45, 2nd Floor, Maharishi Dayanand Marg, Corner Market, Malviya Nagar, New Delhi - 110017 Tel : 49842349 / 49842350 No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried and made our best efforts to provide accurate up-to-date information in this book. All Right Reserved © Copyright Disha Corporate Office DISHA PUBLICATION Typeset by Disha DTP Team www.dishapublication.com www.mylearninggraph.com Books & ebooks for School & Competitive Exams Etests for Competitive Exams Write to us at feedback_disha@aiets.co.in Contents S.no. Chapter Name Page no. Questions Solutions 1. SETS 1-2 75-77 2. RELATIONS & FUNCTIONS-1 3-4 77-81 3. TRIGONOMETRIC FUNCTIONS 5-6 81-84 4. PRINCIPLE OF MATHEMATICAL INDUCTION 7-8 84-88 5. COMPLEX NUMBERS AND QUADRATIC EQUATIONS 9-11 88-92 6. LINEAR INEQUALITIES 12-13 92-96 7. PERMUTATIONS AND COMBINATIONS 14-15 96-98 8. BINOMIAL THEOREM 16-17 99-101 9. SEQUENCES AND SERIES 18-19 102-106 10. STRAIGHT LINES 20-22 107-112 11. CONIC SECTIONS 23-25 113-118 12. LIMITS & DERIVATIVES 26-27 118-123 13. MATHEMATICAL REASONING 28-29 123-124 14. STATISTICS 30-31 124-127 15. PROBABILITY-1 32-33 128-131 16. RELATIONS & FUNCTIONS-2 34-35 132-135 17. INVERSE TRIGONOMETRIC FUNCTIONS 36-38 136-140 18. MATRICES 39-41 141-145 19. DETERMINANTS 42-44 146-152 20. CONTINUITY AND DIFFERENTIABILITY 45-47 153-158 21. APPLICATION OF DERIVATIVES 48-50 159-165 22. INDEFINITE INTEGRATION 51-53 165-170 23. DEFINITE INTEGRATION 54-56 171-175 24. APPLICATION OF INTEGRALS 57-59 176-182 25. DIFFERENTIAL EQUATION AND ITS APPLICATIONS 60-62 182-187 26. VECTOR ALGEBRA 63-65 188-193 27. THREE DIMENSIONAL GEOMETRY 66-68 194-200 28. PROBABILITY-2 69-71 201-207 29. PROPERTIES OF TRIANGLE 72-74 208-212 1 SETS MCQs with One Correct Answer 1. 2. 3. 4. 5. 6. 7. If A and B are non-empty sets such that A É B, then (a) B' – A' = A – B (b) B' – A' = B – A (c) A' – B' = A – B (d) A' Ç B' = B – A Let A = {q : sin(q) = tan(q)} and B = {q : cos(q) = 1} be two sets. Then : (a) A = B (b) A Ë B (c) B Ë A (d) A Ì B and B - A ¹ f A set S contains 3 elements, the number of subsets of which of following sets is 256. (a) S (b) P(S) (c) P(P(S)) (d) None of these In a college of 300 students every student reads 5 newspapers and every newspaper is read by 60 students. The number of newpapers is (a) at least 30 (b) at most 20 (c) exactly 25 (d) None of these Let A and B be two sets then (A È B)'È (A 'Ç B) is equal to (a) A¢ (b) A (c) B¢ (d) None of these The set (A \ B) È (B \ A) is equal to (a) [ A \ ( A Ç B)] Ç [ B \ ( A Ç B)] (b) ( A È B) \ ( A Ç B) (c) A \ ( A Ç B ) (d) A Ç B \ A È B Let A = { x : x ÎR, | x | < 1} B = { x : x ÎR, | x – 1| ³1} and AÈB = R – D, then the set D is : 8. 9. 10. 11. 12. 13. 14. (a) { x : 1< x £ 2} (b) {x : 1£ x < 2} (c) {x : – 2 £ x £ 2} (d) None of these If aN = {ax : x Î N} and bN Ç cN = dN, where b, c Î N are relatively prime, then (a) d = bc (b) c = bd (c) b = cd (d) None of these The set (A È B È C) È (A Ç B' Ç C')' Ç C' is equal to (a) B Ç C' (b) A Ç C (c) B' Ç C' (d) None of these A dinner party is to be fixed for a group of 100 persons. In this party, 50 persons do not prefer fish, 60 prefer chicken and 10 do not prefer either chicken or fish. The number of persons who prefer both fish and chicken is (a) 20 (b) 22 (c) 25 (d) None of these Let U be the universal set and A È B È C = U. Then {(A – B) È (B – C) È (C – A)}¢ is equal to: (a) A È B È C (b) A È (B Ç C) (c) A Ç B Ç C (d) A Ç (B È C) If n(A) = 1000, n(B) = 500 and if n(A Ç B) ³ 1 and n(A È B) = p, then (a) 500 £ p £ 1000 (b) 1001 £ p £ 1498 (c) 1000 £ p £ 1498 (d) 1000 £ p £ 1499 In a battle 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x % lost all the four limbs. The minimum value of x is (a) 10 (b) 12 (c) 15 (d) None of these The number of students who take both the subjects mathematics and chemistry is 30. This represents 10% of the enrolment in mathematics and 12% of the enrolment in chemistry. How MATHEMATICS 2 15. 16. 17. 18. 19. 20. 1 2 3 many students take at least one of these two subjects? (a) 520 (b) 490 (c) 560 (d) 480 At a certain conference of 100 people, there are 29 Indian women and 23 Indian men. Of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. How many foreigners and women doctors are attending the conference? (a) 48, 1 (b) 34, 3 (c) 46, 4 (d) 42, 2 Each student in a class of 40, studies at least one of the subjects English, Mathematics and Economics. 16 study English, 22 Economics and 26 Mathematics, 5 study English and Economics, 14 Mathematics and Economics and 2 study all the three subjects. The number of students who study English an d Math ematics but not Economics is (a) 7 (b) 5 (c) 10 (d) 4 In a market research project, 20% opted for 'Nirma' detergent whereas 60% opted for 'Surf blue' detergent. The remaining individuals were not certain. If the difference between those who opted for 'Surf blue' and those who were uncertain was 720, how many respondents were covered in the survey (a) 1100 (b) 1150 (c) 1800 (d) None of these In a school 80 students like chocolate, 40 like coffee if the number of students doesn’t like any of them is equal to the number of students who like both of them then what is the total number of students in the school? (a) 115 (b) 90 (c) 120 (d) None of these In a school there are 100 students 60 of them don’t like Chocolate and 50 don’t like Biscuit and 10 of them like none then how many of them like both? (a) 20 (b) 30 (c) 40 (d) None of these A survey was conducted of 100 people whether they have read recent issues of ‘Golmal', a (a) (b) (c) 4 5 6 (c) (a) (b) 7 8 9 (b) (a) (a) 10 11 12 monthly magazine. Summarized information is presented below: Only September: 18 September but not August: 23 September and July: 8 September: 28 July: 48 July and August:10 None of the three months: 24 What is the number of surveyed people who have read exactly for two consecutive months? (a) 7 (b) 9 (c) 12 (d) 14 Numeric Value Answer 21. 22. 23. 24. 25. An investigator interviewed 100 students to determine their preferences for the three drinks : milk (M), coffee (C) and tea (T). He reported the following : 10 students had all the three drinks M, C and T; 20 had M and C; 30 had C and T; 25 had M and T; 12 had M only; 5 had C only; and 8 had T only. If number of students who did not n take any of the three drinks is n, then is 5 In a class of 80 students numbered 1 to 80, all odd numbered students opt of Cricket, students whose numbers are divisible by 5 opt for Football and those whose numbers are divisible by 7 opt for Hockey. If the number of students who do not opt any of the three games is n, then n is equal to 4 A survey shows that 61%, 46% and 29% of the people watched “3 idiots”, “Rajneeti” and “Avatar” respectively. 25% people watched exactly two of the three movies and 3% watched none. What percentage of people watched all the three movies? If n(A) = 4 and n(B) = 7, then the difference between maximum and minimum value of n(A È B) is Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The value of m – n is ANSW ER KEY (a) 13 (a) 16 (c) 14 (a) 17 (d) 15 (a) 18 (b) (c) (c) 19 20 21 (d) (b) (4) 22 23 24 (7) (7) (4) 25 (3) 2 RELATIONS & FUNCTIONS-1 MCQs with One Correct Answer 1. 2. x, y Î R If f (x) . f (y) = f (x) + f (y) + f (xy) – 2 and if f (x) is not a constant function, then the value of f (a) is (a) 1 (b) 2 (c) 0 (d) – 1 The domain of the function f ( x ) = x14 - x11 + x 6 - x3 + x 2 + 1 is (a) (-¥, ¥) (a) (c) 7. (b) [0, ¥) 3. 4. üï 1 ìï æ x ö f (x)f (y) – í f ç ÷ + f ( xy ) ý is equal to : 2 ïî è y ø ïþ (a) 0 (b) 1 (c) –1 (d) None of these The domain of the function (d) 8. 5. 6. a + 2b a 2 - b2 a - b2 (d) None of these (a) [1, p) (b) (c) æ pö çè 0, ÷ø - {1} 2 (d) (0, 1) ( 0, 2p ) - [1, p) 1 æp x ö Let f ( x) = - tan ç ÷ , -1< x <1 & 2 è 2 ø (a) é - 21, 21 ù ë û 9. æ x + 59 ö 3f(x) + 2f ç ÷ = 10x + 30 for all real x ¹ 1. è x -1 ø The value of f(7) is (a) 8 (b) 4 (c) –8 (d) 11 10. é1 ù êë 2 ,1úû (b) ö é1 ê 2 , -1 ÷ ë ø é 1 ù é 1 ö (d) ê - , -1ú ê - 2 ,1÷ 2 ë û ë ø The domain of function f(x) = log4[log5{log3 (18x – x2 – 77)}] is: (a) (7, 11) (b) (8, 10) (c) (8, 11) (d) (7, 10) Let f(x) = ax(a > 0) be written as f(x) = g(x) + h(x), where g(x) is an even function and h(x) is an odd function. Then the value of g(x + y) + g(x – y) is (a) 2g(x) g(y) (b) 2g(x + y) g(x – y) (c) 2g(x) (d) None of these (c) (c) éë -5, - 21ùû È éë 21, 5 ùû È {0} (d) (–¥, –5) The function f satisfies the functional equation æ 1 ö If af(x + 1) + bf ç ÷ = x , x ¹ –1, a ¹ b, then è x +1 ø f(2) is equal to ) a 2 g ( x) = 3 + 4 x - 4 x 2 , then dom (f + g) is given by f ( x ) = 10 - x 4 - 21x 2 is (a) [5, ¥) (b) 2 a2 - b 2 (b) The domain of f(x) = cos(sin x ) + log x {x} ; {.} denote the fractional part, is R - [0,1] (c) (- ¥, 0] If f(x) = cos (log x) then ( 2a + b MATHEMATICS 4 11. 12. 13. 14. 15. 16. 17. 1 2 3 Let f be a real valued function such that for any real x, f (l + x) = f (l – x) and f (2l + x) = –f (2l – x) for some l > 0. Then (a) f is even and non-periodic (b) f is odd and periodic (c) f is odd and non-periodic (d) f is even and periodic Let f(x) = ([a]2 – 5[a] + 4)x3 – (6{a}2 – 5{a} + 1) x – (tan x) sgn x, be an even function for all x Î R, then sum of all possible values of 'a' is (where [ ] and { } denote greatest integer function and fractional part functions respectively) 17 53 31 35 (a) (b) (c) (d) 6 6 3 3 The set of all integer values of n for which the 5x function f(x) = cos nx . sin is periodic with n period 2p is equal to (a) {1, 5, 10} (b) {1, 5} (c) {±1, ±5} (d) None of these If f : ¡ ® ¡ & g: ¡ ® ¡ be two given functions, then 2 min {f (x) – g(x), 0} equals (a) f(x) + g(x) – |g(x) – f(x)| (b) f(x) + g(x) + |g(x) – f(x)| (c) f(x) – g(x) + |g(x) – f(x)| (d) f(x) – g(x) – |g(x) – f(x)| The domain of f(x) is (0, 1), therefore the domain of y = f(ex) + f(ln | x |) is : æ1 ö (a) ç , 1÷ (b) (–e, –1) èe ø 1ö æ (d) (–e, –1) È (1, e) (c) ç -1, - ÷ eø è Suppose that f is a periodic function with period 1 æ 9ö and that f(2) = 5 and f ç ÷ = 2 then 2 è 4ø f(–3) – f æç 1 ö÷ has the value equal to è 4ø (a) 2 (b) 3 (c) 5 (d) 7 2 x + 3 ; x £ 1 é Let f ( x ) = ê 2 . If the range of f(x) êë a x + 1 ; x > 1 = R (set of real numbers) then number of integral value(s), which a may take is (a) 2 (b) 3 (c) 4 (d) 5 (b) (a) (a) 4 5 6 (c) (b) (a) 7 8 9 (d) (c) (b) 10 11 12 18. 19. 20. Let f(x) = sin x – cos x and g(x) = log 5x ; then the range of g ( 2 f ( x) + 3) is (a) [0, 1] (b) [0, 2] 3ù é (d) None of these (c) êë 0, 2 úû 1 Let f(x) = 1 + and g(x, y) = log y, then the 4 x æ1 ö domain of g ç , - g (2, f ( x)) - 1÷ is è2 ø (a) 0 < x < 1 (b) 0 < x £ 1 (c) x ³ 1 (d) Null set Let f (x) be defined as |x| 0 £ x <1 ì ï f ( x) = n í| x - 1| + | x - 2 | 1 £ x < 2 ï | x -3| 2£ x<3 î The range of function g(x) = sin (7(f (x)) is : (a) [0, 1] (b) [–1, 0] 1 1 é ù (c) ê - , ú (d) [–1, 1] ë 2 2û Numeric Value Answer 21. 22. 23. 24. 25. If f ( x ) = 1 and S = f(5) + f(4) + f(3) +....+ 1 + e- x f(–3)+ f(–4) + f(–5), then the value of S is If the period of f(x) satisfying the condition: f(x + p) = 1 + {1 – 3f(x) + 3f 2(x) – f 3(x)}1/3 is lp, then evaluate l. If f(x) is an odd function, f(1) = 3, and f(x + 2) = f(x) + f(2), then the value of f(3) is Let f(x, y) be a function satisfying the functional equation: f(x, y) = f(2x + 2y, 2y – 2x) for all real numbers x, y. Define g(x) by g(x) = f(2x, 0). Also given that g(x) is a periodic function with period k k, then find value of . 2 Number of elements in the range set of é x ù é 15 ù f ( x) = ê ú ê - ú " x Î (0, 90) ; (where [.] ë15 û ë x û denotes greatest integer function) ANSW ER KEY (a) 13 (c) 16 (b) 19 (d) 22 (2) 25 (6) (b) 14 (d) 17 (c) 20 (d) 23 (9) (d) 15 (b) 18 (b) 21 (5.5 24 (6) 3 TRIGONOMETRIC FUNCTIONS 6. MCQs with One Correct Answer 1. 2. 1 , then the value of 5 1 1 2 4 + + + cos2 a 1 + sin 2 a 1 + sin 4 a 1 + sin8 a is equal to (a) 2 (b) 4 (c) 6 (d) 10 If a, b, g, d are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity k, then the a b g d value of 4 sin + 3 sin + 2 sin + sin is 2 2 2 2 equal to (a) 2 1 - k (b) 2 1 + k 16 If sin (c) 3. a= 2 k (d) none of these If cos a = 2cos b - 1 (0 < a < b < p) , then 2 - cos b tan a / 2 is equal to tan b / 2 (a) 1 4. 5. (b) 2 a= (a) 3 (d) 1 3 Let 0 £ a, b, g, d £ p whrere b and g are not complementary such that 2 cos a + 6 cos b + 7 cos g + 9 cos d = 0 and 2 sin a – 6 sin b + 7 sin g – 9 sin d = 0 cos (a + d ) m If where m and n are = cos (b + g ) n relatively prime positive numbers, then the value of (m + n) is equal to: (a) 11 (b) 10 (c) 9 (d) 7 If sin q + sin 2q + sin 3q = sin a and cos q + cos 2q + cos 3q = cos a, then q is equal to (a) a/2 (b) a (c) 2a (c) a/6 p is equal to: 5 5 -1 4 (b) 5 +1 4 5 +1 5 -1 (d) 2 2 sin q sin (3q) sin(9q) sin(27 q) + + + = cos (3q) cos(9q) cos(27q) cos(81q) (c) 7. 8. (c) The value of the expression sin 7a + 6sin 5a + 17 sin 3a + 12 sin a where sin 6a + 5 sin 4a + 12 sin 2a (a) sin(81 q) (b) 2 cos(80 q) cos q sin(80 q) 2 cos(81 q) cos q (c) sin(81 q) cos(80 q) cos q sin(80 q) cos(81 q) cos q If cosec q = (d) p+q æp qö , then cot ç + ÷ = p-q è4 2ø p q (b) (c) pq (d) pq q p 9. If sin 2q + sin 2f = 1/2 and cos 2q + cos 2f = 3/2, then cos2 (q – f) = (a) 3/8 (b) 5/8 (c) 3/4 (d) 5/4 10. If u = (1 + cos q) (1 + cos 2q) – sin q × sin 2q, v = sin q (1 + cos 2q) + sin 2q (1 + cos q), then u2 + v2 = (a) 4(1 + cos q) (1 + cos 2q) (b) 4(1 + sin q) (1 + sin 2q) (c) 4(1 – cos q) (1 – cos 2q) (d) 4(1 – sin q) (1 – sin 2q) (a) MATHEMATICS 6 11. The product 19. ìï 2 p üï ìï 2 p üï ì 2p ü cos í 64 ý cos í 64 ý ...cos í 64 ý î 2 - 1þ ïî 2 - 1ïþ îï 2 - 1 ïþ 2 64 16. 1 (d) 32 16 8 32 6464 16 8 Let x, y Î R satisfy the condition such that sin x sin y + 3 cos y + 4 sin y cos x = 26. The value of tan2 x + cot2 y is equal to (a) 9 × 17 (b) 205 1 9 + (c) (d) None of these 16 17 The product xö æ xö æ xö x ö æ æ ç cos ÷ × ç cos ÷ × ç cos ÷ ........ × ç cos ÷ 2ø è 4ø è 8ø 256 ø è è is equal to: sin x sin x (a) (b) x x 128 sin 256 sin 256 256 sin x sin x (c) (d) x x 128 sin 512 sin 128 512 The maximum value of log20 (3 sin x – 4 cos x + 15) is equal to: (a) 1 (b) 2 (c) 3 (d) 4 The value of cos 25° cos 70° cos 85° + + sin 70° sin 85° sin 25° sin 85° sin 25° sin 70° is (a) 1/2 (b) 1 (c) 2 (d) 3/2 The number of solutions of the equation 17. 32sec x + 1 = 10.3tan x in the interval [0, 2p] is (a) 8 (b) 6 (c) 4 (d) 2 Number of solutions of the equation 18. 4 (cos2 2x + cos 2x + 1) + tan x (tan x – 2 3 ) = 0 in [0, 2p] is (a) 0 (b) 1 (c) 2 (d) 3 The number of solutions of the equation 1 (a) 12. 13. 14. 15. 1 (b) 2 (a) 2 (d) (b) (c) 20. 2cos 2 x - 3cos x + 1 (b) 3 4 5 6 (b) (a) (c) = 1 in [0, p] is (c) 4 7 8 9 (b) (b) (b) æ x ö æ 2 xö çè sin çè 2 tan 2 cos 2 ÷ø - 3÷ø + 2 = 0 in [0, 2p] is : (a) 0 (b) 1 (c) 2 (d) 4 If the equation x2 + 12 + 3 sin (a + bx) + 6x = 0 has atleast one real solution, where a, b Î [0, 2p], then the value of a – 3b is (n Î Z) (a) 2np (b) (2n + 1) p p p (c) (4n – 1) (d) (4n + 1) 2 2 Numeric Value Answer 21. The number of solutions of the equation 22. + log15 cos x) for 51/2 + 51/2+ log5 (sinx) = 15 2 x Î [0, 100 p] is Number of solution(s) of th e equation sin x sin 3x sin 9x + = 0 in the interval + cos 3x cos 9x cos 27x 1 23. 24. æ pö ç 0, 4 ÷ is _______. è ø Number of integral value(s) of m for which the 4m - 6 has equation sin x – 3 cos x = 4-m solutions, x Î [0, 2p], is _______. The number of solutions of the equation pö æ cos 2 ç x + ÷ + cos2 x – 2 è 6ø pö p p æ cos ç x + ÷ × cos = sin 2 in interval 6ø 6 6 è 2 2 sin x - 3 1 2 3 1 (c) The number of solutions of the equation æ æ sin x ö ö æ æ sin x ö ö çè 2 sin çè 2 ÷ø ÷ø çè cos çè 2 ÷ø ÷ø 25. æ -p p ö ç 2 , 2 ÷ is _______. è ø The number of solutions of the equation 1 + cos x + cos 2x + sin x + sin 2x + sin 3x = 0, which satisfy the condition ___________. (d) 5 ANSW ER KEY 10 (a) 13 (b) 16 (c) 19 11 (a) 14 (a) 17 (c) 20 12 (c) 15 (c) 18 (b) 21 (a) (c) (50) 22 23 24 p p < 3x - £ p is 2 2 (6) (4) (2) 25 (2) 4 PRINCIPLE OF MATHEMATICAL INDUCTION MCQs with One Correct Answer 1. 5. 1 1 1 1 + + + ...... + n n +1 n + 2 2n - 1 If P(n) = 2 + 4 + 6 + .....+ 2n, n Î N , then P(k) = k (k + 1) + 2 = 1- we can conclude that P(n ) = n(n + 1) + 2 for (a) all n Î N (b) for even values of n (c) for odd values of n (d) not true for any n For all n ³ 1, find (b) n > 1 (c) n > 2 (d) nothing can be said The greatest positive integer, which divides 6. 1 1 1 1 + + + ..... + 1.2 2.3 3.4 n( n + 1) n (n + 1)(n + 2)(n + 3) for all n Î N , is (a) 2 3. (b) 6 (c) 24 (d) 120 (a) n n +1 (b) (c) 1 n(n + 1) (d) None of these For any n Î N , the value of the expression 2 + 2 + ..... + 2 is n - roots 4. 1 1 1 1 + - + ...... + holds for 2 3 4 2n - 1 Þ P(k + 1) = (k + 1)(k + 2) + 2 for all k Î N . So (a) all n Î N 2. If n Î N , then the result 7. (a) æ p ö 2 cos çç ÷÷ è 2 n +1 ø (c) 2 cos( 2 n +1 p) (d) none of these (b) æ p ö 2 sin çç ÷÷ è 2 n +1 ø For a positive integer n, Let a(n) = 1 + 1 1 1 1 + + +…+ n . Then 2 3 4 (2 ) - 1 (a) a(100) £ 100 (c) a(200) £ 100 (b) a(100) > 100 (d) a(200) < 100 1 n +1 For all natural numbers n, find æ 3 ö æ 5 ö æ 7 ö æ 2n + 1 ö ç1 + ÷ ç1 + ÷ ç1 + ÷ ..... ç1 + 2 ÷ è 1 øè 4 øè 9 ø è n ø 8. 9. (a) (n + 1)2 (b) (n – 1)2 (c) n(n + 1) (d) None of these 2n > n2 when n Î N such that (a) n > 2 (b) n > 3 (c) n < 5 (d) n ³ 5 If 4n (2n)! , then P(n) is true for < n + 1 (n!)2 (a) n ³ 1 (b) n > 0 (c) n < 0 (d) n ³ 2 MATHEMATICS 8 10. 11. 12. 13. 14. 15. If P(n) : 3n < n!, n Î N, then P(n) is true (a) for n ³ 6 (b) for n ³ 7 (c) for n ³ 3 (d) for all n (a) If n is a positive integer, then 2 . 42n + 1 + 33n + 1 is divisible by : (a) 2 (b) 7 (c) 11 (d) 27 For every natural number n, n(n2 – 1) is divisible by (a) 4 (b) 6 (c) 10 (d) None of these If p is a prime number, then n p – n is divisible by p when n is a (a) Natural number greater than 1 (b) Irrational number (c) Complex number (d) Odd number If 49n + 16n + l is divisible by 64 for all n Î N, then the least negative value of l is (a) –2 (b) –1 (c) –3 By mathematical induction, (c) 16. 17. 1 2 3 4 ( n + 2 ) ( n + 3) n (n + 2) 4 (n + 1) (n + 3) (b) is n (n + 3) 4 (n + 1) (n + 2) 4 5 6 (a) (a) (a) 7 8 9 (a) (d) (d) 10 11 12 (2n + 1) 3n + 3 4 (d) (2n – 1) 3n + 1 + 3 4 (2n – 1) 3n + 1 + 1 4 For all n Î N, 1+ 1 1 1 + + ..... + 1+ 2 1+ 2 + 3 1 + 2 + 3 + ..... + n is equal to (a) 3n (b) n +1 n (c) n +1 2n (d) n –1 2n n +1 19. When 2301 is divided by 5, the least positive remainder is (a) 4 (b) 8 (c) 2 (d) 6 20. By the principle of induction " n Î N, 32n when divided by 8, leaves remainder (a) 2 (b) 3 (c) 7 (d) 1 21. If n Î N, then 11n + 2 + 122n+1 is divisible by ___________. 22. For all n Î N, 41n – 14n is a multiple of ___________. If m, n are any two odd positive integers with n < m, then the largest positive integer which divides all the numbers of the type m2 – n2 is ___________. For every natural number n, 32n + 2 – 8n – 9 is divisible by ___________. The remainder when 599 is divided by 13, is ___________. 