (C) → (r)..(ii) 4 − sin q + cos q = 1 9 5 4 Required minimum area = 2 ∫ 2 y dy 0 Solving (i) and (ii). 18. we get cosq – sinq = 0 ⇒ ip a = i ± 2e 4 \ x = ± 2 ⇒ [| x |] = 1 be (−5 sin q. . .. 98}. (2) : sin(2 sin x ) = sin − 2 cos x 2 p p2 ⇒ 1 + sin 2 x = 4 16 p2 − 16 sin 2 x = 16 2 2 × 16 32 \ tan x + cot x = = = 2 2 sin 2 x p − 16 p − 16 a +b+c \ a = 32. 15. n(Y5) = 14 Y6 = {6. 12. 100}.. . BE and CF are concurred and BD a CE b if = and = DC b AE g AF g = Then FB a ⇒ BD · CE · AF = DC · AE · FB 3 s −b s −c 1− (D) 1 − 2 s − a s − a 65. (2) : Let P be (5 cos q. a = 2eiq + i . c = 2 ⇒ =2 25 69. 4 cos q) h − 0 h→0 Equation of tangent at P is f ′( x ) y x dx = ∫ dx ⇒ 2 f (x ) = x + c . 96}. (A) → (p)...... (i) ∫ cos q + sin q = 1 f (x ) 5 4 Equation of tangent at Q is x2 \ When a = 0 area is minimum f (x ) = y x ... 94}... . 16. 4(sin q + cos q)) 3/2 0 \ m : n is 1 : 1 ⇒ m + n = 2 67.... .. 8. we get 9 y 3/2 ⇒ 4 = 72 sq.. (B) → (r). . 99}. put it in the equation of the second line. . n(Y4) = 14 0 0 46 MatheMatics tODaY | march ’15 ... 4 sin q) Qand f (h) − f (0) ⇒ lim + f (x ) ⇒ f ′(5xcos ) = q. units ⇒ R = (5(cos q − sin q).. n(Y6) = 14 The largest Y will consist of (i) an element of Y0 (ii) Y1 (iii) Y2 (iv) Y3 or Y4 ⇒ The maximum possible number of elements in Y = 1 + 15 + 15 + 14 = 45. 17... 10.. n(Y2) = 15 1 1 Y3 = {3... 9. 14. .....⇒ 1 73. + f (a x ) 20 = 7 A0 + ∑ Ak x k (1 + a k + ... 74. (0) : Since [ 2046 ] = [ 2047 ] = [ 2048 ] = [ 2049 ] = 45 I = −28 I + 28 ∫ (1 − x 4 )6 dx 0 ⇒ 1 [ 2046 ] = [ 2047 ] = [ 2048 ] = [ 2049 ] = 45 \ 2003rd term is 2003 + 45 = 2048 Hence remainder is 0.. then 1 + ak + a2k + .x5 ) = 7 3/2 3/2 a sin px + bx ∫ (a cos px + b)dx = p 1/2 1/2 −a 3b a b = + − + p 2 p 2 −2a 2 + b = +1 ⇒ b = 1 p p −12 −12 p So. (sin−1 (−1) + cos −1 1) = − + 0 = 6 p p 2 ⇒ g(x1) ... g(x2) = Product of roots = –23 2 6 75... (6) : f ′(x) = –apsin(px) \ p 1 ⇒ f ′ = −ap sin = −ap = p ⇒ a = −1 2 2 x1x2 x3 x 4 x5 = −1 ⇒ g (x1x2 .. g (x5 ) − 30 g (x1x2 .. f (x ) + f (ax ) + .x5 ) = −1 ⇒ g (x1 ) g (x2 ). y = x2 – 2 4 6 29 I − 28 ∫ (1 − x ) dx = 0 0 1 29 ∫ (1 − x 4 )7 dx ⇒ 0 1 4 ∫ (1 − x 4 )6 dx =7 x = y +2 ⇒ ( y + 2 )5 + ( y + 2 )2 + 1 = 0 ⇒ y 5 + 10 y 4 + 40 y 3 + 79 y 2 + 74 y + 23 = 0 0 72... (7) : Let us form that equation having roots y = g(x) i....e.. (7) : f (x ) + f (ax ) + f (a x ) + . + a6k ) k =1 but when k ≠ 7 and k ≠ 14.... + a6k = 0 Hence.......... + f (a6 x ) = 7 A0 + 7 A7 x 7 + 7 A14 x14 = 7( A0 + A7 x 7 + A14 x14 ) \ k=7 nn MatheMatics tODaY | march ’15 47 ... C → R. (Each entry of column I 9. 5. Which of the following homogeneous 7. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE (main & advanced). Shyam Bhushan* 10 Best Problems math archives. Match the column. 1. tan2 By : Prof. the following is/are correct? Then.0). B → R. If y = cos −1 . Narayana IIT Academy. as the title itself suggests. D → R (d) none of these (c) 44 (b) A → P. Mob. C → Q. Director. In every issue of mT. If f (x ) = e . D → S sin( x −[ x ])cos p x (c) A → R. B → S. then f(x) is ([x] (d) A → Q. B → P. lim . then which of sina + sin(a + b) + sin(a + b + g) = 0. 0 and 0. lim x →0 p x →0 x x → cos x ln(cos x ) 2 (B) lim (Q) –1/2 x x →0 where [·] denotes the greatest integer function. D → P denotes the greatest integer function) (a) non-periodic cos 3x 6. : 09334870021 48 MatheMatics tODaY | march ’15 . p) such that x2 + y2 cosa + cos(a + b) + cos(a + b + g) = 0 and 4. C → P. p 1 Column I Column II x − 2 tan x x (A) lim log sin x sin 2 x (P) –1 (0. p 2p 3p 5p 6p 7p x 1 + tan2 + tan2 + tan2 + tan2 + tan2(C) lim − (R) 1 16 16 16 16 16 16 x →1 ln x ln x p 2p 3p 5p 6p 7p x − sin x + tan2 + tan2 + tan2 + tan2 tan2 + tan2 (D) lim (S) 0 16 16 16 16 16 16 is equal to x →0 x − tan x (a) 24 (b) 32 (a) A → S. The readers' comments and suggestions regarding the problems and solutions offered are always welcome. where x →g sin q =1 (a) (b) q < sinq < tanq 1 + sin x − cos x q f (x) = sin2x(1 + cos2x)–1 and g (x ) = 1 + sin x + cos x tan q sin q (c) (d) none of these > (Here f ′(x) denotes derivative of f with respect q q to x). = dx x ⋅ 3 x cos cos 3. Jamshedpur. Let a ∈ R. B → Q. is a collection of various challenging problems related to the topics of JEE (main & advanced) Syllabus. xy (c) (d) all of the above 8. g ∈ (0. then prove that a function f : R → R is functions are of degree zero? differentiable at a if a function f : R → R satisfies x y y x x(x − y ) (a) ln + ln (b) f (x) – f (a) = f(x)(x – a) ∀ x ∈ R and f is y x x y y(x + y ) continuous at a.M th rchives 10 Best Problems Prof. then prove that (b) periodic with no fundamental period cos3 x (c) periodic with period 2 dy 3 (d) periodic with period p . D → Q 2. If q is small and positive number. challenging problems are offered with detailed solution. If b. C → S. evaluate f ′(b) and lim g (x ). Shyam Bhushan. Find the area of the triangle formed with vertices matches with exactly one entry of column II). p) ⇒ [cosa + cos(a + b)]2 + [sina + sin(a + b)]2 = 1 ⇒ 2 + 2[cos(b)] = 1 \ cosb = –1/2 Similarly.10. Let O = (0.. (a) 5.. (d) 2.c’s of two lines then roots of (3) are l1 l and 2 m1 m2 l l bf Product of the roots = 1 ⋅ 2 = m1 m2 ag ll mm \ 12 = 1 2 f /a g /b \ ll mm nn \ 12 = 1 2 = 1 2 f /a g /b h/c Q lines are perpendicular \ l 1l 2 + m 1m 2 + n 1n 2 = 0 ⇒ f g h + + =0 a b c MatheMatics tODaY | march ’15 nn 49 . Given cosa + cos(a + b) + cos(a + b + g) = 0 sina + sin(a + b) + sin(a + b + g) = 0 where b. (i) = 4 − 3 sec2 x cos2y = 4 – 3sec2x = 4 – 3(1 + tan2x) = 1 – 3tan2x sin2y = 3tan2x ⇒ sin y = 3 tan x dy = 3 sec2 x dx dy 3 \ = [ from (i)] dx cos x ⋅ cos 3x 7. m2.. 8.. if l1. m1. if f g h + + = 0. (d) : f (x. Q f : R → R is continuous at x = a and satisfies f (x) – f (a) = f(x)(x – a) ∀ x ∈ R f (x ) − f (a) ⇒ = f(x ) x −a f (x ) − f (a) ⇒ lim = lim f(x ) x −a x →a x →a ⇒ cos y ⇒ f ′(a) = f(a) lim f(x ) = f(a) x →a ⇒ f is differentiable at x = a. Prove that the straight lines whose direction cosines are given by the relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular..(2) 2 l l ⇒ ag + (af + bg − ch) + bf = 0 . cos y = ⇒ ⇒ cos 3x cos3 x = 4 cos3 x − 3 cos x cos3 x . n2 are d. lim = (0. g ∈ (0.(1) . 1) x →0 x 1 \ Area of ∆OAB = 0 − 2 − 0 = 1 square unit 2 10. 0) p cos x x→ 2 1 tan x x B = 0. (c) : f (x) = esin(x – [x])cospx sin(x – [x]) = sin{x}. y) is homogeneous function of degree n ∈ R in x. cosg = –1/2 2p \ b=g= 3 sin 2 x x = tan x and g (x ) = tan But f (x ) = 1 + cos 2 x 2 2p 2p \ f ′ = sec2 =4 3 3 p and lim g (x ) = tan = 3 2p 3 x→ 3 9. ky) = knf (x. 0) p x − 2 A = lim . 4.. al + bm + cn = 0 fmn + gnl + hlm = 0 Eliminating n. a b c sOlutiOns 1. period is 1 cospx . (c) 6... Period is 2 Hence period of f(x) is 2 3. we get . y if f (kx.(3) m m Now. 0 = (−2. n1 and l2. y) . where k > 0. he trains IIt and olympiad aspirants. sin hx x 2 Lt = x →0 x (a) e1/2 (b) 1 (c) e1/6 (d) e1/3 8. 2a] and left hand derivative at x = a is 0. The value of c in Lagrange’s Theorem for the 1 x cos . f(2) = 4. zx and xy planes respectively. 50 MatheMatics tODaY | March ’15 . 2a] → R is an odd function such that f(2a – x) = f(x). b. x ∈ [2. IIT Kanpur x y z + + = 1 meets the axes at A. (d) none of these k k k 2. 3) (c) f ′′(x) = 3 for " x ∈ (2. ∞) where a > 0. b > 0. then which of the following is definitely true? (a) f ′′(x) = 2 for " x ∈ (1. " x ∈ [a. in the interval function f (x ) = x 0. The function f defined by 1 . 3) (b) f ′′(x) = f ′(x) = 5 for some x ∈ (2. B. then equation of the plane ABC must be (a) bcx + acy + abz = 3abc (b) bcx + acy + abz =2abc (c) bcx + acy + abz = abc x y z + + =0 (d) a b c 3. x ≠ 0 . a b c B.Tech. then find left hand derivative at x = –a. The global maximum value of f(x) = log10 (4x3 – 12x2 + 11x – 3). where k4 = a2b2 + b2c2 + c2a2 ab2c 2 ba 2c 2 ca 2b2 (a) 4 . assumes its minimum value only at one point if (a) a ≠ b (b) a ≠ c (c) b ≠ c (d) a = b = 0 6. C be images of P in yz. f(3) = 9. if x is rational f (x ) = 2 is 1 . B. x = 0 [–1. If a function f : [–2a. 3] is 3 (a) − log10 3 (b) 1 + log10 3 2 3 (c) log10 3 (d) log 3 2 10 4. 3) 1. 1] is 1 (a) 0 (b) 2 1 (c) − (d) none of these 2 1 7. c > 0. 2 k k k a 2 b2 c 2 (c) . Let A. 4 . The plane 5. Let f(x) be a twice differentiable function for all real value of x and satisfies f(1) = 1. 1 (a) 0 (b) –1 (c) (d) a 2 * Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). 3) (d) f ′′(x) = 2 for some x ∈ (1. if x is irrational 3 (a) discontinuous for all x (b) continuous at x = 2 1 (c) continuous at x = 2 (d) continuous at x = 3 9.* ALOK KUMAR. " x ∈(–∞. The function f(x) = |ax – b| + c|x|. . 4 k k k a 3 b3 c 3 (b) 2 . 2 . The orthocentre of the triangle must be at. P is a point (a. C. c). his chance of winning is 31 30 (a) (b) 61 61 60 15 (c) (d) 61 61 MatheMatics tODaY | March ’15 51 .10. 7) 29 (c) a ∈( −∞. units (c) (d) sq. (1 − x )} . ∫ x ln x dx (a) x2 x2 ln x − +c 2 4 is equal to (x ≠ 0) 1 1 x x ln x + x x + c 2 4 2 x x2 (c) − ln x + +c 2 4 1 1 x x ln x − x x + c (d) 2 4 (b) −1 y x ∫ 17. If A starts the game. The differential equation representing the family of the curves y 2 = 2c(x + c ). −1 d 2 y dy dx 2 dx d2 y − (b) 2 dx −2 −1 dy dx −3 d 2 y dy −3 (d) − dx 2 dx 11. degree 3 (d) order 4. 7 (d) a ∈(3. degree 3 (b) order 2. then −p/2 (b) ae (d) a −p/2 e 2 dx is equal to 1 1 log( x 3 − 1) − log(x 6 + x 3 + 1) 9 18 2x 3 + 1 1 tan −1 − +c 2 3 3 1 1 (b) − log( x 3 − 1) + log(x 6 + x 3 + 1) 9 18 2x 3 + 1 1 − tan −1 +c 3 3 3 1 1 (c) − log( x 3 + 1) + log(x 6 − x 3 + 1) 9 18 2x 3 + 1 1 − tan −1 +c 3 3 3 (d) none of these log x 2 0 1− x p (a) − log 2 2 p (c) − log 2 8 dx is equal to p (b) − log 2 4 p log 2 (d) 4 15. A and B throw alternatively with a pair of dice. Let { f (x ) = min⋅ x + 1. unit 6 6 11 7 sq. degree 4 19. –3) 12. units 6 6 18. −3) ∪ 3. 1] is 1 5 (a) (b) sq. d2x dy 2 2 (a) d y dx 2 (c) 1 = 14. where c is a positive parameter. If the function f(x) = x3 + 3(a – 7)x2 +3(a2 – 9)x – 1 has a point of maximum at positive values of x. (a) x5 ∫ 1− x 9 tan . unit sq. ∫ (x − 3){sin −1 (ln x ) + cos −1 (ln x )}dx is equal to p (a) (b) 0 (x − 3)3/2 + c 3 (c) does not exist (d) none of these 16. degree 2 (c) order 3. then 29 (a) a ∈ −∞. y(0) > 0. A wins if he throws a sum 6 before B throws 7 and B wins if he throws a 7 before A throws 6. a > 0. is (a) order 1. then area bounded by f(x) and x-axis in [–1. 7 (b) a ∈ (–∞. If x2 + y2 = a e y′′(0) = a p/2 (a) e 2 −2 − p/2 e (c) a 13. ∞) ∪ (–∞. If the probability 2 that no two adjacent persons are selected is . Unit vector ^c is inclined at an angle ‘q’ to unit ^ vectors a^ and b.. b and c are three non-coplanar unimodular vectors. 4 2 4 4 29. 3 persons are selected at random. b and c is (a) (c) 30. Two integers x and y are chosen from the set {0. the probability that |x – y| ≤ n. each inclined with other at an angle 30°. 3.. then ∠C = (a) 30° (b) 150° (c) 30° or 150° (d) none of these . If in the triangle ABC. with replacement.. then t = 4 (a) (b) a + b + c (a + b − c)abc (c) abc (d) abc(a + b + c) 26. C of an acute angled DABC to the opposite sides meet the EF circumcircle at D. then = BC (a) sinA (b) cosA (c) 2sinA (d) 2cosA 33. 2. then volume of tetrahedran whose edges are a . For ^c = m(a^ + b) + n (a^ × b). 3sinA + 4cosB = 6 and 4 sinB + 3cosA = 1. which are perpendicular to each ^ ^ other. (n ∈ N) is (a) (c) 3n2 + 3n + 1 (2n + 1)2 3n2 + 1 (2n + 1)2 (b) (d) 3n2 + 3n (2n + 1)2 n2 + n + 1 2 (2n + 1) 25. 2. If D is the area of a triangle whose sides are t a. In DABC. From a group of n persons arranged in a circle. (d) . B. 3 3 −5 12 (b) 5 2 +3 12 ∞ (d) 3 3 +5 12 5 2 −3 13 3 ∑ cot −1 r 2 + 4 = __________ r =1 (a) tan–1(2) (c) tan–1(1) (b) cot–1(2) p 1 + tan −1 (d) 2 2 31. b. 2n}. the probability that exactly one pair of shoes is obtained is 16 4 2 8 (a) (b) (c) (d) 21 21 21 21 22. 1. If a . 3. m. A box contains 5 pairs of shoes. p 2 2 p p p 3p (c) .The probability that at least 7 consecutive heads show up is 9 7 1 1 (a) (b) (c) (d) 256 256 64 128 23. Six fair dice are thrown independently . 28.. is 2 (a) (a – b)2 (c) (a + b)2 (b) a2 + b2 (d) a2 – b2 52 MatheMatics tODaY | March ’15 AB 3 and D is a = AC 4 point lying on BC such that AB ^ AD and D divides BC externally in the ratio 3 : 2. then 7 n= (a) 6 (b) 7 (c) 8 (d) 9 21. . 6) is 5 5 125 25 (a) (b) (c) (d) 36 72 144 72 24. F respectively. 5. q lies in p p (a) 0. The least value of ‘c’ for which a2 b2 + =c sin x 1 − sin x p (a > b > 0) has at least one solution in 0. c and D ≤ . The areas of a circle and a regular polygon of n sides and of equal perimeter are in the ratio p p p p (a) cot : (b) cos : n n n n (c) tan p : p (d) none of these n n 32. E. An unbiased coin is tossed 12 times . (b) . The altitudes from the vertices A. n real.20. then ∠A = (a) 30° (b) 75° (c) 45° (d) 60° 27. 1..The probability that there are exactly 2 different pairs (A pair is an ordered combination like 2. If 6 shoes are selected at random.. MatheMatics tODaY | March ’15 53 . b.. The smallest positive integer ‘n’ such that 1 1 + + sin 45° sin 46° sin 47° sin 48° 1 1 . A circle is drawn through P and Q so that the origin ‘O’ is outside. –c) ⇒ Equation of plane ABC bcx + acy + abz = abc 3. 