23. (d) None of these For all n Î N, 3.52n + 1 + 23n + 1 is divisible by (a) 19 (b) 17 (c) 23 (d) 25 For all n Î N, 1.3 + 2.32 + 3.33 + ..... + n.3n is equal to (d) (c) (a) 4 (b) Numeric Value Answer equal to n (n + 1) 18. (d) – 4 1 1 1 + + ..... + 1× 2 ×3 2 × 3× 4 n (n + 1)(n + 2) (a) (c) (2n + 1) 3n + 1 + 3 24. 25. ANSW ER KEY (b) 13 (a) 16 (c) 14 (b) 17 (b) 15 (b) 18 (b) (b) (d) 19 (c) 22 20 (d) 23 21 (133) 24 (27) (8) (16) 25 (8) 5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS the point in the complex plane given by wk = MCQs with One Correct Answer 1. 2. (b) 16(x12 + y12) (a) 32(x12 + y12) (d) 32 (c) 4(x12 + y12) Let z and w be two non-zero complex numbers such that | z | = | w | and Arg z + Argg w = p, then z equals (a) 3. cos 2a k + i sin 2a k for k = 1, 2, 3. The origin, O zk Let x1 and y1 be real numbers. If z1 and z2 are complex numbers such that |z1| = |z2| = 4, then |x1z1 – y1z2|2 + |y1z1 + x1z2|2 = w (b) – w (c) w 7. the line b z + b z = c, b ¹ 0 in the argand plane, (d) – w If | z - 1| + | z + 3 | £ 8, then the range of values of | z - 4 | is 4. (a) (0, 7) (b) (1, 8) (c) [1, 9] (d) [2, 5] Which of the following is/are value of 2 is the (a) incentre of DA1A2A3 (b) orthocentre of DA1A2A3 (c) circumcentre of DA1A2A3 (d) centroid of DA1A2A3 If a point z1 is the reflection of a point z2 through 8. 2 then b z2 + b z1 is equal to (a) 4c (b) 2c (c) c (d) None of these z1 and z2 lie on a circle with centre at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by sin ln (ii )i + cos ln (ii )i ? (a) – 1 (b) 1 (c) 0 (d) None of these (a) 1 ( z1 + z2 ) 2 (b) 2z1 z2 z1 + z2 5. If n1, n2 are positive integers, then (1+ i ) n1 + (c) (d) z1 + z2 z1 + z2 6. (1 + i 3 )n1 + (1+ i 5 )n2 + (1+ i 7 )n2 is real number if and only if: (a) n1 = n2 + 1 (b) n1 + 1 = n2 (c) n1 = n2 (d) n1,n2 are any positive integers Let rk > 0 and zk = rk (cos ak + i sin ak) for k = 1, 1æ 1 1 ö + 2 çè z1 z2 ÷ø 2, 3 be such that 1 1 1 + + = 0 Let Ak be z3 z2 z1 9. If z1, z2 are two complex numbers such that z1 – z2 = 1 and i z1 = Kz2, where K Î R, then z1 + z2 the angle between z1 – z2 and z1 + z2 is æ 2K ö (a) tan– 1 ç 2 ÷ è K + 1ø æ 2K ö (b) tan– 1 ç ÷ è 1– K 2 ø (c) – 2 tan– 1 K (d) 2 tan– 1 K MATHEMATICS 10 10. Let 'z' be a complex number and 'a' be a real parameter such that z2 + az + a2 = 0, then which is of the following is not true? (a) locus of z is a pair of staight lines (b) |z| = |a| 16. 2p 3 (d) None of these Let P denotes a complex number z = r(cos q + i sin q) on the Argand's plane, and Q denotes a 17. (c) arg (z) = ± 11. æ p pö complex number 2| z |2 çcosæçq+ ö÷ +i sinæçq+ ö÷÷ . è 4øø è è 4ø 12. y = 2(a - x) ( x + x 2 + b2 ) is 18. If 'O' is the origin, then DOPQ is (a) isosceles but not right angled (b) right angled but not isosceles (c) right isosceles (d) equilateral If w ¹ 1 and w3 = 1, then a w + b + c w2 + 19. 2 is equal to (d) 2w2 1 1 = 1 and a = z2017 + 2017 and b is z z 13. If z + 14. the last digit of the number 22 – 1, when the integer n > 1, the value of a2 + b2 is (a) 23 (b) 24 (c) 26 (d) 27 If y1 = max ||z – w| – |z – w2||, where |z| = 2 and n 20. 1 y2 = max ||z – w| – |z – w2||, where |z| = and w and 2 w2 are complex cube roots of unity, then 15. (a) y1 = 3 ; y2 = 3 (b) y1 < 3 ; y2 = 3 (c) y1 = 3 ; y2 < 3 (d) y1 > 3; y2 < 3 If a and b are the roots of the equation ax2 + bx + c = 0, (c ¹ 0) , then the equation whose roots 1 1 and is a b +b aa + b (a) acx2 – bx + 1 = 0 (b) x2 – acx + bc + 1 = 0 (c) acx2 + bx – 1 = 0 (d) x2 + acx – bc + 11 = 0 are (a) a 2 + b2 (b) (c) a 2 + 2b 2 (d) None of these a2 - b2 If a, b are the roots of x 2 + px + q = 0, and x 2 n + p n x n + q n = 0 and a w2 + b + cw a w2 + b w + c a + b w + c w (a) 2 (b) w (c) 2w If the roots of the equation x2 + px + c = 0 are 2, –2 and the roots of the equation x2 + bx + q = 0 are –1, –2, then the roots of the equation x2 + bx + c = 0 are (a) –3, –2 (b) –3, 2 (c) 1, – 4 (d) –5, 1 If x Î R, then the maximum value of a is a root of b x n + 1 + ( x + 1)n = 0, a n ¹ bn , the n must be (a) any integer (b) an even integer (c) an odd integer (d) None of these If 0 < a < b < g < p/2, then the equation (x – sin b) (x – sin g) + (x – sin a) (x – sin g) + (x – sin a) (x – sin b) = 0 has (a) real and unequal roots. (b) non-real roots. (c) real and equal roots. (d) real and unequal roots greater than 2. The set of values of a for which inequation (a – 1) x2 – (a + 1)x + a – 1 ³ 0 is true for all x ³ 2 (a) é 7ù ê1, 3 ú ë û (b) (– ¥, 1) (c) é7 ö ê 3 , ¥÷ø ë (d) None of these Numeric Value Answer 21. a, b, c are integers, not all simultaneously equal and w is cube root of unity (w ¹ 1), then minimum value of |a + bw + cw2| is 22. If 2 – i is a root of the equation ax2 + 12x + b = 0 (where a and b are real), then the value of ab is equal to 23. If the equations ax 2 + bx + c = 0 and cx2 + bx + a = 0, a ¹ c have a negative common root, then the value of a - b + c is Complex Numbers and Quadratic Equations 24. If a, b are the roots the quadratic equation x2 – (3 + 2 log2 3 –3 log 3 2 ) x –2(3log3 2 – 2log2 3 ) 11 28. If z and w are two complex numbers having non-negative imaginary parts such that æ z - 2ö æ w - 1ö = arg ç = p / 2, then arg ç è z + 2 ÷ø è w + 1÷ø = 0, then the value of a2 + ab + b2 is equal to 25. Let f (x) = x4 + ax3 + bx2 + cx + d be a polynomial with real coefficients and real zeroes. If |f (i)| = 1, (where i = -1 ) then a + b + c + d is equal to 26. If w and w2 be the non-real cube roots of unity 1 1 1 + + = 2w 2 and a+w b+w c+w and 1 a + w2 + 1 b + w2 + 1 30. 1 1 1 + + is a + 1 b +1 c + 1 equal to : 27. For any real x, the maximum value of 2 k2 1 2 3 4 ( (x - k) x + (a) (d) (c) (b) 5 6 7 8 (d) (d) (c) (b) 2 x +k 9 10 11 12 29. If z = = 2w, where a, b, c c + w2 are real then the value of | w- z |< k; evaluate k. (Here k is least upper bond) 2 ) æ 1 - pi p p -i ö (1 + i )4 çç + ÷÷ , where 4 è p + i 1 + pi ø æ |z| ö i = -1, then ç amp( z ) ÷ equals to è ø If p, q, r are positive and are in A.P., then the roots of the quadratic equation px2 + qx + r = 0 are real for r - 7 ³ k 3 then find the value p of k. is equal to (d) (d) (c) (d) 13 14 15 16 ANSW ER KEY (c) 17 (a) 21 (c) 18 (b) 22 (a) 19 (a) 23 (c) 20 (c) 24 (1) (45) (0) (7) 25 26 27 28 (0) (2) (2) (3) 29 30 (4) (4) MATHEMATICS 12 6 LINEAR INEQUALITIES MCQs with One Correct Answer 1. The set of all x satisfying the inequality 1 2x - 1 - 3 ³0 2 x - x +1 x +1 x +1 (a) (–¥, 2] (b) [1, 2] (c) (–¥, –1) È (–1, 2] (d) (2, ¥] The set of all x satisfying the inequality (2x + 1) (x – 3) (x + 7) < 0 2 2. - (a) (–¥, 7) 3. 4. 5. 6. æ 1 ö (b) ç - ,3 ÷ è 2 ø æ 1 ö (c) (–¥, 7) È ç - ,3 ÷ (d) (–¥, 3) è 2 ø The set of real values of x satisfying | x - 1| £ 3 and | x - 1| ³ 1 is (a) [2, 4] (b) (-¥, 2] È [4, + ¥) (c) [-2, 0] È [2, 4] (d) none of these The system of equation | x - 1 | +3y = 4, x - | y - 1 |= 2 has (a) No solution (b) A unique solution (c) Two solutions (d) More than two solutions The number of real roots of the equation | 2- | 1- | x |||= 1 is (a) 1 (b) 3 (c) 5 (d) 6 The solution set of the inequality | x + 2 | - | x - 1 |< x - 3 is 2 7. 8. 9. (a) ö æ9 ç , ¥÷ 2 ø è (c) 3ö æ ç - 2, - ÷ 2ø è 12x ³ 1 for all real values of x, the 4x 2 + 9 inequality being satisfied only if | x | is equal to 2 1 3 1 (a) (c) (d) (b) 3 3 2 2 The equation | x - 1| +a = 4 can have real solutions for x if ‘a’ belongs to the interval (a) (-¥, + ¥) (b) (-¥, 4] If (c) (4, + ¥) (d) [–4, 4] The interval(s) that satisfy the equation x 2 - 8x + 12 x 2 - 10x + 21 10. 11. 3ö æ (b) ç - ¥, ÷ 2ø è 3ö æ (d) ç - 1, ÷ 2ø è =- x 2 - 8x + 12 x 2 - 10x + 21 is /are (a) (-¥, 2] (b) [2, ¥) (c) [6, 7) (d) [3, 6] È [7, ¥) The set of all real x satisfying the inequality 3- | x | ³ 0 is 4- | x | (a) [– 3, 3] È (– ¥, – 4) È (4, ¥) (b) (– ¥, – 4) È (4, ¥) (c) (– ¥, – 3) È (4, ¥) (d) (– ¥, – 3) È (3, ¥) If x, y, z be any three positive real number and x +z x+y y+z A= 2 + 2 + 21 2 2 2 x +y y +z x +z and B = 1 1 1 + + ; x y z Linear Inequalities 13 then which of the following is true? (a) A ³ B (b) A £ B (c) A = B (d) A < B 12. If x, y, z are arbitrary real numbers satisfiying the condition xy + yz + zx < 0 and if x 2 + y2 + z 2 then only one of the xy + yz + zx following statements is always correct. Which one is it? (a) – 1 £ u < 0 (b) u takes all negative real values (c) – 2 < u £ – 1 (d) u £ – 2 13. If a1,a2.....,an are positive real numbers whose product is a fixed number c, then the minimum value of a1 + a2 + .......+an-1 + 2an is (a) n(2c)1/n (b) (n + 1)c1/n 1/n (c) 2nc (d) (n + 1)(2c)1/n 14. The set of real values of x for which u = log 0.2 (a) æ 3p ö æ 3p ö (a) x Î (3, p) È ç p, ÷ È ç , 5 ÷ è 2 ø è 2 ø (b) x Î (3, p) È (p, 5) æ 5p ö (c) x Î ç 3, ÷ è 2 ø (d) None of these 20. If x is real, the maximum value of 3x 2 + 9 x + 17 3x 2 + 9 x + 7 x+2 £ 1 is x 1 4 (a) 5ù æ ç - ¥, - ú È (0, ¥) (b) 2û è é5 ö ê 2 , ¥÷ ë ø is (b) 41 (c) 1 (d) 17 7 Numeric Value Answer (-¥, - 2) È [0, ¥) (d) None of these 21. If ‘k’ any integer such that set of all real values of x, satisfy the equation (sin3 x ) (cos3 x ) x 2 + 6x + 9 < - log 2 ( x + 1), then x lies 2(x + 1) in the interval 3x + 3- x ; then find the minimum value k. k 22. For any x, y Î R, xy > 0. Then the minimum value (c) 15. If log1 / 2 (a) (-1, - 1 + 2 2 ) (b) 1 + log 5 ( x 2 + 1) £ log 5 (ax 2 + 4x + a ) is true for all x Î R is (a) 6 (b) 7 (c) 10 (d) 1 17. The set of all real numbers x for which x2 – [x + 2] + x > 0, is (a) (- ¥,-2) È (2, ¥ ) (b) (c) (d) (c) (c) (c) (- ¥,- 2 )È ( 2 , ¥) 7 8 9 2x y + 3 (a) (b) (c) 10 11 12 x 3 y 4y 2 + 4 is. 3 9x 23. Number of integeral solution satisfying x + 3 - x ³ 3 - x + 3 is x+3 + x > 1 and then find x+2 modulus of smallest integral member of the interval. The number of integers satisfying the equation 24. Solve for x, |x|+ ( 2 , ¥) (b) (c) (a) of 25. (- ¥,-1) È (1, ¥ ) 4 5 6 £ (1 - 2 2 , 2) (c) (-1, ¥) (d) None of these 16. The least integer a, for which 1 2 3 18. Solution set of the in equality log102x – 3 (log10 x) (log10 (x – 2)) + 2log102 (x – 2) < 0, is : (a) (0, 4) (b) (– ¥, 1 ) (c) (4, ¥) (d) (2, 4) 19. The solution set of log|sin x| (x2 – 8x + 23) 3 > contains log 2 | sin x | ANSW ER KEY (a) 13 (a) 16 (b) 14 (a) 17 (d) 15 (a) 18 4 - x2 4 - x2 = x+ is x x (b) (b) (c) 19 20 21 (a) (b) (4) 22 23 24 (2) (1) (4) 25 (4) MATHEMATICS 14 PERMUTATIONS AND COMBINATIONS MCQs with One Correct Answer 1. 2. 3. 4. Ten different letters of an alphabet are given words with five letters are formed from three given letters. Then the number of words which have at least one letter repeated are (a) 69760 (b) 30240 (c) 99748 (d) None of these How many different nine digit numbers can be formed from th e number 223355888 by rearranging its digits so that the odd digits occupy even positions ? (a) 16 (b) 36 (c) 60 (d) 180 The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is (a) 360 (b) 192 (c) 96 (d) 48 The total number of 5-digit numbers of different digits in which the digit in the middle is the largest is (a) 9 å n p4 (b) 4563 n =4 5. 6. (c) 2688 (d) 5292 All possible 120 permutations of WDSMC are arranged in dictionary order, as if each were an ordinary five-letter word. The last letter of the 86th word in the list, is : (a) W (b) D (c) M (d) C If m be the number of different words that can be formed with the letters of the word BHARAT in which B and H are never together and n be number of different words that can be formed with the letters of the words BHARAT in which 7. 8. 9. 10. 11. 7 words always begin with B and end with T. Then m/n is (a) 10 (b) 20 (c) 1 (d) 2 Anil have tiled his square bathroom wall with congruent square tiles. All the tiles are red, except those along the two diagonals, which are all blue. If he used 121 blue tiles, then the number of red tiles used are (a) 900 (b) 1800 (c) 3600 (d) 7200 The number of distinct natural numbers up to a maximum of four digits and divisible by 5, which can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, each digit not occurring more than once in each number, is (a) 1246 (b) 952 (c) 1106 (d) None of these 6 white and 6 black balls are distributed among ten identical urns, so that there is atleast one ball in each urn. Balls are all alike except for the colour and each box can hold any number of balls. The number of different distributions of the balls is: (a) 26250 (b) 132 (c) 12 (d) 10 Number of ways in which two Americans, two British, one Chinese, one Dutch and one Egyptian can sit on a round table so that persons of the same nationality are separated is (a) 48 (b) 240 (c) 336 (d) None of these In an examination of 9 papers a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is (a) 255 (b) 256 (c) 193 (d) 319 Permutations and Combinations 15 12. During a draw of lottery, tickets bearing numbers 1, 2, 3, ..., 40, 6 tickets are drawn out and then arranged in the descending order of their numbers. In how many ways, it is possible to have 4th ticket bearing number 25? (a) 15C3 × 24C2 (b) 12C3 × 20C2 (c) 15C3 + 24C2 (d) None of these 13. A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not take the same three children to the zoo more than once he finds than she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is (a) 12 (b) 10 (c) 60 (d) None of these 14. Messages are conveyed by arranging four white, one blue, and three red flags on a pole. Flags of the same colour are alike. If a message is transmitted by the order in which the colours are arranged, the total number of messages that can be transmitted if exactly six flags are used is (a) 45 (b) 65 (c) 125 (d) 185 15. There were two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women. The number of participants is (a) 6 (b) 11 (c) 13 (d) None of these 16. In a class tournament, all participants were to play different games with one another. Two players fell ill after having played three games each. If the total number of games played in the tournament is equal to 84, the total number of participants in the beginning was equal to (a) 10 (b) 15 (c) 12 (d) 14 17. The number of ways in which 5 X's can be placed in the squares of the figure so that no row remains empty is (a) 97 1 2 3 (a) (c) (c) (b) 44 4 5 6 (d) (b) (b) 7 8 9 (c) 100 (d) (c) (c) (d) (c) (b) (a) 10 11 12 18. There are three coplanar parallel lines. If any p points are taken on each the lines, the maximum number of triangles with vertices at these points is (a) 3p2(p – 1) + 1 (b) 3p2(p – 1) 2 (c) p (4p – 3) (d) None of these 19. From the vertices of a regular polygon of 10 sides, the number of ways of selecting three vertices such that no two vertices are consecutive is (a) 10 (b) 30 (c) 50 (d) 40 20. 5 different objects are to be distributed among 3 persons such that no two persons get the same number of objects. Number of ways this can be done, is (a) 60 (b) 90 (c) 120 (d) 150 Numeric Value Answer 21. If a, b, c are three natural numbers in AP such that a + b + c = 21 and if possible number of ordered triplet (a, b, c) is then the value of (l – 5) is 22. In a single correct match the column question, column I contain 10 questions and Column II contain 10 answers written in some arbitrary order. If the number ways a student can answer this question so that exactly 6 of his matching are correct is k, then (sum of digits of k)/2 is equal to 23. There are 720 permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged from smallest to largest numerical values, beginning from 1 2 3 4 5 6 and ending with 6 5 4 3 2 1. Then the digit in unit place of number at 267th position is ...... . 24. If N is the number of ways in which a person can walk up a stairway which has 7 steps if he can take 1 or 2 steps up the stairs at a time, then the value of N/3 is ...... . 25. In an international convention participants from 10 different countries were arranged in a row such that all the participants from the same country were together. Each country has different number of participants with maximum 10 participants from a country. If K is the number of ways that they can be arranged in a row then find the highest power of 10 in K. 126 ANSWER KEY 13 (b) 16 14 (d) 17 15 (c) 18 (b) (b) (c) 19 20 21 (c) (b) (8) 22 23 24 (9) (6) (7) 25 (9) MATHEMATICS 16 8 BINOMIAL THEOREM MCQs with One Correct Answer 1. 2 60 when divided by 7 leaves the remainder (a) 1 (b) 6 (c) 5 (d) 2 2. If 7 divides 32 32 , the remainder is (a) 1 (b) 0 (c) 4 (d) 6 3. 4. 5. 8. 32 æ r -1 ö n r p÷ ç C C 2 is equal to å r p å ç ÷ r=1 è p =1 ø (a) 4n – 3n + 1 (b) 4n – 3n – 1 n n (c) 4 – 3 + 2 (d) 4n – 3n 2 6 If (1 + x – 2x ) = 1+ a1x + a2x2 + ..... + a12 x 12, then the value of a2 + a4 + ..... + a12 is : (a) 1024 (b) 64 (c) 32 (d) 31 The last term in the binomial expansion of n 1/ 3 (2 n 1/ 3 - 1/ 2) is (1/ 3.9 9. 10. n 11. ) log 3 8. Then the If (1 + ax) = 1+ 8x +24x2 + ......... ; than a – n is a+n equal to (n being a positive integer) (a) 3 (b) – 3 n (c) – 7. 1 3 (d) 1 3 The unit digit of 17 2009 + 112009 – 7 2009 is (a) 1 (b) 2 (c) 3 (d) 0 1 ö æ 1/3 ç 3 + 1/3 ÷ . If y = 12x then the value of n is : 4 ø è (a) 9 (b) 8 (c) 10 (d) 11 The coefficient of the term independent of x in æ ö x +1 x –1 ÷ ç – the expansion of ç 2 1 1÷ çè 3 ÷ x – x3 +1 x – x2 ø 5th term from the beginning is (a) 10 C6 (b) 2.10C4 10 (c) 1/2. C4 (d) None 6. Let N = 21224 – 1, a = 2153 + 277 + 1 and b = 2408 – 2204 + 1. Then which of the following statement is correct ? (a) a divides N but b does not (b) b divides N but a does not (c) a and b both divides N (d) neither a nor b divides N The number N = 20C7 – 20C8 + 20C9 – 20C10 + ..... – 20C20 is not divisible by : (a) 3 (b) 7 (c) 11 (d) 19 Let x be the 7th term from the beginning and y be the 7th term from the end in the expansion of 12. 10 is (a) 70 (b) 112 (c) 105 (d) 210 If the middle term of (1 + x)2n (x > 0, n Î N) is the greatest term of the expansion. Then the interval in which x lies, is (a) én +1 n + 2ù êë n , n úû (b) é n –1 n + 1ù êë n , n úû (c) n + 1ù é n êë n + 1 , n úû (d) None of these Binomial Theorem 17 13. The number of irrational terms in the expansion ( of 8 ) 19. The value of 5 + 6 2 100 is (a) 97 (b) 98 (c) 96 (d) 99 14. The greatest value of the term independent of x in the expansion of ( x sin a + x 5 (a) 2 (c) 1 –1 (b) 10! (5!)2 (d) None of these . 2 5 (5!) 2 25 The largest term in the expansion of (2 + 3x ) where x = 2 is its. (a) 13th term (b) 19th term th (c) 20 term (d) 26th term If the middle term in the expansion of 15. 16. 10 æ1 ö 7 ç + x sin x ÷ equals to 7 then x is equal to x è ø 8 (n Î I) p p (a) 2np ± (b) n p + 6 6 np n 5p (c) n p + (-1) (d) n p + ( -1) 6 6 17. The sum of the coefficients of all the integral 40 powers of x in the expansion of (1 + 2 x ) is (a) 3 40 + 1 (b) 3 40 – 1 1 40 1 40 (3 – 1) (3 + 1) (c) (d) 2 2 18. If p (n) denotes product of all binomial coefficients in (1+ x ) n , then ratio of p (2002) to p (2001) is 2001 (a) 2002 (c) 1 2 3 (a) (c) (d) (b) (2001) 2002 (2002)! 