3) x <0 b − (a + c)x . we have b 1 × y ′ + 2 × y0 = 3 3 b3 (a 2 + c 2 ) ba 2c 2 ⇒ y′ = b − = a 2b 2 + b 2 c 2 + c 2 a 2 k4 2. then 2r is equal to PQ + RS PQ ⋅ RS (a) (b) 2 2 2 2PQ ⋅ RS (c) (d) PQ + RS PQ + RS 2 40. Now if y coordinate of orthocentre is y′ then we know that centroid divide the line segment joining the orthocentre and circumcentre internally in the ratio 2 : 1.. Then the length of PQ is (a) 5 2a (b) 4 2a (c) 3 2a (d) 2 2a 54 MatheMatics tODaY | March ’15 sOlutiOns 1. + = is sin133° sin134° sin n° (a) 2 (b) 1 (c) 3 (d) 4 35. Consider the parabolas y2 = 4ax and x2 = 4ay (a > 0) have a common tangent at the points P and Q respectively. (–a.(i) + + =1 a b c and (x – a)2 + y2 + z2 = x2 + (y – b)2 + z2 = x2 + y2 + (z – c)2 From these relations we can easily get x= y= a 2 − b2 + 2by c 2 − b2 + 2by . (a) : Let (x.z = 2a 2c b(a 2b2 + b2c 2 ) 2k 4 where k4 = a2b2 + b2c2 + c2a2 b3 (a 2 + c 2 ) = y0 2k 4 We can write similar expressions for x and z. Angle between k and the line of intersection of the planes r ⋅ (i + 2 j + 3k ) = 0 and r ⋅ (3i + 3j + k ) = 0 is equal to 2 (a) cos −1 3 −1 3 (b) cos 122 7 1 (c) cos −1 (d) cos −1 122 3 38. (c) : Since A is image of P(a. A must be the point. (b) : Clearly f(x) is increasing in [2. g′(x) = 0 for at least one x ∈ (2. then min {A + B. we get g′(x) = 0 for at least one x ∈(1. b . b 5. then the expression r be any arbitrary (a × b ) × (r × c ) + (b × c ) × (r × a ) + (c × a ) × (r × b ) is equal to (a) [a b c]r (b) 2[a b c]r (c) 3[a b c]r (d) none of these 37. B + C.. C + A} (a) ≤ 30° (b) < 30° (c) < 60° (d) ≤ 60° 36. 3) \ f ′′(x) = 2 for some x ∈ (1. RQ intersect at a point X on the circumference of the circle. (b) : f (x ) = b + (c − a)x . 0 ≤ x < a b x≥ (a + c)x − b . z) be the circumcentre of space triangle ABC then x y z . Then the length of the tangent to the circle from ‘O’ is (a) 15 (b) (c) (d) 8 8 15 39. RS are the tangents of the extremities of a diameter PR of a circle of radius ‘r’ such that PS. a . 2) Also. b. c) in yz plane. 3) \ g′′(x) = 0 for at least one x ∈ (1. c be three non coplanar vectors and vector. Similarly B and C be the points (a. Let a . Let PQ. b. c).. The parabola y = x2 – 8x + 15 cuts the x-axis at P and Q... y. In DABC.34. sinA + sinB + sinC ≤ 1. c) and (a. (d) : Let g(x) = f(x) – x2 Using Rolle’s Theorem. 3] f(x)max = f(3) = log10 30 = 1 + log10 3 4. –b. 3 n→∞ f ( − a ) − f ( − a − h) h h→0 9. we get. (c) : Let l = Lt x →0 x 1 log l = Lt x →0 x 2 sin hx log by L-Hospitals Rule x ⇒ l = e1/6 8. (d) : = dy dx Differentiating with respect to y. Similarly in case of irrational. when x = 0 ⇒ y = ae d2 y −2 −2 − p/2 = = e 2 a dx at x =0 y 13. x = 1 ⇒ q = 2 p /2 log sin q cos qdq \ I= ∫ 2 0 1 − sin q z Let 2 = −1 dx dy 10. (c) : f(x) = x3 + 3(a – 7)x2 + 3(a2 – 9)x – 1 f ′(x) = 3x2 + 6(a – 7)x + 3(a2 – 9) The roots of f ′(x) = 0 are positive and distinct which is possible if (i) b2 – 4ac > 0 ⇒ 36(a – 7)2 – 4(3)(3)(a2 – 9) > 0 29 ⇒ a< 7 (ii) Product of Roots > 0 ⇒ a2 – 9 > 0 (iii) Sum of Roots > 0 ⇒ a – 7 < 0 ⇒ a < 7 29 ⇒ From (i). (iii) a ∈(−∞. (ii). 16. + ∫ dz 3 3 1− z 3 1+ z + z2 1 1 = − log( x 3 − 1) + log(x 6 + x 3 + 1) 18 9 2x 3 + 1 1 − tan −1 +c 3 3 3 14. (c) : Differentiating with respect to x two times. 11. −3) ∪ 3. 1 1 x x ln x − x x + c 2 4 MatheMatics tODaY | March ’15 55 . [3. (d) : Case I : If x > 0. (d) : \ Clearly f(x) is not differentiable at x = 0. e = φ e Hence the given integral does not exist. then |x| = – x ⇒ ⇒ ∫ x ln x dx = ∫ x ln xdx = ∫ x ln x dx = − ∫ x ln(− x )dx = − x2 x2 ln x + +c 2 4 ⇒ Combining both cases. then 3x2dx = dz \ A Bz + C + (1 − z )(1 + z + z ) 1 − z 1 + z + z 2 1 (z − 1) 1 1 dz 1 3 I= ∫ . "n ∈N n n n 1 ⇒ Lt f (xn ) = .6. we get d2 y dx 2 = 2(x 2 + y 2 ) (x − y )3 1 zdz 3 ∫ (1 − z )(1 + z + z 2 ) I= using p /2 a ∫ 0 ≡ log sin qdq = a p /2 ∫ 0 p log sin − q dq 2 ∫ f (x)dx = ∫ f (a − x)dx 0 0 p \ I = − log 2 2 15. then |x| = x x2 x2 +c ln x − 2 4 Case II : If x < 0. 1 sin hx x 2 7. (b) : Putting x3 = z. If x is rational \ " n ∈ N $ an irrational number xn such that 1 1 1 x − < x n < x + ⇒ x n − x < . (a) : Put x = sinq then dx = cosqdq p And x = 0 ⇒ q = 0. 7 12. (a) : There lie infinitely many rationals as well as infinitely many irrationals between any two rational or irrational. (c) : x − 3 is defined only when x ≥ 3 and sin–1(lnx) + cos–1(ln x) is defined only when. 1 −1 ≤ ln x ≤ 1 ⇒ ≤ x ≤ e e 1 Then. ∞ ) ∪ . (a) : f ′(−a − ) = Lt − f (a) + f (a + h) − f (a) + f (2a − (a − h) = Lt h h h→0 h→0 − f (a) + f (a − h) = Lt =0 h h→0 = Lt p /2 From given eqn. units 6 1 ∫ 1 − x dx 0 18. 3. Then find minimum value of c.. after which again a head or tail can show up. degree = 3 5 19. n + 1 .. of outcomes = 66. (d) : \ \ P. (a) : x and y can be any one of (2n + 1) numbers given |x – y| ≤ n ⇒ x – n ≤ y ≤ x + n Hence number of possibilities of y. 2. (c) : Solve as >1 = 0and get sin x = a+b a −b dx is not possible..17.. 36 5 B’s chance of losing in a throw = 6 A can winning the game = 5 31 5 5 31 5 31 5 5 + × × + × × × × + . 36 36 6 36 36 6 36 6 36 2 30 5 155 155 = = 1 + .V. 1.. + + 36 216 216 61 20. + ⋅ 7 2 2 2 27 2 1 5 7 = 1 + = 27 2 28 23. n + 3. 2n.. .... (d) : The sequence of consecutive heads may starts with 1st toss or 2nd toss or 3rd toss .. 5 × 16 8 Hence probability = = 10 C6 21 22.. 2(n + 1 + n + 2 + . 27. n +1 respectively. of D = 3c − 2b 2b 1 AB ⋅ AD = 0 ⇒ cos A = = \ A = 60° 3c 2 .. 2n. + 2n) + 2n + 1 \ Probability = (2n + 1)2 \ Probability = 1 7 = 25.. (c) : From 10 shoes.. 36 1 B’s chance of winning in a throw = 6 31 A’s chance of losing in a throw = ...... or at 6th toss. (c) : P(no two adjacent persons are selected) n(n − 4)(n − 5)/6 (n − 4)(n − 5) 2 = = = n(n − 1)(n − 2)/6 (n − 1)(n − 2) 7 ⇒ 5n2 – 57n + 136 = 0 ⇒ n = 8 21. for x = 0. 56 MatheMatics tODaY | March ’15 1 1 1 1 + ⋅ + . Other 4 shoes can be selected in 4C4(24) = 16 ways. In any case . (a) : y 2 = 2cx + 2c 3/2 ⇒ 4y(y′)3 + 4xyy′ – y2 – 4x2(y′)2 = 0 ⇒ order =1...... (d) : 3n2 + 3n + 1 (2n + 1)2 (s − a) + (s − b) ≥ (s − a)(s − b) 2 c ≥ (s − a)(s − b) Similarly others 2 abc D 2 (a + b + c)abc \ ≥ ⇒ D≤ s 8 4 a a dc 26.if it starts with rth throw.. n..... 2n +1. (c) : x + 1.. n –1.. n + 2. Number of ways of choosing 4 other different numbers is 6C2 and choosing 2 out of remaining 4 can be done in 4C2 ways..... the first (r – 2) throws may be head or tail but (r – 1)st throw must be tail.. 2n –1. 6 can be selected in 10C6 ways and if there is to be one pair among them it can be selected in 5C1 ways. 0 < x ≤ 1 \ required area = 0 ∫ (x + 1)dx + −1 7 = sq....... (b) : Total no. 2n are n + 1... .. (1 − x )} = 1 − x . . of which 2 are alike and 2 are alike is 2! 2! 6! 6C × 4 C × 2 2 2 ! 2 ! = 25 Required probability = 6 72 6 24.. (a) : A’s chance of winning in a throw = . Also number of ways of arranging 6 numbers 6! . − 1 ≤ x < 0 f (x ) = min{x + 1.... (c) : The points P and Q are (3. (a) : −1 ≤ cos q ≤ 2 a ⋅c b ⋅c c ⋅c 1 2 3 1 1 cot −1 r 2 + = tan −1 r + − tan −1 r − 2 2 4 \ ∞ 3 p 1 ∑ cot −1 r 2 + 4 = 2 − tan −1 2 r =1 1 = cot −1 = tan −1 (2) 2 31. (d) : } The equation of the circle is x2 + y2 – 8x + 2fy + 15 = 0. (b) : sin(1°) = sin((x + 1)° – x°) sin(1°) \ = cot x ° − cot(x + 1)° sin x ° sin(x + 1)° Using this. we can arrive n = 1. (c) : ^c ⋅ a^ = cos q ⇒ m = cos q Also ^c ⋅ ^c = 1 ⇒ 2m2 + n2 = 1 ⇒ n2 = 1 – 2cos2q ≥ 0 ⇒ a ⋅ a a ⋅b 1 29.Q(–2a.e.. we get 1 24sin(A + B) = 12 ⇒ sinC = 2 ⇒ either C = 30° or 150° But if C = 150°. A < 30° so that 3 3sinA + 4cosB < +4 < 6. (b) : Line of intersection of the given planes will be parallel to (i + 2 j + 3k ) × (3i + 3j + k ) { 38. i. (b) : Given expression can be written as 3 a b c r − a b r c + b c r a + c a r b = 3 a b c r − a b c r = 2 a b c r 37. 0). (c) : Points of contacts are ends of latusrectum i. = tan q 2 PR PR 39. (a) : V 2 = b ⋅a b ⋅b 36 c ⋅a c ⋅b 30. (c) : ‘s’ – side of polygon. a contradiction. (a) : Squaring and adding the given equations. P(a. (5. Length of tangent from origin = c = 15 PQ RS p = tan − q = cot q . a) ⇒ PQ = 3 2 a nn IMPORTANT EXAMINATION DATES 2015 4 april JEE Main (Offline) 29-30 april Karnataka cET 10 – 11 april JEE Main (Online) 3 May aIPMT 8 – 19 april VITEEE 10 May cOMED K 18-19 april WB JEE 16 May UPSEaT 19 april MGIMS 24 May JEE advanced 20-21april Kerala PET 1 June aIIMS 22-23 april Kerala PMT 7 June JIPMEr MatheMatics tODaY | March ’15 57 .28. 2 40. – 2a). (b) : \ PQ ⋅ RS PR 2 = 1 ⇒ PR 2 = PQ ⋅ RS \ 2r = PQ ⋅ RS \ ∠ADE = ∠ABE = 90° – A ∠ADF = ∠ACF = 90° – A ∠D = 180° – 2A EF 2R sin D = = 2 cos A BC 2R sin A 33. 2pr = ns. 0) \ p p : n n 32. r – radius of the circle. \ Required ratio = pr 2 p ns 2 cot n 4 = tan 34. 35. B + C < 30° 36.e. (b) : Without loss of generality assume A ≥ B ≥ C We have b + c > a ⇒ sinB + sinC > sinA ⇒ sinA + sinB + sinC = 2sinA 1 \ It follows sinA < and A ≥ A + B + C = 60° 2 3 which gives A > 150°. ( gof ) π = g 5 = 1 8 4 A letter is to come from either LONDON or CLIFTON. The word CLIFTON contains 6 types of consecutive letters (CL. ND.P.Nihal Pandey (Bihar) 58 MatheMatics tODaY | March ’15 1 x3 + x + 1 ∫ x 2 + 2 x + 1 dx −1 π π f (x ) = sin2 x + sin2 x + + cos x cos x + 3 3 1 2π = 2 sin2 x + 1 − cos 2 x + 2 3 Evaluate : Ans.Niyati Shukla (U. Now 1 P (E1 ) = = P (E2 ) 2 2 P (E | E1 ) = and P (E | E2 ) = 1 5 6 By Bayes’ theorem. From the serious to the silly. π If f (x ) = sin2 x + sin2 x + 3 π + cos x cos x + 3 5 and g = 1 . ON) of which there is one “ON”. π π + cos 2 x + + cos 3 3 π 3 1 2π = 2 sin2 x − cos 2 x + + cos 2 x + + 2 3 3 2 1 3 = 2 sin2 x + cos 2 x + 2 2 1 3 5 = 1 + = 2 2 4 Therefore f(x) = 5/4 for all x. The word LONDON contains 5 types of consecutive letters (LO. We have.) 1 1 x 3 + ( x + 1) x3 + x + 1 = dx dx ∫ x2 + 2 x + 1 ∫ 2 −1 −1 ( x + 1) = 1 x3 ∫ dx + 2 −1 ( x + 1) =0+ 1 1 ∫ ( |x|2 = x2) 1+ x 2 −1 ( x + 1) dx 1 ∫ 1 + x dx −1 x3 is an odd function 2 ( x + 1) 1 = 2∫ 1 dx 1+ x = 2∫ 1 1 dx 1+ x is an even function 1+ x 0 1 0 1 = 2[log |1 + x|]0 = 2(log 2 – log 1) = 2 log 2 nn . TO. Hence 2. 4 π then find the value of ( gof ) . The postal mark on the letter legibly shows consecutive letters “ON”. E2 : Letter coming from CLIFTON. ON) of which there are two ON’s.Richa Kaushik (Hyderabad) Ans. What is the probability that the letter has come from LONDON ? . E : Two consecutive letters ON. Let the event be defined as E1 : Letter coming from LONDON. Ans. the controversial to the trivial. LI. the team will tackle the questions. The best questions and their solutions will be printed in this column each month. .Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. 1. FT. easy and tough. IF. 8 . DO. 1 2 × 12 2 5 = P (E1 | E ) = 1 2 1 1 17 × + × 2 5 2 6 3. ON. Who is the worst player? (a) The woman (b) Her son (c) Her brother (d) Her daughter (e) No solution is consistent with the given information. d – s = 1. III. Of the three equations 2 2 I.Exam on 10th May * ALOK KUMAR. In the unit circle shown in the figure to the right. N is the mid-point of BC. her brother. Given an equilateral triangle with side of length s. and AN and CM intersect at O. consider the locus of all points P in the plane of the triangle such that the sum of the squares of the distances from P to the vertices of the triangle is a fixed number a. The ratio of the area of AOCD to the area of ABCD is C D O A M N B (a) 5/6 (e) (b) 3/4 (c) 2/3 (d) 3 /2 ( 3 − 1) 2 3. M is the mid-point of AB. Chords MP. MatheMatics tODaY | march ’15 59 . her son and her daughter are chess players (all relations by birth). The worst player’s twin (who is one of the four players) and the best player are of opposite sex. he trains iit and Olympiad aspirants. ds = 1. B. II. What is the smallest integer larger than ( 3 + 2 )6 ? (a) 972 (e) 968 (b) 971 (c) 970 (d) 969 * alok Kumar is a winner of iNDiaN NatiONal MatheMatics OlYMpiaD (iNMO-91). A woman. In the adjoining figure AB and BC are adjacent sides of square ABCD. The worst player and the best player are of the same age. This locus (a) is a circle if a > s2 (b) contains only three points if a = 2s2 and is a circle if a > 2s2 (c) is a circle with positive radius only if s2 < a < 2s2 (d) contains only a finite number of points for any value of a (e) none of these 5. 4. II and III 2. IIT Kanpur PART-A Multiple choice test 1. d − s = 5 those which are necessarily true are (a) I only (b) II only (c) III only (d) I and II only (e) I. PQ and NR are each s units long and chord MN is d units long.Tech. chords PQ and MN are parallel to the unit radius OR of the circle with centre at O. 1/2). 5 4 3 2 8. where P is on edge AB and Q is on edge CD.6. Find the least possible distance between a pair of points P and Q. The number of real solutions to the equation x = sin x is 100 (a) 61 (b) 62 (c) 63 (d) 64 (e) 65 15. is the word CONTEST spelled out as the path is traversed from beginning to end ? (a) 63 (b) 128 (c) 129 (d) 255 (e) None of these C COC CONOC CONTNOC CONTETNOC CONTESETNOC CONTESTSETNOC D Q A C P B (a) 1 2 (b) 3 4 (c) 2 2 (d) 3 2 3 3 11. For how many paths consisting of a sequence of horizontal and/or vertical line segments. A second circle is tangent internally to the circumcircle at T and tangent to sides AB and AC . the polynomial g(x) ? 5 5 (a) 6 (b) 5 – x cx 3 . x ≠ − . B. then the measure of ∠A44A45A43 equals (e) Not uniquely determined by the given A3 information A1 (a) 30° (e) 120° (b) 45° 13. 0). (d) . with each segment connecting a pair of adjacent letters in the diagram below. What is the 5 5 3 3 3 5 3 3 remainder when the polynomial g(x12) is divided by 2 4 (e) . z) satisfy the equations x + 2y + 4z = 12 . (c) . each time making a 90° turn counter clockwise and travelling half as far as in the previous move. then c is mid-point of line segment An An + 1 for all positive (a) –3 (b) –3/2 (c) 3/2 (d) 3 integers n. First it moves 1 unit right to (1. x≠− . C and D each have length one. If it continues in this fashion. Then it makes a 90° turn counterclockwise and travels 1/2 a unit to (1. (b) . 60 MatheMatics tODaY | march ’15 (e) 14. xy + 4yz + 2xz = 22 xyz = 6? (a) None (b) 1 (c) 2 (d) 4 (e) 6 7. A bug (of negligible size) starts at the origin on the coordinate plane. 12. The polynomial x2n + 1 + (x + 1)2n is not divisible by x2 + x + 1 if n equals (d) 64 (a) 17 (b) 20 (c) 21 (e) 65 A5 A6 A4 (c) 60° A2 (d) 90° 10. The edges of a regular tetrahedron with vertices A. y. Let g(x) = x + x + x + x + x + 1. to which of the following points will it come closest ? 4 2 2 2 2 1 2 4 (a) . c a constant. satisfies f(f(x)) = x for all real 2x + 3 2 9. If the function f defined by f (x ) = (c) 4 – x + x2 (d) 3 – x + x2 – x3 2x + 3 2 cx 3 (e) 2 – x + x2 – x3 + x4 f (x ) = . Equilateral DABC is inscribed in a circle. How many distinct ordered triples (x. If DA1A2A3 is equilateral and An + 3 is the number x except –3/2. T is the point such that OM = MT. Find the number of quadratic polynomials. where a is a rational number. 24. In the adjoining diagram.. ∠A = 30°. Prove it. On the line OM. and AC = 18. G and F AC are concyclic. At each of the eight corners of a cube write + 1 or – 1 arbitrarily. If side BC has length 12. There are two urns each containing an arbitrary number of balls.0 (c) 3.5 (b) 3. 2. 28. and MN is parallel to BC.. the converse is also true.answer type test 19. Is it possible to arrange the numbers + 1 and –1 at the corners initially so that this final sum is zero ? MatheMatics tODaY | march ’15 61 . FB EC PD 27. show that the triangles ABC and GAP are similar.0 (e) 4. which satisfy the following conditions: (i) a. If. 1999} and (iii) x + 1 divides ax2 + bx + c. y) satisfying 3x2 + 3y2 – 4xy + 10x – 10y + 10 = 0. then the perimeter of DAMN is A M (a) 30 (e) 42 B (b) 33 O N (c) 36 C (d) 39 PART-B short . The equation of the line is x = (a) 2. If 11 + 11 11a2 + 1 is an odd integer. 0). Show that AT = 2BC. Show that a2 = b(b + c). E. Both are non-empty to begin with.5 (d) 4. CA. Show that after performing these operations finitely many times. in base 10) ends exactly in 1993 zeroes. where 0 < a < . BO bisects ∠CBA. 25. AB in AF AE AP D. Then. 1) in the xy-plane into two regions of equal area. Let G be the centroid of the triangle ABC in which the angle at C is obtuse and let AD and CF be medians from A and C respectively onto the sides BC and AB. c are distinct. If the four points B. Find the real roots of the equation 1 1 1 x 2 + 2ax + = −a + a2 + x − . D. P is BC a point on the line BG extended such that AGCP is a parallelogram. .) 26. on each of the six faces of the cube write the product of the numbers written at the four corners of that face.at points P and Q. 16 16 4 20. ax2 + bx + c. F respectively. 1) and (9. further. c ∈ {1. O is the orthocentre and M is the mid-point of BC. 21. 22. 23. b. ∠A is twice ∠B. (1. Find the real points (x. 29. In a triangle ABC. then c is (a) 0 (b) 1 (c) 3 (d) 4 (e) Not uniquely determined 17. Suppose P is an interior point of a triangle ABC and AP. Show that + = . Add all the fourteen numbers so written down. (In fact. show that > 2 . If the base 8 representation of a perfect square is ab3c. A vertical line divides the triangle with vertices (0. CO bisects ∠ACB. CP meet the opposite sides BC. then segment PQ has length A Q P C B T (b) 6 3 (a) 6 (e) 9 (d) 8 3 (c) 8 16.. Show that there is a natural number n such that n! when written in decimal notation (that is. In an acute-angled triangle ABC. BP. 3. prove that a is a perfect square. (ii) a. b. BC = 24. where a ≠ 0. (ii) Double the number of balls in any one of them.5 18. both the urns can be made empty. We are allowed two types of operations: (i) Remove an equal number of balls simultaneously from both urns. If AB = 12. s = 2sin18° and d = 2sin54° d = s +1 = 62 MatheMatics tODaY | march ’15 It follows from this and trigonometric identities that d = 2sin54° = 2cos36° = 2( 1– 2sin218°) = 2 – s2 and s = 2sin18° = 2cos72° = 2(cos236° – 1) = d2 – 2 Adding d = 2 – s 2 and s = d 2 – 2. Then 2 5 +1 d = s +1 = . have base angles of measure 1 (180° − 36°) = 72° . In parallelogram ORNT. TN of intersecting chords MN and PL satisfy (PT)(TL) = (MT)(TN). s. chords QN and KM have length s. In the adjoining figure.sOlUtiONs 1. we find that s2 + s – 1 = 0. d − s = (d + s)(d − s) = d + s = 5 . The area of AOCD 6 is obtained by subtracting the areas of triangles AOB and COB from that of the square. Since PT = OP – OT = 1 – s. and TL = OL + OT = 1 + s. d + s = 5 2 2 2 and d − s = 5 . 1/3 the side length. TN = OR = 1.e. So DPMT is isosceles with MT = MP = s. Since O is the intersection of the medians of DABC. so. area of DCOB = s 2 . d – s = 1. (c) : 1st Solution: In the adjoining figure diagonals AC and DB are drawn. of the square. We saw that ∠MPO = 72°. II and III are all true by giving geometric arguments for I and II and then showing algebraically that III is a consequence of I and II. 1 s = ( 5 − 1) . The segments PT. because MT || KO. TL and MT. so area of 1 2 AOCD = s 2 − s 2 = s 2 3 3 2nd Solution : Introduce coordinates with respect to which AB is the unit interval on the positive x-axis. the altitude of DAOB from O is 1/3 the altitude of DABC from C. goes into chord NR. Then chord PQ. 2. Hence 1 area of DAOB = (area of DABC ) 3 1 1 2 1 2 = s = s 3 2 6 D C P O A M N B 1 Similarly. and TO = NR = s. which has one positive root. and MN of length d subtends an angle 3 . Substituting s + 1 for d in d = 2 – s 2 . Each chord of length s subtends an angle 36° at the centre O. Therefore. 2 Now rotate the entire configuration clockwise through 72° about O. now ∠MTP = 72° also. parallel to diameter PL. whose positive root is s = ( 5 − 1). i. 36° = 108°. AD is the unit interval on the positive y-axis. (e) : We shall show that I. this equation takes the form (1 – s)(1 + s) = s · 1 or 1 – s2 = s Multiplying equation s2 + s = 1 is equivalent to 1 s2 + s – 1 = 0. ds = 1. parallel to diameter KR. Therefore d = MT + TN = s + 1. Now. 2 Therefore 5 1 + . The five equal chords of length s in the semicircle with diameter KR subtend central angles of 180°/5 = 36° each. Area of AOCD = Area of DACD+ Area of DAOC 2 1 1 1 2 = + 3 3 = 2 2 1 1 3 . and 2 2 2 2 III. so that I. Denote by T the intersection of MN and PL. d – s = 1. we obtain d + s = d2 – s2 – (d + s)(d – s). The five isosceles triangles. each with base s and opposite vertex at the centre O. But the woman and her daughter cannot be of the same age. If the brother is the worst player. (c) : Instead of trying to compute ( 3 + 2 )6 directly. 4. 2 . 5. v. Consider the polynomial p(t) = (t – u)(t – v)(t – w). Since the change of variables (1) is one-to-one. This is consistent with the given information. the son must be the daughter’s twin. uv + vw + uw = 11. When (a + b)2k and (a – b)2k are expanded by the binomial theorem. and u. their even-powered terms are identical. and hence the woman’s twin. her brother. 2 . the original system has 6 distinct solutions (x. the locus is a single point if a = s2. But the woman and her son cannot be of the same age. 4. 2. 2. in fact. we compute something slightly larger and easier to compute. Those of O are . y) in the coordinate system in which the vertices of the s s 3 equilateral triangle are (0. y. Remark: Since the area of DABN is 1/4 of the area of the square. 2 2 Then P belongs to the locus if and only if 2 2 s s 3 2 . this triple is a solution to system (2). p(t) = (t – 1)(t – 2)(t – 3). 6. But the woman and her daughter cannot be of the same age. + b2k . (6. 2 4 3. 0) and . 970 is the smallest integer larger than ( 3 + 2 )6 . b = 2 . the woman must be his twin. namely we compute ( 3 + 2 )6 + ( 3 − 2 )6 . If it is not hard to see that p(t) = 0 has three distinct solutions. (a) : Let point P have coordinates (x. 1). . Conversely. 2 . w) is a solution of the system (2). But the woman and her daughter cannot be of the same age. 2 This substitution yields the transformed system (2) u + v + w = 6. equivalently. If the brother is the worst player. cannot be of the same age as the son. because many terms cancel. The assumption that any of the other players is worst leads to a contradiction: If the woman is the worst player. her brother must be her twin and her daughter must be the best player. So the triple (1. if and only if a = (3x 2 − 3sx ) + (3 y 2 − s 3 y ) + 2s 2 a − 2s 2 3 2 2 s s 3 s2 = x − + y − − . The best player is then the son. Then (3) p(t) = t3 – 6t2 + 11t – 6. 2 6 3 a − s2 2 2 s s 3 = x − + y − 2 3 6 Thus the locus is the empty set if a < s2. 2 This principle. (s. The best player must then be the brother. 2 . it is clear that the desired ratio r satisfies 1 1 < r < . Only (c) fulfills this condition. k = 3 yields ( 3 + 2 )6 + ( 3 − 2 )6 = 2[27 + 15(18 + 12) + 8] = 970 Since 0 < 3 − 2 < 1 .. if the roots of p(t) = 0 are listed as a triple in any order. 0). 1. 1. 3. the woman must be his twin. y = v. a = x2 + y2 + (x − s ) + y2 + x − + y − 2 2 or. z): (2. If the daughter is the worst player. 1) or 6. 4.The rows of the determinant are the coordinates 2 1 of O and C. z = w . w are the solutions of p(t) = 0. because O is 3 3 the intersection of medians of DABC. applied with a = 3 . MatheMatics tODaY | march ’15 63 . 3) and each of its permutations satisfies the system (2). uvw = 6. since the brother and the son could be of the same age. v. 3 3 1 1 2. the daughter must be his twin. where (u. and the locus is a circle if a > s2. (e) : We observe that we can find a system of symmetric equations by the change of variables 1 (1) x = 2u. 2. (b) : If the son is the worst player. The best player is then the son. and their odd-powered terms differ only in sign. The best player must be the woman.. so their sum is 2k 2 a2k + a2k −2b2 + . 3. that is. (x – 1)g(x) = x6 – 1. for example. and so does the line through C and D. Now M is the foot of the perpendicular to AB from any point Q on CD. Since g(x)(x – 1) = x6 – 1. 30°. 120°. if P is only point on AB. We claim that M and N are the unique choices for P and Q which minimize the distance PQ. MQ < PQ unless P = M. we have g(x12) = (x12)5 + (x12)4 + (x12)3 + (x12)2 + x12 + 1 = (x6)10 + (x6)8 + (x6)6 + (x6)4 + (x6)2 + 1 Subtracting 1 from each term on the right yields the equation g(x12) – 6 = [(x6)10 – 1] + [(x6)8 – 1] + . . if we spell TSETNOC. respectively.e. + [(x6)2 – 1] Each expression on the right is divisible by x6 – 1. The next cycle yields four triangles. they lie in S. Therefore. DA5A6A7 is again equilateral. We may therefore write g(x12) – 6 = (x6 – 1)P(x) where P(x) is a polynomial in x6. Since the degree of the remainder is less than that of the divisor. each similar to the corresponding triangle in the previous cycle. Expressing x6 – 1 in terms of g(x). if R(x) = 6 for all x.. starting at the bottom centre and traversing sequences of horizontally and/or vertically upward directed segments. The count becomes easier still if we take advantage of the symmetry of the figure and distinguish those paths whose horizontal segments are directed to the lift (see figure) from those whose horizontal segments are directed to the right. Since there are 6 steps. 90° respectively. (c) : Let M and N be the mid-points of AB and CD. respectively. To show this we consider the set S of all points equidistant from A and B. Starting at the bottom corner “T” in our figure.. . the five zeros of g(x) are –1 and the other four (complex) sixth roots of unity. respectively. we have at each stage of the spelling the two choices of taking the next letter from above or from the left. i.7. 90°. Since we have counted the central column twice.. Since ∠A1A2A3 = 60° and A2A4 and A2A5 have the same length. then a6 = 1. Therefore DAn An + 1 An + 2 ~ DAn + 4 An + 5 An + 6 with An and An + 4 as corresponding vertices. we arrive at g(x12) = (x – 1)g(x)P(x) + 6. 30°. g(a12) = g(a) Q (a) + R(a). (e) : Triangle A2A3A4 has vertex angles 60°. since ∠A4A5A6 = 60° and A5A6 and A5A7 have the same length. C CO CON CONT CONTE CONTES CONTEST 8. where 64 MatheMatics tODaY | march ’15 Q(x) is a polynomial and R(x) is the remainder we are seeking. Then DA 4A 5A6 has vertex angles 30°. DA3A4A5 has vertex angles 30°.. Our count is easier if we go from the end to the beginning of each path. this leads 10 26 paths in this configuration..6 = R(a). By definition of the function g. (e) : All admissible paths end at the centre “T” in the bottom row of the diagram. DA2A4A5 is equilateral. and this holds for five distinct values of a.. 2nd Solution: Write g(x12) = g(x)Q(x) + R(x). 