4 5 6 (d) (a) (c) (a) 2n + 1 n +1 (b) 2n n +1 (c) 2n + 1 n -1 (d) 2n - 1 n +1 cos a )10 , a Î R, is 10! (2002) (2001) ! C1 C 3 C 5 + + + ....... is equal to 2 4 6 n 20. If I is integral part of (2 + 3) and f is its fractional part. Then (I + f ) (1 – f ) is (a) I + 1 (b) 1 (c) n (d) 2n Numeric Value Answer 21. The largest real value for x such that æ 34 - k öæ x k ö 32 ÷ç ÷ = å çç ÷ç ÷ 3 is k = 0 è ( 4 - k )! øè k! ø 4 22. Find the coefficient of x13 in the expansion of (1 – x)5 (1 + x + x2 + x3 )4 is 23. If (1 + x)n = C0 + C1x + C2x2 + ..... + Cnxn, then find the value of 3C0 – 5C1 + 7C2 + ...+ (–1)n (2n+3)Cn 24. If 1 1 1 1 2k + + + ... + = 1!10! 3!8! 5!6! 11!1! 11! then k is equal to 2 25. (1 – 2x + 5x2 – 10x3) (1 + x)n = 1 + a1x + a2x2 + ... and that a21 = 2a2, then the value of n is____ (d) 2001 7 8 9 (a) (c) (c) 10 11 12 (a) (d) (c) ANSWER KEY 13 (a) 16 14 (c) 17 15 (c) 18 (c) (d) (b) 19 20 21 (d) (b) (1) 22 23 24 (4) (0) (5) 25 (6) MATHEMATICS 18 9 SEQUENCES AND SERIES 1. MCQs with One Correct Answer Sum to n terms of the series (a) x, y, z are in A.P. 3 13 + 3. 23 + 33 + 3. 43 + 5 + ................ is (n is even) (a) n(n 2 + 1) (2 n + 1) 3 (b) n(n3 + 4n2 + 10n + 8) 8 (c) n(n3 + 1) 8 7. (d) 2. 3. (a) n (4n2 –1) c 2 6 (b) n (4n2 +1) c 2 3 n (4n2 +1) c 2 n (4n2 –1) c 2 (d) 6 3 If a1, a2, ..., an are in A.P. with common difference d ¹ 0, then (sin d) [sec a1 seca2 + sec a2 sec a3 + ... + sec an–1 sec an] is equal to (a) cot an – cot a1 (b) cot a1 – cot an (c) tan an – tan a1 (d) tan an – tan an–1 1 1 1 1 3 1 3 If x = 2 + 2 + 2 + ..., y = 2 + 2 + 2 + 2 1 3 5 1 2 3 4 1 1 1 1 + ... and z = 2 – 2 + 2 – 2 +..., then 1 2 3 4 8. 9. (c) 4. 5. y x z , , are in A.P.. 6 3 2 y x z , , are in A.P.. (d) 6y, 3x, 2z are in A.P. 6 3 2 a +b b+c , b, are in For a, b, c Î R – {0}, let 1– ab 1– bc A.P. If a, b are the roots of the quadratic equation 2ac x2 + 2abc x + (a + c) = 0, then the value of (1 + a)(1 + b) is (a) 0 (b) 1 (c) – 1 (d) 2 An A.P. consist of even number of terms 2n having middle terms equal to 1 and 7 respectively. If n is the maximum value which satisfy t1t2n + 713 ³ 0, then the value of the first term of the series is (a) 17 (b) – 15 (c) 21 (d) – 23 If a, b, c are in G. P., x and y be the A. M.’s between (c) 6. n2 (2n 2 + 6 n + 5) 4 If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P., then x equals (a) log3 4 (b) 1 – log3 4 (c) 1 – log4 3 (d) log4 3 In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is (b) 10. 11. æ a cö æ b bö a, b and b, c respectively, then ç + ÷ ç + ÷ è x yø è x yø is equal to (a) – 2 (b) – 4 (c) 2 (d) 4 Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. 3 if a < b < c and a + b + c = , then the value of a is 2 1 1 (a) (b) 2 3 2 2 1 1 1 1 (c) 2 (d) 2 3 2 An infinite G.P. has first term ‘x’ and sum ‘5’, then x belongs to (a) x < – 10 (b) – 10 < x < 0 (c) 0 < x < 10 (d) x > 10 In the quadratic equation ax2 + bx + c = 0, D = b2 – 4ac and a + b, a2 + b2, a3 + b3, are in G.P. where a, b are the root of ax2 + bx + c = 0, then (a) D ¹ 0 (b) bD = 0 (c) cD = 0 (d) D = 0 Sequences and Series 19 12. The sum of an infinite geometric series is 2 and the sum of the geometric series made from the cubes of this infinite sereis is 24. Then the series is 3 3 3 3 3 3 (a) 3 + - + - .... (b) 3 + + + + .... 2 4 8 2 4 8 3 3 3 (c) 3 - + - + ... (d) None of these 2 4 8 13. If a, b, c are in G. P. and log a – log 2b, log 2b – log 3c and log 3c – log a are in A. P., then a, b, c are the sides of a triangle which is (a) Acute angled (b) Obtuse angled (c) Right angled (d) None of these 14. Ar ; r = 1, 2, 3, ........... , n are n points on the parabola y 2 = 4x in the first quadrant. If Ar = ( xr , yr ) , where x1 , x2 , x3 , ..............., xn are in G. P. and x1 = 1, x2 = 2, then yn is equal to n +1 n +1 –2 2 (b) 2 n +1 n (d) 2 2 (a) (c) ( 2) 15. If three successive terms of a G..P. with common ratio r (r > 1) form the sides of a D ABC and [r] denotes greatest integer function, then [r] + [– r] = (a) 0 (b) 1 (c) – 1 (d) None of these 16. If a, b, c, are in A.P. and p, p¢ are respectively A.M. and G.M. between a and b while q, q¢ are respectively AM.and G.M. between b and c, then (a) p 2 + q 2 = p '2 + q '2 (b) pq = p ' q ' (c) p 2 - q 2 = p '2 - q '2 (d) p 2 + p¢2 = q 2 + q '2 17. The sum of the series 1 + 2.2 + 3.2 2 + 4.2 3 + 5.2 4 + ... + 100.2 99 is (a) 99.2100 – 1 (b) 100.2100 (c) 99.2100 (d) 99.2100 + 1 18. If a, b, c, d are positive real number such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation: (a) 0 < M £ 1 (b) 1 £ M £ 2 (c) 2 £ M £ 3 (d) 3 £ M £ 4 2 3 3 1 4 æ1ö 5 æ1ö . + .ç ÷ + .ç ÷ 1 .2 2 2.3 è 2 ø 3.4 è 2 ø + ......... to n terms is equal to 19. The sum of 1 2 3 (d) (b) (c) 4 5 6 (c) (b) (b) 7 8 9 (d) (d) (d) 10 11 12 (c) (c) (c) 13 14 15 (a) 1 – (c) 1- 1 ( n + 1)2 n 1 (b) n - 1 2 n +1 (d) None of these n.2n+1 b ³ c for all positive x, where a < 0 20. Let ax2 + x and b < 0. The value of the expression 27ab2 cannot be less than (a) 4c3 (b) 4c2 (c) 8c3 (d) c3 Numeric Value Answer 21. Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is 22. Three numbers a, b, c are in GP. If a, b, c – 64 are in AP and a, b – 8, c – 64 are in GP, then the sum of the numbers may be 23. a, b, c are positive integers forming an increasing G.P. and b – a is a perfect cube and log6 a + log6 b + log6 c = 6, then a + b + c = 24. The sum to infinite term of the series 25. 26. 27. 28. 2 6 10 14 1 + + 2 + 3 + 4 + ... is 3 3 3 3 The 20th term of the series 2 + 3 + 5 + 9 + 16 +.......is Two consecutive numbers from 1, 2, 3,.........., n are removed. If the arithmetic mean of the n remaining numbers is 105/4 then is equal to 10 Let a, b, c, d be four distinct real numbers in A.P. Then half of the smallest positive value of k satisfying 2(a – b) + k(b – c)2 + (c – a)3 = 2(a – d) + (b – d)2 + (c – d)3 is ..... . Let x1, x2, ... Î (0, p) denote the of values of x satisfying the equation 2 3 27(1 + |cos x| + cos x + |cos x| + ....upto ¥) = 93, find 1 the value of ( x1 + x2 + ...) p 29. For a, b > 0, let 5a – b, 2a + b, a + 2b be in A.P. and (b + 1)2, ab + 1, (a – 1)2 are in G.P., then the value of (a–1 + b– 1) is ..... . 30. If one geomteric mean G and two Arithmetic means P and q be inserted between two quantities, then G2 = (kp – q)(kq – p) then find k. ANSWER KEY (b) 16 (c) (c) 17 (d) (c) 18 (a) 19 (a) 22 (124) 25 (990) 28 20 (a) 23 (189) 26 (5) 29 21 (0.60) 24 (3) 27 (8) 30 (1) (6) (2) MATHEMATICS 20 10 STRAIGHT LINES MCQs with One Correct Answer 1. 5. Through the point P (a, b), where ab > 0 , the x y + = 1 is drawn so as to form a b with coordinate axes a triangle of area S. If ab > 0, then least value of S is straight line (a) (c) 2. 3. 2ab ab (b) x y + = 1 meets the axis of x and y at A a b and B respectively and the line y = x at C so that area of the trinagle AOC is twice the area of the triangle BOC, O being the origin, then one of the positions of C is The line (a) (a, a) 1 ab 2 (c) (d) None of these 1ö æ The vertices of a triangle are ç ab, ÷ , è abø 6. The range of values of b such that (0, b ) lie on or inside the triangle formed by the lines y + 3x + 2 = 0, 3y – 2x – 5 = 0, 4y + x – 14 = 0 is 1ö 1ö æ æ çè bc, bc ÷ø and ç ca, ca ÷ where a, b, c are the è ø (a) 5 < b £ 7 roots of the equation x3 – 3x2 + 6x + 1 = 0. The coordinates of its centroid are. (a) (1, 2) (b) (2, – 1) (c) (1, – 1) (d) (2, 3) Consider points A (3, 4) and B (7, 13). If P be a point on the line y = x such that PA + PB is minimum, then coordinates of P are 5 7 £ b £ (d) None of these 3 2 The intercepts on the straight line y = mx by the lines y = 2 and y = 6 is less than 5, then m belongs to (a) æ 12 12 ö ç , ÷ è7 7ø æ 13 13 ö (b) ç , ÷ è7 7ø æ 31 31 ö ç , ÷ (d) (0, 0) è 7 7ø If the straight lines 2x + 3y – 1 = 0, x + 2y – 1 = 0 and ax + by – 1 = 0 form a triangle with origin as orthocentre, then (a, b) is given by (a) (6, 4) (b) (– 3, 3) (c) (– 8, 8) (d) (0, 7) (b) (c) 7. (a) æ 4 4ö çè - , ÷ø 3 3 (b) æ 4 3ö çè , ÷ø 3 8 (c) 4ö æ4 ö æ ç -¥, - ÷ È ç , ¥ ÷ 3ø è3 ø è (d) æ4 ö çè , ¥÷ø 3 (c) 4. æ b bö çè 3 , 3 ÷ø æ 2a 2a ö , 3 3 ÷ø æ 2b 2b ö (d) ç , ÷ è 3 3ø (b) ç è 1 £ b£1 2 Straight Lines 8. 21 If three distinct points A, B, C are given in the 2-dimensional coordinate plane such that the ratio of the distance of each one of them from the point (1, 0) to the distance from (– 1, 0) is equal 1 to , then the circumcentre of the triangle ABC 2 is at the point (a) æ5 ö ç ,0÷ è3 ø (a) ax + by + 2a = 0 (c) bx + ay – 2b = 0 (d) ay – bx + 2b = 0 13. If the point (a, 2) lies between the lines x – y – 1 = 0 and 2 (x – y) + 5 = 0, then the set of values of ‘a’ is æ9 ö (a) (– ¥ , 3) È ç , ¥ ÷ è2 ø (b) (0, 0) æ1 ö (d) (3, 0) ç ,0÷ è3 ø 9. Let A (–3, 2) and B (–2, 1) be the vertices of a triangle ABC. If the centroid of this triangle lies on the line 3x + 4y + 2 = 0, then the vertex C lies on the line : (a) 4x + 3y + 5 = 0 (b) 3x + 4y + 3 = 0 (c) 4x + 3y + 3 = 0 (d) 3x + 4y + 5 = 0 10. The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points (a2 + 1, a2 + 1) and (2a, – 2a), a ¹ 0. Then for any a, the orthocentre of this triangle lies on the line: (a) y – 2ax = 0 (b) y – (a2 + 1)x = 0 (c) y + x = 0 (d) (a – 1)2x – (a + 1)2y = 0 11. The line parallel to the x- axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0, where (a, b) ¹ (0, 0) is (c) (a) below the x - axis at a distance of 3 from it 2 2 (b) below the x - axis at a distance of from it 3 (c) above the x - axis at a distance of 3 from it 2 2 from it 3 The straight line y = x – 2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then its equation is (b) æ 9ö ç 3, ÷ è 2ø (c) (– ¥ , 3) (d) æ 1 ö çè - , 3÷ø 2 14. If two vertices of a triangle are (5, –1) and (–2, 3) and its orthocentre is at (0, 0), then the third vertex is (a) (4, – 7) (b) (– 4, – 7) (c) (– 4, 7) (d) (4, 7) 15. The base of an equilateral triangle is along the line given by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then the length of a side of the triangle is: (a) 2 3 15 (b) 4 3 15 (c) 4 3 5 (d) 2 3 5 16. The equation of bisector of that angle between the lines x + y + 1 = 0 and 2x – 3y – 5 = 0 which contains the point (10, – 20) is (a) x ( 13 + 2 2 ) + y ( 13 – 3 2 ) + ( 13 – 5 2 ) = 0 (b) x ( 13 – 2 2 ) + y ( 13 + 3 2 ) + ( 13 + 5 2 ) = 0 (d) above the x - axis at a distance of 12. (b) ax – by – 2a = 0 (c) x ( 13 + 2 2 ) + y ( 13 + 3 2 ) + ( 13 + 5 2 ) = 0 (d) None of these MATHEMATICS 22 17. The bisector of the acute angle formed always pass through a fixed point P for all possible values of q. If the maximum value of the difference of distances of P and B (3, 4) from a between the lines 4 x - 3 y + 7 = 0 and 3x - 4 y + 14 = 0 has the equation : (a) x+ y+3 = 0 (b) x - y -3 = 0 (c) x- y+3 = 0 (d) 3x + y - 7 = 0 The straight lines (3sec q + 5cosec q)x + (7 sec q - 3cosec q) y + 11(sec q - cosec q) = 0 L . If the equation of the side BC is ax + by – 5 A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. If the point O divides the segment PQ in the ratio 19. 20. m , then n m + n is ________. The vertex of an equilateral triangle is (2, –1), and the equation of its base is x + 2y = 1. If the 23. 24. length of its sides is 2 / K , then value of K is ____. If (sin q, cos q), q Î [0, 2p] and (1, 4) lie on the same side or on the line 3x – y + 1 = 0, then the maximum value of sin q will be ______. 21. k2 is 10 equal to . The straight line L º x + y + 1 = 0 and L1 º x + 2y + 3 = 0 are intersecting. m is the slope of the straight line L2 such that L is the bisector of the anlge between L1 and L2. The unit digit of 812m2 + 3 is equal to If tana, tanb, tanl are the roots of the equation t3 – 12t2 + 15t – 1 = 0; then the centroid of triangle having vertices (tana, cota); (tanb, cotbb); (tanl, cotl) is given by G(h, k); then evaluate (h + k)/(k – h). Consider a DABC whose sides AB, BC, and CA are represented by the straight lines 2x + y = 0, x + py = q, and x – y = 3, respectively. The point P(2, 3) is the orthocenter. The value of (p + q)/10 is ..... . In DABC, the vertex A = (1, 2), y = x is the perpendicular bisector of the side AB and x – 2y + 1 = 0 is the equation of the internal angle bisector of 22. Numeric Value Answer 18. point on the line x - y + 3 = 0 is k then 25. = 0, then the value of a – b is ..... . 1 2 3 (a) (b) (c) 4 5 6 (c) (d) (c) 7 8 9 (c) (a) (b) 10 11 12 ANSW ER KEY (d) 13 (d) 16 (a) 14 (b) 17 (d) 15 (b) 18 (a) (c) (7) 19 20 21 (15) (0) (4) 22 23 24 (1) (9) (5) 25 (4) 11 CONIC SECTIONS MCQs with One Correct Answer 1. 2. 3. The line 4 x + 3 y - 4 = 0 divides the circumference of the circle centered at (5, 3), in the ratio 1 : 2. Then the equation of the circle is (a) x 2 + y 2 - 10 x - 6 y - 66 = 0 (b) x 2 + y 2 - 10 x - 6 y + 100 = 0 (c) x 2 + y 2 - 10 x - 6 y + 66 = 0 (d) x 2 + y 2 - 10 x - 6 y –100 = 0 Let A(– 4, 0) and B(4, 0). Then the number of points C = (x, y) on the circle x2 + y2 = 16 lying in first quadrant such that the area of the triangle whose vertices are A, B and C is a integer is (a) 14 (b) 15 (c) 16 (d) None of these If (a, b) is a point on the circle whose centre is on the x-axis and which touches the line x + y = 0 at (2, –2), then the greatest value of a is (a) 4 – 4. 5. 2 2 y = | x | + c and x + y 2 – 8| x | – 9 = 0 have no solution is (a) (– ¥ , – 3) È (3, ¥ ) (b) (– 3, 3) (c) (–¥, – 2 ) È (5 2 , ¥) (d) (5 2 –4, ¥) (a) 2x2 + 2 y 2 + gx + fy = 0 2 (b) x + y 2 + gx + fy = 0 6. (c) x 2 + y 2 + 2gx + 2fy = 0 (d) None of these A ray of light incident at the point (– 2, – 1) gets reflected from the tangent at (0, –1) to the circle x 2 + y 2 = 1. The reflected ray touches the circle. The equation the line along which the incident ray moved, is 7. (b) 6 (c) 4 + 2 2 (d) 4 + 2 The set of values of ‘c’ so that the equations Tangents are drawn from O (origin) to touch the circle x2 + y2 + 2gx + 2fy + c = 0 at points P and Q. The equation of the circle circumscribing triangle OPQ is 8. (a) 4 x - 3 y + 11 = 0 (b) 4 x + 3 y + 11 = 0 (c) 3 x + 4 y + 11 = 0 (d) 4x + 3 y + 7 = 0 If the line y = mx + 1 meets the circle x2 + y2 + 3x = 0 in two points equidistant from and on opposite sides of x-axis, then (a) 3m + 2 = 0 (b) 3m – 2 = 0 (c) 2m + 3 = 0 (d) 2m – 3 = 0 If a circle passes through the point (a, b) and cuts the circle x 2 + y 2 = 4 orthogonally, then the locus of its centre is (a) 2ax - 2by - (a 2 + b2 + 4) = 0 (b) 2ax + 2by - (a 2 + b2 + 4) = 0 (c) 2ax - 2by + (a 2 + b2 + 4) = 0 (d) 2ax + 2by + (a 2 + b2 + 4) = 0 MATHEMATICS 24 9. 10. 11. 12. 13. 14. 15. The set of all real values of l for which exactly two common tangents can be drawn to the circles x2 + y2 – 4x – 4y + 6 = 0 and x2 + y2 – 10x – 10y + l = 0 is the interval: (a) (12, 32) (b) (18, 42) (c) (12, 24) (d) (18, 48) A circle bisects the circumference of the circle x2 + y2 – 2y – 3 = 0 and touches the line x = y and the point (1, 1). Its radius is : 3 9 (a) (b) (c) 4 2 (d) 3 2 2 2 Let L1 be the length of the common chord of the curves x2 + y2 = 9 and y2 = 8x, and L2 be the length of the latus rectum of y2 = 8x, then: (a) L1 > L2 (b) L1 = L2 L1 = 2 (c) L1 < L2 (d) L2 If the tangent at the point P (x1, y1) to the parabola y2 = 4ax meets the parabola y2 = 4a (x + b) at Q and R, then the mid-point of QR is (a) (x1 + b, y1 + b) (b) (x1 – b, y1 – b) (c) (x1, y1) (d) (x1 + b, y1 – b) Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (a) (–6, –11) (b) (–9, –13) (c) (–10, –15) (d) (–6, –7) A circle is drawn with centre at the focus S of the parabola y2 = 4x so that a common chord of the parabola and the circle is equidistant from the focus and the vertex. Then the equation of the circle is 9 9 (b) (x – 1)2 = – y2 (a) (x – 1)2 + y2 = 16 4 9 9 (c) (x – 1)2 + x2 = (d) (y – 1)2 + x2 = 16 4 Locus of all such points so that sum of its distances from (2, – 3) and (2, 5) is always 10, is (a) ( x - 2)2 ( y - 1)2 + =1 25 9 (b) ( x - 2)2 ( y - 1)2 + =1 25 16 (c) ( x - 2)2 ( y - 1)2 + =1 16 25 ( x - 2)2 ( y - 1)2 + =1 9 25 The radius of the circle passing through the foci (d) 16. 2 2 of the ellipse x + y = 1 , and having its 16 9 centre at (0, 3) is 7 1 (d) 2 2 Equation of the line passing through the points of intersection of the parabola x2 = 8y and the (a) 4 17. (b) 3 (c) x2 + y 2 = 1 is : 3 (a) y – 3 = 0 (b) y + 3 = 0 (c) 3y + 1 = 0 (d) 3y – 1 = 0 Equation of the largest circle with centre (1, 0) that can be inscribed in the ellipse x2 + 4y2 = 16, is ellipse 18. 2 2 (a) 2 x + 2 y – 4x + 7 = 0 (b) x 2 + y 2 – 2x + 5 = 0 2 (c) 3 x + 3 y 2 – 6x – 8 = 0 (d) None of these æ 3ö ç 2, ÷ to the ellipse, è 2ø 19. The normal at 20. x2 y 2 + = 1 touch es a par abola, whose 16 3 equation is (a) y2 = – 104 x (b) y2 = 14 x 2 (c) y = 26x (d) y2 = – 14x The angle subtended by the common tangent of the two ellipse ( x - 4)2 y 2 + 25 4 = 1 and ( x + 1)2 y 2 + = 1 at the origin is 1 4 (a) p 2 (b) p 4 (c) p 3 (d) p 6 Conic Sections 25 Numeric Value Answer 21. Two equal chords AB and AC of the circle x2 + y2 – 6x – 8y – 24 = 0 are drawn from the point A ( 33 + 3, 0) . Another chord PQ is drawn intersecting AB and AC at points R and S, respectively given that AR = SC = 7 and RB = AS = 3. The value of PR/QS is 22. If p and q be the longest and the shortest distance respectively of the point (–7, 2) from any point (a, b) on the curve whose equation 27. S1 and S2 be the foci of the hyperbola whose transverse axis length is 4 and conjugate axis length is 6, S3 and S4 be the foci of the conjugate hyperbola. If the area of the quadrilateral S1 S3 A . 13 28. If the ratio of the area of equilateral triangles made of the common chord of the circles x2 + y2 = 4 and x2 + y2 – 8x + 4 = 0 and their respective pairs of tangents drawn from points on the S2 S4 is A, then find positive x- axis is 57 + 24 3 : k then k is ________. is x 2 + y 2 - 10 x - 14 y - 51 = 0 and G.M. of p and q is 2 k , then value k is _______. 23. The straight line y = mx + c (m > 0) touches the 29. are drawn through P touching the coordinate axes, such that the length of common chord of these circle is maximum. If possible values of parabolas y 2 = 8 (x + 2) then the minimum value taken by c is 24. Two tangents are drawn from a point (–2, –1) to the curve, y2 = 4x. If a is the angle between them, then |tan a| is equal to: 2 y x2 25. Tangents are drawn to the ellipse + =1 5 9 at ends of latus rectum. The area of quadrilateral so formed is 26. A trapezium is inscribed in the parabola y2 = 4x such that its diagonal pass through the point 25 . If the area of (1, 0) and each has length 4 éPù trapezium be P then ê ú is equal to ë4û P(a , b) is a points in the first quadrant. Circles a/b is k1 ± k2 2 then k1 + k2 is equal to______. 30. C is the centre of the hyperbola and ' A ' is any point on it. The tangent at A to the hyperbola meets the line x - 2 y = 0 and x + 2 y = 0 at Q and R respectively. The value of CQ.CR is equal to ANSW ER KEY 1 2 3 4 (a) (b) (c) (d) 5 6 7 8 (b) (b) (b) (b) 9 10 11 12 (b) (b) (c) (c) 13 14 15 16 (d) (a) (d) (a) 17 18 19 20 x2 y 2 = 1, 4 1 (d) (c) (a) (a) 21 22 23 24 (1) (11) (4) (3) 25 26 27 28 (27) (4) (2) (9) 29 30 (5) (5) MATHEMATICS 26 12 LIMITS & DERIVATIVES 5. MCQs with One Correct Answer 1. Given a real valued function 'f ' such that lim (d) 3. 6. 