9. But the polynomial R(x) – 6 of degree less than 5 can vanish at 5 places only if R(x) – 6 = 0 for all x. the remainder is 6. A3 A5 A6 A7 A1 A4 A2 Finally. + x + 1) = xn + 1 – 1 Thus. Therefore g(a12) = g[(a6)2] = g(1) = 6 On the other hand. (a) : 1st Solution: We shall use the identity (x – 1)(xn + xn – 1 + . there are altogether 2·26 – 1 = 127 distinct paths. Therefore. These two sets have the central vertical column in common and contain an equal number of paths. Since C and D are equidistant from A and B. we know that the degree of R(x) is at most 4. Thus ∠A44A45A43 = ∠A4A5A3 = 120° 10. S is the plane perpendicular to AB through M. 60°. When this is divided by g(x). We get 26 paths also in the symmetric configuration.. so if a is a zero of g(x). 5 = 2nd Solution: The line segments may be regarded as a complex geometric sequence with a1 = 1 and r = i/2. Indeed. This proves the claim. 4 8 These inequalities alone prove that (b) is the correct choice. x(2y – c) = –3y. 2 11. 1 − + − . . the limit is the point 4 2 5 . Therefore. y > . MN < PQ unless P = M and Q = N. indeed if −3x 3 f (x ) = =− . 1 . y) of the bug satisfies x > . (c) : 1 st Solution: Let f(x) = x 2 + x + 1 and g n (x) = x 2n + 1 + (x + 1) 2n .. −3 y = f ( y ) . thus MN < MQ unless Q = N.. (b) : 1st Solution: If the bug travels indefinitely. Its sum is ∞ 4 + 2i 4 2 a 2 = = + i ∑ ai = 1 = 1− r 2 − i 5 5 5 i =1 In coordinate language. the result will be x = f(y). To compute the length of MN. the algebraic sum of the vertical components of its moves approaches 2 1 1 1 = − + − . In particular. 12. then f (x) ≠ – 3/2.. and x = 2y −c Remark: In order to form f [f(x)]. if cx y= 2x + 3 is solved for x as a function of y. By transitivity. This assumption can be justified expost factor now that we have found c = – 3. Thus. cx c 2x + 3 c2 x x = f ( f (x )) = = . 2 2 i2p 1 3 w′ = − − i = cos 240° + i sin 240° = e 3 = w2 2 2 MatheMatics tODaY | march ’15 65 . = 4 16 1 1− − 4 Similarly. Thus. c = –3. 13. MN ^ CD. Note that (x – 1)f(x) = x3 – 1. 2x + 3 2 then 6x = 6x + 9.D N S C A 1/4 1/2 1 M B Similarly. which implies c = –3.. we note that MN is the altitude of isosceles DDMC with sides of 3 3 . 5 2 8 32 Therefore. 2nd Solution: The condition f [f(x)] = x says that the function f(x) is its own inverse. (a) : 1st Solution: For all x ≠ –3/2.... so that the zeros of f(x) are the complex cube roots of 1: ip 1 3 w=− + i = cos 120° + i sin 120° = e 3 . a contradiction. 2c + 6 = 0 and 9 – c2 = 0. we obtain 2xy + 3y – cx = 0. 2cx + 6 x + 9 cx 2 +3 2x + 3 which implies (2c + 6)x + (9 – c2) = 0. we assumed that if x ≠ –3/2. the algebraic sum of the horizontal components of its moves approaches 4/5. the plane through N perpendicular to CD contains AB. 5 Remark: The figure shows that the limiting 3 3 position (x. the limit of the geometric series 1 1 1 . The Pythagorean theorem 2 2 now yields lengths 2 1/8 2 MN = (MC ) − (NC ) = 3 1 − 4 4 2 = minimal distance PQ. the bug will get arbitrarily close to 4 2 5 . Remark: The second solution. so 22n + 1 + 32n leaves a remainder of 3 upon division by 7. (c) : Let O and H be the A points at which PQ and BC. since 26k = 64k = (7·9 + 1)k. 66 MatheMatics tODaY | march ’15 Also. whenever n is a multiple of 3. We note that ip ip 1 3 w +1 = + i = e 6 . 2nd Solution: Both of the given polynomials have integer coefficients. However if n is divisible by 3. then Q(x) has integer coefficients. let n = (de)8. and (c) is a correct answer since 3 divides 21. (6631)8 and (2531)8. there are three choices for n: (33)8.5 ≤ and ≤ 16 . The total number of solutions is. is 1. all positive solutions are less than or equal to 100. which is the units digit of e2 . we obtain 22n + 1 + 32n = Q(2)·7.) . respectively. Since DAPQ is an equilateral triangle with a side in common with DPQT. Sides AB and AC form a O Q P portion of the equilateral H C triangle circumscribing the B smaller circle and tangent T to the smaller circle at T. DPQT is an equilateral triangle. the 3 in ab3c is the first digit (in base 8) of the sum of the eights digit of e2 (in base 8) and the units digit of (2de) (in base 8). so the former is odd. therefore. The entire table of base 8 representations of squares of base 8 digits appears below. which shows that 7 divides 22n + 1 + 32n. and g3k + 1(w) = w2 + w + 1 = –1 + 1 = 0. Furthermore. w2n = w 6k + 2 = w2 . 100 −x = sin(− x ) . so (w + 1)2 = e 3 = w 2 2 Thus. (w + 1)2n = wn = w 3k + 1 = w. in either case c. (b) : 1st Solution: If n2 = (ab3c)8. unlike the first. Therefore. DAPQ @ DPQT. Thus there is one solution to the given equation between 0 and p and two solutions in each of the intervals from (2k – 1)p to (2k + 1)p. If n = 3k + 1. 1 ≤ k ≤ 15. intersect diameter AT. (c) : We have. and 36k = (729)k = (7·104 + 1)k. 1 + 2(1 + 2·15) = 63. 15. (w + 1)2n = w 3k + 2 = w2 and g3k + 2(w) = w + w2 + 1 = 0 Thus gn(w) ≠ 0 if and only if n is a multiple of 3. Since w and w′ are complex conjugates. respectively. Thus AO = OT and O is the centre of the larger circle. Setting x = 2 in (1). thus. it suffixes to determine those n for which gn(w) = w2n + (w + 1)2n + 1 = 0. say n = 3k. (In fact. and x2 + x + 1 has leading coefficient 1. so that PQ = (BC ) = 8 3 3 16. Suppose n is not a multiple of 3. The latter is even. This implies 2 2 ( AH ). x = 0 is a solution. Thus. the given equation has an 100 equal number of positive and negative solutions. e 1 2 3 4 5 6 7 AO = e 2 1 4 11 20 31 44 61 The eights digit of e2 is odd only if e is 3 or 5. Hence if (1) x2n + 1 + (x + 1)2n = Q(x)(x2 + x + 1). x = sin x if and only if 14. and gn(x) fails to be divisible by f(x) only in that case. the graphs of 100 (2 p) sinx are as shown in the figure. does not tell us what happens when n is not divisible by 3. w2n = w 6k + 4 = w. We know that only one listed answer is correct. both 22n and 32n leave remainders of 1 upon division by 7. Then n2 = (8d + e)2 = 64d2 + 8(2de) + e2. (73)8 and (45)8. we have g3k(w) = w6k + (w + 1)6k + 1 = 1 + 1 + 1 = 3 ≠ 0. If n = 3k + 2. Now g n (x) is divisible by f(x) if and only if gn(w) = gn(w′) = 0. The squares are (1331)8. but it does enable us to answer the question.Note that w3 = (w′)3 = 1. since |x| = 100|sinx| ≤ 100 x 100 Since 15. and its remainder upon division by 8 is 0 or 4. n2 has remainder 1 upon division by 8. then n2 = 4(8L + 7) another impossibility since no odd squares have the form 8L + 7. (b) : In the adjoining figure. n2 is divisible by 4. a + 2aa + = −a + a2 + a − . then n2 = 4(k2 + k) + 1. ∠MOB = ∠CBO = ∠OBM. ABC is the given triangle and x = a is the dividing line. If c = 0. Let us show that the real roots of 1 1 . Thus in all cases. 16 16 From (3) and (1). Let 1 . 2 9 2 or (9 – a) = 36 Then 9 – a = ±6. 4 3 1 3 2 a − a + = − a − a > 0 16 4 4 Therefore the roots are real and distinct. and a = 15 or 3.. Since the line x = a must intersect DABC. . 1 3 . in fact. Let a be any root of (2). say n = 2k + 1. Hence AM + MO + ON + AN = (AM + MB) + (AN + NC) = AB + AC = 12 + 18 = 30. (4) ⇒ b2 + 2ab + = a 16 MatheMatics tODaY | march ’15 67 . 17.2nd Solution: We are given n2 = (ab3c)8 = 83a + 82b + 8·3 + c If n is even. the two regions must each 2 have area 2. Thus c = 1. it is not difficult to see that they are positive. 9 Thus 1 a Area DDEC = 2 = 1 − (9 − a) . Conversely suppose a is a real root of (1). 16 16 i. a is a root of (1)... Note that the given value BC = 24 was not needed. then n 2 = 8(8k + 3). (2) 16 Note that the roots of (2) can be easily computed. If n is odd. an impossibility since 8 is not a square.. Then 1 1 (a + a)2 = a2 + a − ⇒ (a + a) = a2 + a − 16 16 1 ⇒ a = −a + a2 + a − 16 1 1 2 Therefore. the line x = a is indeed right 2 of A as shown.. and ∠CON = ∠OCB = ∠NCO A 12 B M O N 18 C Therefore MB = MO and ON = NC. we get 1 b = −a + a 2 + a − 16 1 ⇒ (b + a)2 = a2 + a − 16 1 ⇒ b2 + a2 + 2ab = a2 + a − 16 1 .. Since the equation of line BC is x y= the vertical line x = a intersects BC at a 9 a point E : a. they are − a ± a2 − a + 2 16 1 Since 0 < a < . 19. 18. and since k2 + k = k(k + 1) is always even. (a) : Since MN is parallel to BC. Since area 1 DABC = (1)(8) = 4 .e. Since the portion of DABC to the vertical line through vertex A has area less than 1 area DABF = .. the only possible values of c are 0. If c = 4. (3) a2 + 2aa + = b 16 1 1 Then b = (a + a)2 − a2 + ≥ − a2 > 0 .. x = 3. 1 or 4. (1) x 2 + 2ax + = −a + a2 + x − 16 16 1 are precisely the roots of x 2 + 2ax + = x . only the sum of the lengths of the other two sides was used.. a = b. y are real. Since T divides the segment 2 OM in the ratio 2 : –1. i. This and two similar expression yield 2(b2 + c 2 ) = 4 AD 2 + a2 . Since M is the mid-point of BC. 1 st Solution: The given equation can be considered as a quadratic in x : 3x2 + (10 – 4y)x + (3y2 – 10y + 10) = 0.S. Also BC = 2Rsin30° = R. So the triangles GAP AB BC CA 3 and ABC are similar.e. 3 A Since G lies on SO such that SG : SO = 1 : 3. 21.. (x. Solving for x. a is a root of (2). Thus S B 68 O T P C MatheMatics tODaY | march ’15 2 23.(3) Since ∠C is obtuse. D. i. z3 be complex numbers representing A. (1) 2(c 2 + a2 ) = 4 BE 2 + b2 and 2(a2 + b2 ) = 4CF 2 + c 2 Since B. of the given equation = 3(x2 + 2x + 1) + 3(y2 – 2y + 1) + 4(1 + x – y – xy) = 3(1 + x)2 + 3(1 – y)2 + 4(1 + x)(1 – y) = (1 + x)2 + (1 – y)2 + 2[(1 + x)2 + (1 – y)2 + 2(1 + x)(1 – y)] = (1 + x)2 + (1 – y)2 + 2[(1 + x) + (1 – y)]2 Since x. Therefore by (4). Let l = 11 + 11 11a + 1 Then (l – 11)2 = 112(11a2 + 1) .. its coordinate is z + z 1 −1(z1 + z2 + z3 ) + 2 2 3 = − z1 2 −1 + 2 \ AT = |–2z1| = 2R. 1). Let z1. Therefore BC is equal to the radius of the circle. Since CO ^ AB and CO || TB. Then the centroid z +z +z G is 1 2 3 . Hence BTCO is a parallelogram. B. By appolonius theorem. we get 3 4AD2 = 3c2 . This is the circumcircle of triangle ABC. 1 3 1 3 − a + a2 − a + and − a − a2 − a + 2 16 2 16 20. Now using these two relations and (2).e.(2) Therefore the first relation of (1) gives 2b2 = c2 + a2 . we get 4BE2 = 3b2 and 4CF2 = 3a2. Hence the real roots of (1) are the roots of (2). Therefore AT = 2BC.H. Then ∠BSC = 2∠BAC = 60° and ∠SBC = ∠SCB = 60°.. a – b = 0 . Similarly TC ^ AC.e.Therefore a > 0 and (3) – (4) ⇒ (a – b)(a + b + 2a + 1) = 0 Since a. y = 1 and therefore x = –1. y are real (10 – 4y)2 – 12(3y2 – 10y + 10) ≥ 0 which on simplification gives (y – 1) 2 ≤ 0. its z +z coordinate is 2 3 . Let S be the circumcentre. we get 1 x = 4 y − 10 ± (10 − 4 y )2 − 12(3 y 2 − 10 y + 10) 6 Since x. i. 1 + x = 0 = 1 – y. z2.. GP = 2GE = BE = 3 3 3 3 2 a and AP = GC = CF = 3 3 2 2 2 2 2b > 2a + b ⇒ b > 2a2 ⇒ GA AP PG 1 = = = . . a are positive. . 2nd Solution: L. we get 2 b 2 c AG = AD = . now 2 using AG = AD .. A E F G B D C AC > 2 BC Using (3) in the second and third relations of (1). therefore by (3).. therefore DSBC is equilateral.. F are concyclic AG·AD = AF·AB. 1st Solution : The diagonals of the quadrilateral BTCO bisect each other at M. Thus the circle having AT as diameter passes through B and C. b.. 2 nd Solution : Let the circumcentre of DABC be the origin.. C respectively. c2 > a2 + b2 . we have AB2 + AC2 = 2(AD2 + BD2). we have TB ^ AB.. y) = (–1.. viz.. g. O is z1 + z2 + z3. 22. that is we have to solve for n in the equation: n n n 1993 = + + + .. s ∈ N. in triangles ACB and BCD.. Thus we have l = 11m = 112n2. as desired. it follows that s = 1. Since 7972 7972 7972 7972 7972 5 + 25 + 125 + 625 + 3125 = 1989 We have to arrange for four more multiples of 5. We refer to the same figure. (b − d ) (b + d ) MatheMatics tODaY | march ’15 69 . Produce CA to D such that A AD = AB. s) = 1. n n n n n n 5 + 25 + 125 + . A . So DABC is similar to DBDC. as a consequence of the assumption a2 = b(b + c). Writing l = 11m. Thus 7985 is the required number. = BC DC It follows that a2 = b(b + c) Now we prove the convers e. (1) ∠ADB = 2 On the other hand. in all 4 more multiples of 5) than 7972!... Assume that a2 = b(b + c). we have ∠ABD = ∠ADB. We also use some simple properties of equal fractions. ≤ 5 + 25 + 125 + . AC BC = ... If 11 | m – 2. such that s (r. n ≥ 7972. a c Specifically. Therefore 11|m and hence m = 11n 2 for some n ∈ N. By construction. 5 25 125 Now. Since multiples of 2 occur far more often than multiples of 5 it is enough to find an n such that 51993 is the maximum power of 5 that divides n!. r Putting | a | = w here r... n 1 1 n5 n ≤ 1 + + + .Simplifying we get l(l – 22) = 113a2. 2 nd Solution: We may use the sine rule for a triangle to dispose of both the implications simultaneously. We are to find an n such that 101993 is the highest power of 10 dividing n!. then m is a square of the form 11n + 2 which is not possible. BC DC So the two triangles are similar and the result follows: ∠CDB = ∠CBA = ∠B . A = 2B. Join BD. 11 |l. Thus we have m(m – 2) = 11r2. Since m – 2 and m are consecutive odd integers. ∠ADB = ∠ABD = = B 2 In triangles ABC and BDC we have ∠ABC = ∠BDC and ∠C is common. it follows that B = A/2.. Thus. we have. Note that 7975! will have one more multiple of 5 and one more multiple of 25 than 7972! and 7985! will have 3 more multiples of 5 and 1 more multiple of 25 (that is.. they are relatively prime.. AC BC .. Since 112 |s because otherwise 11 would divide r. and ∠C is common. in particular.. ≤ = 5 5 25 5 4 4 i. 1st Solution: First assume that in the triangle ABC. 1st Solution: We use the fact that the areas of two triangles having the same height are in the ratio of their bases. A = 2B ⇔ A – B = B ⇔ sin(A – B) = sinB ⇔ sin(A – B)sin(A + B) = sinB sinC ⇔ sin2 A – sin2B = sinB sinC ⇔ (2RsinA)2 – (2RsinB)2 = (2RsinB)(2RsinC) ⇔ a2 – b2 = bc ⇔ a2 = b(b + c) 26. we get m(m – 2)s2 = 11r2. So each of these angles is equal to half of their sum which is A.