7. lim f ( x ) exist lim f ( x ) does not exists 8. x®0 4. (c) –3 (a) å 4k e (b) æ 4ö çè ÷ø e 1/ k æ 4ö (c) ç ÷ è eø k (d) æ eö çè ÷ø 4 k a p (b) 2a e p (c) - ) + (1/ x)sin x , (d) 2 The limit lim æç 2 - ÷ xø x ®a è is equal to - 2 p (d) 1 (a) e e If a and b are the roots of the quadratic lim x® (d) –1/3 log ( n + r ) - log n 1ö æ = 2 ç log 2 - ÷ , è 2ø n r =1 1 k k k 1/ n then lim k [(n + 1) (n + 2) ...(n + n) ] is n®¥ n equal to lim ( (d) 4 1/ x æ px ö tan a ö èç 2a ø÷ 1 a 9. 1 - cos(cx 2 + bx + a ) 2(1 - ax )2 = (a) c æ 1 1ö 2a èç a bø÷ (b) (c) c æ 1 1ö ab èç a bø÷ (d) None of these n n®¥ (c) 3 equation ax 2 + bx + c = 0 , then æ x 2 + 5x + 3 ö lim ç ÷ x ®¥ çè x 2 + x + 2 ÷ø (a) e4 (b) e2 (c) e3 (d) 1 2 x x3 £ f ( x) £ 3 + for all x ¹ 0, then the If 3 – 12 9 value of lim f ( x ) is equal to h®0 is a finite non-zero The value of lim (sin x) x ®0 where x > 0 is (a) 0 (b) –1 (c) 1 - x®0 (a) 1/3 (b) 3 Given that xn number is (a) 1 (b) 2 x 2. (cos x - 1)(cos x - e x ) x®0 ì tan 2 {x} for x > 0 ï 2 2 ï x - [ x] ï f ( x) = í 1 for x = 0 ï ï {x} cot{x} for x < 0 ï î then (a) LHL = 1 (b) RHL = cot 1 (c) The integer n for which c æ 1 1ö 2b èç a bø÷ cos2 (1- cos2 (1- cos2(............ - cos2(x))))..........) x®0 ìï æ x + 4 - 2ö üï sin ípç ÷ý x ø ïþ ïî è is equal to p p 4 2 (a) (b) (d) (c) p 2 4 p lim Limits & Derivatives 27 10. The value of 1/ cos2 x lim [1 x®p / 2 1/ cos 2 x +2 1/ cos 2 x cos 2 x + ..... + n ] is (a) 0 11. (c) ¥ (b) n (d) lim {log n -1 (n) log n (n + 1)log n +1( n + 2)... n(n+1) 2 n®¥ 17. Let f(x) be a polynomial function satisfying æ1ö æ1ö f ( x ) × f ç ÷ = f ( x ) + f ç ÷ . If f(4) = 65 and è xø è xø l1, l2, l3 are in GP, then f '(l1 ), f '(l2 ), f '(l3 ) are in (a) AP (b) GP (c) HP (d) None of these Numeric Value Answer ...log k (n x )} is equal to n -1 (a) n (b) k (c) ¥ (d) None of these 12. If [x] denotes the greatest integer £ x , then lim 1 n®¥ n 3 (a) {[12 x] + [22 x] + [32 x] + ... + [n 2 x]} equals x / 2 (b) (c) x/3 (d) 0 x/6 If lim f ( x ) exist, then l = x ®3 9 (b) 2 14. The value of 2 9 (c) sec 2 ì æ p öü lim ísin 2 ç ÷ý x ®0 î è 2 - ax ø þ (a) e-a / b p 2 -bx (b) (d) None é ù 20. The value of lim ê x + x + x - x ú is x ®¥ ë û f (x) =1 21. Let f (x) be a function such that lim x®0 x x (1 + a cos x ) - b sin x and lim = 1, then b – 3a x ®0 { f ( x )3} is equal to 22. The largest value of non-negative integer a for is equal to 2 e-a / b x ü is 23. The value of lim ì ï ï 3 x ®¥ ï x ï íx+ ý 3 x ï x+ ....¥ ï ïî ïþ x+3 x 24. If a, b are two distinct real roots of the 2 g (1 - cos 2 x) x4 x ®0 equation ax3 + x - 1 - a = 0, (a ¹ -1, 0), none of which is equal to unity, then the value of is 4a a a (b) (c) (d) None b 4b b Let f (x) be a polynomial function of second degree. If f (1) = f ( -1) and a1, a2 , a3 are in A.P.. then f '(a1 ), f '( a2 ), f '( a3 ) are in (a) A. P. (b) G. P. (c) H. P. (d) None of these (a) 16. p 2 1- x 4a / b (where b ¹ 0 ), then lim 19. ìï - ax + sin ( x - 1) + a üï1- x 1 which lim í = is ý 4 x ®1 î ï x + sin ( x - 1) - 1 þï (d) e a 2a / b f (1 - cos x ) f ( x) =b If lim 2 = a and lim x ®0 g ( x ) sin 2 x x ®0 x (c) 15. 2 3 g ( x) f ( a ) - g ( a ) f ( x ) is x-a x®a The value of lim (sinx)tanx is then the value of lim x® x ì ï (e( x+ 3) ln 27 ) 27 - 9 ; x<3 ï 3x - 27 13. If f ( x) = í ï 1 - cos( x - 3) ; x>3 ïl î ( x - 3) tan( x - 3) (a) 18. If f (a ) = 2, f ' (a) = 1 , g (a) = -1 , g ' (a) = 2 , lim (1 + a) x 3 - x 2 - a (e1-ax - 1)( x - 1) the value of kl. x ®(1/ a ) 25. is al ( k a - b) . Find a 3x + 33- x - 12 is equal to _______. x ®2 3- x /2 - 31- x lim ANSW ER KEY 1 2 3 (d) (a) (b) 4 5 6 (c) (c) (c) 7 8 9 (c) (a) (b) 10 11 12 (b) (b) (b) 13 14 15 (c) (a) (c) 16 17 18 (a) (b) (5) 19 20 21 (1) (0.50) (6) 22 23 24 (2) (1) (1) 25 (36) MATHEMATICS 28 13 MATHEMATICAL REASONING MCQs with One Correct Answer 1. 2. For the statement “17 is a real number or a positive integer”, the “or” is (a) Inclusive (b) Exclusive (c) Only (a) (d) None of these Let p and q be any two logical statements and 7. 8. r : p ® (: p Ú q) . If r has a truth value F, then the truth values of p and q are respectively : (a) F, F (b) T, T (c) T, F (d) F, T 3. If p : Ashok works hard q : Ashok gets good grade 9. The verbal form for (~ p ® q) is 5. (a) If Ashok works hard then gets good grade (b) If Ashok does not work hard then he gets good grade (c) If Ashok does not work hard then he does not get good grade (d) Ashok works hard if and only if he gets grade If p is false and q is true, then (a) p Ù q is true (b) p Ú ~ q is true (d) p Þ q is true (c) q Ù p is true ~ p Ù q is logically equivalent to 6. (a) p ® q (b) q ® p (c) ~ (p ® q) (d) ~ (q ® p) Which of the following is a contradiction? 4. (a) (p Ù q)Ù ~ (p Ú q) (b) p Ú (~ p Ù q) (c) (p Þ q) Þ p (d) None of these 10. (p Ù ~ q) Ù (~ p Ù q) is (a) A tautology (b) A contradiction (c) Both a tautology and a contradiction (d) Neither a tautology nor a contradiction The false statement in the following is (a) p Ù (~ p) is contradiction (b) (p Þ q) Û (~ q Þ ~ p) is a contradiction (c) ~ (~ p) Û p is a tautology (d) p Ú (~ p) Û p is a tautology The conditional ( p Ù q) Þ p is (a) A tautology (b) A fallacy i.e., contradiction (c) Neither tautology nor fallacy (d) None of these If p and q are two statements, then ( p Þ q) Û (-q Þ~ p) is a 11. 12. (a) contradiction (b) tautology (c) neither (a) nor (b) (d) None of these Which of the following is false? (a) p Ú ~ p is a tautology (b) ~ (~p) « p is a tautology (c) p Ù ~ p is a contradiction (d) ((p Ù q) ® q) ® p is a tautology In the truth table for the statement ( p ® q) « (~ p Ú q), the last column has the truth value in the following order is (a) TTFF (b) FFFF (c) TTTT (d) FTFT Mathematical Reasoning 29 13. If p Þ (~ p Ú q) is false, then truth values of p and q are respectively (b) F, F (a) F. T (c) T, T (d) T, F 14. Negation of “2 + 3 = 5 and 8 < 10” is (a) 2 + 3 ¹ 5 and < 10 (b) 2 + 3 = 5 and 8 </ 10 21. The negation of (p Ú q)Ù (p Ú ~ r) is (a) (~ p Ù ~ q) Ú (q Ù ~ r) (b) (~ p Ù ~ q) Ú (~ q Ù r) (c) (~ p Ù ~ q) Ú (~ q Ù r) (d) (p Ù q) Ú (~ q Ù ~ r) 22. Let p, q and r be any three logical statements. Which of the following is true? (c) 2 + 3 ¹ 5 or 8 </ 10(d) None of these 15. If the compound statement p ® (~ p Ú q) is false then the truth value of p and q are respectively (a) T, T (b) T, F (c) F, T (d) F, F 16. The contrapositive of p ® (~q ® ~r) is (a) (~ q Ù r) ® ~ p (b) (q ® r) ® ~p (c) (q Ú ~r) ® ~ p (d) None of these (c) ( p Ú ~ q ) Ú ~ p (d) None of these ( p Ù ~ q) Ú ~ p 18. The inverse of the statement (p Ù ~ q) ® r is (a) ~ (p Ú ~q) ® ~ r (b) (~p Ù q) ® ~ r (c) (~p Ú q) ® ~ r (d) None of these 19. ~ ( (~ p) Ù q ) is equal to (a) p Ú (~ q) (b) p Ú q (c) p Ù (~ q) (d) ~ p Ù ~ q 20. Which of the following is true? (a) p Þ q º ~ p Þ ~ q (b) (c) (d) (b) ~ [( p Ú q ) Ù (~ r ) º (~ p) Ú (~ q) Ú (~ r ) (c) ~ [ p Ú (~ q)] º (~ p) Ù q (d) ~ [ p Ú (~ q)] º (~ p)Ù ~ q (b) [p Ú q] Ú (~ p) is a tautology (c) [p Ù q) Ù (~ p) is a contradiction p Ú (~ p Ú q) is ( p Ù ~ q ) Ù ~ p (b) ~ [ p Ù (~ q)] º (~ p) Ù q 23. Identify the false statements (a) ~ [p Ú (~ q)] º (~ p) Ú q 17. The negation of the compound proposition (a) (a) (d) ~ [p Ú q] º (~ p) Ú (~ q) 24. Negation of the statement (p Ù r) ® (r Ú q) is (a) ~ (p Ù r) ® ~ (r Ú q) (b) (~p Ú ~r) Ú (r Ú q) (c) (p Ù r) Ù (r Ù q) (d) (p Ù r) Ù (~ r Ù ~q) 25. Let A, B, C and D be four non-empty sets. The contrapositive statement of “If A Í B and B Í D, then A Í C ” is: (a) If A Í C, then A Í B and B Í D (b) If A Í C, then B Ì A or D Ì B ~ ( p Þ ~ q) º ~ p Ù q ~ (~ p Þ ~ q) º ~ p Ù q ~ (~ p Û q) º [~ ( p Þ q)Ù ~ (q Þ p)] (c) If A Í C, then A Í B and B Í D (d) If A Í C, then A Í B or B Í D ANSW ER KEY 1 2 3 (a) (c) (b) 4 5 6 (d) (d) (a) 7 8 9 (b) (b) (a) 10 11 12 (b) (b) (c) 13 14 15 (d) (c) (b) 16 17 18 (a) (a) (c) 19 20 21 (a) (c) (c) 22 23 24 (c) (d) (d) 25 (d) MATHEMATICS 30 14 STATISTICS MCQs with One Correct Answer 1. n.2n-1 2n - 1 (b) 6. The median of 100 observations grouped in classes of equal width is 25. If the median class interval is 20 - 30 and the number of observations less than 20 is 45, then the frequency of median class is (a) 10 (b) 20 (c) 15 (d) 12 If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to : (a) 20.0 (b) 10.1 (c) 20.2 (d) 10.0 In a series of 2 n observations, half of them equal a and remaining half equal –a. If the standard deviation of the observations is 2, then | a | equals 3n (n + 1) 2 (2n + 1) (n + 1) (2 n + 1) n (n + 1) (c) (d) 6 2 The A.M. of n observations is M. If the sum of n – 4 observation is a, then the mean of remaining 4 observation is (a) nM - a 4 (b) nM + a 2 nM – a (d) nM + a 2 The mean income of a group of persons is ` 400. Another group of persons has mean income ` 480. If the mean income of all the persons in the two groups together is ` 430, then ratio of the number of persons in the groups is (c) 3. The mean of n items is X . If the first item is increased by 1, second by 2 and so on, the new mean is : x (a) X + (b) X + x 2 n +1 (d) none of these (c) X + 2 The weighted mean of first n natural numbers whose weights are equal to the number of selections out of n n atural numbers of corresponding numbers is (a) 2. 5. (a) n1 5 = n2 3 (b) n1 2 = n2 5 n1 7 = (d) None of these n2 4 The mean of six numbers is 30. If one number is excluded, the mean of the remaining numbers is 29. The excluded number is (a) 29 (b) 30 (c) 35 (d) 45 7. 8. 1 2 (b) 2 (c) 2 (d) n n The standard deviations of two sets containing 10 and 20 members are 2 and 3 respectively measured from their common mean 5. The SD for the whole set of 30 members is (a) 9. (c) 4. (a) (c) 2 3 æ 22 ö ç 3 ÷ è ø (b) 6 (d) 3 Statistics 31 10. If M. D. is 12, the value of S.D. will be (a) 15 (b) 12 (c) 24 (d) None of these 11. Suppose values taken by a variable x are such that a £ xi £ b, where xi denotes the value of x in ith case for i = 1, 2, ... n. Then (a) a £ Var(x) £ b (b) a2 £ Var(x) £ b2 a2 £ Var( x) (d) (b – a)2 ³ Var(x) 4 12. If the median and the range of four numbers {x, y, 2x + y, x – y}, where 0 < y < x < 2y, are 10 and 28 respectively, then the mean of the numbers is : (a) 18 (b) 10 (c) 5 (d) 14 (c) 13. Let X and M.D. be the mean and the mean deviation about X of n observations xi, i = 1, 2, ........, n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are : (a) (b) X, M.D. the following statistical measures will not change even after the grace marks were given ? (a) mean (b) median (c) mode (d) variance 17. Coefficient of variation of two distribution are 60 and 70, and their standard deviations are 21 and 16, respectively. What are their arithmetic means? (a) 35, 22.85 (b) 22.85, 35.28 (c) 36, 22.85 (d) 35.28, 23.85 Numeric Value Answer 18. Variance of the data 2, 4, 5, 6, 8, 17 is 23.33. Then, variance of 4, 8, 10, 12, 16, 34 will be 19. Coefficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25, respectively. Then, difference of their standard deviations is 20. In an experiment with 15 observations on x, the following results were available: å x2 = 2830, å x = 170 X + 5, M.D. (c) X, M.D. + 5 (d) X + 5, M.D. + 5 14. The mean and variance for first n natural numbers are respectively 21. 2 (a) mean = n +1 n -1 , variance = 2 12 (b) mean = n -1 n2 + 1 , variance = 2 12 22. n +1 n2 - 1 , variance = 2 12 2 n -1 n +1 (d) mean = , variance = 2 2 15. The variance of 20 observations is 5. If each observation is multiplied by 2, then the new variance of the resulting observation is (a) 23 × 5 (b) 22 × 5 (c) 2 × 5 (d) 24 × 5 16. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of (c) mean = 23. 24. 25. One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is Consider the following data 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 If 1 is added to each number, then variance of the numbers so obtained is Coefficient of variation of two distribution are 50% and 60% and their standard deviation are 10 and 15, respectively. Then, difference of their arithmetic means is The mean of 5 observation is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, then difference of the other two observations is If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to _______. If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, then x × y is equal to ____. ANSW ER KEY 1 2 3 (a) (a) (a) 4 5 6 (c) (c) (a) 7 8 9 (b) (c) (c) 10 11 12 (a) (d) (d) 13 14 15 (b) (a) (b) 16 (d) 19 17 (a) 20 18 (93.32) (0) (78) 21 (8.25) 24 22 (5) 25 23 (5) (18) (52) MATHEMATICS 32 15 PROBABILITY-1 1. MCQs with One Correct Answer If two numbers p and q are choosen randomly from the set {1, 2, 3, 4} with replacement, then 6. the probability that p 2 ³ 4q is equal to 1 3 1 7 (b) (c) (d) 4 16 2 16 A natural number x is chosen at random from the first 100 natural numbers. Then the probability, 100 for the equation x + > 50 is x 1 11 1 3 (b) (c) (d) (a) 20 20 3 20 A person throws two dice, one the common cube and the other regular tetrahedron with numbers 1, 2, 3, 4 on its faces, the number on the lowest face being taken in the case of a tetrahedron.The chance that the sum of numbers thrown is not less than 5 is 5 1 3 4 (a) (b) (c) (d) 6 4 5 4 A is a set containing n element. A subset P of A is chosen at random, and the set A is reconstruced by replacing the elements of P. Another subset Q of A is now chosen at random. The probability that P È Q contains exactly r elements, 1 £ r £ n, is (a) 2. 3. 4. n (a) 4n n (c) 5. Cr 3r Cr 3n - r n (b) 3n 4n n (d) C r 3n 4 4n 10 persons sit around a circular table with 10 numbered chairs. The probability that the two particular persons A and B are always together is 2 1 1 2 (b) (c) (d) 9 5 9 5 5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is 15 21 1 1 (a) (b) (c) (d) 64 64 4 4 If three squares are chosen at random on a chess board, then find the probability that they should be in a diagonal line (a) 9/743 (b) 11/743 (c) 13/743 (d) None of these If the letters of the word MATHEMATICS are arranged arbitrarily, the probability that C comes before E, E before H, H before I and I before S is 1 1 1 1 (b) (c) (d) (a) 75 120 720 24 In a game of chance a player throws a pair of dice and scores points equal to the difference between the numbers on the two dice. Winner is the person who scores exactly 5 points more than his opponent. If two players are playing this game only one time, then the probability that neither of them wins is 1 1 53 107 (a) (b) (c) (d) 54 108 54 108 A box contains 2 fifty paise coins, 5 twenty five paise coins and 15 ten paise coins. Five coins are taken out of the box at random. Probability that the value of these five coins is less than one rupee and fifity paise is 150 170 (a) 22 (b) 1 - 22 C5 C5 (a) 7. 8. 9. 10. (c) 170 1 - 22 C5 (d) None of these Probability-1 11. 17. From a well-shuffled pack of 52 cards, four cards are selected at random. The probability of drawing exactly 2 spades and exactly 2 aces is 1494 1594 (a) (b) 270725 270725 1296 1396 (c) (d) 270725 270725 Numeric Value Answer 18. The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’, the two I’s does not come together is 19. A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is 20. Two numbers x and y are chosen at random without replacement from amongst the numbers 1, 2, 3, .........,3n. The probability that x3 + y3 is divisible by 3 is 21. 5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15. If the probability that the end seats are occupied by the girls and between any two girls odd number of boys take 2 2 3p + 2p p + 3p (a) (b) n 20 , then seat is 2 4 is equal to 1001 n p + 3 p2 3 p + 2 p2 22. A quadratic equation is chosen from the set of (c) (d) all quadratic equations which are unchanged by 2 4 squaring their roots. If the chance that the chosen There are three events E1, E2 and E3. one of which must, and only one can happen. The odds are 7 p equation has equal roots, is , then p + q = to 4 against E1 and 5 to 3 against E2. The odds q against E3 is (a) 4 : 11 (b) 3 : 8 (c) 23 : 88 (d) 65 : 23 23. In a multiple choice question, there are five alternative answers of which one or more than The chance of an event happening is the square one are correct. A candidate will get marks on the of the chance of a second event but the odds question, if he ticks all the correct answers. If he against the first are the cube of the odds against decides to tick answers at random, then the least the second. The chances of the events are number of choice should he be allowed so that 1 1 1 1 the probability of his getting marks on the , , (a) (b) 1 9 3 16 4 question exceeds is 8 1 1 , 24. A die is rolled three times, if p be the probability (c) (d) None of these 4 2 of getting a large number than the previous Ashmit and Bishmit are two students who number, then the value of 54p is appeared in Class X exam, Probability that 25. A class contains 20 boys and 20 girls of which Ashmit will pass is 0.5 while probability that half the boys and half the girls have cat eyes. If both of them will pass is at least 0.1 and at most one student is selected from the class & if the 0.3, and probability that Bishmit will pass is P(B) probability that either the student is a boy or then find the range of P(B) a (a) (0.2, 0.8) (b) (0.15, 0.25) has cat eyes is , then a 2 + b2 = (c) (0.1, 0.8) (d) None of these b ANSWER KEY (d) (b) (d) 10 (c) 13 (a) 16 (c) 4 7 19 (0.50) 22 (3) 25 (5) (b) (a) (c) 11 (b) 14 (d) 17 (a) 5 8 20 (0.33) 23 (4) (b) (b) (c) 12 (d) 15 (a) 18 (0.80) 21 (3) 24 (5) 6 9 Two distinct numbers a and b are chosen randomly from the set {2, 22, 23, .... 225}. Then the probability that logab is an integer is 31 131 (a) (b) 300 300 21 62 (c) (d) 200 300 12. There are two vans each having numbered seats, 3 in the front and 4 at the back. There are 3 girls and 9 boys to be seated in the vans. The probability of 3 girls sitting together in a back row on adjacent seats, is 1 1 1 1 (a) (b) (c) (d) 13 39 65 91 13. For the three events A, B, and C, P (exactly one of the events A or B occurs) = P (exactly one of the two events B or C occurs) = P(exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is 14. 15. 16. 1 2 3 33 16 RELATIONS & FUNCTIONS-2 MCQs with One Correct Answer 1. 2. Let P be the relation defined on the set of all real numbers such that P = {(a, b) : sec2a – tan2b = 1}. Then P is: (a) reflexive and symmetric but not transitive. (b) reflexive and transitive but not symmetric. (c) symmetric and transitive but not reflexive. (d) an equivalence relation. Consider the following relations. R = {(x, y) | x, y are real numbers and x = wy for some rational 3. æm pö number w} S = {ç , ÷ | m, n, p and q are èn qø integers such that, q ¹ 0, n ¹ 0 and qm = pn}. Then (a) Neither R nor S is an equivalence relation (b) S is an equivalence relation but R is not an equivalence relation (c) R and S both are equivalence relations (d) R is an equivalence relation but S is not an equivalence relation For real numbers x and y, we define xRy iff 4. x - y + 5 is an irrational number. Then, relation R is (a) reflexive (b) symmetric (c) transitive (d) None of these If f : R ® S , defined by 5. f ( x) = sin x - 3 cos x + 1, is onto, then the interval of S is (a) [ –1, 3] (b) [–1, 1] (c) [ 0, 1] (d) [0, 3] If the function f : (– ¥ ,¥) ® B defined by f (x) = – x2 + 6x – 8 is bijective, then B = (a) [1,¥) (b) (– ¥, 1] (c) (– ¥, ¥) (d) None of these 6. 7. 8. Let f : f : {x, y, z} ® {1, 2, 3} be a one-one mapping such that only one of the following three statements is true and remaining two are false : f (x) ¹ 2, f (y) = 3, f (z) ¹ 1, then – (a) f (x) > f (y) > f (z) (b) f (x) < f (y) < f (z) (c) f (y) < f (x) < f (z) (d) f (y) < f (z) < f (x) Let A º {1,2,3,4} B º {a, b, c}, then number of functions from A ® B, which are not onto is (a) 8 (b) 24 (c) 45 (d) 6 A quadratic polynomial maps from [–2, 3] onto [0, 3] and touches X-axis at x = 3, then the polynomial is 3 2 3 2 ( x - 6 x + 16) (b) ( x - 6 x + 9) (a) 16 25 3 2 3 2 ( x - 6 x + 16) (d) ( x - 6 x + 9) (c) 25 16 9. Let f : R + ® {-1, 0, 1} defined by 10. f ( x ) = sgn( x - x 4 + x 7 - x8 - 1) where sgn denotes signum function. Then f(x) is (a) many-one and onto (b) many-one and into (c) one-one and onto (d) one-one and into Let f : R ® R be a function defined by f ( x) = x + x2 , then f is (a) injective (b) surjective (c) bijective (d) None of these 11. If f (x) = sin x + cos x, g (x) = x2 – 1, then g (f (x)) is invertible in the domain (a) (c) é pù ê0 , 2 ú ë û é p pù ê- 2 , 2 ú ë û (b) é p pù ê- 4 , 4 ú ë û (d) [0, p] Relations & Functions-2 35 12. Let g : R ® R be given by g (x) = 3 + 4x. -n n If g ( x) = gogo....... og (x), then g (x) = (where g–n (x) denotes inverse of g n ( x) ) (a) (4n - 1) + 4n x (b) ( x + 1)4- n - 1 (c) ( x + 1)4n - 1 (d) (4- n - 1) x + 4n 13. If f (x) = 3 | x | - x - 2 and g (x) = sin x, then domain of definition of fog (x) is pü ì (a) í2np + ý 2 þ n ÎI î (b) (c) 7p 11p ö æ U çè 2np + 6 , 2np + 6 ÷ø n ÎI 7p ü ì í 2 np + ý 6 þ n ÎI î p 7p 11p ù é (d) {(4m + 1) : m ÎI} U ê2np + , 2np + 2 6 6 úû nÎI ë pö pö æ æ 14. If f ( x) = sin 2 x + sin 2 ç x + ÷ + cos x cos ç x + ÷ è è 3ø 3ø æ5ö and g ç ÷ = 1, then graph of y = g [f (x)] is è4ø Numeric Value Answer 18. Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is 19. If for x > 0, f ( x ) = (a - x n )1/ n , g ( x) = x 2 + px + q; p, q Î R and the equation g (x) – x = 0 has imaginary roots, then number of real roots of equation g (g (x)) – f (f (x)) = 0 is ì -1, x < 0 ï 20. Let g(x) = 1 + x - [x] and f ( x ) = í 0, x = 0 . ï 1, x > 0 î Then for all x, f (g(x)) is equal to 21. Let g ( x) = (a) a circle (b) a straight line (c) a parabola (d) None of these 15. Let f (x) = x2 and g(x) = sin x for all x Î R. Then the set of all x satisfying (f o g o g o f) (x) = (g o g o f) (x), where (f o g) (x) = f (g(x)), is (a) ± np , n Î{0,1, 2,....} (b) 17. Let f : [0, 1] ® R be an injective continuous function that satisfies the condition –1 < f (0) < f (1) < 1. Then, the number of functions g : [–1, 1] ® [0, 1] such that (gof ) (x) = x for all x Î [0, 1] is (a) 0 (b) 1 (c) more than 1, but finite (d) infinite ± np , n Î{1, 2,....} p + 2np, n Î{... - 2, -1, 0,1, 2....} 2 (d) 2np, n Î{... - 2, -1, 0,1, 2,....} 16. Let f : [0, 1] ® [–1, 1] and g : [–1, 1] ® [0, 2] be two functions such that g is injective and gof : [0, 1] ® [0, 2] is surjective. Then, (a) f must be injective but need not be surjective (b) f must be surjective but need not be injective (c) f must be bijective (d) f must be a constant function (c) e x - e- x and g(f(x)) = x, then 2 æ e22 -1ö f ç 11 ÷ - 5 =. ç 2e ÷ è ø x£0 ìï - x + 1, , then the number 22. If f ( x) = í 2 ïî-( x - 1) , x ³ 1 of solutions of f ( x) - f -1 ( x) = 0 is æp ö æp ö 23. If f (x) = cos2 x + cos2 ç + x ÷ - cos x cos ç + x ÷ è3 ø è3 ø and g(3/4) = 2, then find the value of (gof) (1). 24. If a non-zero function f(x) is symmetrical about y = x, then the value of p (constant) such that f 2 ( x) = ( f -1 (x))2 - px × f ( x) × f -1( x) + 2x2 f ( x) for all x Î R + is ....... . 25. Let f be defined on the natural numbers as follows : f (1) = 1 and for n > 1, f (n) = f [f (n – 1)] + f [n – f (n – 1)], the value of 1 20 å f (r ) is 30 r =1 ANSWER KEY 1 2 3 (d) (b) (a) 4 5 6 (a) (b) (c) 7 8 9 (c) (b) (b) 10 11 12 (d) (b) (b) 13 14 15 (d) (b) (a) 16 17 18 (b) (d) (14) 19 20 21 (0) (1) (6) 22 23 24 (4) (2) (2) 25 (7) 17 INVERSE TRIGONOMETRIC FUNCTIONS 4. MCQs with One Correct Answer 1. ac (a) a 2 + c2 2 ac (c) 2. f (x) = sin -1 (log[ x ]) + log(sin -1[ x]); (where [.] denotes the greatest integer function) is (a) R (b) [1, 2) If ax + b (sec (tan–1 x)) = c and ay + b (sec (tan–1 y)) = c, then 2 2 x+ y = 1 - xy (b) 2 ac a-c (d) a+c 1 - ac a -c The range of the function pü ì (d) {– sin 1} ílog ý 2þ î If x1, x2, x3, x4 are roots of the equation x4 – x3 sin 2b + x2 cos 2b – x cos b – sin b = 0, then (c) 5. 4 å tan -1 xi is equal to i =1 -1 é 2 1 ù -1 é 2 1 ù f (x) = sin ê x + ú + cos ê x - ú , where 2 2û ë û ë [.] is the greatest integer function, is (a) p – b (b) p – 2b (a) ìp ü í , pý î2 þ (b) (c) {p} æ pö (d) ç 0, ÷ è 2ø p -b (d) 2 The maximum value of f(x) æ 12 - 2 x 2 ö ÷ is –1 = tan çç 4 x + 2x2 + 3 ÷ è ø If å cos-1 br = 2 for any k ³ 1 where (a) 18º (b) 36º (c) 22.5º (d) 15º k 3. The range of the function kp 1ü ì í0, - ý 2þ î (c) 6. r =1 k 7. br ³ 0"r and A = å (b r ) r . Then r =1 lim (1 + x 2 )1/ 3 - (1 - 2 x )1/ 4 x + x2 x® A (a) 1 2 (b) 0 (c) ( A 2 (d) p 2 ) The number of solutions of the equation tan -1 x = = p - 2b 2 (a) 2 (c) 4 ( x2 + 1) - 4x2 is 2 (b) 3 (d) none of these Inverse Trigonometric Functions 8. 9. 37 æ ö 1 1 ö -1 æ S = tan-1 ç ÷ + tan ç 2 ÷ + ... 2 è n + n +1ø è n + 3n + 3 ø æ ö 1 + tan -1 ç ÷ , then tan S is 1 + ( n + 19)( n + 20) è ø equal to : 20 n (a) (b) 2 401 + 20 n n + 20n + 1 20 n (c) (d) 2 401 + 20 n n + 20n + 1 (a) a = –3 & b = 1 (b) a = 1 & b = - 1 1 &b= 6 2 (d) None of these (c) -1 x- 1 = cos -1 x- 1 is cos -1 x (a) 0 (b) 1 (c) 2 (d) 3 10. The minimum integral value of a for which the quadratic equation (cot–1 a)x2 – (tan–1 a)3/2 x + 2(cot–1 a)2 = 0 has both positive roots (a) 1 (b) 2 (c) 3 (d) 4 11. The sum of series sin -1 x sec -1 2 + sec-1 sec -1 a= cot–1 7 19 39 + cot –1 + cot –1 + ..... then 4 4 4 (a) S n = tan –1 n 2n + 5 (b) Sn = cot –1 n+5 2n (c) S n = cot –1 4n 2n + 5 1 2 –1 15. Find : tan (1/3) + tan–1(1/7) + tan–1 (1/13) (d) S¥ = cot –1 10 50 + sec -1 + . . . . + 3 7 ( n2 + 1)( n2 - 2n + 2) is 2 ( n2 - n + 1) 1 3 14. If Sn denotes the sum to n terms of the series The number of roots of the equation sin æ ö ÷ +....¥ è n + n +1ø +..... + tan -1 ç 1 2 p p p p (b) (c) (d) 2 3 4 6 16. If the equation x3 + bx2 + cx + 1 = 0 (b < c) has only one real root a. Then the value of 2 tan –1 (cosec a) + tan –1 (2 sin a sec2 a) is: (a) (a) tan–1 1 (b) tan–1 n (c) tan–1 (n + 1) (d) tan–1 (n – 1) 12. The value of x satisfying the equation (sin -1 x)3 - (cos-1 x)3 + (sin -1 x)(sin -1 x - cos -1 x ) = p3 16 is: 13. when (a) p cos 5 (b) p cos 4 (c) cos p 8 (d) cos - (c) p 2 p 2 (b) –p (d) p 17. Solution of equation cos -1 x 3 + cos -1 x = is p 12 æ ö a 2 a3 sin -1 ç a + + .... ÷ ç ÷ 3 9 è ø + cos -1 (1 + b + b2 + ...) = (a) p 2 1 2 (a) x= (c) x= ± (b) 1 2 x= - 1 2 (d) None of these p 2 MATHEMATICS 38 18. The sum of infinite series Numeric Value Answer æ 1 ö -1 æ 2 - 1 ö -1 æ 3 - 2 ö sin -1 ç ÷ + sin çç ÷ + sin çç ÷ ÷÷ è 2ø è 6 ø è 2 2 ø æ n - n -1 ö ÷ + ... +... + sin -1 ç is ç n ( n + 1) ÷ è ø p (b) p 2 The value of (a) 19. p 3 (c) 21. 22. 3 a aö b æ1 æ1 æ b öö cosec2 ç tan-1 ÷ + sec2 ç tan -1 ç ÷ ÷ 2 2 b 2 2 è a øø è ø è is equal to 20. (a) ( a - b ) ( a 2 + b2 ) (b) ( a + b ) ( a2 - b2 ) (c) ( a + b ) ( a2 + b2 ) (d) None of these cot–1(2.12) + cot–1(2.22) + cot–1(2.32) +... is equal to (a) p 4 (b) p 3 (c) p 2 p p < A< 4 2 The number of ordered triplets (x, y, z) that satisfy the equation p2 + (sec–1 y)2 + (tan–1 z)2 is 4 Find the number of solution of cos (2 sin–1(cot(tan–1(sec (6 cosec–1x))))) + 1 = 0; where x > 0. If the domain of the function (sin–1 x)2 = 24. 25. f(x) = p 5 (d) equals when 23. 3 1/ x æ1 ö tan -1 ç (tan 2 A) + tan -1 (cot A) + tan -1 (cot 3 A) ÷ , 2 è ø p 4 (d) lim æ π - tan -1 x ö ÷ø x ®¥ çè 2 Find the value of 3cos -1 ( 4x ) - p is [a, b], then the value of (4a + 64b) is _______. ANSWER KEY 1 2 3 (c) (c) (a) 4 5 6 (c) (c) (d) 7 8 9 (c) (c) (c) 10 11 12 (b) (b) (c) 13 14 15 (b) (d) (c) 16 17 18 (b) (a) (a) 19 20 21 (c) (a) (1) 22 23 24 (0) (2) (3) 25 (7) Matrices 39 18 MATRICES MCQs with One Correct Answer 1. The product of the matrices é cos 2 q A=ê êë cos q sin q cos q sin q ù ú and sin 2 q úû 4. (b) A (z) = A (x) [A (y)]–1 p 2 (c) 3. (d) A (z) = A (x) – A (y) (b) np (d) n p 2 5. é - 3 2 ù é1 0 ù é2 1ù A ê If ê ú= ê ú , then the ú ë 5 - 3û ë0 1 û ë 3 2û matrix A equals (a) (d) None of these (c) A (z) = A (x) A (y) (c) 2np 2. (c) I + P For each real x : – 1 < x < 1. Let A (x) be the matrix (a) A (z) = A (x) + A(y) q-f = (2n + 1) (b) 2I + P x+ y é 1 – xù (1 – x)–1 ê then and z = ú 1 + xy ë–x 1 û é cos2 f cos f sin f ù B=ê ú is a null matrix if êëcos f sin f sin 2 f úû (a) (a) I + 2p é1 1 ù ê1 0ú ë û (b) é1 0ù ê1 1 ú ë û (d) é1 1ù ê ú ë0 1û é0 1ù ê1 1ú ë û - tan (q / 2) ù é 0 If matrix P = ê ú , then find tan q / 2 0 ë û é cos q - sin q ù (I – P) ê ú ë sin q cos q û 6. é1 0 ù If A = ê1/ 2 1ú , then A50 is ë û é1 0 ù ê50 1 ú ë û (a) é1 0 ù ê0 50 ú ë û (b) (c) é1 25 ù ê0 1 ú ë û (d) None of these If A = [aij]m × n and aij = (i2 + j2 – ij) (j – i), n is odd, then which of the following is not the value of Tr(A) (a) 0 (b) | A | (c) 2| A | (d) None of these MATHEMATICS 40 7. é 3 ê If P = ê 2 ê 1 êë 2 1 ù ú é1 1ù 2 ú and A = ê ú and Q = 3ú ë 0 1û ú 2 û 10. 11. é u1 ê P = ê u2 êë u 3 PAPT and x = PTQ2005P then x is equal to 8. (a) é1 2005ù ê0 1 úû ë (b) é 4 + 2005 3 6015 ù ê ú 4 - 2005 3 ûú ëê 2005 (c) 1 é2 + 3 1 ù ú ê 4 êë - 1 2 - 3 úû (d) 1 é 2005 2 - 3 ù ú ê 4 ëê2 + 3 2005 ûú (a) 9. ± (b) ± (c) ± (d) ± 1 2 1 3 1 2 1 2 ,± 12. 13. ,± ,± ,± 1 6 1 2 1 6 1 2 ,± ,± ,± ,± 1 v3 é1ù é u1 v1 w1 ù é x ù ê ú ú êyú ê the equation ê u 2 v 2 w 2 ú ê ú = ê 1 ú , êë 5 úû êë u 3 v 3 w 3 úû êë z úû the value of y comes out to be –3. Then the value of m is equal to (a) 27 (b) 7 (c) – 27 (d) – 7 If A, B are two square matrices such that AB = A and BA = B, then (a) only B is idempotent (b) A, B are idempotent (c) only A is idempotent (d) None of these If A1, A3, ..., A2n–1 are n skew – symmetric matrices of same order, then (a) (b) (c) (d) 1 6 1 3 2 v2 2 1ù é 2 w1 ù 1 ê ú ú w 2 ú ; Q = ê 13 – 5 m ú 9 êë – 8 1 5 úû w 3 úû X = A1 + 3 A33 + ...(2 n - 1)(A 2n -1 )2 n -1 will be 2 1 v1 are such that PQ = I, an identity matrix. Solving é 0 2b c ù The value of a, b, c when êê a b - c úú is êë a - b c úû orthogonal, are Matrix A is such that A2 = 2A – I when I is unit matrix then An is equal to (a) nA – (n – 1) I (b) nA – I (c) 2n–1 A – I (d) none of these The matrices 14. symmetric skew-symmetric neither symmetric nor skew symmetric depends on ‘n’ is even or odd éa b c ù ê ú For A = ê d e f ú , a, b, c, d, e, f, g, h, i Î C, we ëê g h i ûú é 0 1 0ù If A = ê 0 0 1ú then A3 – rA2 – qA = ê ú êë p q r úû éa d g ù ê ú say A = êb e h ú and we say that A is the êc f i ú ë û (a) p I (c) r I Hermitian matrix if A = Aq. Suppose A is the Hermitian matrix such that A2 = O then (a) A = – A' (b) A = A' (c) A = O (d) A = I (b) q I (d) None of these q Matrices 41 15. A, B, C are three matrices of the same order such that any two are symmetric and the 3rd one is skew symmetric. If X = ABC + CBA and Y = ABC – CBA, then (XY)T is (a) symmetric (b) skew symmetric (c) I – XY (d) – YX écos q sin q ù é 1 0ù T 16. If A = ê ú , B = ê –1 1ú , C = ABA , sin q – cos q ë û ë û then ATCnA equals to (n Î I+) (a) é – n 1ù ê 1 0ú ë û (b) é1 – n ù ê0 1 ú ë û (c) é0 1 ù ê1 – n ú ë û (d) é 1 0ù ê – n 1ú ë û 21. éa b c ù If matrix A = ê b c a ú where a, b, c are real ê ú êë c a b úû positive numbers, abc = 1 and ATA = I, then find the value of a3 + b3 + c3. 22. Let z = -1 + 3i , where i = 2 -1 , and r, s Î é (- z)r {1, 2, 3}. Let P = ê 2s êë z z 2s ù ú and I be the z r úû identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P2 = –I is 17. Let A be a 3 × 3 matrix given by A = (aij)3 × 3. If for every column vector X satisfies X 'AX = 0 and a12 = 2008, a13 = 1010 and a23 = – 2012. Then the value of a21 + a31 + a32 = (a) – 6 (b) 2006 (c) – 2006 (d) 0 Numeric Value Answer é 1 0 0ù ê ú 23. Let P = ê 4 1 0 ú and I be the identity matrix êë16 4 1 úû of order 3. If Q = [qij] is a matrix such that P50 – Q =I, then q 31 + q32 equals 103 q 21 24. Let X be the solution set of the equation Ax = I; 18. If B, C are square matrices of order n and if A = B + C, BC = CB, C2 = 0, then for any positive integer N, A N + 1 = BK[B + ( N + 1) C], then K/N is 19. A and B are two square matrices such that A2B = BA and if (AB)10 = AkB10, then k is 20. If the product of n matrices é1 1 ù é1 2 ù é1 3 ù é1 n ù ê 0 1ú ê 0 1ú ê 0 1ú ... ê 0 1ú is equal to the ë ûë ûë û ë û é1 378 ù matrix ê ú , then the value of n is equal to ë0 1 û é2 where A = ê ë3 –1 ù and I is the corresponding – 2 úû unit matrix and x Î ¥ , then find the minimum ì np ü value of S(cos x q + sin x q), qΡ – í ; n ΢ ý . î2 þ é0 a ù éa b ù and (A + I)50 – 50A = ê 25. Let A = ê ú ú, ë0 0 û ëc d û find a + b + c + d. ANSWER KEY 1 2 3 (a) (a) (c) 4 5 6 (c) (d) (d) 7 8 9 (a) (c) (a) 10 11 12 (a) (d) (b) 13 14 15 (b) (c) (d) 16 17 18 (d) (c) (1) 19 20 21 (1023) 22 (27) 23 (4) 24 (1) (1) (2) 25 (2) 19 DETERMINANTS 1. MCQs with One Correct Answer The value of the determinant 5. 1 a a2 cos (n –1) x cos nx cos (n + 1) x is zero, if sin (n –1) x sin nx sin (n + 1) x (a) sin x = 0 2. Q = (cos(b - a ),sin b) and R = (cos(b - a + q),sin(b - q)) , where (b) cos x = 0 p 0 < a, b, q < . Then, 4 (a) P lies on the line segment RQ (b) Q lies on the line segment PR (c) R lies on the line segment QP (d) P, Q, R are non-collinear 1 + a2 (c) a = 0 (d) cos x = 2a If a, b, c, are the sides of a triangle ABC such a2 b2 (a) 0 (b) positive (c) negative (d) can not be determined Consider three points P = ( - sin(b - a ), - cos b), c2 that (a + 1) 2 (b + 1)2 (c + 1) 2 = 0 , then (a - 1)2 3. (c - 1)2 (a) DABC is non-isosceles right-angled triangle (b) D ABC is an equilateral (c) D ABC is an acute angled triangle with no two angles being equal (d) D ABC is an isosceles If a, b, c, are the sides of a triangle A B C and A, B, C are angles opposite to a, b, c, and a2 D = b sin A c sin A 4. (b - 1)2 6. 3r - 2 n -1 a 1 (n - 1)(3n - 4) 2 n -1 then the value of å D r r =1 cos A then, 1 (a) D = area of triangle (b) D = perimeter of the triangle (c) D = Sa2 (d) None of these If P, Q and R represent the angles of an acute angled triangle, no two of them being equal then 1 1 + sin P sin P(1 + sin P) 1 1 + sin Q sin Q(1 + sin Q) the value of is 1 1 + sin R sin R(1 + sin R) If D r = 2r - 1 1 n(n - 1) (n - 1)2 2 b sin A c sin A 1 cos A r n 2 7. (a) depends only on a (b) depends only on n (c) depends both on a and n (d) is independent of both a and n If a2 + b2 + c2 + ab + bc + ca £ 0 " a, b, c Î R, then value of the determinant, ( a + b + 2)2 a 2 + b2 1 (b + c + 2) c2 + a2 1 equals (a) 65 (c) 4(a2 + b2 + c2) 1 2 2 b + c2 (c + a + 2) 2 (b) a2 + b2 + c2 + 31 (d) 0 Determinants 8. 43 If A is matrix of order 3 such that |A| = 5 and B = adj A, then the value of |A 9. –1 T | ( AB) is equal to (where |A| denotes determinant of matrix A.AT denotes transpose of matrix A, A– 1 denotes inverse of matrix A. adj A denotes adjoint of matrix A) 1 (a) 5 (b) 1 (c) 25 (d) 25 If f(x), h(x) are polynomials of degree 4 and f ( x ) g ( x ) h( x ) a b c = mx4 + nx3 + rx2 + sx + t be p q r an identity in x, then f "'(0) – f "(0) g "'(0) – g "(0) h "'(0) – h "(0) a b c is p q r (a) 2(3n – r) (b) 2(2n – 3r) (c) 3(n – 2r) (d) None of these 1 1 1 10. If a b c = (a – b) (b – c) (c – a) (a + b + c), a 3 b3 c 3 where a, b, c are all different, then the determinant 1 1 1 ( x – a)2 ( x – b) 2 ( x – c) 2 ( x – b)( x – c ) ( x – c )( x – a) ( x – a)( x – b) vanishes when (a) a + b + c = 0 (b) x = 1 (a + b + c) 3 1 (a + b + c) (d) x = a + b + c 2 2 If t is real and l = t - 3t + 4 , then the number t 2 + 3t + 4 (c) x = 11. of solutions of the system of equations 3x – y + 4z = 3, x + 2y – 3z = – 2, 6x + 5y +lz = –3 is : (a) one (b) two (c) zero (d) infinite 12. The set of homoegeneous equations : tx + (t + 1) y + (t - 1) z = 0; (t + 1) x + ty + (t + 2) z = 0; (t - 1) x + (t + 2) y + tz = 0 has non-trivial solution for (a) three values of t (b) two values of t (c) one value of t (d) no value of t 13. If a, b, c are non-zeros, then the system of equations : (a + a) x + ay + az = 0; ax + (a + b) y + az = 0; ax + ay + (a + c ) z = 0 has a non-trivial solution if : (a) a -1 = -(a -1 + b-1 + c -1 ) (b) a -1 = a + b + c (c) a + a + b + c = 1 (d) None of these 14. If the system of equations ax + y + z = 0, x + by + z = 0 and x + y + cz = 0 (a, b, c ¹ 1) has a non-trivial solution, then the value of 1 1 1 + + is 1– a 1– b 1– c (a) –1 (b) 0 (c) 1 (d) None of these 15. If a, b, c are non-zeros, then the system of equations (a + a)x + ay + az = 0, ax + (a + b)y + az = 0, ax + ay + (a + c)z = 0 has a non-trivial solution if (a) a– 1 = – (a– 1 + b– 1 + c– 1) (b) a– 1 = a + b + c (c) a + a + b + c = 1 (d) None of these 16. If c < 1 and the system of equations x + y – 1 = 0, 2x – y – c = 0, and –bx + 3by – c = 0 is consistent, then the possible real values of b are 3ö æ (a) b Î ç – 3, ÷ è 4ø 3 (b) b Î æç – , 4 ö÷ è 2 ø 3 (c) b Î æç – ,3ö÷ è 4 ø (d) None of these 1 x+3 np ( -1)n n ! then cos 2 3n +1 3 a a5 x n -1 cos x 17. If f(x) = dn 0 a [f ( x )] x =0 = dx n (a) 1 (c) 0 a1 18. Suppose D = a2 a3 (b) –1 (d) None of these b1 c1 b2 c2 and b3 c3 a1 + pb1 b1 + qc1 c1 + ra1 D¢ = a2 + pb2 b2 + qc2 c2 + ra2 . Then a3 + pb3 b3 + qc3 c3 + ra3 MATHEMATICS 44 19. 20. (a) D¢ = D (b) D¢ = D(1 – pqr) (c) D¢ = D(1+p+q+r) (d) D¢ = D(1 + pqr) é x 2 –1ù –11ö ê ú æ If A = ê –1 1 2 ú ; çè x ¹ ÷ and det (adj(adj 3 ø êë 2 –1 1úû A)) = (14)4. Then the value of x is (a) 2 (b) – 2 (c) 0 (d) None of these 3 2 x é ù ê1 y 4 ú Matrix A = ê ú , if x y z = 60 and 8x + 4y + 3z êë 2 2 z úû = 20, then A(adj A) is equal to é88 0 0ù é 64 0 0ù ê ú ê ú (a) (b) ê 0 88 0ú 0 64 0 ê ú êë 0 0 88úû ëê 0 0 64ûú (c) é 68 0 0 ù ê 0 68 0 ú ê ú êë 0 0 68úû (d) 24. D = log an + 6 log an + 8 log an + 10 is equal to: 25. value of (1 + aq)2 (1 + bq)2 (1+ cq)2 x 22. 23. 1 2 3 4 The value of the determinant æ 3p ö æ 5p ö æ 7p ö sin2 ç x + ÷ sin2 ç x + ÷ sin2 ç x + ÷ è è è 2ø 2ø 2ø æ 5p ö sin ç x + ÷ è 2ø 7p ö æ sin ç x + ÷ is è 2ø æ 3p ö sin ç x - ÷ è 2ø æ 5p ö sin ç x - ÷ è 2ø 7p ö æ sin ç x - ÷ è 2ø ìæ a ü a ö Let S = íç 11 12 ÷ : aij Î{0,1, 2}, a11 = a22 ý îè a21 a22 ø þ Then the number of non-singular matrices in the set S is : é3; when $i = $j Let A = [aij]3 × 3 be such that aij = ê $i ¹ $j , ëê 0; ì det (adj(adj A)) ü then í ý equals: (where {×} 5 î þ denotes fractional part function) (a) (d) (d) (d) 5 6 7 8 (d) (d) (a) (b) 9 10 11 12 (a) (b) (a) (c) 13 14 15 16 (a) (c) (a) (c) is (1 + ar )2 (1 + br )2 (1 + cr )2 27. æ 3p ö sin ç x + ÷ è 2ø log an + 12 log an + 14 log an + 16 Area of triangle whose vertices are (a, a2) (b, b2) 1 (c, c2) is , and area of another triangle whose 2 vertices are (p, p2) , (q, q2) and (r, r2) is 4, then the (1 + ap)2 (1 + bp)2 (1+ cp)2 Numeric Value Answer 21. log an + 2 log an + 4 log a n 26. é34 0 0 ù ê 0 34 0 ú ê ú êë 0 0 34 úû If a1, a2, a3, ..., an are in G.P. and ai > 0 for each i, then the determinant 28. 