(2) From (1) and (2). 24. Therefore. gives l(l – 22)s2 = 113r2. it is clear c b that ABC is an isosceles triangle and so B C a ∠ADB = ∠ABD But ∠ADB + ∠ABD = ∠BAC (the external angle) A Hence.. Note that 7986. 7987.. in the isosceles triangle ABD.e. As before. 7988 and 7989 also satisfy the requirement. D 25. then each fraction is also b d (a − c) (a + c) equal to as well as . if = . Since 11 |s for otherwise we would have 11| r. A F B P D E C 27.. relation. 1999} is thus equal to 1997 + 1995 + 1993 + . be zero only if some seven terms are + 1 each and whose sides are cut by the line FPC.. AP [ ABP ] [CAP ] [ ABP ] + [ ACP ] . then for each a with 1 ≤ a ≤ 999. 2. c) with a < c. [ ACF ] = AF = [ APF ] . i =1 x + x x x x + x + x1x 4 x5 x8 + x2 x3 x6 x7 + x1x2 x5 x6 + x3 x 4 x7 x8 x x x ∑ i 1 2 3 4 5 6 7 8 AF AE AF BD AF BC . + 1 = 9992. BE.(1) + = 1 + i =1 = . x3. the sum will Now applying ‘Menelaus’ Theorem to triangle ABD.) We now double the number of balls of the urn which contains only one ball and remove one ball from each of the urn. 1999} is 9992 ...(2) ⋅ = FB DC PD of –1’s in the above sum.. and so on.(2) = = EC [CBE] − [CPE] [CBP ] From (1) and (2). AF BC DP But.. (If m and n denote the number of balls in the urns.. Similarly the number of pairs (a..(4) = = = PD [DBP ] [DCP ] [BCP ] From (3) and (4). nn 70 MatheMatics tODaY | march ’15 .. giving 1995 pairs (a. 28.. Thus we have to count the number of triples (a. Continuing this way we reach a stage when both the urns contain one ball each whence we can empty the urns removing one ball from each of the two urns. . by addition... it is impossible to have an odd number Consequently. then we can empty both the urns in one operation.. c can take values from a + 1 to 1999 – a. 3. 2. CF which are concurrent at P. Because there are fourteen terms in the above sum FB AC FB DC FB DC and each of the terms is + 1 or – 1. The number of ordered pairs (a. b. Hence the required number of polynomials is 2·9992 = 1996002. Then. x7 and x8 be the numbers written at the corners.(3) + = FB EC [BCP ] Again. Comparing (1) and (2).... Else. b. c runs from 3 to 1997.. (1) FB [BCF ] − [BPF ] [BCP ] Similarly from [ ABE] AE [ APE] = = [CBE] EC [CPE] one obtains. c) with a < c and a + c lying in the set {1. If both the urns have the same number of balls... c runs from 2 to 1998 giving 1997 ordered pairs (a. . then take out n – 1 balls from each. . c) with a < c.. Since x + 1 divides ax2 + bx + c.. c lie in the set {1. we have AF BC CE ⋅ ⋅ =1 FB DC EA AE AF BD = ⋅ Therefore. we have the desired We conclude that the desired arrangement is not possible. This process decreases the number of balls in the other urn by 1.. we get AF AE [ ACP ] + [ ABP ] . 29. we have the remaining seven terms are – 1 each.. x4... c) with c < a and a + c lying in the set {1.. 3. c) with the condition that a. the product of the fourteen terms is ⋅ ⋅ =1 FB DC PA (x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 )4 = (±1)4 = +1. 2. 8 EC FB DC ∑ xi + x1x2 x3 x 4 + x5 x6 x7 x8 + x1x 4 x5 x8 + x2 x3 x6 x7 + x1x 8 Hence... Thus for a = 1. AE [ ABE] − [ APE] [ ABP ] . AF BC AP Therefore. for a = 2.. the final sum is given by Now. [BCF ] FB [BPF ] AF [ ACF ] − [ APF ] [ ACP ] = = So. Let x1. x5. a ≠ c and a + c = b. we remove the same number of balls from each of the urns so that one of the urns contains exactly one ball. 1999}. and say m > n.. we have the desired result... 3. . 2 nd Solution: Applying Ceva’s theorem to the Cevians AD. If we take a < c. x2. x6. we must have a + c = b. . § dx ¶ ^ ^ ^ OR 4/3 Show that f(x) = x is differentiable at x = 0. Prove that 2a 3a 2b 4a 3b 2c a3 3a 6a 3b 10a 6b 3c Find two positive numbers whose sum is 24 and whose product is maximum. Prove that: a2 + b2 + c2 + 2abc = 1. one sentence or as per the exact requirement of the question. If the E E E E vectors a b and a b are orthogonal to each other. E ^ ^ ^ ^ ^ ^ E 3. 6. Find the distance of the plane 2x – y + 2z + 1 = 0 from the origin. What is the value of tan–12 + tan–13. 2 · ¨x 1 0 © A © 1 2 x 3¸¸ ©ª 2 0 4 ¸¹ 2. if required. ¥ dy ´ 4. section B comprises of 13 questions of four marks each and section C comprises of 7 questions of six marks each. if n is even ® 2 is a many-one onto function. Show that f : N N. z = bx + ay where. then find the value of . a a b a b c 8. and hence find f (0). z are not all zero. However. y = az + cx. OR ^ ^ 5. x. Section A comprises of 6 questions of one mark each. if n is odd ®® f (n) ¬ 2 ® n . (ii) Th e question paper consists of 26 questions divided into three sections A. 15 Marks : 100 GENERAL INSTRUCTIONS (i) All questions are compulsory. 9. Prove that tan 1 ¦ § 1 sin x µ¶ ¦§ 4 2 µ¶ 10. (iii) All questions in Section A are to be answered in one word. Solve the differential equation sin ¦ µ a . internal choice has been provided in 5 questions of four marks each and 2 questions of six marks each. Two schools A and B decide to award prizes to their students for three values honesty (x). (v) Use of calculators is not permitted. SECTION-B 7. (iv) Th ere is no overall choice. Evaluate: ° cos x x x´ ¥ ¦§ cos sin µ¶ 2 2 OR 4 2 (x x ) Evaluate: ° dx 2x 1 2 3 dx 12. Let a 5 i j 8 k and b i j L k . You may ask for logarithmic tables. You have to attempt only one of the alternatives in all such questions. If A is a singular matrix.SAMPLE PAPER CBSE Board 2 Time : 3 hrs. Write the value of (k s j) i j k . ¥ cos x ´ ¥ P x ´ . Given x = cy + bz. y. B and C. defined by «n 1 . find x. 11. SECTION-A 1. MATHEMATICS TODAY | MARCH ’15 71 . Find a point on the curve y = (x – 2)2.3 and 0. Find the equation of the plane passing through the intersection of the planes ^ ^ ^ ^ ^ ^ r ⋅ (2 i + j + 3 k ) = 7. The male-female ratio of a village increases continuously at the rate proportional to the ratio at any time.2. x 17. Physics and Chemistry in the examination. respectively. A and B. 15. The machine A requires an area of 1000 m2 and 12 skilled men for running it and its daily output is 50 units. (ii) is it possible to solve the system of equations so obtained using matrices? (iii) which value you prefer to be rewarded most and why? 13. whereas the machine B required 1200 m2 area and 8 skilled men. then (i) represent the above situation by a matrix equation and form linear equations using matrix multiplication. Then. 4 and 3 students respectively while school B wishes to award ` 10700 for the three values to 4. 0. 16. then find dy . 3). x = 2.5. If all the three prizes together amount to ` 2700. If y = xx . Find the probability that he gets (i) grade A in no subject. 23. for his factory. If an area of 7600 m2 and 72 skilled men be available to operate the machine. what is the ratio of the areas of these parts? . the x-axis and the abscissa x = –1.punctuality (y) and obedience (z). then what will be the ratio in 2019? (i) Why gender equality is value for society? (ii) What should society do to reduce the male-female ratio to 1? 22. 19. (ii) grade A in two subjects. x ≥ 0. 0) and (4. r ⋅ (2 i + 5 j + 3 k ) = 9 and the point (2. Find the adjoint of the matrix A = and 3 4 verify that A[adj (A)] = |A| I. Find the area bounded by the line y = x. dx 1 2 18. Prove that a −b θ b + a cos θ 2 tan −1 tan = cos −1 2 a + b cos θ a +b 21. how many machines A and B respectively should be purchased to maximize the daily output? OR Find the shortest distance between the lines r = (8 + 3λ)i − (9 + 16 λ)j + (10 + 7 λ)k and r = 15i + 29 j + 5k + µ(3i + 8 j − 5k ) . Find the probability that the balls drawn are black and blue. at which the tangent is parallel to the chord joining the points (2. If the ratio of male-female of the villages was 1000 : 980 in 1999 and 1000 : 950 in 2009. OR 2 The parabola y = 2x divides the circle x2 + y2 = 8 in two parts. x = cos −1 1 + x 2 OR Solve the equation : x −1 x +1 π tan −1 + tan −1 = x − 2 x + 2 4 72 MatheMatics tODaY | March ’15 sectiOn-c 20. Coloured balls are distributed in three bags as shown in the following table : Bag I II III Colour of ball Green Black Blue 1 2 3 2 4 1 3 3 2 A bag is selected at random and then two balls are drawn from the selected bag. The probabilities of getting grade A in these subjects are 0. 3 and 5 students respectively. School A desires to award a total of ` 11000 for the three values to 5. Prove that tan −1 1 1− x . OR Solve: (x x 2 + y 2 − y 2 )dx + xy dy = 0 14. 4). 1. and its daily output is 40 units. X is taking up subjects : Mathematics. A factory owner wants to purchase two types of machines. if x +1 x −1 tan −1 + tan −1 = tan −1 (−7). \ f is onto. 2 × 3 = 6 > 1 2+3 tan–12 + tan–13 = π + tan −1 1 − 2 ⋅ 3 π 3π = p + tan–1(–1) = π − = 4 4 3. ^ ^ ^ ^ ^ ^ ^ ^ ^ (k × j) ⋅ i + j⋅ k = (− i ) ⋅ i + j⋅ k = – (1) + 0 = –1 6.. 25.H. we get D = a × (a2 – 0) = a3 Hence proved. x − 1 x 3 cos x + 2 26. a a +b a +b+c 8. R3 → R3 – 3R1. = tan −1 1 + sin x = tan π sin 2 − x 1 + cos π − x 2 −1 MatheMatics tODaY | March ’15 73 . cos x 9. consider an arbitrary element n ∈ N. A is singular \ |A| = 0 |A| = (x – 1)(–8 – 0) + 2(4) = 0 ⇒ –8x + 8 + 8 = 0 ⇒ x = 2 2. \ f is many-one. Evaluate : ∫ dx sin x + 2 cos x + 3 sOLutiOns 1.S. Given. we get a a +b a +b+c ∆= 0 a 2a + b 0 3a 7a + 3b Applying R3 → R3 – 3R2. We have. Given plane is 2x – y + 2z + 1 = 0 . Find the volume of the largest cylinder that can be inscribed in a sphere of radius r cm.. f(1) = f(2) while 1 ≠ 2. Find x. We have . for each n ∈ N (whether even or odd) there exists its pre-image in N.(i) Distance of (i) from origin (2 × 0) − (1 × 0) + (2 × 0) + 1 1 = = 3 (2)2 + (−1)2 + (2)2 7. If n is odd. L. f (0 + h) − f (0) f (h) − f (0) = lim h h h→0 h→0 4 /3 4 /3 h −0 h lim = lim = lim h1/3 = 0 h h h→0 h→0 h→0 and f (0 − h) − f (0) f ( − h) − 0 = lim Lf ′(0) = lim −h −h h→0 h→0 Rf ′(0) = lim (−h)4/3 − 0 (−h)4/3 = lim = lim (−h)1/3 = 0 ( − h) h→0 h → 0 ( − h) h→0 = lim Thus. we get y = x sin–1 (a) + c 5. Let ∆ = 2a 3a + 2b 4a + 3b + 2c 3a 6a + 3b 10a + 6b + 3c Applying R2 → R2 – 2R1. then 2n is even and 2n f (2n) = =n 2 Thus. We have. Rf ′(0) = Lf ′(0) = 0 Hence f(x) = x4/3 is differentiable at x = 0 and f ′0) = 0. (a + b ) ⋅ (a − b ) = 0 2 ⇒ a 2 − b = 0 ⇒ 90 = 2 + l2 ⇒ λ = 88 \ dy dy = sin −1 (a) 4. Here.24. OR We have. (1 + 1) 2 2 f (1) = = = 1 and f (2) = = 1 2 2 2 Thus. f is many-one onto. we get a a +b a +b+c ∆= 0 a 2a + b 0 0 a Expanding along C1. Hence. In order to show that f is onto. sin = a ⇒ dx dx ⇒ dy = sin–1 (a) dx On integrating both sides. then (2n – 1) is odd and (2n − 1 + 1) 2n f (2n − 1) = = =n 2 2 If n is even. We have. x + y + z = 2700 5 4 3 (ii) Since 4 3 5 = –3 ≠ 0 1 1 1 therefore.e. z = bx + ay. OR Let the two positive numbers be x and 24 – x and let y = x(24 – x). . π x π x 2 sin 4 − 2 cos 4 − 2 −1 = tan π x 2 cos2 − 4 2 π x π x = tan −1 tan − = − = R. (iii) Preference should be given to honesty as an honest person is expected to follow all ethical values such as punctuality. i. 4 2 4 2 \ cos x π x = − tan −1 1 + sin x 4 2 =∫ = cos( x /2) − sin(x /2) x x cos + sin 2 2 2 dx = 2 ∫ 1 t2 dt . the required parts are 12 and 24 – 12 = 12. = (60 − 6 5 ) − ∫ (2 x + 1)3/2 dx 1 −c −b ⇒ c −1 a = 0 ⇒ 1(1 – a2) – (– c)(– c – ab) + (– b)(ca + b) = 0 b a −1 ⇒ 1 – a2 – c2 – abc – abc – b2 = 0 ⇒ a2 + b2 + c2 + 2abc = 1 Hence proved..(ii) dx For maxima/minima.r. 4x + 3y + 5z = 10700. x x where t = cos + sin 2 2 −2 −2 +C = +C t cos( x /2) + sin(x /2) OR 10.H..t. (i) We can represent the given problem in matrix form as follows: 5 4 3 x 11000 4 3 5 y = 10700 1 1 1 z 2700 \ 5x + 4y + 3z = 11000. So. it is possible to solve the system of equations obtained in (i) by using matrices. justice. dy = 0 ⇒ 24 – 2x = 0 ⇒ x = 12 dx Differentiating (ii) w. y = az + cx.. x. 4 2 4 4 (x + x ) (x 2 + x ) ⋅ 2 x + 1 − (2 x + 1) ⋅ 2 x + 1 dx This means that the system of homogeneous = dx 2 ∫ ∫ 2x + 1 equations 2 2 4 4 2 x – cy – bz = 0 4 (x + x ) 2 cx – y + az = 0 ∫ 2x + 1 dx = (x + x) ⋅ 2x + 12 − ∫ (2x + 1) ⋅ 2x + 1 dx 2 2 bx + ay – z = 0 4 has non-trivial solution. 12 and 12. 0 < x < 24 ⇒ y = 24x – x2 . We are given that there is a non-zero solution of Integrating by parts. We have cos x ∫ x x cos + sin 2 2 3 dx = ∫ cos2 (x /2) − sin2 (x /2) {cos(x /2) + sin(x /2)}3 74 MatheMatics tODaY | March ’15 dx 2 4 1 = (60 − 6 5 ) − ⋅ (2 x + 1)5/2 2 5 243 = (60 − 6 5 ) − −5 5 5 57 57 − 5 5 = − 5 = 5 5 12.. 11. x.r.(i) Differentiating (i) w. we get dy = 24 − 2 x . we get the equation x = cy + bz. equality and obedience etc..t.S. we get d2 y = −2 < 0 at x = 12 dx 2 ⇒ y has a local maximum value at x = 12. Given equation of planes in cartesian form are 2x + y + 3z – 7 = 0 .3 and P(C) = 0.8 × 0.28 (ii) P(grade A in two subjects) = P (M ) ⋅ P (P ) ⋅ P (C ) + P (M ) ⋅ P (P ) ⋅ P (C ) + P (M ) ⋅ P (P ) ⋅ P (C ) [∵ M .2 × 0. C .