29. 30. sin x cos x If f ( x) = x 2 - tan x 2 x sin 2 x - x3 then lim x ®0 5x f ¢( x) x is equal to If a, b are the roots of the equation x2 – 2x + 4 = 0, find the value of a+b a2 + b2 a 3 + b3 a 2 + b2 a 3 + b3 a4 + b4 a 3 + b3 a 4 + b 4 a 5 + b5 Find the co-efficient of x in the expansion of the determinant (1 + x) 2 (1 + x )4 (1 + x)6 (1 + x )3 (1+ x)6 (1 + x) 9 . (1 + x) 4 (1 + x )8 (1 + x)12 éa b ù Suppose A = ê ú is a real matrix with nonëc d û zero entries, ad – bc = 0 and A2 = A. Then, a + d equal to The number of integers x satisfying é1 x -3x 4 + det ê1 x 2 ê1 x3 ë ANSWER KEY 17 (c) 21 (0) 25 18 (d) 22 (20) 26 19 (d) 23 (0.20) 27 20 (c) 24 (0) 28 x2 ù x 4 ú = 0 is equal to x 6 úû (16) (4) (0) (0) 29 30 (1) (2) 20 CONTINUITY AND DIFFERENTIABILITY MCQs with One Correct Answer 1. 2 The function f ( x ) = (sin 2 x ) tan 2 x is not p p defined at x = . The value of f æç ö÷ so that f 4 è 4ø is continuous at x = (a) (c) 2 2. e 4. æ ç ç f ( x) = ç ç ç ç è p is 4 (b) 1 e (d) None of these integer function] (a) Continuous at x = –1 (b) Continuous at x = 0 1 (c) Discontinuous at x = 2 (d) All of these Suppose 'f ' is continuous function from R to R and f ( f (a)) = a for some a Î R then the equation f (x) = x has (a) no solution (b) exactly one solution (c) at most one solution (d) atleast three solutions Let g (x) be a polynomial of degree one and f (x) ìï g ( x), x £ 0 . If f (x) is be defined by f ( x) = í sin x ,x>0 ïî x , p <x<0 2 x=0 , 0< x< (cos x - sin x)cosec x , a 1/ x +e ae 2/ x e 2/ x +e + be 3/ x 3/ x p 2 If f (x) is continuous at x = 0, then (a, b) = ì1 - [ x] ï , x ¹ -1 If f (x) = í 1 + x , then the value of ïî1 , x = -1 f ( 2 k ) will be (where [] shows the greatest 3. 5. continuous satisfying f '(1) = f (–1), then g (x) is (a) (1 + sin 1) x +1 (b) (1– sin1) x + 1 (c) (1– sin 1) x –1 (d) (1 + sin 1) x – 1 Let f (x) be defined as follows : (a) 6. æ 1ö çè e, ÷ø e æ1 ö e (b) ç , e÷ è ø (c) (e, e) (d) The function f defined by ( e -1 , e -1 ) ìï (1 + sin p x )t - 1 üï f ( x) = lim í ý is t t ®¥ ï x (1 + sin p ) + 1 î þï (a) every where continuous (b) discontinuous at all integer values of x (c) continuous at x = 0 (d) None of these 7. Let f : R ® R be a function defined by f (x) = min {x + 1, x + 1} ,then which of the following is true? (a) f (x) is differentiable everywhere (b) f (x) is not differentiable at x = 0 (c) f (x) ³ 1 for all x Î R (d) f (x) is not differentiable at x = 1 MATHEMATICS 46 8. Let g ( x) = ( x – 1) n log cos m ( x – 1) ; 0 < x < 2, m and n 14. æ x ®1+ 9. 10. 11. 12. 15. Let f: R ® R be a function such that f ( x ) £ x 2 , for all x Î R . Then, at x = 0, f is: (a) continuous but not differentiable. (b) continuous as well as differentiable. (c) neither continuous nor differentiable. (d) differentiable but not continuous. æ xx – x – x ö If f (x) = cot–1 ç ÷ , then f ¢ (1) is 2 è ø (a) – 1 (b) 1 (c) log 2 (d) – log 2 1 If g is the inverse of f and f ' (x) = , then 1 + x3 find g' (x). (a) 1 + [g(x)]2 (b) 1 – [g(x)]2 3 (c) 1 + [g(x)] (d) 1 – [g(x)]3 æp 2ö If f ( x) = sin ç [ x] - x ÷ , where 2 < x < 3 and 3 è ø [ . ] represetns greatest integer function, then 13. 5p 3 (c) -2 5p 3 (b) 2 5p 3 (d) - 5p 3 16. (a) 1 (c) x n -1 y n -1 (b) x/y (d) None of these f ''(0) f n (0) + .... + is 2! n! (a) n (b) 2n (c) 2n– 1 (d) None of these If a and b are any two roots of equation ex cos x = 1, then the equation ex sin x – 1 = 0 has (a ) exactly one root in (a, b) (b) exactly two roots in (a, b) (c) at least one root in (a, b) 17. 18. (d) no root in (a, b) Let f(x) = x |sin x|, x Î R. Then (a) f is differentiable for all x, except at x = np, n = 1, 2, 3, ...., (b) f is differentiable for all x, except at x = np, n = ±1, ±2, ±3, ... (c) f is differentiable for all x, except at x = np, n = 0, 1, 2, 3 (d) f is differentiable for all x, except at x = np, n = 0, ± 1, ± 2, ± 3, ... f : [–1, 1] ® R be a function defined by ì 2 æp ö ï x cos ç ÷ f ( x) = í è xø ï 0 î for x ¹ 0 for x = 0 The set of points where f is not differentiable is (a) {x Î [–1, 1] : x ¹ 0} (b) If 1 - x 2n + 1 - y 2n = a( x n - y n ) , then 1 - x 2 n dy is equal to 1 - y 2 n dx (a) f (x, y) = y /x (b) f (x, y) = y2 / x2 2 2 (c) f (x, y) = 2y / x (d) f (x, y) = x2 / y2 If f (x) = (1 + x)n, then the value of f (0) + f '(0) + æ 5p ö f ¢ç is equal to ç 3 ÷÷ è ø (a) 6ö è1- x ø hand derivative of |x – 1| at x = 1. If lim g(x) = p, (b) n = 1, m = – 1 (d) n > 2, m = n dy 3 3 (1 - x 6 ) + (1 - y 6 ) = a(x – y ), and dx = f (x, y) ç 1 - y6 ÷ , then are integers, m ¹ 0, n > 0 , and let p be the left then (a) n = 1, m = 1 (c) n = 2, m = 2 If 2 ì ü , n ÎZ ý í x Î[ -1,1] : x = 2n + 1 î þ (d) [–1, 1] Let f : R ® R be a continuous function such that f (x2) = f (x3) for all x Î R. Consider the following statements. (c) 19. 2 ì ü , n ÎZý í x Î[-1, 1] : x = 0or x = 2 n + 1 î þ Continuity and Differentiability 47 I. f is an odd function. II. f is an even function. III. f is differentiable everywhere Then (a) I is true and III is false (b) II is true and III is false (c) both I and III are true (d) both II and III are true 20. Let f : R ® R be function defined by 26. If y = (1 + 1/ x ) x then find the value of 2 y2 (2) + 1/ 8 log 3/ 2 - 1/ 3 27. If y = ( x - sin x ) + ( x - sin x ) + ..., then 2 ( ) if x ¹ 0, ì sin x 2 ï f (x) = í x ï î 0 dx - 2p = ......... dy x = p 2 if x = 0, Then, at x = 0, f is (a) not continuous (b) continuous but not differentiable (c) differentiable and the derivative is not continuous (d) differentiable and the derivative is continuous Numeric Value Answer 28. ìa cot x b ïï x + x 2 , 0 <| x |£ 1 Let f ( x) = í ï1 x=0 ïî 3 If f (x) is continuous at x = 0, then the value of a2 + b2 is 29. Number of points where function f (x) defined as f :[0, 2p ] ® R, f ( x) 21. If f is a real valued differentiable function satisfying | f (x) – f (y) | £ ( x - y )2 , x, y Î R and f (0) = 0, then f (1) is equal to 22. If x = cosec q - sin q ; y = cosecn q- sin n q , ì 1 1 , | sin x |< ï3 - cos x 2 2 ï =í ï 2 + cos x + 1 , | sin x |³ 1 ï 2 2 î 2 æ dy ö and ( x 2 + 4) ç ÷ - n 2 y 2 = kn2, then value of è dx ø k is 23. If f (x) = cos x cos 2x cos 22 x cos 23 x ..... æ pö cos 2 n–1 x and n > 1, then f ' çè 2 ÷ø is non-differentiable is 30. If the derivative of the function ì 1 ü f ( x ) = cos-1 í (2 cos x - 3sin x) ý î 13 þ ì 1 ü + sin -1 í (2 cos x + 3sin x) ý î 13 þ 24. If f '' (x) = – f (x) and g (x) = f ' (x) and 2 25. 1 2 3 4 2 æ æ x ö ö æ æ x öö F ( x) = ç f ç ÷ ÷ + ç g ç ÷ ÷ and given that è è 2 ø ø è è 2 øø F (5) = 5, then F (10) is equal to Number of functions f : [0, 1] ® [0, 1] satisfying | f (x) – f (y) | = |x – y| for all x, y in [0, 1] is (b) (d) (d) (b) 5 6 7 8 (b) (b) (a) (c) 9 10 11 12 (b) (a) (c) (b) 13 14 15 16 (c) (d) (b) (c) is 3 10 w.r.t. 1 + x 2 at x = is , then b = 4 b ANSWER KEY 17 (b) 21 18 (c) 22 19 (d) 23 20 (d) 24 (0) (4) (1) (5) 25 26 27 28 (2) (3) (3) (2) 29 30 (4) (3) 21 APPLICATION OF DERIVATIVES MCQs with One Correct Answer 1. The function f (x) = tan–1(sin x + cos x) is an 5. increasing function in 2. (a) æ pö çè 0, ÷ø 2 æ p pö (b) ç - , ÷ è 2 2ø (c) æ p pö çè , ÷ø 4 2 p p (d) æç - , ö÷ è 2 4ø 6. The greatest of the numbers 1, 21 / 2 , 31 / 3 , 41 / 4 , 51 / 5 , 61 / 6 and 71 / 7 is (a) 3. 21 / 2 (b) 31 / 3 (d) All but 1 are equal (c) 71 / 4 The interval of decrease of the function 7. f ( x) = x 2 log 27 - 6 x log 27 2 + (3x - 18 x + 24) log( x - 6 x + 8) is 4. (3 - 1 + 1/ 3e , 2) È (4,3 + 1 + 1/ 3e ) (b) (3 - 1 + 1/ 3e , 3 + 1 + 1/ 3e ) (c) (-¥, 3 - 1 + 1/ 3e ) È (3, 4 + 1 + 1/ 3e ) (d) None of these. The set of positive values of the parameter ‘a’ for each of which the function f ( x) = sin 2 x - 8(a + 1)sin x - (4a 2 + 8a - 14) x is monotonic increasing in R and has no critical points are (a) (0, 6 - 2) (b) (-2 - 6, 6 - 2) (c) (-2 - 6, 0) (d) None of these æ p pö Let the function g : ( -¥, ¥) ® ç - , ÷ be è 2 2ø p -1 u given by g (u ) = 2 tan (e ) - . Then, g is 2 (a) even and is strictly increasing in (0, ¥ ) (b) odd and is strictly decreasing in ( -¥, ¥) (c) odd and is strictly increasing in ( -¥, ¥) (d) neither even nor odd, but is strictly increasing in ( -¥, ¥) The largest term in the sequence, an = 2 (a) x x and g(x) = , where 0 < x £ 1, sin x tan x then in this interval (a) both f(x) and g(x) are increasing functions (b) both f(x) and g(x) are decreasing functions (c) f(x) is an increasing function (d) g(x) is an increasing function. If f (x ) = n2 n3 + 200 (a) a 6 (c) a 8 8. 9. is (b) a 7 (d) None of these x2 + 1 [ x] for all x Î [1, 4], where [.] denotes the greatest integer function. then, f (x) is monotonically. (a) increasing on [1, 4) (b) decreasing on [1, 4) (c) increasing on [1, 2) (d) decreasing on [2, 3) If A > 0, B > 0 and A + B = p/3, then the maximum value of tan A tan B is 1 1 (a) (c) 3 (d) (b) 3 3 3 Let f (x) be a function given by f ( x) = Application of Derivatives 49 10. The set of values for which the function x x f ( x) = (4a - 3)( x + ln 5) +2( a - 7) cot sin 2 2 2 does not possess critical points is (a) 4ö æ ç -¥, - ÷ È (2, ¥) 3ø è (b) (-¥, 2) (c) [1, ¥) (d) (1, ¥) 11. A given right circular cone has a volume p, and the largest right circular cylinder that can be inscribed in the cone has a volume q. then p : q is (a) 9 : 4 (b) 8 : 3 (c) 7 : 2 (d) None of these 12. 2 x ¹ -2 ïì 2 - | x + 5 x + 6 |; f ( x) = í 2 ; x = -2 ïî a + 1 then the range of a is so, that f(x) has maxima at x = – 2 is (a) | a |³ 1 (b) | a |< 1 (c) a >1 (d) a <1 p p <x< . 2 2 In order that f ( x) has exactly one minimum, 15. Two men P and Q start with velocities v at the same time from the junction of two roads inclined at 45o to each other. If they travel by different roads, the rate at which they are being separated is (a) v 2 (b) v 2+ 2 (c) v 2- 2 (d) v/ 2 16. The general value of a such that the line x cos a+ y sin a = p is a normal to the curve (x + a) y = c2 is p 3p (a) æç 2np + ,(2n +1)p ö÷ È æç 2np + ,(2n + 2)p ö÷ 2 2 è ø è ø (b) 3p ö æ ç 2np + p , 2np + ÷ 2 ø è (c) 3p æ ö ç 2np + ,(2n + 2)p ÷ 2 è ø (d) p ö æ ç 2np , 2np + , ÷ È 2 ø è 13. Let f ( x) = sin 3 x + l sin 2 x , - 3p ö (n Î I ) æ ,÷ ç (2n + 1)p , 2np + 2 ø è l should belong to (a) (–1, 1) æ 3 ö æ 3ö (b) ç - , 0 ÷ È ç 0, ÷ è 2 ø è 2ø æ 3 1 ö æ 1 3ö (d) ç - , - ÷ Èç , ÷ è 2 2 ø è 2 2ø p 14. The function f (x) = 1 + x (sin x) [cos x], 0 < x £ 2 (where [ . ] is G.I.F.) (c) (0, ¥) (a) is continuous on æç 0, p ö÷ è 2ø æ pö (b) is strictly increasing in ç 0, ÷ è 2ø (c) is strictly decreasing in æç 0, p ö÷ è 2ø (d) has global maximum value 2 17. Let P be a point on the hyperbola x 2 - y 2 = a 2 , where a is a parameter, such that P is nearest to the line y = 2x. Then the locus of P is (a) y = 2x (b) y = x (c) 2y = x (d) x + y = 0 18. For the curve y = 3 sinq cosq, x = eq sin q, 0 £ q £ p, the tangent is parallel to x-axis when q is: (a) 3p 4 (b) p 2 (c) p 4 (d) p 6 19. Let C be the curve y3 – 3xy + 2 = 0. If H is the set of points on the curve C, where the tangent is parallel to x-axis and V is the set of points on C where the tangent is parallel to y-axis, then MATHEMATICS 50 20. (a) H = {(x, y): y = 0, x Î R}, V = {(1, 1)} (b) H = {(x, y) : x = 0, y Î R}, V = {(1, 1)} (c) H = f, V = {(1, 1)} (d) H = {(1, 1)} , V = {(x, y) : y = 0, x Î R} The number of polynomials p : R ® R satisfying p(0) = 0, p(x) > x2 for all x ¹ 0 and p¢¢(0) = 25. 26. f ( x ) = (b2 - 3b + 2)(cos 2 x - sin 2 x) 1 is 2 27. (a) 0 (b) 1 (d) infinite 28. Numeric Value Answer 22. The fuel charges for running a train are proportional to th e square of the speed generated at 16 miles per hour and costs ` 48 per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to ` 300 per hour is The curve x + y – ln (x + y) = 2 x +5 has a vertical tangent at the point (a, b). Then a + b is equal to 23. 24. 1 2 3 +(b - 1) x + sin(b + 1) does not possess stationary point is If the curves ax2 + by2 = 1 and a1x2 + b1y2 = 1 may cut each other orthogonally such that æ1 1 ö 1 1 - = l ç - ÷ then l is equal to a a1 è b b1 ø (c) more than 1, but finite 21. Let P be a non-zero polynomial such that P(1 + x) = P(1- x) for all real x, and P(l) = 0. Let m be the largest integer such that (x - l)m divides P(x) for all such P(x). Then m is equal to The integral value of b for which the function Two ships A and B are sailing straightaway from a fixed point O along routes such that ÐAOB is always 120°. At a certain instance, OA = 8 km, OB = 6 km and the ship A is sailing at the rate of 20 km/hr while the ship B sailing at the rate of 30 km/hr. If the distance between A and B is k changing at the rate km/hr, then value of k 37 is _______. Tangent at P1 (2,3) on the curve 3 y = x3 + 1 meets the curve again at P2 . The tangent at P2 meets the curve at P3 and so on. If the sum of the ordinates for P1, P2 , P3 ,.....P60 be S then æ 2183 - 8 ö S +ç ÷ is equal to 5k, when k is equal to è 27 ø 29. A conical vessel is to be prepared out of a circular sheet of metal of unit radius. In order that the vessel has maximum volume, the sectorial area that must be removed from the sheet is A1 and the area of the given sheet is A. If A = m + n , then m + n is equal to A1 30. An arch way is in the shape of semi-ellipse, the road level being the major axis. If the breadth of the road is 30 m and the height of the arch is 6 m at a distance of 2 m from the side, the greatest height of the arch in metres, is A rectangle with its sides parallel to the X-axis and Y-axis is inscribed in the region bounded by the curves y = x2 – 4 and 2y = 4 – x2. The maximum possible area of such a rectangle is (d) (b) (a) 4 5 6 (a) (c) (c) 7 8 9 (b) (c) (b) 10 11 12 (a) (a) (a) ANSWER KEY 13 (b) 16 (a) 14 (a) 17 (c) 15 (c) 18 (c) 19 20 21 (c) 22 (1) 25 (a) 23 (260) 26 (40) 24 (9.22) 27 (2) (2) (1) 28 29 30 (4) (9) (6) 22 INDEFINITE INTEGRATION MCQs with One Correct Answer 1. If then value of (A, B) is (a) (- cos a, sin a) (b) (cos a, sin a) (- sin a, cos a) (sin a, cos a) (c) 2. sin x ò sin( x - a) dx = Ax + B log sin( x - a) + C , (d) If f ( x ) = lim [2 x + 4 x3 + ......... + 2nx 2 n -1 ] ; n®¥ (0 < x < 1), then ò ( f ( x )) dx is equal to (a) (c) 3. 4. - 1 - x2 + c 1 x2 - 1 (b) +c (d) 1 1 - x2 1 1 - x2 +c (a) 1 + nx n -1 - x 2 n (1– x n ) 1 - x 2 n (a) æ 1 - xn ö ex ç ÷ +C è 1 + xn ø (c) æ 1+ xn ö ex ç ÷ + C (d) çè 1 - x n ÷ø (b) 1 æ g ( x) ö ç ÷ +C 2 è f ( x) ø (c) æ g ( x) ö ö 1æ ç log ç ÷÷ +C 2è è f ( x) ø ø (d) æ æ g ( x ) ö2 ö log ç ç ÷ ÷+C ç è f ( x) ø ÷ è ø 2 (b) æ 1 + xn ö ex ç ÷ +C è 1 - xn ø æ 1 - xn ö ex ç ÷ +C çè 1 + x n ÷ø If f ( x ) = lim n 2 ( x1/ n - x1/( n +1) ) , x > 0 then n ®¥ 6. (a) x2 + ln x + C 2 (b) - x2 x2 ln x + +C 4 2 (c) x3 + xl n x + C 3 (d) x2 x2 ln x +C 2 4 If f ( x ) = lim xn - x -n n®¥ x n + x - n ò (sin 2 dx = ò xf ( x) dx is equal to +c æ f ( x ) g '( x ) - f '( x ) g ( x ) ö ÷ (log ( g ( x )) ò çè f ( x) g ( x) ø - log ( f ( x))) dx is equal to æ g ( x) ö log ç ÷+C è f ( x) ø 5. x òe -1 , 0 < x < 1 , n Î N then x ) f ( x)dx is equal to (a) -[ x sin -1 x + 1 - x 2 ] + C (b) x sin -1 x + 1 - x 2 + C (c) x2 +C 2 (d) 1 (sin -1 x )2 + C 2 MATHEMATICS 52 7. x æ ln a a x / 2 ln bb ö ç x + ò ç 3a5 x / 2b3 x 2a 2 x b 4 x ÷÷ dx (where è ø (c) + log{log e e 3 x 3 } + C + a, b Î R ) is equal to (a) (b) (c) (d) 8. (d) a 2 x b3 x 2 x 3x a b +k ln e 6 ln a 2b3 1 1 1 2 3 2 x 3x 6 ln a b a b 1 1 ln 1 2x 3x ea b +k 11. ln(a 2 x b 3 x ) + k 6 ln a 2b 3 a 2 x b3 x 1 1 ln(a 2 x b3 x ) + k 2 3 2 x 3x 6 ln a b a b n If I n = ò (sin x + cos x ) dx (n - 1) (sin x + cos x )n -1 I n -2 , (sin x – cos x) +2 = n n then I5 equals (t = (sin x + cos x)) (sin x - cos x) 5 [t - 3t 3 + 8t - 32] + C (a) 15 (b) 12. 10. + log{loge e 4 x} + C (b) 1 log{loge x} - log{loge x} 2 + log{loge x} + C (a) 2 x -a +C a -b b - x (b) 2 ( x - a) (b - x) + C a -b If æp ö tan ç - x ÷ è4 ø ò cos2 x tan 3 x + tan 2 x + tan x dx. = –2 tan–1 u + c then u is equal to (sin x - cos x) 4 [3t + 8t 2 + 32] + C 15 (d) None of these sin 2q æ cos q + sin q ö 1 = ´ loge ç ÷ - f ( x) + c 2 è cos q - sin q ø 2 where f (x) is (a) sec 2q (b) log e (sec 2q) (c) 2 tan–1q (d) tan 2q 1 ò x{logex e × log e2 x e × loge3 x e} dx is equal to 1 (a) log{log e e 2 x} - log{loge e3 x} 2 dx ò (x - b) (x - a)(b - x ) is a-b ( x - a) b - x + C 2 (d) None of these (sin x - cos x ) 5 [t + 3t 4 + 8t 3 + 32t + 1] + C 15 æ cos q + sin q ö ò cos 2q loge çè cos q - sin q ÷ø d q 1 log{loge ex} - log{loge e 2 x} 2 1 + log{log e e 3 x} + C 2 (c) (c) 9. 1 log{log e ex} - log{loge e 2 x 2 } 2 13. ( a) 1 + tan x + cot x (b) (c) tan x + cot x If 1 + tan x + tan 2 x (d) tan -1 (tan x + cot x) dx ò (a + bx2 ) b - ax 2 = K tan–1 ( L tan q) + M , M being constant of integration then KL is equal to 1 1 (a) 1 (b) (c) a (d) a a a 14. If ò f ( x)sin x cos xdx = 1 2(b - a 2 ) then f (x) is (a) (b) (c) (d) 1 a sin x + b cos x 1 a 2 sin 2 x + b2 cos2 x 1 a 2 sin x + b2 cos x 1 a sin 2 x + b cos2 x 2 l n f ( x) + c, Indefinite Integration 53 15. The integral æ sec 6α sec18α sec54α ö ò çè cosec 2α + cosec 6α + cosec18α ÷ø dα is equal to ln | sec 54a | ln | sec 2a | +c (a) 108 4 ln | sec 6a | ln | sec18a | ln | sec54a | + + (b) +c 6 18 54 1 sec 6α 1 sec18α ln + ln 6 cosec 2α 18 cosec 6α (c) 1 sec 54α + ln +c 54 cosec18α cos x + ax + b l n | 2 cos x - sin x | + c, 2 cos x - sin x then value of |a + b| is ______. = 20. If ò 21. 16. If I n = ò (sin x + cos x)n dx (n ³ 2), then 22. n (a) (sin x + cos x) (sin x - cos x ) (b) (sin x + cos x) n -1 (sin x - cos x) (c) (sin x + cos x)n +1 (sin x - cos x) x ®0 f ( x) x2 exists finitely 3 = e , and ò f ( x) log e x dx is equal to ax 3 ( log e x - b ) + c , then value of a + b is _______. 1 2 3 cos 2 x + sin 2 x (2 cos x - sin x )2 (b) (d) (c) 4 5 6 (c) (d) (a) 1 log e | (sin x ) k + (cos x ) k | + c k then k is equal to q sin3 dq 2 If q 3 cos cos q + cos 2 q + cos q 2 ò = tan -1 f (q) + c then the least value of f(q) for allowable values of q is equal to Let f(x) be a function such that f ( xy ) = f ( x) × f ( x ) " x, y Î R. If f (1 + x) = 1 + x(1 + g(x)) where f ( x) xk dx = + C, f '( x ) k t ®0 then evaluate the numerical quantity k. 23. 99 ò {sin (101x) × sin x}dx = then 24. If ò sin(100 x)(sin x) l , m l is equal to .......... m dx 3 cos x - sin3 x = A tan -1 f ( x ) + B ln 2 + f ( x) +C 2 - f ( x) where f (x) = sin x + cos x find the value of (12 A + 9 2 B) - 3. 1/ x f ( x) ö æ lim ç1 + x + x ÷ø x ®0 è 19. If ò dx lim g ( x) = 0. Further if ò (d) (sin x - cos x )n -1 (sin x + cos x ) 17. Let ln x denote the logarithm of x with respect to the base e. Let S Ì R be the set of all points where the function ln (x2 –1) is well-defined. Then, the number of functions f : S ® R that are differentiable, satisfy f ¢ (x) = ln (x 2 – 1) for all x Î S and f (2) = 0, is (a) 0 (b) 1 (c) 2 (d) infinite Numeric Value Answer 18. If lim tan x + (cot x )2011 = (d) None of these nI n - 2(n - 1) I n- 2 is equal to 1 - (cot x )2010 dx 7 8 9 (b) (c) (b) 10 11 12 25. If ò sin8 x sin 7 x sin 5 x dx = + cos x a b sin 3 x sin x æp xö + + ln tan ç + ÷ + C , c d è 4 2ø then evaluate (a + b + c + d). ANSWER KEY (d) 13 (c) 16 (a) 14 (b) 17 (a) 15 (a) 18 - (b) (d) (1) 19 (0.60) 22 20 (2012) 23 (3) 21 24 (2) (1) (8) 25 (4) 23 DEFINITE INTEGRATION MCQs with One Correct Answer 1. Let p (x) be a polynomial of least degree whose graph has three points of inflection (–1, –1), (1, 1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissa at an ±1 (c) ±2 p/2 5. If 1 minimum value of a cos x + b sin x is (a) – 4 (b) – 8 (c) – 2 (d) -2 2 0 3 3 +4 14 (c) 3+ 7 14 p 2. (b) 3 3 7 (d) 3+2 7 p/3 6. Let I = 2 sin x dx , then I belongs to x æ 3 2ö ç ÷ , ç 8 6 ÷ è ø æ 2 (b) ç , ç 2 è 3ö ÷ 2 ÷ø ò (cos px - sin qx) dx; where p, q are integers; (a) is equal to (a) – p (b) 0 The value of æ1 2ö (d) None of these çç , ÷ 2 ÷ø è2 Let f(x) be positive, continuous and differentiable on the interval (a, b) and 2p ò (c) p (d) 2p the definite integral (a) 0 (b) p (c) 2p (d) ep If p, q, r, s are in arithmetic progression and p + sin x q + sin x f ( x) = q + sin x r + sin x p - r + sin x -1 + sin x r + sin x s - q + sin x s + sin x 2 such that ò f ( x )dx = -4 , then the common 0 difference of the progession is (c) 7. ecos q cos q(sin q)d q is 0 4. ò p/4 -p 3. p dx ò a2 cos2 x + b2 sin2 x = 16 . Then 0 angle of 60°. Then ò p ( x)dx is equal to (a) 1 2 (d) None of these (b) (a) lim f ( x ) = 1 , lim f ( x ) = 31/ 4 . If x®a + x ®b – f '( x ) ³ f 3 ( x ) + 1 then the greatest value f ( x) of b – a is (b) 31 / 4 (a) 1 (c) (31/ 4 - 1 ) p 24 (d) p 24 Definite Integration 8. 55 Let a, b, c be non-zero real numbers such that 13. Suppose the limit L = lim 1 n®¥ 8 2 ò (1 + cos x)(ax + bx + c) dx 0 exists and is larger than 2 = ò (1 + cos8 x)(ax2 + bx + c) dx . ax 2 + bx + c = 0 has (a) no root in (0, 2) (b) at least one root in (0, 2) (c) a double root in (0, 2) (d) two imaginary roots x 0 2 x | f '(w) | where w is a complex cube root of unity is (b) 3 3 3 2 3 (c) (d) 4 3 sec 2 x lim p x® 4 (a) (c) 2 p2 x 16 equals 2 8 f (2) p 2 æ 1ö fç ÷ p è 2ø Then f (l) equals (a) e (b) e 2 (c) e 4 (d) e 6 1 x dx. 8 0 1+ x Consider the following assertions: 15. Let J = ò I. J> 1 4 II. J< p 8 Then (a) only I is true (b) only II is true (c) both I and II are true (d) neither I nor II is true ò f (t )dt 10. 1 . Then, 2 f (x) = 2ò tf (t )dt + 1 for all x ³ 0. Let f ( x ) = ò (t - 1)dt . Then the value of (a) dx 1 <L<2 (b) 2 < L < 3 2 (c) 3 < L < 4 (d) L > 4 14. Suppose a continuous function f : [0, ¥) ® R satisfies Then the quadratic equation x 1 1 0 (1 + x 2 ) n (a) 0 9. nò 2 f (2) p (b) (d) 4 f (2) p 11. The value of the integral ò0 (1- | sin8 x |)dx is (a) 0 (b) p –1 (c) p –2 (d) p –3 12. Let S be the set of real numbers p such that there is no non-zero continuous function f : R ® R satisfying ò0 f ( t )dt = p f (x) for all x Î R. Then, x S is (a) the empty set (b) the set of all rational numbers (c) the set of all irrational numbers (d) the whole set R. x 16. For x ÎR, let f (x) = | sin x | and g (x) = ò f (t ) dt. 0 2 Let p(x) = g(x) - x. Then p (a) p(x + p) = p(x) for all x (b) p(x + p) ¹ p(x) for at least one but finitely many x (c) p(x + p) ¹ p(x) for infinitely many x (d) p is a one-one function 17. Let f: [0, 1] ® [0, 1] be a continuous function such that 1 x2 + (f(x))2 £ 1 for all x Î [0, 1] and ò f ( x ) dx = 1 2 Then 0 f ( x) ò 1 - x2 dx equals 1 2 p . 4 MATHEMATICS 56 (a) (c) p 12 2 -1 p 2 (b) p 15 (d) p 10 22. x x ò ò x (1 - t ) f (t )dt = t f (t )dt "x Î (0, ¥) and 0 Numeric Value Answer 18. Let f : (0, ¥) ® R be a differentiable function such that Let f : R ® R be a differentiable function and 23. f ( x) ò 2t dt f (1) = 4. Then the value of lim x ®1 4 f(1) = 1. The value of the limit lim f ( x ) is x ®¥ equal to If p and q are different roots of the equation 1 tan x = x then , if x -1 0 0 f ' (1) = 2 is 19. If 3 24. 2 sin x sec x x -1 f ( x ) = cosec x x sin x cos x tan x x tan x x2 + 1 ò f (x)dx= -p / 3 é1 x 2 1 1 ù lim ê 5 ò e-t dt - 4 + 2 ú = x®0 ê x 3x úû x ë 0 If I (n) = x Î [–1, 2] and G(x) = 4 5 6 ò q sin qd q, n Î N , n > 3, then n (a) (a) (a) 7 8 9 (d) (b) (c) 10 11 12 ò t | f ( f (t)) | dt for all F ( x) 1 = , then the value of x ®1 G ( x ) 14 xÎ[–1, 2]. If lim 1 [2010 I (2010) -2009 I (2008)]-1 is 1005 equal to (a) (d) (c) –1 –1 0 1 2 3 25. greatest integral value of x. Let f : ¡ ® ¡ be a continuous odd function, which vanishes exactly at one point and x 1 f (1) = . Suppose that F(x) = ò f (t )dt for all 2 x p/2 21. Find the value ò éë x ùû dx , when [x] is the 0 then p/3 20. ò 2 sin( pt )sin(qt )dt is equal to æ 1ö f ç ÷ is è 2ø ANSWER KEY (a) 13 (a) 16 (c) 14 (a) 17 (d) 15 (a) 18 (a) 19 (0) 22 (a) 20 (0.30) 23 (16) 21 (2) 24 0 0 (2) 25 (7) 24 APPLICATION OF INTEGRALS 5. MCQs with One Correct Answer 1. points ( x1, y1 ) and (x2, y2) where (x1 < x2), then The area bounded by the x-axis, the curve y = f(x) and the lines x = 1, x = b, is equal to x2 2ö æ the value of m for which ò ç mx + 2 - x ÷ dx is 2ø x è b 2 + 1 - 2 for all b > 1, then f(x) is (a) x -1 (b) (c) 2 (d) x +1 1 x +1 minimum, is x 1+ x 2 2 If the line y = mx + 2 cuts the parabola 2y = x2 at 6. 2 (b) (a) 2 The area 8 1 (c) (d) 0 3 3 bounded by the curve 2. Area of region bounded by [ x ] = [ y ] f ( x) = x + sin x and its inverse function between the ordinates x = 0 to x = 2p is 3. if x Î [1, 5] (where [.] represents the greatest integer function), is (a) 10 sq. unit (b) 8 sq. unit (c) 6 sq. unit (d) 5 sq. unit If a point P moves such that its distance from 7. line y = 3x - 7 is same as its distance from (2 3, -1), then area of the curve, described by P, enclosed between the coordinate axes is 3 (b) 2 3 (c) 6 (d) 3 2 The area bounded by the curve f (x) = ||tan x + cot x| – |tan x – cot x|| between the lines x = 0, x= p and the x-axis, is 2 (a) ln 2 (c) ln 4 (b) 3pa 2 sq. unit 2 3pa 2 sq. unit 4 (d) 6pa 2 sq. unit 5 8. The area bounded by the curves y = xe x , y = xe - x and the line x = 1 is 2 2 1 1 (b) 1 (c) (d) 1 e e e e Area enclosed by the curve | x + y – 1| + |2x + y – 1| = 1 is (a) 2 sq. units (b) 3 sq. units (c) 6 sq. units (d) 7 sq. units (a) (b) 2 ln 2 (d) (a) 3pa2 sq. unit (c) (a) 4. (a) 4p square units (b) 8p square units (c) 4 square units (d) 2 square units The area bounded by the curve y2 (2a – x) = x3 and the line x = 2a is 2ln 2 9. MATHEMATICS 58 10. The area bounded by the curve (a) f ( x) = max.{4 - x 2 ,| x - 2|, ( x - 2)1/ 3} for x Î[-2, 4] and x-axis is (a) 45 sq. units 4 (b) 23 sq. units 4 15. (a) ò 0 t (1 + t ) 1 - t 2 2 -1 (b) ò 0 4t (1 + t 2 ) 1 - t 2 2 +1 (c) ò 0 4t (1 + t 2 ) 1 - t 2 2 +1 (d) ò 0 12. t (1 + t 2 ) 1 - t 2 dt {( x, y ) Î ¡ : y ³ x + 3 , 5y £ x + 9 £ 15} 16. 1 4 5 3 (b) (c) (d) 6 3 3 2 The maximum possible area bounded by the parabola y = x2 + x + 10 and a chord of the parabola of length 1 is 17. 1 1 1 1 (b) (c) (d) 12 6 3 2 The area of the region bounded by the lines x = l, x = 2 and the curves x(y - ex) = sin x and 2xy = 2sin x + x3 is dt dt (a) e2 - e - 1 6 (b) e2 - e - 7 6 (c) e2 - e + 1 6 (d) e2 - e + 7 6 Numeric Value Answer 18. dt If [x] is the greatest integrer £ x , then 2 ò min{x - [ x], - x - [ - x ]} dx = 3 13. The area enclosed by the curves y = x , y = x , x = 0 and x = p, where p > 1, is 1/6. Then p equals (a) 8/3 (b) 16/3 (c) 2 (d) 4/3 If a straight line y – x = 2 divides the region 14. x2 + y 2 £ 4 into two parts, then the ratio of the area of the smaller part to the area of the greater part is (a) 3p – 8 : p + 8 (b) p – 3 : 3p + 3 (c) 3p – 4 : p + 4 (d) p – 2 : 3p + 2 The area of the region above the x-axis bounded p by the curve y = tan x, 0 £ x £ and the tangent to 2 p is: 4 1 (1 + log 2 ) 2 (a) 2 the curve at x = (d) (a) p is 4 2 1æ 1ö ç log 2 + ÷ 2è 2ø is equal to 1 - sin x 1 + sin x y= and y = bounded by cos x cos x 2 -1 (b) 2 59 53 sq. units (d) sq. units 4 4 The area of the region between the curves the lines x = 0 and x = 1 (1 - log 2 ) 2 Area of the region (c) (c) 11. 1æ 1ö ç log 2 - ÷ 2è 2ø -2 19. Let for all n Î N , n - 2 ³ 2 log e n then the minimum value of the area bounded by the curve | y|+ 20. 1 £ e -| x| is n ïì {x}, If f ( x ) = í ïî1, x ÏI x ÎI 2 and g ( x ) = {x} , (where {x} is fractional part of x), then area bounded between f (x) and g(x) "x Î[0,10] is . Application of Integrals 59 21. P (x, y) is a point, which moves in the xy plane such that 2 [ y ] = 3 [ x ] where [ . ] denotes the greatest integer function and -2 £ x £ 5 and -3 £ y £ 6 . The area of the region containing the point P (x, y) is equal to 22. If the value of a ( a ³ 1) for which the area of the figure bounded by pair of straight lines 2 y 2 - 3 y + 2 = 0 and the curves y = [ a ] x , 1 [ a ] x2 is greatest, is [a, b), then 2 (b – a) = where [.] denotes the greatest integer function. If the area bounded by y = f (x) and the curve y= 23. y= 2 1 + x2 24. Let f ( x) be a polynomial of degree 3 such that the curve y = f ( x) has relative extremes at x = ±2 / 3 and passes through (0, 0) and (1, –2) dividing the circle x2 + y2 = 4 in two parts. Find out the difference of areas of these two parts. 25. Area of the region bounded by the curve y = {x 2 }, where {.} denotes fractional part function 7ö æ x Î [–2, 2] is p ç q + r - ÷ . Find è 3ø the value of p + q + r. where f is a continuous function satisfying the conditions f (x) · f (y) = f (xy). " aö æ x, y, Î R and f ¢ (1) = 2, f (1) = 1is ç p - ÷ , then 3ø è a= 1 2 3 (d) (b) (a) 4 5 6 (c) (d) (d) 7 8 9 (b) (a) (a) 10 11 12 ANSWER KEY (c) 13 (d) 16 (b) 14 (a) 17 (d) 15 (c) 18 (b) (b) (1) (2) 19 22 20 (3.33) 23 (4) 21 24 (1) (2) (0) 25 (7) 25 DIFFERENTIAL EQUATION AND ITS APPLICATIONS MCQs with One Correct Answer 1. 2. Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the co-ordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of (a) circles (b) pair of straight lines (c) parabolas (d) rectangular hyperbolas Lef f (x) be a positive, continuous and differentiable function on the interval (a, b). If lim f ( x ) = 1 an d x®a + f ¢ (x) ³ f 3 (x) + 5. lim f ( x) = 31/ 4 . Also x ®b - 1 then f ( x) 6. p p (b) b – a £ 4 4 p (c) b – a £ (d) None of these 24 Let f be a non-negative function defined on the interval [0, 1]. If ò x 0 x 1 - ( f '(t )) 2 dt = ò f (t ) dt , 0 £ x £ 1 and 0 f (0) = 0, then (a) f (1/ 2) < 1/ 2 and f (1/ 3) > 1/ 3 (b) f (1/ 2) > 1/ 2 and f (1/ 3) > 1/ 3 (c) f (1/ 2) < 1/ 2 and f (1/ 3) < 1/ 3 (d) f (1/ 2) > 1/ 2 and f (1/ 3) < 1/ 3 The solution of the differential equation dy x3 = y 3 + y 2 y 2 - x 2 is dx (a) y + y 2 - x2 = cxy (b) y - y 2 - x 2 = cxy (c) y y 2 - x2 = cx + y (d) x y 2 - x 2 = cx + y The solution of the differential equation, 2x2 y b–a³ (a) 3. 4. dy = tan( x 2 y 2 ) – 2 xy 2 given y (1) = dx is (a) sinx2y2 = ex–1 (b) sin(x2y2) = x 2 2 (c) cosx y + x = 0 (d) sin(x2y2) = e.ex Solution of the differental equation æ yö æ yö x cos ç ÷ ( ydx + xdy ) = y sin ç ÷ ( xdy - ydx ) x è ø èxø is (a) æxö y = cx cos ç ÷ è yø y (b) sec æç ö÷ = cxy èxø æ yö æ yö ç ÷ sec ç ÷ = c (d) None of these è xø èxø The family of curves whose tangents form an p with the hyperbolas xy = C are angle of 4 (a) pair of straight lines (c) 7. p 2 (b) y 2 - xy - x 2 = C (c) y 2 - 2 xy - x 2 = C (d) y 2 + 2 xy - x 2 = C Differential Equation and its Applications 8. 9. Solution of the differential equation ìï 1 ìï x 2 1 üï y 2 üï - ý dy = 0 is dx + í í 2ý 2 y þï îï x ( x - y ) þï îï ( x - y ) x xy + =c (a) ln y x- y xy =c (b) ln | xy | + ( x - y) xy = ce x / y (c) ( x - y) xy = c e xy (d) ( x - y) (where c is arbitrary constant) The equation of the curve passing through the points (3a, a) (a > 0) in the form x = f(y) which satisfy the differential equation; a 2 dx x y × = + - 2, is xy dy y x æ 1 + e y -k ö x = y + a çç (a) y-k ÷ ÷ è 1 - 2e ø y k æ1+ e ö (b) x = y + a çç y-k ÷ ÷ è 1- e ø y k æ1+ e ö (c) y = x + a çç y-k ÷ ÷ è 1- e ø (d) None of these 10. The solution of the differential equation ( y + x xy ( x + y ))dx + ( y xy ( x + y) - x)dy = 0, is (a) x2 + y 2 x + 2 tan -1 =C 2 2y (b) x2 + y 2 + 2 tan -1 2 x =C y x2 + y 2 x =C y + 2 tan -1 2 (d) None of these (c) 11. dy Solution of dx = (a) e -2 x 2 y3 e2 x + y 2 y + 2 ln | y |= c is 61 (b) e2 x y 2 - 2 ln | y |= c (c) e x + ln | y |= c (d) None of these 12. The orthogonal trajectory of the curve an–1 y = xn is (a) n2 + x2 = constant (b) ny2 + x2 = constant (c) x2 + y2 = constant (d) y2 + nx2 = constant 13. A curve f (x) passes through the point P (1, 1). The normal to the curve at point P is a( y - 1) + ( x - 1) = 0 . If the slope of the tangent at any point on the curve is proportional to the ordinate at that point, then the equation of the curve is (a) y = e ax - 1 (b) y - 1 = eax (d) y - a = e ax (c) y = e a ( x -1) 14. A tangent and a normal to a curve at any point P meet the x and y axes at A, B and C, D respectively. If the centre of circle through O, C, P and B lies on the line y = x (O is the origin) then the differential equation of all such curves is : (a) (c) dy y - x = dx y + x dy x - y = dx xy (b) dy y 2 - x 2 = dx y 2 + x 2 (d) None of these dy - y log e 2 = 2sin x (cos x - 1) log e 2, then y = dx (a) 2sin x + c 2 x (b) 2cos x + c 2 x 15. If (c) 2sin x + c 2- x (d) 2cos x + c 2- x 16. Solution of differential equation æ d ö t ç ( g ( x )) ÷ - t 2 dt dx ø is = è dx g ( x) (a) t= g ( x) +c x (b) t = g ( x) (c) t= g ( x) x+c (d) t = g ( x) + x + c x2 +c MATHEMATICS 62 17. (a) 1 2 x2 + 3x 3 (b) 1 4 x2 - + 3x 3 (c) 1 2 - + 2 x x (d) 1 x 22. æ 1 + x 2 - y2 ö x dx - y dy = ç ÷ be x dy - y dx è x2 - y 2 ø æ f ( x, y ) + 1 + f ( x, y ) = c ç è Find the sum of the order ‘O’ and degree D of the differential equation n®¥ 23. 24. where c is an arbitrary constant then f (3, 2) is equal to A curve y = f (x) is such that f ( x) ³ 0 and equal to 10 k, where k equals Let y = f(x) is a polynomial function satisfying æ1ö æ1ö æ1ö x2 f '( x) f ç ÷ - f ( x) f ' ç ÷ = x2 f '( x) - f ' ç ÷ . x x è ø è ø èxø If f(1) = 2 and f(5) = 26, then find f(6) – 30. e ò g ( x) dx equals to. 1/ e 25. If the solution of the differential equation x+ y ö ÷ f ( x, y ) ø f (0) = 0 and bounds a curvilinear trapezoid with the base [0, x] whose area is proportional to (n + 1)th power of f (x). If f (1) = 1, then {f(10)}n is n 2 x n æ dy ö æ dy ö x æ dy ö y = 1 + x ç ÷ + ç ÷ + ..... + ç ÷ n! è dx ø è dx ø 2! è dx ø +........¥ . Let y = f (x) be a curve passing through (e, ee), which satisfy the differential equation (2ny + xy logex) dx – x logex dy = 0, x > 0, y > 0. If g ( x) = lim f ( x), then dy 1 = is 2 dx xy[ x sin y 2 + 1] If the ar ea bounded by y = f ( x), x = 1 , 2 3 and the X-axis is A sq. units where 2 2 2 4 2 4 6 f ( x ) = x + x3 + × x 5 + × × x 7 + 3 3 5 3 5 7 ....¥, | x |< 1, then the value of [4A] is (where [.] is G..I.F) x= 2 21. 0 æ 1ö then y ç ÷ is equal to è 2ø If the solution of the differential equation 2 20. 0 t 2 f ( x ) - x 2 f (t ) lim = 1 for each x > 0, then f(x) t®x t-x Numeric Value Answer 19. x x ò y (t )dt = ( x + 1) ò ty (t )dt , x > 0 , and y(1) = e, is 18. x Let f(x) be continuously differentiable on the interval (0, ¥ ) such that f(1) = 1, and x 2 (cos y 2 - sin y 2 - 2Ce - y ) = k , then value of k is ______. If the equation of a curve y = y(x) satisfies the differential equation ANSWER KEY 1 2 3 (c) (c) (c) 4 5 6 (a) (a) (b) 7 8 9 (c) (a) (b) 10 11 12 (b) (a) (b) 13 14 15 (c) (a) (a) 16 17 18 (c) (a) (2) 19 20 21 (0) (2) (8) 22 23 24 (5) (1) (7) 25 (1) 26 VECTOR ALGEBRA 1. 2. 3. 4. MCQs with One Correct Answer r The vectors a ( x) = cos xiˆ + sin xjˆ and r b ( x) = xiˆ + sin xjˆ are collinear for p (a) Unique value of x, 0 < x < 6 p p (b) Unique value of < x < 6 3 (c) No value of x p (d) Infinitely many values of x, 0 < x < 2 If cos a ¹ 1, cos b ¹ 1 and cos g ¹ 1, then the r r vector a = iˆ cos a + ˆj + kˆ, b = iˆ + ˆj cos b + kˆ , r c = iˆ + ˆj + kˆ cos g are (a) coplanar vectors (b) coplanar vectors if cos a = cos b = cos g ¹ 1 (c) coplanar vectors if os a ¹ cos b ¹ cos g (d) never coplanar Let a, b, c be distinct non-negative numbers. If the vectors aiˆ + ajˆ + ckˆ, iˆ + kˆ and ciˆ + cjˆ + bkˆ lie in a plane, then c is (a) the Arithmetic Mean of a and b (b) the Geometric Mean of a and b (c) the Harmonic Mean of a and b (d) equal tor zero r Let a and b are two vectors making angle q with each other, then which of the following r represents unit vectors along bisector of a and r b is: aˆ + bˆ aˆ + bˆ (a) ± (b) ± 2 cos q 2 ˆ aˆ + b (aˆ + bˆ) (c) ± (d) 2cos q / 2 | aˆ + bˆ | 5. uuur uuur In a parallelogram ABCD, | AB |= a,| AD |= b and uuur uuuur uuur | AC |= c , the value of DB . AB is (a) 3a 2 + b2 - c 2 2 (b) a 2 + 3b2 - c 2 2 a 2 + 3b2 + c 2 a 2 - b2 + 3c 2 (d) 2 2 r r r a and c are unit vectors and | b | = 4 . The angle (c) 6. r r r æ 1ö r r between a and c is cos -1 ç ÷ . If b - 2c = la , è 4ø then l is equal to 1 3 , 4 4 1 3 (c) - 3,4 (d) - , 4 4 The angles of a triangle, two of whose side are r represented by the vectors 3(aˆ ´ b) and r r r ˆ ˆ where b is a non-zero vector and aÌ‚ b - (a.b).a is a unit vector, are (a) 3, – 4 7. (b) (a) æ 3+2ö æ 1 ö æ 1ö tan -1 ç ; tan -1 ç ÷ ; tan -1 ç ÷ ÷ è 2ø è 3ø è1- 2 3ø (b) æ 1ö tan-1( 3); tan -1 ç ÷ ;cot -1(0) è 3ø (c) æ 3+2ö tan -1 ( 3); tan -1 (2); tan -1 ç ÷ è 2 3 - 1ø (d) æ 2 +3ö tan -1 ( 3); tan -1 (2); tan -1 ç ÷ è 3 2 - 1ø MATHEMATICS 64 8. uur If the vectors a = (c log 2 x)iˆ - 6 ˆj + 3kˆ and uur b = (log 2 x)iˆ + 2 ˆj + (2c log 2 x)kˆ make an 12. vector 2iˆ - ˆj + 2kˆ and is coplanar with the vectors iˆ + ˆj - kˆ and 2iˆ + 2 ˆj - kˆ is obtuse angle for any x Î(0, ¥) , then the interval to which ‘c’ belongs is (0, ¥) (a) 9. (a) (-¥, 0) 4 (c) (- , 0) (d) ( -1, 0) È (0, 2 ) 3 3 ur r Let p = ai$+ b$j + ck$ and q = bi$+ c$j + ak$, ur where a, b, c Î R. If 'q' be the angle between p r and q then, q Î (0, p / 2) (a) 10. (b) 13. (b) q Î[0, 2p / 3] (c) q Î (2p / 3, p] (d) q Î[p / 2, p] The value of ‘a’ so that th e volume of parallelopiped formed by iˆ + ajˆ + kˆ, ˆj + akˆ and aiˆ + kˆ becomes minimum is (a) –3 11. (b) 3 (c) 1 3 (d) ® ® ® 3 14. If a , b , c are three non-zero, non-coplanar vectors and ® ® ® ® ® ® b× a ® ® ® b× a ® b1 = b a , b2 = b + a, ® ® 2 2 |a| |a| ® ® ® ® |a| |c| ® ® c× a ® b× c ® c1 = c a+ b1 , ® ® 2 2 ® ® ® ® c× a ® b × c ® 1 c2 = c - ® a - ® b1 , 2 2 |a| | b1 | ® ® ® ® ® ® c× a ® b× c ® c3 = c a+ b1 , ® ® 2 2 ® ® ® ® ® (a) ( a , b1, c3 ) (b) ( a , b , c ) 1 2 (c) ( a , b1, c1) (d) ( a , b , c ) 2 2 ® ® ® ® ® ® 5 ˆ 3i + 2 ˆj + 2kˆ (b) 3iˆ + 2 ˆj - 2 kˆ r iˆ ˆj kˆ v = + + , then a b c r r (a) u and v are parallel vectors r r (b) u and v are orthogonal vectors r r (c) u × v = 1 r r (d) u ´ v = iˆ + ˆj + kˆ If the two adjacent sides of two rectangles are represented by vectors r r r r r r r r r p = 5a - 3b ; q = -a - 2b an d r = -4a - b ; r r r s = - a + b , r espectively, then the angle r 1 r r r between the vectors x = ( p + r + s ) and 3 r 1 r r y = ( r + s ) is 5 æ 19 ö (a) - cos -1 ç ÷ è 5 43 ø æ 19 ö cos -1 ç ÷ è 5 43 ø æ 19 ö p cos -1 ç ÷ è 5 43 ø (d) Cannot be evaluated r r r a, b and c be three non-coplanar vectors and r d be a non-zero vector, which is perpendicular r r r to (a + b + c ). Now if r r r r r r r d = sin x(a ´ b ) + cos y (b ´ c ) + 2(c ´ a ), then 2 2 minimum value of x + y is equal to (a) p2 (b) 0 (c) p2/4 (d) 5p2/4 (c) 15. 