(ii) 4−2 2 Since. Let the event of getting grade A in Mathematics. the required point is (3. Physics and Chemistry be represented by M.3 × 0.. P (P ) = 0. We know that y = x is the line passing through the origin. dy \ = Slope of the chord dx ⇒ 2(x – 2) = 2 ⇒ x – 2 = 1 ⇒ x = 3 On putting x = 3 in (i).5 = 0. 1).03 + 0.. dy dv Putting y = vx ⇒ ...5 P (M ) = 0..5) + (0. the tangent is parallel to the chord joining the points.7 and P (C ) = 0.22 15.5) + (0.5 (i) P(grade A in no subject) = P (M ) ⋅ P (P ) ⋅ P (C ) = 0. we get dy = 2(x − 2) dx Slope of the chord joining the points (2.5) = 0.7 × 0.. P . B y= x Y X O (–1. Given curve is y = (x – 2)2 .t. 3) \ (4 + 1 + 9 – 7) + l(4 + 5 + 9 – 9) = 0 −7 ⇒ 7 + 9l = 0 ⇒ l = 9 On putting the value of l in (iii).(i) and 2x + 5y + 3z – 9 = 0 .0) X A Y Required area = (area DDBO) + (area DACO) 2 = ∫ y dx + 0 2 = ∫ x dx + 0 2 0 ∫ (− y)dx −1 0 [∵ area OACO is below the x-axis] ∫ (− x)dx −1 0 x2 −x2 1 5 = + = 2 + = sq.07 + 0. 1. Now. P(M) = 0.(iii) Plane (iii) passes through the point (2. P and C are independent events] =(0. we get −7 (2x + y + 3z – 7) + (2x + 5y + 3z – 9) = 0 9 18 x + 9 y + 27 z − 63 − 14 x − 35 y − 21z + 63 ⇒ =0 9 ⇒ 4x – 26y + 6z = 0 ⇒ 2x – 13y + 3z = 0 Equation of plane in vector form is ^ ^ ^ ^ ^ ^ (x i + y j + z k ) ⋅ (2 i − 13 j + 3 k ) = 0 ^ ^ ^ \ r ⋅ (2 i − 13 j + 3 k) = 0 16.8 × 0. 0) and 4−0 4 (4.8. M . which is clearly = dx xy homogeneous...r. 4) = = =2 .0)C D(2. P and C. we have to find the area of the shaded region.(i) On differentiating (i) both sides w. units 2 0 2 −1 2 2 OR The given equation may be written as 2 2 2 dy y − x x + y . MatheMatics tODaY | March ’15 75 .12 = 0.2 × 0.7 × 0. respectively. we get =v+x dx dx dv v 2 x 2 − x x 2 + v 2 x 2 v+x = dx vx 2 ⇒ x dv v 2 − 1 + v 2 = −v dx v ⇒ x dv − 1 + v 2 = ⇒ dx v ∫ ⇒ 1 + v 2 = − log x + C ⇒ x 2 + y 2 + x log x = Cx v 1+ v 2 dv = − ∫ dx x 14. we get y = 1 Hence.(ii) Equation of plane passing through the intersection of planes (i) and (ii) is (2x + y + 3z – 7) + l(2x + 5y + 3z – 9) = 0.3 × 0.2.. x. P(P) = 0..13. Given.S.(i) On differentiating (i) both sides w.r. a −b θ 20. = = L. x. y = xxx Taking log on both sides.(i) 3 4 Now. we get 1 1 dy 1 1 1 ⇒ ⋅ ⋅ = x ⋅ + log x ⋅1 + ⋅ log y y dx x log x x dy 1 1 ⇒ ⋅ = 1 + log x + y log y dx x log x dy 1 ⇒ = y log y 1 + log x + dx x log x 1 = x ( x ) log x ( x ) 1 + log x + x log x 1 2 18. We have. 1 2 4 −2 4 − 6 −2 + 2 A[adj (A)] = = 3 4 −3 1 12 − 12 − 6 + 4 19.t.17. we get log (log y) = log(xx log x) ⇒ log (log y) = log xx + log (log x) ⇒ log (log y) = x (log x) + log (log x) .S..H. we get a −b θ tan2 x = tan2 a +b 2 θ a −b 1− tan2 a + b 1 − tan2 x 2 Now.. R.H. = cos −1 1 + cos2 θ 2 76 MatheMatics tODaY | March ’15 OR Given equation is x −1 x +1 π tan −1 + tan −1 = x − 2 x + 2 4 x −1 x +1 + x −2 x +2 = π 1 − x − 1 + x + 1 4 x −2 x +2 −1 ⇒ tan ⇒ (x − 1)(x + 2) + (x + 1)(x − 2) π = tan 4 (x − 2)(x + 2) − (x − 1)(x + 1) ⇒ x 2 − x + 2x − 2 + x 2 + x − 2x − 2 (x 2 − 4) − (x 2 − 1) =1 ⇒ 2x4 – 4 = –3 ⇒ 2x2 = 1 1 2 1 ⇒ x = ⇒x=± 2 2 ⇒ 4 −2 \ adj (A) = and −3 1 1 2 A= = 4 − 6 = −2 . Let x = tan −1 tan 2 a +b C11 = (–1)2(4) = 4 C12 = (–1)3(3) = –3 C21 = (–1)3(2) = –2 C22 = (–1)4(1) = 1 −2 0 1 0 = = −2 = A I 0 −2 0 1 1 1 = cos −1 (cos 2θ) = × 2θ = θ = tan −1 x 2 2 [from (i)] tan x = a −b θ tan a +b 2 On squaring both sides. A = 3 4 C11 C12 ′ adj(A) = where Cij’s are the C21 C22 x x cofactors of aij’s 1 1− x cos −1 1 + x 2 Put x = tan2q ⇒ q = tan −1 x 1 − cos2 θ 1 \ R... cos 2 x = = 2 1 + tan x 1 + a − b tan2 θ 2 a +b 2θ 2 θ a + b − a tan 2 + b tan 2 = 2θ 2 θ a + b + a tan 2 − b tan 2 θ θ a 1 − tan2 + b 1 + tan2 2 2 = θ θ a 1 + tan2 + b 1 − tan2 2 2 . we get log y = xx log x Again. taking log on both sides.S.H. we get r= t 50 10k 10 ⋅ (e ) = 49 In the year 2019. Let male-female ratio at any time be r. 1000 50 So in 1999. we get 20 50 10k 98 = e ⇒ e10k = 19 49 95 Substituting in (ii).085 ≈ 1085 : 1000. (c) Provide special opportunities to women to come at par with men in all spheres of life. MatheMatics tODaY | March ’15 77 . (b) Empower women to realise their rights.20 2 θ 1 − tan 2 a +b θ 1 + tan2 2 = 2 θ 1 − tan 2 a +b θ 1 + tan2 2 a cos θ + b ⇒ cos 2x = a + b cos θ \ = b + a cos θ ⇒ 2 x = cos −1 a + b cos θ a −b θ b + a cos θ \ 2 tan −1 tan = cos −1 2 a + b cos θ a +b 21. t = 0 and r = = 980 49 Substituting in (i). 95 × 95 Thus in the year 2019.. (i) Gender equity promotes economic growth. we get log r = kt + log c (where log c is the constant of integration) r ⇒ log r – log c = kt ⇒ log = kt c \ r = cekt . y ≥ 0 Mathematical formulation of the LPP is Maximize z = 50x + 40y subject to constraints : 5x + 6y ≤ 38 3x + 2y ≤ 18 x ≥ 0. t = 10 and r = Substituting in (ii). r = e kt 49 Also in the year 2009.. x and y must satisfy the following conditions : (Area) 1000x + 1200y ≤ 7600 ⇒ 5x + 6y ≤ 38 (Man power) 12x + 8y ≤ 72 ⇒ 3x + 2y ≤ 18 x ≥ 0. dr dr Given ∝r ⇒ = kr dt dt dr = kdt We have r Integrating both sides.(i) \ Let us start reckoning time from the year 1999 for the problem.. fertility. y ≥ 0 Now. we get 50 50 = c ⋅ e0 ⇒ c = 49 49 \ 50 (i) becomes.. Now given information can be summarized as : A (x) Area (m2) B (y) Maximum available capacity 1000 1200 Man power 12 8 Output 50 40 7600 72 According to question. child mortality and under nutrition. Let z be the daily output.. Let the number of machine A be x and number of machine B be y.(iii) l2 : 3x + 2y = 18 l3 : x = 0 and l4 : y = 0 Lines l1 and l2 meet at E(4. 3). t = 20 t 50 98 10 49 95 50 98 10 50 98 98 r= = × × 49 95 49 95 95 . 22. the male-female ratio will be 1085 : 1000. (ii) (a) Stop female-foeticide.(ii) 1000 20 = 950 19 100 × 98 ≈ 1. we draw the lines l1 : 5x + 6y = 38 .. 0).. Let E1. 3) and B 0. a parabola with vertex (0. Vertices of the feasible region are O(0. E(4. E3 and A be the events defined as follows : E1 = bag I is selected E2 = bag II is selected E3 = bag III is selected A = one black and one blue ball has been drawn from the selected bag. b2 = 3i + 8 j − 5k a2 − a1 = 7i + 38 j − 5k i j k b1 × b2 = 3 −16 7 3 8 −5 = i(80 − 56) − j(−15 − 21) + k (24 + 48) = (24i + 36 j + 72k ) \ = (a2 − a1 ) ⋅ (b1 × b2 ) Shortest distance. C(6.e.. a circle with centre (0. 19 3 Maximize z = 50x + 40y At O. 3).= The shaded region OCEB is the feasible region which is bounded. z = 50 × 4 + 40 × 3 = 320 19 At B. find A2 . E2. we get P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3) 1 2 1 4 1 3 1 2 4 3 = ⋅ + ⋅ + ⋅ = + + 3 5 3 21 3 14 3 5 21 14 1 169 169 = ⋅ = 3 210 630 OR We have y 2 = 2x .(i).e. the maximum output = 320 is at E(4.. equations of lines r = (8 + 3λ)i − (9 + 16 λ)j + (10 + 7 λ)k i. As the bags are selected at random. when 4 machines A and 3 machines B are purchased. OR Given. therefore the probability of their selection is equally likely. We have to A1 ...(ii).33 3 Clearly. i. 0). r = 8i − 9 j + 10k + λ(3i − 16 j + 7k ) and r = 15i + 29 j + 5k + µ(3i + 8 j − 5k ) Here. z = 50 × 0 + 40 × 0 = 0 At C. z = 50 × 6 + 40 × 0 = 300 At E.. 1 \ P (E1 ) = P (E2 ) = P (E3 ) = 3 Total number of balls in bag I = 1 + 2 + 3 = 6 2 C × 3C1 2 × 3 2 P(A|E1) = 1 = = 6 15 5 C2 Total number of balls in bag II = 2 + 4 + 1 = 7 4 C1 × 1C1 4 × 1 4 = = P(A|E2) = 7 21 21 C2 Total number of balls in bag III = 3 + 3 + 2 = 8 3 C1 × 2C1 3 × 2 3 = = P(A|E3) = 8 28 14 C2 By using law of total probability. b1 = 3i − 16 j + 7k a2 = 15i + 29 j + 5k . d = | b1 × b2 | (7i + 38 j − 5k ) ⋅ (24i + 36 j + 72k ) (24)2 + (36)2 + (72)2 78 MatheMatics tODaY | March ’15 168 + 1368 − 360 1176 1176 = = = 14 576 + 1296 + 5184 84 7056 23. 0) and radius 2 2 Let the area of the smaller part of the circle be A1 and that of the bigger part be A2. a1 = 8i − 9 j + 10k . 0) and x 2 + y 2 = 8 . z = 50 × 0 + 40 × = 253. MatheMatics tODaY | March ’15 79 . On solving (i) and (ii). Then. –4 x = –4 is not possible as both the points of intersection have the same positive x-coordinate.. units 2 3 4 3 [neglecting –ve value of R] R= r. we get x = 2. A2 = 8 π − A1 = 6 π − given by 3 2 2r 4 πr 3 Then. Let h be the height and R be the radius of the base of the inscribed cylinder. A1 = 2[Area(OBCO) + Area(CBAC)] 2 2 2 = 2 ∫ y1 dx + ∫ y2 dx 0 2 2 2 2 ⇒ A1 = 2 ∫ 2 x dx + ∫ 8 − x 2 dx 0 2 For maximum or minimum values of V. V is maximum when h = 2 . units \ The maximum volume of the cylinder is 4 Hence. the required ratio is V = πR2h = π r 2 = 3 3 3 3 4 A1 3 + 2 π 2 + 3π 25. = − 3πr < 0 dh2 h = 2r 2 2 2 3 2 8 x x ⇒ A1 = 2 2 ⋅ x 3/2 + 2 8 − x 2 + sin −1 2r 3 0 2 2 2 Thus. 1 1 − × x V = pR2h . Now. we obtain 4 3 16 π 4 = + 2 2 π − 2 + 4 × = + 2 π sq. We have.(i) From DOCA. we must have 3πh2 4r 2 2r dV = 0 ⇒ πr 2 − = 0 ⇒ h2 = ⇒h= 4 3 dh 3 2 −2 dV r∵ > 0 neglecting h = 3 dh2 h = −2 r 3 d 2V Now. we have 2 B′ C′ h2 h/2 h r 2 = + R2 ⇒ R2 = r 2 − 2 4 O 2 2 2 h 2 2 2 h r h/2 = +R ⇒R =r − 2 4 B C h2 ∴ V = π r2 − h 4 π ⇒ V = πr 2h − h3 4 d 2V 3πh dV 3πh2 =− and ⇒ = πr 2 − 2 2 dh 4 dh 80 MatheMatics tODaY | March ’15 A′ x −1 1 ⇒ tan −1 (1) + tan −1 + tan −1 + tan −1 7 = 0 x x 1 x −1 + 1+ 7 x =0 ⇒ π + tan −1 + tan −1 x x −1 1− 7 1− x2 { } A x2 4 + tan −1 = −π 2 3 x − x +1 ⇒ − tan −1 ⇒ x2 4 − 2 3 tan −1 x − x + 1 = −π 2 4x 1 + 3x 2 − 3x + 3 . 2 3 2 2 2 2 8 x x ⇒ A1 = 2 2 ⋅ x 3/2 + 2 8 − x 2 + sin −1 2r h2 3 0 2 2 2 2 2 Putting h = in R2 = r 2 − . Let V be the 1 1+ volume of the cylinder and r be the radius of x + tan −1 x − 1 = − tan −1 (7) ⇒ tan −1 1 x sphere. 2 3 Area of circle = π(2 2 ) = 8 π sq. = = 4 9π − 2 A2 x −1 x +1 6π − = tan −1 (−7) + tan −1 tan −1 3 x x − 1 24.. we get l – 2m = 0. cos x and constant term on both sides. 2l + m = 3. µ = and v = − 5 5 5 λ(sin x + 2 cos x + 3) + µ(cos x − 2 sin x ) + ν ∴ I=∫ dx sin x + 2 cos x + 3 ⇒ I = λ ∫ dx + µ ∫ cos x − 2 sin x dx sin x + 2 cos x + 3 + ν∫ 1 dx sin x + 2 cos x + 3 ⇒ I = lx + m log |sin x + 2 cos x + 3| + n I1.⇒ 3x 2 − 4 x 2 + 4 x − 4 = tan(–p) = –tanp = 0 3x 2 − 3x + 3 + 4 x 2 ⇒ – x2 + 4x – 4 = 0 ⇒ x2 – 4x + 4 = 0 ⇒ (x – 2)2 = 0 ⇒ x = 2 Putting. cos x = 1 − tan2 (x / 2) 1 + tan2 (x / 2) 1 2 tan(x / 2) 1 + tan2 (x / 2) 26. I=∫ 2 tan(x / 2) + 2(1 − tan2 (x / 2)) 1 + tan2 (x / 2) dx +3 1 + tan2 (x / 2) 2 tan(x /2) + 2 − 2 tan2 (x /2) + 3(1 + tan2 (x /2)) ⇒ I1 = ∫ . We have. I = l x + m log |sin x + 2 cos x + 3| x tan + 1 + ν tan −1 2 +C 2 6 3 ⇒ I = x + log sin x + 2 cos x + 3 5 5 8 tan(x / 2) + 1 − tan −1 + C 5 2 nn MatheMatics tODaY | March ’15 81 . sin x = we get I1 = ∫ 3 cos x + 2 dx sin x + 2 cos x + 3 ⇒ I1 = ∫ Let 3 cos x + 2 = l (sinx + 2 cos x + 3) + m(cosx – 2 sin x) + n Comparing the coefficients of sin x. where I1 = ∫ 1 + tan2 (x / 2) 1 dx sin x + 2 cos x + 3 . 3l + n = 2 6 3 8 ⇒ λ = . dx sec2 (x / 2) dx tan2 (x / 2) + 2 tan(x / 2) + 5 x x 1 Putting tan = t ⇒ sec2 dx = dt 2 2 2 dt 2dt ⇒ I1 = ∫ = 2∫ 2 t + 2t + 5 (t + 1)2 + 22 x tan + 1 2 t + 1 − 1 2 +C = tan −1 + C = tan 2 2 2 Hence. t = 1. 3. 999 x = abc 999∑ x = 111 ⋅ 9 ⋅ 8(0 + 1 + . Khokon Kumar Nandi : W. N. (a) : x = et ⇒ I = ∫ t 3e 3t dt = 0 2 (2e 3 + 1) 27 1 3. (b) : 2ab cos C = a2 + b2 – c2 ∴ 2 8. Replacing x by p – x and adding with I π ⇒ 2 I = ∫ f (x )dx = 7. Shreyam Maity : W... Devjit acharjee : W.B.Solution Set-146 1.B. nn .. x = 1 + 3 . 4. + 19)2 2 –(12 + 32 + 52 + . 5. Khokon Kumar Nandi : W. + 9) ∴ ∑ x = 360. − 2i − j + 5k 16 0 = (x 2 + x + 1)2 + (x 2 − 1)2 − (2 x + 1)2 Solving.. (d) : PQ2 = 5 cos2q + (1 + sinq)2 = f(q) 1 5 f ′(θ) = 0 ⇒ sin θ = . Replacing x by p – x and adding with I ⇒ 2I = π π ∫ f (x )dx = π ⋅ 0 16 π 2 ⇒I= 8 π Solution Sender of Maths Musing set-146 ⇒ (2 − 3 )x 2 + (2 − 3 )x − (1 + 3 ) = 0 x = −(2 + 3 ) ⇒ 2 x + 1 = c < 0 ∴ x = 1 + 3 6.. tan α = 2 3 4 5! 5 = 2 + 1 + 1 + 1 ⇒ ⋅ = 720 words 1 3 2! 5 +1 2 1 2.. PQmax = 2 2 82 MatheMatics tODaY | March ’15 2 ⇒I= 8 π π2 S. chirayata Bhattacharyya : W.. (c) : N = [(−1 + 3 − 5 + 7 − . k ∈R –(cosa + isina)k2 + ki + 1 = 0 ⇒ k2cosa = 1. Gouri Sankar adhikari : W. rintu Giri. 3.. 5. set-145 1. (c) : P. 5 −1 cos α = . Gouri Sankar adhikari. 4.B. PQmax = 4 2 9. 2. Jayanthi : hyderabad : W. with digit sum 9. Eliminating k.B. (b) : PQ2 = 2cos2q + (1 + sin q)2 = f(q) f ′(θ) = 0 ⇒ θ = π . abc. |N| = 615.B. with digit sum 12 4. I = – I ⇒ I = 0 R.. + 192)] = – 615.. chandan Pramanik 2.. (a. (c) : x = 0. (d) : z = ki. I = ∫ 2x − π 10.B. − 3 3 |a | The vectors are 2i + 3j − 3k . (3) : CC LL UU AS 3 3 5! = 270 words 5 = 2 + 2 +1 ⇒ ⋅ 2 1 2!2! 2 3 (x + x + 1)(x − 1) 5 5 = 1 + 1 + 1 + 1 + 1 ⇒ ⋅ 5 ! = 120 words 5 Total = 1110 words Digit sum of N = 3 0 Replacing x by p – x.B. c): A vector in the plane of b and c is b + λc = (1 + λ)i + (2 + λ)j − (1 + 2 λ)k 2 (b + λc ) = ⋅ a ⇒ λ = 1. Sattwik Sadhu : W..B. ksina = 1. −(2 + 3 ) π /2 16 16 π cos x ⇒ I = t sin t dt = 2 ∫ 2 π 0 π2 π π sin 2 x sin cos x dx 2 Q. PaPer-1 section-i Multiple CorreCt ChoiCe type this section contains 10 multiple choice questions. 1) and satisfying the differential dy equation + y cos x = cos x is such that it is dx (a) a constant function (b) periodic (c) neither an even nor an odd function (d) continuous and differentiable for all x 6. The solution of satisfying = dx 2 xy y(1) = 1 is (a) a hyperbola (b) a circle (c) y2 = x(1 + x) – 1 (d) (x – 2)2 + (y – 3)2 = 5 x is x +1 x − x + fog (x ) + c. each question has four choices (a). The value of the integral π/ 4 dx is ∫ 2 2 2 2 0 a cos x + b sin x 7.: (022) 24306367 mathematics today | march ‘15 83 . g (x ) = x 2. The value of the integral ∫ e x dx is (a) less than e (c) less than 1 (d) 2a 0 (b) greater than e (d) greater than 1 4. Senapati Bapat Marg. The graph of the function y = f (x) passing through the point (0. Dadar (W). b = 1) 4 1 a 1 tan −1 + (d) b ab ab (c) 5. dy 2 =1 3 d2 y dy =0 + 2 dx dx π/2 1 + sin 3x ∫ dx is 0 1 + 2 sin x (a) 1 b tan −1 (a > 0. [Correct answer 3 marks and wrong answer no negative mark] 2 2 dy x + y + 1 1. f (x ) = tan x π (a = 1. b > 0) (a) a ab (b) 1 b tan −1 (a < 0. The differential equation of all parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis is (a) of degree 2 and order 1 (b) of order 2 and degree 3 −1 (c) f (x ) = tan x .28. (c) and (d) out of which oNe or More may be correct. g (x ) = x (d) none of these 1 (c) 2a 2 3. then x sin −1 x +1 (a) f(x) = sin–1x. Tel. Pearl Centre. If the antiderivative of sin −1 −1 (b) g (x ) = x + 1. (b). Mumbai . b < 0) a ab (c) (b) d2 x π/2 1 − cos 3x dx ∫ 0 1 + 2 cos x π/2 cos 3x + 1 dx ∫ 2 cos x − 1 0 π 2 (d) 1 By : Vidyalankar Institute. when worked out will result in one integer from 0 to 9 (both inclusive). Orthogonal trajectories of family of parabolas y2 = 4a(x + a). If ∫ f (sin 2 x )sin xdx =A 2 ∫ f (cos 2 x )cos xdx . 3 dx y(0) = 4/3 is PaPer-2 section-i SiNgle CorreCt optioN this section contains 10 multiple choice questions. then ∫ f (x )dx = −1 1 1 1 tan −1 tan x + c (a) − tan x + 2 2 2 1 1 1 tan −1 tan x + c (b) − cot x + 2 2 2 12. Find the area enclosed by the curve [x] + [y] = 4 in the 1st quadrant (where [·] denotes greatest integer function). The order of the differential equation whose general solution is given by y = C1 + C2 cosx + C3 + C4ex + C5. where a is an arbitrary constant is (a) ax2 = 4cy (b) x2 + y2 = a2 − x 2a 14. The integral of 2 2 sin x + tan x 2 must be 11. The area bounded by the curves f (x) = sin–1(sinx) and g(x) = [sin–1(sinx)] in the interval [0. The value of y 8 − if 1 + x 2 = x(1 − y ). out of which oNly oNe is correct. where k is 2 2 ln x − 1 10. C4. The value of 0 . ∫ dx is equal to (ln x )2 + 1 ln x x (a) (b) +c +c (ln x )2 + 1 x2 + 1 (c) x 2 (ln x ) + 1 +c 1x ∫ {3t − 2 g ′(t )}dt . C2. (c) and (d) for its answer. The value of ∫ | 1 − x 2 | dx is (d) e x x + c 2 x +1 −2 π /2 π/ 4 0 0 19.1 8. each question. If f (x) = max{2 – x. ∫ sin 2 x tan (sin x )dx (c) y = ce (d) axy = c2 where c is a constant. section-ii oNe iNteger value CorreCt type this section contains 10 questions. C5 arbitrary constants are π /2 −1 16. where C1. 2. [Correct answer 3 marks and wrong answer –1 mark] 1. 15. where [·] denotes the greatest integer function. is 84 mathematics today | march‘15 π −1 (b) 2 (a) p/2 (c) p 2 π (d) − 1 4 2 ∫ [x − 1]dx 2 2 2. then 2g′(2) = x2 e37 π sin(π ln x ) dx is x 1 17. [Correct answer 3 marks & wrong answer no negative mark] then the value of A is 1 dy 20. x-axis and x = e is 13. C3. is given by (a) 3 − 3 − 2 (b) 2 − 3 (c) 4 − 3 − 2 (d) none of these . p]. 0 π must be equal to − k. If g (x ) = 1 1 (c) − cot x − tan −1 tan x + c 2 2 1 1 1 (d) − cot x − cot −1 tan x + c 2 2 2 9. each question has four choices (a). where [·] is a greatest integer function. 1 + x}. The area bounded by the curves y = lnx. The value of integral. The value of ∫ 2 18. (b). Periodicity of its publication 3.3 – 4I3. The anti-derivative of f (x ) = 6. Based upon each of the paragraphs 2 multiple choice questions have to be answered. Mahabir Singh Publisher mathematics today | march ‘15 85 . then there exists a polynomial f (x) of degree (n – 1) and a constant such that ∫ φ(x )dx (ax 2 + bx + c) +c 4 (b) 2 = f (x ) ax 2 + bx + c + D ∫ dx ax 2 + bx + c Form IV 1. If ∫ tan7 xdx = f (x ) + log | cos x |. Gurgaon Haryana . [Correct answer 3 marks & wrong answer –1 mark] Paragraph for Q. then (a) f (x) is a polynomial of degree 8 in tanx (b) f (x) is a polynomial of degree 5 in tanx (c) f (x) is a polynomial of degree 6 in tanx (d) f (x ) = 1 tan6 x 1 − tan 4 x + tan2 x 6 4 2 + log|cosx| + c −1 (c) 0 (d) 1 section-ii paragraph type this section contains 3 paragraph. where f(x) (ax 2 + bx + c) is a polynomial in x. 406. then the value of A + B is 0 cos x − sin x B= ∫ (b) p/4 (a) 3p/2 (d) p (c) 0 1 . If f (x) is an even and differentiable function. Anil Ahlawat Indian Mathematics Today. If f (x) is a polynomial of degree n. 3 + 5 sin x + 3 cos x whose graph passes through the point (0. (c) and (d) out of which oNly oNe is correct. 11 and 12 φ(x )dx Consider the integral ∫ . New Delhi . Place of Publication 2. Taj Apartment New Delhi 5. 19. Printer’s and Publisher’s Name Nationality Address : : : : : 4. then 7I4. 0) is 5. If A = ∫ (a) 6 3 π /2 sin x dx. Mahabir Singh.2 = (a) constant (b) –cos2x + c (c) –cos4x cos3x + c (d) cos7x – cos4x + c 10. each of these questions has four choices (a). (b). ∞ x log xdx is equal to ∫ (1 + x 2 )2 (a) 1 (c) 2 0 (b) 0 (d) none of these 3 π /2 9. Taj Apartment. If Im.110029. (a) 1 5 x ln 1 − tan 5 3 2 (b) (c) 1 5 x ln 1 + cot 5 3 2 (d) none of these ∫ xe x (1 + x )2 (a) dx is ex +c x +1 (c) − 1 5 x ln 1 + tan 5 3 2 ex (x + 1)2 (b) ex(x + 1) + c +c (d) ex 1 + x2 dy 2 dy 7. A solution of y = 2 x + x is dx dx 2 2 (a) y = 2 c1/2 x1/4 + c (b) y = 2 cx + c (c) y = 2 c (x + 1) 2 (d) y = 2 cx + c 8. National Media Centre.3. Editor’s Name Nationality Address : : : New Delhi Monthly Mahabir Singh Indian Mathematics Today. hereby declare that particulars given above are true to the best of my knowledge and belief. cos x dx and 0 cos x − sin x 1 then the value of ∫ x 3 f (x ) + xf ′′(x ) + 3 dx = 4. Name and address of : individuals who own the newspapers and partners or shareholders holding more than one percent of the total capital I. n = ∫ cosm x sin nx dx .122002 Mahabir Singh 406. No. Now apply this method to evaluate the given integral. we get 1 2 (a) φ(x ) = f ′(x )(ax + bx + c) + (2ax + b) f (x ) + D 2 multiply by (a) 1 2 (b) φ(x ) = f ′(x )(ax 2 + bx + c) − (2ax + b) f (x ) + D (b) 1 1 2 (c) φ(x ) = f ′(x )(ax + bx + c) + (2ax + b) f (x ) + D 2 2 (d) none of these (c) C0 n C1 n C2 − + − . matching List tyPe this section contains 4 questions.. Column I (P) x Also.. Choices for the correct combination of elements from list-i and list-ii are given as options (a). (R) Area contained between 3. = ( Ax 2 + Bx + C ) (5 + 6 x − x 2 ) + D sin −1 14 then 2 (a) A = 3 9 (b) B = 5 227 1 (c) C = (d) C = 6 6 Paragraph for Q..11.. out of which one is correct. Function g(x) is (a) odd (b) even (c) neither even nor odd (d) can’t be determined 14. to (n + 1) terms 3 4 5 n C1 n C2 n C3 − + − . The area of a loop of the above curve is (Q) The area of the curve a2y2 = x2(a2 – x2) is 2.. [Correct answer 3 marks & wrong answer –1 mark] 17... to (n + 1) terms 3 4 5 (d) none of these 12. 4a2 y2(a2 + x2) = x2(a2 – x2). to (n + 1) terms 3 4 5 n C3 n C4 n C5 − + − . Value of g(2) in terms of A is (a) 2A (b) A/2 (c) 4A (d) A/4 Paragraph for Q. each having two matching lists. Then ∫ f (x )dx is equal to (a) 3ln3 – 5ln2 (c) 5ln2 + 3ln3 86 mathematics today | (b) 5ln2 – 3ln3 (d) none of these march‘15 π 2a2 1 + 4 π a 2 − 1 2 4 2 a 3 . 0 13.. function g (x ) = ∫ f (t )dt and g (1) = A. Match the following.. If ∫ n Section-iii (x 3 + 4 x 2 − 6 x + 3)dx 5 + 6x − x 2 3− x . the curve y2(a – x) = x2(a + x) and its asymptotes is (S) The area enclosed by the 4. (c) and (d). No. 15 and 16 2 Let f (x ) = x 3 + 6 x 2 + 11x + 6 Column II The equation of curve 1. Differentiating both sides with respect to x and 16. f (n) is equal to ax 2 + bx + c . No. (b). 13 and 14 An even function f is defined and integrable everywhere and is periodic with period 2. parabola ay = 3(a2 – x2) and the x-axis is P Q R S 1 (a) 1 2 3 4 0 (b) 3 4 2 1 (c) 3 1 2 4 (d) 3 4 1 2 15. 11.18. ∫ | ln x | dx − 1/e P 2 1 3 2 Q 3 2 4 1 R 4 3 1 3 1 2 π2 6 S 1 4 2 4 1. 2 1 − e 0 (1 + sin x )(2 + sin x ) ∫ (S) 19. (a. 19. b. c) (a. 7. 20. 10. 13. 5.vidyalankar. d) (b. (b) (a) (a) (a) 2. 16. 8. (a) 15. ex ln(x + 2) + c 2 2. Q 4 1 1 1 1 +c 1+ x π x ln tan − + c 2 8 2 1 2 R 1 2 4 2 tan −1 2 x −1 x 2 (S) dx 2 − 3x − x 5 x ∫ 2 dx x +1 (a) (b) (c) (d) P 3 4 1 3 Q 4 1 2 1 x 4 x2 − 4 2 2. 15. 12. (c) (c) (a) (a) 5. b. 13.2 5 sin x sin x − +c 3 5 4. c) (a. 7. 9. (c) (a. (b) 10. (a) (d) (c) (b) 3. 19. 3 3. www.org nn S 2 3 4 4 mathematics today | march ‘15 87 . ln4 – ln3 ansWer Keys ex 2x + 3 [(1 + (x + 2) 1. 8. 17. Match the following: 20. (a) For detailed solution to the Practice Paper. 18. d) (c) (3) (1) (1) 2. (a. sin −1 +c x +2 17 ln(x + 2)] dx ∫ ∫ | cos x | dx 0 (a) (b) (c) (d) 1 + ln(x 2 + 1) + c 2 (R) 41π/ 4 1 (R) +c Column II (Q) ∫ sin2 x cos3 x dx (Q) S 2 4 2 3 Column I (P) cos x 1 dx 1. Match the following: ∫ π /2 (P) 1 2 ln x P 3 3 3 4 2. d) (c. 20 + e 4. d) (7) (5) (2) (3) 3. (b) 20. b. 6. 18. 11. tan–1(tan2x) + c ∫ sin x − cos x (Q) ∫ (R) Column II (a) (b) (c) (d) sin(2x ) dx x ln x e 1 ln 1+ x2 x2 + 1 x 3. 14. d) (c) (1) (3) (4) PaPer . dx ln x − ln(1 + x ) + dx 3 (1 + x ) 4. 14. 17. 6. R 1 2 3 2 ln(1 − x ) dx x 0 ∫ PaPer-1 3. 12. 9. 4. 16. visit our website. Match the following: Column I (P) ∫ ∫ (S) Column I 1 dx 1. (b) (d) (a) (c) 4. Column II 1. . CD and BE meet AB and AC at P and Q respectively. Let AB = c. and according to Menelaos' Theorem applied to DACD. (a) Let g(n) = f(f(n)). Then from (1).OLYMPIAD CORNER 1. In a group of nine mathematicians each speaks at most three languages and any two of them speak at least one common language. 2 2 2 PA + PB + PC = 5. and AD : AB = AE : AC = x : 1. PB. [ACE] : [ABD] = b2 : c2 = (PC)2 : (DP)2 = [ABC]2 : [ABD]2 .. we get collinear points B*. so that ∠ABD = ∠ACE and ∠BAD = ∠CAE. it is possible to construct a triangle of sides PA. AP = AQ if and only if xb ⋅ DP ⋅ (c − b) b ⋅ DP 1 = AE * ⋅DP ⋅ CB * = = . (b) Show that for all points P of Γ . Similar triangles ABD and ACE are drawn outwardly on the sides AB and AC of DABC. Q and E around the internal bisector of ∠BAC. B* C Q A Q* E* P D B Reflecting B. (b) For any n ≥ 1. so that AE* = AE = xb and AD = xc. 5. etc). with area 3 / 4. Two externally tangent circles of radii R1 and R2 are internally tangent to a semicircle of radius 1. sOlutiOns E 1. let r(n) be the least integer r such that f r (n) = 0 (where f 2 (n) = f(f(n)). 88 MatheMatics tODaY | march ’15 (a) Show that for all points P of Γ . i. Show that at least five of them share a common language. Now. ABC is a triangle with AB ≠ AC.. AB and AD respectively. Prove that R1 + R2 ≤ 2( 2 − 1). PC. . it follows that B* ≠ C and E* ≠ D. DE * ⋅CP ⋅ AB * (xc − xb) ⋅ PC ⋅ c c ⋅ PC that is. Compute n lim inf r (n) ⋅ n→∞ 2 3. where [XYZ] denotes the area of DXYZ. f 3 (n) = f(f 2 (n)). f(2n) = 2f(n) + 1 for n ≥ 1. AC = b. Q* and E* on the lines AC. AP = AQ if and only if P = Q*. 2. Show that g(n – g(n)) = 0 for all n ≥ 0. The function f is defined on non-negative integers by. P and E* are collinear. AP = AQ ⇔ B*. f(0) = 0 and f(2n + 1) = 2f(n) for n ≥ 0. as in the figure. In an equilateral triangle ABC (of side length 2) consider the incircle Γ . since A will lie between P and B and between Q and C if and only if ∠BAE = ∠DAC > 180°.(1) From AB ≠ AC.e. Prove that AP = AQ if and only if [ABD]·[ACE] = [ABC]2. 4. ... 101 . . In other words. Thus..0 ..e.. M1 and M2 have L1 and L2 in common. t..... To prepare for this.. and s ≥ 1. 10.. and since n + f(n) = L(n) – 1 by the lemma.... Furthermore. Suppose now that each speaks three languages.... where x denotes the greatest integer ≤ x. say.. Since . we claim that L(2n + 1) = L(2n).. g (n) = f ( f (n)) = 1 * .. .. Then f(2n + 1) = 2f(n) = 2L(n) – 2n – 2 = L(2n) – 2n – 2 = L(2n + 1) – (2n + 1) – 1 and f(2n) = 2f(n) + 1 = L(2n) – 2n – 1...... Since g(0) = f(f(0)) = f(0) = 0. Three cases arise: (i) Two mathematicians have all the three languages in common. Suppose M3 speaks (L1.. So the limit inferior of n/2r(n) is the limit of the sequence (in base 2) 1 .. we first prove the following lemma... unless n is in the sequence (in base 2) 1. Since f(1) = 0 and L(1) = 2. 1 10 1 if m is odd.. 10 .... we have g(n – g(n)) = 0 for all n ≥ 1... n/2r(n) will exceed 1 unless n has exactly r(n) digits....1 . 10 * .that is.. we see that f acts on n by interchanging all 1's and 0's in the binary expansion of n.. s1 s2 sm n = . u ≥ 0. . [ABD]·[ACE] = [ABC]2.... it is necessary to examine more closely the blocks of binary digits of n. 10 s t +u Since the last expression contains no final block of digits after the initial block of 1's and 0's.. L5). If any one of them speaks less than three languages then by the pigeon hole principle one of the languages he (or she) speaks should be spoken by four others and we are done..0 if m is even n.0 . * .. *′ ′ . we have log2(2n) < k + 1 ≤ log2(2n + 1) which implies 2n < 2k + 1 ≤ 2n + 1 or n < 2k ≤ n + 1/2.0 1 * ... 10 1 s2 sm s1 and since f always wipes out the leading block of 1's..... f (n) = 1 . L2. = 2 .. i. g(0 – g(0)) = 0. 2 8 32 3 3..... (b) The answer is 2/3...... f(n) = L(n) – n – 1 holds for n = 1. it is clear that r(n) = m... clearly an impossibility.. n ≥ 1.. n − g (n) = 1 .. which is (in base 10) 10 102 103 1 + 1 + 1 + . Proof : Since 1 + log 2 (2n) = 1 + log 2 (2n) = log 2 (4n) we have 2L(n) = L(2n)... assume the formula holds up to 2n – 1 for some n ≥ 1.......... Since L(n) – 1 is just a string of 1's in binary notation.. Let L3 and L4 be the third languages of M1 and M2 respectively.... If there is one more mathematician having L1 and L2 as two of his languages then we are done as follows. 2.. 101.... Of the remaining six mathematicians if there are MatheMatics tODaY | march ’15 89 .e. .... Since n has atleast r(n) digits and 2r(n) has 1+ r(n) digits....... 1010.... ... 0 . LeMMa : Let L(n) = 2 log 2 (2n) . The binary representation of n can be written in the form n = 1 .... This completes the proof of the lemma. as a result.. Then letting log 2 (2n) = k. To see this..0 ... . (ii) Some two mathematicians have two languages in common... 10101....... u i.... g(n) is the original final block of u digits in n. Here again we are through since one of the three languages should be spoken by at least three of the remaining seven mathematicians.. 1 0* t u where each *' is 1 – *. assume henceforth that n ≥ 1. Then f(n) = L(n) – n – 1. (a) Most of the solution can be worked out suitably in binary notation. Suppose log 2 (2n) < log 2 (2n + 1) . s t u where each * is either 0 or 1. O2 and O denote the centres of the circles. Now consider the function f(x) = 2x( 2 – x) f(x) is increasing on the interval (0.. Squaring. 90 MatheMatics tODaY | march ’15 1 B 1 r r rO 1 C 1 1 Take O as origin. C have coordinates . A2. 4. O A2 1 Then O1O2 = R1 + R2. so that OO1 = 1 – R1 and OO2 = 1 – R2. with the x-axis parallel to BC.(2) ≤ (R1 + R2 )( 2 − R1R2 ) and therefore R1R2 ≤ 2 − 1. and O2A2 = R2. say M5 has L2 in common with M1..Therefore A1A2 = OA1 + OA2 more than two who speak L1 or L2 then we are through. there are five who do not speak L3 and L4 simultaneously. O1A1 = R1. equality holds when R1 = R2. then dividing each term by 2 and rearranging the terms. so A1 A2 = (O1O2 )2 − (O1 A1 − O2 A2 )2 2 2 = (R1 + R2 ) − (R1 − R2 ) = 2 R1R2 . B. M4 also. O1 A1 . = (OO1 )2 − (O1 A1 )2 + (OO2 )2 − (O2 A2 )2 = (1 − R1 )2 − R12 + (1 − R2 )2 − R22 = 1 − 2R1 + 1 − 2R2 . Let M1 speak (L1. B2 B1 Squaring both sides and simplify 8R1R2 = (2R1R2 + R1 + R2 )2 . O1B1 = R1 and O2B2 = R2. OB2 = 1. Now M5 has to speak to each of M2. We will show that L1 is spoken by all the nine mathematicians. Let O1. as shown in the diagram. M2. Suppose not.. 2 2R1R2 = 2R1R2 + R1 + R2 . L 4 ). So suppose only two have L1 and L2 common. R1 + R2 attains its maximum when R1R2 = 2 − 1. M3.(1) So. there are four mathematicians who do not speak L1 or L2. L3). But then these five should speak either L1 or L2 to be able to converse with both M1 and M2 which in turn implies that either L1 or L2 is spoken by at least five mathematicians. The remaining eight have to speak either L1 or L2 or L3 and so one of these three languages has to be spoken by at least three more mathematicians. Since 0 < R1R2 ≤ 2 − 1 < 1 / 2 and R1 + R2 = f ( R1R2 ) from (2). L 5 to be able to converse with M1. if not. Consider the pair (L 3 . L2. we get (1 − 2R1 )(1 − 2R2 ) = 2R1R2 + R1 + R2 − 1. M4 in a different language other than L2 (since any two mathematicians have only one common language) which is not possible as no mathematician speaks more than three languages and we are done. If some three mathematicians speak both these languages we are done as before. B1 and B2 denote the points of tangency of these circles with the semicircle. L4 and L5. and let A1. M3 and so there will be five mathematicians who speak L3. 1 / 2 ) since f′(x) = 2 2 – 4x > 0 for x in the interval. Hence R1 + R2 ≤ 2( 2 – 1)[ 2 – ( 2 – 1)] = 2( 2 − 1). Thus 1 − 2R1 + 1 − 2R2 = 2 R1R2 . also OB1 = 1. Thus R1 + R2 = 2 R1R2 ( 2 − R1R2 ) . and y-axis along OA. (iii) any two of the nine mathematicians have exactly one language in common. if not. L 4. say L1 is spoken by M2. O2 A 5. Then the incircle has radius r = 1 / 3 and A. M3. But then these four are forced to speak L 3.. 0 ≤ q < 2p given by P 1 cos θ. BP. of this triangle is given by 1 (x + y + z) 1 (–x + y + z)· 1 (x – y + z)· 2 2 2 1 (x + y – z) 2 = 1 [(y + z)2 – x2][x2 – (y – z)2] 16 F2 = +2 ⋅ 1 (5 + 2s + 2 3c)1/2 (c + 2s − 2 3c)1/2 3 (b) Now. 3 y = BP = ( ) y = 1 5 + 4 1 s + 3 c = 1 5 + 4 cos θ − π 2 2 6 3 3 7 ≤y≤ 3 so that and 3 z= 1+ 5 > y and (z + x )min = max = 3 . F. BP. From Heron's formula the area. s = sinq. 3 By reflection and rotational geometry the distances AP. Thus 3 3 MatheMatics tODaY | march ’15 91 . CP will be a permutation of those obtained when p/6 ≤ q ≤ p/2. so that 1 ≤ x ≤ 1. and x. C 1.− − A 0. 1 . 3 = 1 1 (5 + 2s + 2 3c + 5 + 2s − 2 3c − 5 + 4 s) 16 3 = 1 [3(c2 + s2) + 12] = 5 3 also and ( y + z )min = 7 + 5 > x max = 1 3 ⋅ 1 (5 − 4 s − 5 − 2s − 2 3c − 5 − 2s + 2 3c) 3 +2 ⋅ 1 (5 + 2s + 2 3c)1/2 (5 + 2s − 2 3c)1/2 3 = 1 [(5 + 8s) + 2 (5 + 2s)2 − 12c 2 ] 48 [−(5 + 8s) + 2 (5 + 2s)2 − 12c 2 ] = 1 [100 + 80s + 16s2 – 48c2 – 25 – 80s – 64s2] 48 = 1 [75 − 48] = 9 . y. set x = AP = 1 5 − 4 s . 3 3 3 (x + y )min = 1 + 7 > z max = 7 3 3 a point on the incircle has coordinates parameterized by q. 1 sin θ . since c2 + s2 = 1.e. 3 3 (a) Let c = cosq. 2 . AP. 3 1 5 2 2 3 + s+ c . B −1. z (i. 1 . Then AP2 + BP2 + CP2 = 1 c 2 + 1 (2 − s)2 + 1 (c + 3 )2 + 1 (s + 1)2 3 3 3 3 1 (c − 3 )2 + 1 (s + 1)2 3 3 = 1 [c 2 + 4 − 4 s 2 + s 2 + c 2 + 2 3c + 3 + 3 s 2 + 2s + 1 + c 2 − 2 3c + 3 + s 2 + 2s + 1] since c2 + s2 = 1. CP) can form the sides of a triangle. 48 16 Thus F = 3. 4 nn ( ) 1 5 41 3 = 1 5 + 4 sin θ − π + s − c 2 2 3 3 3 so that 5 ≤z≤ 7. and 3 z = CP = 1 5 + 2s − 2 3c. Problem solving is further defined as the complex interplay of cognitive. social work. Peer pressure. It makes sense that effective problem solvers are flexible. adaptable. Obviously. if a goal is readily attainable. problem solving abilities come into the picture. says Vintu Augustine O ur lives are replete with daily hassles and stressful events. You need to acquire the abilities to handle problematic situations and emerge successful. they do have a psychological effect on the young minds. It becomes extremely difficult to search for solutions for all such matters. extreme competition and lack of opportunities make their lives hard and impair their abilities. clinical psychology. Sure.Vital Problem-solving skills for students Pushing towards growth : Students need more than just technical education to survive today. Similarly. everyone is born with some talents and abilities in this world but daily hassles and problematic events test every aspect of an individual. effective problem solvers It has been suggested for some time that ineffective problem solving results in stressful outcomes and psychological maladjustment. and are able to develop . Problem solving abilities fill that void and help students gain better chances of survival. No matter how small or how big the stressors are. we face manifold difficulties. one must go beyond the information given to overcome obstacles and reach the goal. The failure to arrive at a solution to a problem puts us into despair and makes us demotivated. Problem solving has been defined as an overt or cognitive process that makes available a variety of potentially effective response strategies for coping with problematic situations. The problem solving skills of a student depends on his critical personal resources for dealing with stressors. students also face a number of obstacles in their daily lives. health care and so on. The result is stress and disappointment. To ensure students have a well-rounded personality and survive in this world. From the morning crossword puzzle to retrieving keys from a locked car. It is a topic of great significance especially in areas like counselling. and behavioural processes for the purpose of adapting to internal or external demands or challenges. But in problem solving situations. there isn’t a problem. Only talent doesn’t matter anymore. 92 MatheMatics tODaY | march ’15 What is problem solving? Problem solving refers to active efforts to discover what must be done to achieve a goal that is not readily available. affective. suicidal ideation. and evaluation of outcomes. including problem orientation. Students who promote problem-based learning as an learning approach are expected to be more successful in academic performance and to be better prepared for selfdirected. selection of strategies. Given the socio-economic and technological advancements. when education has taken a commercial turn and there is a renewed thrust on enhancing the performance of students. it may become an important predictor of managing daily chaos of life. there has been mounting evidence that problem solving or coping does play a role in adaptive responses to stress. it is vital that we form strategies or prepare intervention programs and necessary tools to improve problem solving abilities of students. it is imperative that we assess students’ problem solving skills and how that create an impact in their academic performance. college students. Likewise. and exams to prepare for. problem solving has been related to depression. To enhance quality. In addition to the link with problems. Therefore. student affairs professionals are often interested in psycho-educational programming to impart specific skills (assertiveness). Such an empirical research should serve as a fundamental template in syllabus formulation and curriculum design. students’ life is the perfect time to assess the problem solving ability and determine if there is a need to improve it before they get in to their career life. a student faces a lot of problems. spatial ability. the advent of problem-based learning has made problem solving central to the educational process. For example. The procedure of overcoming difficulties or barriers which interfere with the satisfaction of wants and fulfillment of goals is called problem solving. hopelessness. adolescents. Educators are often interested not only in imparting knowledge about specific topics but also in increasing students’ problemsolving abilities. physical health. a better understanding of students’ problem solving abilities and training them with strategies to improve their capacities in this regard will definitely go a long way in determining the quality of education today. thus. and resistance to distraction. life in today’s world is becoming more and more complex with excess of problems. school psychologists. Indeed. field independence. and student affairs professionals. verbal and general reasoning ability. at the higher level of evolution. in the last few years. In all. and older adults.suitable methods to solve problems and reach personal goals. divergent thinking. Courtesy: Deccan Herald nn MatheMatics tODaY | march ’15 93 . Relation to academic performance Problem solving is very relevant for educators. They will have papers to write. as well as generic problemsolving skills for preventive purposes. It also depends on the difficulty of the problem and the ability of the solvers. It is important to recognise problem solving ability of students as a tool kit to academic performance and in addition. problem solving is a topic that has a great deal of applicability for a wide array of practitioners as they work to increase the problem-solving effectiveness of a broad range of people like children. They may have to maintain a B average to qualify for scholarship or else an a average to sustain parents’ support. he will have to engage in quite a balancing act in meeting deadlines and fulfilling requirements. This can be achieved only by providing quality education. The nature of approach to problems varies from person to person. most researchers see problem solving as consisting of a sequence of activities. as the end of a semester approaches. there is a large body of empirical research demonstrating the relation between problem solving and other variables. augmenting performance We live in an era. generation of alternatives. coping with stress In a given academic year. positive attitudes. In human beings. and the responsibility of school authorities and teachers especially has become increasingly demanding to develop scientific attitude and skills in students to cope with the issues of the time. the animals also solve their problems by insight. presentations to give. animals solve their problems by habitual behaviour or by following trial and error method. help seeking. lifelong learning. Towards this end. reasoning is the most important method of problem solving. adults. Studies have found problem solving to be related to mathematical achievement. career planning and academic performance. Effective problem solving can decrease negative affective states by enabling a student more effectively managing the problems. Theoretical accounts of problem solving postulates specific cognitive behavioural skills as important aspect involved in academic achievement. 94 MatheMatics tODaY | march ’15 .