2 ˆj + kˆ 17 ˆ 2i + 2 ˆj - 2kˆ (c) (d) 3 17 If a, b and c are pth, qth, rth terms of HP and r u = (q - r )iˆ + (r - p) ˆj + ( p - q)kˆ, (b) |c| |c| uur uur uur uur uur uur c . a uur b . c uur c4 = c - uur a = uur b1 , | c |2 | b |2 then the set of orthogonal vectors is ® ® ® A unit vector which is perpendicular to the Vector Algebra 65 r 16. The vectors a = 2l 2iˆ + 4lˆj + kˆ and r b = 7iˆ - 2 ˆj + lkˆ make an obtuse angle r whereas the angle between b and kÌ‚ is acute and less than p/6, 1 (a) 0 < l < (b) l > 159 2 1 (c) - < l < 0 (d) null set 2 r 17. The angle q between two non-zero vectors a r and b satisfies the relation r r r r cos q = (a ´ iˆ) × (b ´ iˆ) + (a ´ ˆj ) × (b ´ ˆj ) r r +(a ´ kˆ) × (b ´ kˆ), then the least value of |a| + |b| is equal to (where q ¹ 90°) (a) (c) 1 2 (b) 2 a, b, where x1, x2, x3 Î {– 3, – 2, – 1, 0, 1, 2}. r r If the number of possible vectors b such that a r and b are mutually perpendicular is t, then t /5 = 22. Find the absolute value of parameter t for which the area of the triangle whose vertices are A(–1, 1, 2); B(1, 2, 3) and C(t, 1, 1) is minimum. r 23. Given a vector A defined as r r r r r r r A = (a ´ b ) ´ (c ´ d ) + (a ´ c ) r r r r r r ´(b ´ d ) + (a ´ d ) ´ (b ´ c ) , then find the value r r of | A ´ a | . r é1ù r 24. Let v0 be a fixed vector and v0 = ê ú . Then for ë0û n ³ 0 a sequence is defined (d) 4 2 g are scalars. If a= k1 ( Fˆ × aˆ ) - k2 ( Fˆ × bˆ), then the value of 2(k1 + k2) is r r k , b = x1$i + x2 $j + x3 $ k, 21. Let a = $i + $j + $ where Numeric Value Answer 18. If the sum of two unit vectors is a unit vector, and the magnitude of their difference is k , then value of k is ˆ yˆ and zÌ‚ are three unit vectors in three19. If x, dimensional space, then the minimum value of 2 2 2 xˆ + yˆ + yˆ + zˆ + zˆ + xˆ r r æ1ö vn +1 = vn + ç ÷ è2ø 25. 20. Let aÌ‚ and bÌ‚ be two unit vectors such that r 1 aˆ × bˆ = and aˆ ´ bˆ = cˆ. Also F = aaˆ + bbˆ + gcˆ, 3 n +1 é0 -1ù ê ú ë1 0 û n +1 r v0 then éa ù r a lim vn = ê ú . Find . n ®¥ b b ë û If A1, A2, ...., Ag are vertices of a regular octagon, 7 uuur uuur uuur uuur then å (OA j ´ OA j +1 ) = K (OA1 ´ OA2 ) j =1 where the value of K is ........ ANSWER KEY 1 2 3 (b) (d) (b) 4 5 6 (c) (a) (a) 7 8 9 (b) (c) (b) 10 11 12 (c) (b) (d) 13 14 15 (b) (b) (d) 16 17 18 (d) (c) (3) 19 20 21 (3) (3) (5) 22 23 24 (2) (0) (2) 25 (7) 27 THREE DIMENSIONAL GEOMETRY MCQs with One Correct Answer 1. The angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0 , 6nm - 2nl + 5lm = 0 is (a) (c) 2. æ1ö cos -1 ç ÷ è6ø æ2ö cos -1 ç ÷ è3ø (b) (d) æ 1ö cos -1 ç - ÷ è 6ø 4. æ pù ç 0, ú è 4û (b) 5. (b) l1 + l2 m + m2 n +n , 1 , 1 2 2 cos q / 2 2 cos q / 2 2 cos q / 2 1 (d) –2 2 L1 and L2 are two lines whose vector equations r are L1 : r = l[(cos q + 3)ˆi + (b) –1 (c) - r and L2 : r = m(aˆi + bˆj + ckˆ ) where, l and m are é p pù æ p pù (d) ç , ú ê 4 , 2ú ë û è 3 2û If l1, m1, n1 and l2, m2, n2 are DCs of the two lines inclined to each other at an angle q, then the DCs of the internal bisector of the angle between these lines are l1 + l2 m + m2 n +n , 1 , 1 2 2 sin q / 2 2 sin q / 2 2 sin q / 2 If the straight lines x = 1 + s , y = -3 - l s , z = 1 + l s ( 2 sin q)ˆj + (cos q - 3)kˆ ] é p pù ê6 , 3ú ë û (a) l1 - l2 m - m2 n -n , 1 , 1 2 2 cos q / 2 2 cos q / 2 2 cos q / 2 (a) 0 scalars and a is the acute angle between L1 and L2. If the angle a is independent of q, then the value of a is (c) 3. (d) t , y = 1 + t , z = 2 - t , with parameters 2 s and t respectively, are co-planar, then l equals. A line in the 3-dimensional space makes an angle (a) l1 - l2 m - m2 n -n , 1 , 1 2 2 sin q / 2 2 sin q / 2 2 sin q / 2 and x = æ 5ö cos -1 ç - ÷ è 6ø pö æ q çè 0 < q £ ÷ø with both the x and y axes. Then 2 the set of all values of q is the interval: (c) (a) 6. p 6 (b) p 4 (c) p 3 (d) p 2 x + 6 y + 10 z + 14 = = is the 5 3 8 hypotenuse of an isosceles right angled triangle whose opposite vertex is (7, 2, 4). Then which of the following is not the side of the triangle? The line Three Dimensional Geometry (a) x-7 y-2 z-4 = = -3 2 6 (b) x-7 y-2 z-4 = = 3 6 2 ABC satisfies the relation 1 + 1 + 1 = k , x2 y 2 z 2 then the value k is (a) 3 x-7 y-2 z-4 = = 3 5 -1 (d) None of these The vertex A of the triangle ABC is on the line r r = iˆ + ˆj + lkˆ and the vertices B and C have (c) 7. 67 12. The 8. 9. (c) [–2, 2] (d) [-4, - 2] È [2, 4] A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals (a) 1 (b) (d) 2 2 (c) 3 Through a point P (h, k, l) a plane is drawn at right angles to OP to meet the co-ordinate axes in A, B and C. If OP = p, then the area of DABC is : (a) p 2 hk l2 (b) p 3l 3hk p5 p 2t 2 (d) 2hkl 2hk 10. Projection of the line x + y + z – 3 = 0 = 2x + 3y + 4z – 6 on the plane z = 0 is (c) (a) x y-6 z = = -2 1 0 (b) x y-6 z = = 1 -2 0 x y-6 z = = (d) none of these 1 2 0 11. A variable plane at a distance of the one unit from the origin cuts the coordinates axes at A, B and C. If the centroid D (x, y, z) of triangle (c) plane (c) 1 3 containing (d) 9 the line x -1 y - 2 z - 3 = = and parallel to the line 1 2 3 respective position vectors iˆ and jÌ‚ . Let D be é 3 33 ù the area of the triangle and D Î ê 2 , 2 ú , then ë û the range of values l corresponding to 'A' is (a) [-8, - 4] È [4, 8] (b) [–4, 4] (b) 1 13. x y z = = passes through the point: 1 1 4 (a) (1, – 2, 5) (b) (1, 0, 5) (c) (0, 3, –5) (d) (– 1, – 3, 0) The position vectors of points a and b are iˆ - ˆj + 3kˆ and 3iˆ + 3 ˆj + 3kˆ respectively. The equation of a plane is r × (5iˆ + 2 ˆj - 7 kˆ) + 9 = 0. The points a and b (a) lie on the plane (b) are on the same side of the plane (c) are on the opposite side of the plane (d) None of the above 14. The angle between the pair of planes represented by equation 2 x 2 - 2 y 2 + 4 z 2 + 6 xz + 2 yz + 3xy = 0 is (a) æ1ö cos -1 ç ÷ è 3ø (b) æ 4ö cos -1 ç ÷ è 21 ø (c) æ4ö cos -1 ç ÷ è9ø (d) æ 7 ö cos -1 ç ÷ è 84 ø 15. Let P = (–3, 1, 1) and Q = (3, 4, 2). R divides PQ in the ratio PR : PQ = 1 : 3. Then, the equation of uuur the plane perpendicular to PQ at R is (a) 18x + 9y + 3z = 8 (b) 18x + 9y + 3z = 4 (c) 9x + 18y + 3z = 4 (d) 3x + 9y + 18z = 8 16. Let s1, s2, s3 be planes passing through the origin. Assume that s1 is perpendicular to the vector (1, 1, 1), s2 is perpendicular to a vector (a, b, c), and s3 is perpendicular to the vector (a2, b2, c2). MATHEMATICS 68 17. What are all the positive values of a, b and c so that s1 Ç s2 Ç s3 is a single point? (a) Any positive value of a, b, and c other than 1 (b) Any positive values of a, b and c where either a ¹ b, b ¹ c or a ¹ c (c) Any three distinct positive values of a, b, and c (d) There exist no such positive real numbers a, b and c r ˆ ˆ ˆ r r Let a = i + j + k , b = 2iˆ + 2 ˆj + kˆ , and c = 5iˆ + ˆj - kˆ be three vectors. The area of the region formed by the set of points whose position vectors r r r r s art i sfy t h e eq u a t i on s r . a = 5 an d r r r | r - b | + | r - c | = 4 is closest to the integer (a) 4 (c) 14 (b) 9 (d) 19 21. 22. f (-2) = 1, f (1) ¹ 1, f (0) ¹ 2 and the remaining two are false. If the area of the triangle formed by (–2, 1, 0) and (f(–2), f(1), f(0)) and the origin is 19. 20. A line with direction ratios (2, 1, 2) intersects r the lines r = - ˆj + l(iˆ + ˆj + kˆ) and r r = -iˆ + m(2iˆ + ˆj + kˆ) at A and B, respectively,, then length of AB is equal to The shortest distance between the z-axis and the line, x + y + 2z - 3 = 0 , 2x + 3y + 4z - 4 = 0 is The distance of the point (1, 0, –3) from the plane x - y - z = 9 measured parallel to the line k ; then sum of digits of k is. 2 If the equation of the plane through the intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is ax + by + cz + 173 = 0, then b – 9(a + c) is equal to ........ If a line is passing through (a, b, c) and intersecting y = 0, z2 = 4ax lies on the surface given by 23. Numeric Value Answer 18. Let the equation of the plane containing line x – y – z – 4 = 0 = x + y + 2z – 4 and parallel to the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2 be x + Ay + Bz + C = 0. Then the values of |A + B + C – 4| is ........ . Let f be a one-one function with domain {–2, 1, 0} and range {1, 2, 3} such that exactly one of the following statements is true : 24. 25. (bz - cy)2 = k a(b - y)(bx - ay); then find the value of k. If the volume enclosed by the equation | x |£ 8, | y |£ 8, | z |£ 8 and | x + y + z |£ 8 is t, then t = 512 x-2 y+2 z -6 is = = 2 3 -6 ANSW ER KEY 1 2 3 (b) (c) (b) 4 5 6 (d) (a) (c) 7 8 9 (d) (c) (d) 10 11 12 (b) (d) (b) 13 14 15 (c) (c) (b) 16 17 18 (c) (a) (3) 19 20 21 (2) (b) (7) 22 23 24 (7) (6) (4) 25 (4) 28 PROBABILITY-2 1. Let A and B be the sets {1, 2, …10} and {1, 2, …20} respectively. A function is selected randomly from A to B the probability that the function is non-decreasing is 29 (a) (20) 29 3. (b) 10 (c) 2. 29 C10 C19 (20)10 Raj and Sanchita are playing game in which they throw two dice alternately till one of them gets 9. Which one of the following could be the probability that Sanchita win the game? (a) 7/15 or 8/15 (b) 6/11 or 5/11 (c) 8/17 or 9/17 (d) None of these A boy whose hobby is tossing a fair coin is to score one point for every tail and 2 points for every head. The boy goes on tossing the coin, till his score reaches n or exceeds n where n > 2. If pn is the probability that his score attains exactly n, then pn is equal to (b) The odds in favour of a book reviewed by three independent critics are, respectively, 5 : 2, 4 : 3 and 3 : 4. The probability that majority of the critics give favourable remark is (a) (20)10 1 p n -1 + pn -2 2 n 4. C20 (d) None of these (a) pn–1 + pn–2 5. n 2n +1 + ( -1) 2 ( -1) + (c) (d) 3 2n 3.2 n Entries of a 2 × 2 determinant are chosen from the set {–1, 1}. The probability that determinant has zero value is 1 3 (a) 1 4 (b) (c) 1 2 (d) None of these 6. (c) 211 343 (d) 205 343 1 1 + 6 73 1 1 7 (b) 1 9 7 + 1 3 7 - 1 76 1 1 1 (d) + 3 6 7 79 7 7 79 If a Î[– 20, 0], then probability that the graph of the function y =16x2 + 8(a + 5) x – 7a – 5 is strictly above the x-axis is (c) 3 + 13 17 7 3 (c) (d) (b) 20 20 20 20 In a Competitive test, a candidate guesses, copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/ 8. Find the probability that he knew the answer to the question, given that he answered it correctly. (a) 24/29 (b) 26/29 (c) 22/29 (d) None of these (a) 8. 209 343 Matrices of order 3 × 3 are formed by using the elements of the set A = {–3, –2, –1, 0, 1, 2, 3}, then probability that matrix is either symmetric or skew symmetric is (a) 7. 210 (b) 343 MATHEMATICS 70 9. A wire of length l is cut into three pieces. Then the probability that the three pieces form a triangle is (a) 1 2 (b) (a) 14. 2 (d) None of these 3 Given that the sum of two non-negative quantities is 200, the probability that their product is not less than 2 1 , otherwise it is . If he writes 3 4 2 books, the probability that at least one book will be published is 3 times their greatest 4 7 101 9 10 (a) (b) (c) (d) 16 201 16 16 On each evening a boy either watches DOORDARSHAN channel or TEN SPORTS. The 3 1 that he will be asleep, while it is 4 4 when he watches TEN SPORTS. On one day, the boy is found to be asleep. The probability that the boy watched DOORDARSHAN is 15. parameters n = 8 and p = (a) (c) 13. ( 2 - p )k +1 2a . p (2 - p) (b) pk ( 2 - p )k +1 k k +1 (d) None of these A die is thrown 2n + 1 times, n Î N. The probability that faces with even numbers show odd numbers of times, is 121 119 117 115 (b) (c) (d) 128 128 128 128 Ravi and Rashmi are each holding 2 red cards and 2 black cards (all four red and all four black cards are identical). Ravi picks a card at random from Rashmi, and then Rashmi picks a card at random from Ravi. This process is repeated a second time. Let p be the probability that both have all 4 cards of the same colour. Then p satisfies (a) p £ 5% (b) 5% < p £ 10% (c) 10% < p £ 15% (d) 15% < p The probability of men getting a certain disease 1 is and that of women getting the same 2 1 disease is . The blood test that identifies the 5 4 disease gives the correct result with probability . 5 (a) 16. (a) of n children have the same probability. If k ³ 1 , then the probability that a family contains exactly k boys is 1 , then p(|x – 4| < 2) is 2 equal to chance of 5 2 3 4 (b) (c) (d) 7 7 7 7 Let the probability Pn that a family has exactly n children be apn, when n ³ 1 and P0 = 1 – ap (1 + p + p2 + ...). Suppose that all sex distributions 407 411 405 307 (b) (c) (d) 576 576 576 576 If X follows a binomial distribution with (a) 4 . 5 If he watches DOORDARSHAN, there is a 2a 1 (d) None of these 2 A book writer writes a good book with probability be published is probability that he watches TEN SPORTS is 12. 1 2 1 . If it is a good book, the probability that it will 2 product value is 11. (b) less than (c) greater than 1 4 (c) 10. 2n + 1 4n + 3 17. Suppose a person is chosen at random from a group of 30 males and 20 females, and the blood test of the person is found to be positive. What is the probability that the chosen person is a man? (a) 75 107 (b) 3 5 (c) 15 19 (d) 3 10 Probability-2 71 Numeric Value Answer 18. A and B play a game of tennis. The situation of the game is as follows : if one scores two consecutive points after a deuce, he wins, if loss of a point is followed by win of a point, it is deuce. The chance of a server to win a point is 2 . The game is at deuce and A is serving. 3 Probability that A will win the match is, (serves are changed after each game). 22. A number x is selected from the set of first 9 natural numbers (i.e., x = 1, 2, 3, ......., 9). If the probability that f(f(x)) = x where f(x) = x2 – 3x + 3 is 23. A special die is so constructed that the probabilities of throwing 1, 2, 3, 4, 5 and 6 are (1 – k) / 6, (1 + 2k) / 6, (1 – k) / 6, (1 + k) / 6, (1 – 2k) / 6 and (1 + k) / 6, respectively. If two such dice are thrown and the probability of 19. The probablilty that the length of a randomly chosen chord of a circle lies between getting a sum equal to 9 lies between 2 5 and 3 6 of its diameter is 20. In a hurdle race, a runner has probability p of jumping over a specific hurdle. Given that in 5 trials, the runner succeeded 3 times, the conditional probability that the runner had succeeded in the first trial, is 24. A sportsman's chance of shooting an animal at a distance r > a ("a" is constant) is given to be a2 r2 . He fires at r = 2a and if he misses, then again fires at r = 3a. He repeats the same process at r = 4a, 5a and 6a. When he misses at r = 6a, the animal escapes into the jungle. If the odds against the sportsman are p : q, then q – p is _______., 8 or at the point (II) with probability 9 1 . There are 21 shells each of which can be 9 1 2 and , 9 9 then the integral value of k is. 21. An artillery target may be either at point I with probability m , then m is equal to _________. 9 25. A determinant of the second order is made with m be the probability n that the determinant made is non-negative, where m and n are relative primes, then the value of n – m is the elements 0 and 1. If fired either at point I or II. Each shell may hit the target independently of the other shell with 1 . Minimum number of shells that 2 must be fired at point I to hit the target with maximum probability is equal to 2k. Then value of k is. probability ANSW ER KEY 1 2 3 (c) (c) (d) 4 5 6 (c) (b) (d) 7 8 9 (b) (a) (b) 10 11 12 (b) (c) (c) 13 14 15 (d) (a) (b) 16 (a) 19 17 (a) 20 18 (0.50) 21 (0.25) (0.60) (6) 22 23 24 (2) (0) (5) 25 (3) 29 PROPERTIES OF TRIANGLE 5. MCQs with One Correct Answer 1. 2. If the sines of the angles A and B of a triangle ABC satisfy the equation c2x2 – c(a + b) x + ab = 0, then the triangle is (a,b,c, are sides of D) (a) acute angled (b) obtuse angled (c) right angled (d) no such triangle is possible sin 3 A + sin 3 B + sin 3 C (a) 7 3. 4. 6. = 7, then the maximum possible value of a is (b) 49 (c) 3 7 7. (d) 1 7 If in a triangle ABC, sin A, sin B, sin C are in A.P., then (a) the altitudes are in A.P. (b) the reciprocals of altitudes are in A.P. (c) the altitudes are in G.P. (d) the medians are in A.P. Let a, b and A are given and c1 and c2 are two values of c, then the value of c12 + c22 – 2c1c2 cos 2A = (a) a2sin2A (b) a2 cos2 A 2 2 (c) 2(a + b ) (d) 4a2cos2 A (a + b + c )2 , then ab + bc + ca (a) P Î [1, 2] (b) P Î [3, 4) (c) P Î (2, 4] (d) None of these If a, b, c are the sides of a triangle such that b. c = l2, for some positive l, then P= In DABC, a ³ b ³ c . If a 3 + b 3 + c3 If a, b, c be the sides of a triangle and (a) a ³ 2l sin A 2 (b) b ³ 2l sin B 2 (c) c ³ 2l sin C 2 (d) all are correct In DABC, a2 (s – a) + b2 (s – b) + c2 (s – c) = (a) 4R(cos A + cos B + cos C) (b) 4RD(sin A + sin B + sin C) A B Cö æ 4RD ç 1 + 4 sin sin sin ÷ 2 2 2ø è (d) None of these In an obtuse angled triangle, the obtuse angle is (c) 8. 3p and the other two angles are equal to two 4 values of q satisfying a tan q + b sec q = c, 2 2 where | b | £ a + c , then a2 – c2 is equal to (a) ac (c) a c (b) 2ac (d) None of these Properties of Triangle 9. 73 1 In a triangle ABC, 2ac sin ( A - B + C ) = 2 (a) a2 + b2 – c2 (c) b2 – c2 – a2 (b) c2 + a2 – b2 (d) c2 – a2 – b2 10. The sides of a triangle are sin a, cos a and 1 + sin a cos a for some 0 < a < p . Then the 2 greatest angle of the triangle is (a) 150° (b) 90° (c) 120° (d) 60° 11. Given a triangle DABC such that sin2 A + sin2 C =1001 × sin 2 B. Then the value of 2 (tan A + tan C) × tan 2 B tan A + tan B + tan C 1 1 1 1 (a) (b) (c) (d) 2000 1000 500 250 12. In DABC, if sin A sin B sin C c b a + + = + + , then c sin B c b ab ac bc the value of angle A is (a) 120° (b) 90° (c) 60° (d) 30° 13. If the hypotenuse of a right-angled triangle is four times the length of the perpendicular drawn from the opposite vertex to it, then the difference of the two acute angles will be (a) 60° (b) 15° (c) 75° (d) 30° 14. In a DABC, ÐB = p p and ÐC = let D divide 3 4 BC internally in the ratio 1 : 3, then sin (ÐBAD) sin (ÐCAD) is equal to: 1 1 1 2 (b) (c) (d) 3 3 6 3 15. In a triangle ABC, a = 5, b = 4 and cos (A – B) (a) 31 , then the third side is equal to: = 32 (where symbols used have usual meanings) (b) 6 6 (a) 6 (c) 6 (d) (216)1/4 16. In triangle ABC if A : B : C = 1 : 2 : 4, then (a2 – b2) (b2 – c2) (c2 – a2) = l a2b2c2, where l = (where notations have their usual meaning) (a) 1 (b) 2 (c) 4 (d) 9 17. If in a right angle triangle ABC, 4 sinA cosB – 1 = 0 and tan A is real, then (a) angles are in A.P. (b) B2 = AC 2 AC (d) None of these A+ C 18. A tower of height b subtends an angle at a point O on the level of the foot of the tower and at a distance a from the foot of the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is (c) B= (a) æ a 2 - b2 ö bç ÷ ç a 2 + b2 ÷ è ø æ a2 + b2 ö (b) b ç ÷ ç a 2 - b2 ÷ è ø (c) æ a 2 - b2 ö aç ÷ ç a2 + b2 ÷ è ø (d) æ a2 + b2 ö aç ÷ ç a 2 - b2 ÷ è ø 19. An observer on the top of a tree, finds the angle of depression of a car moving towards the tree to be 30°. After 3 minutes this angle becomes 60°. After how much more time, the car will reach the tree? (a) 4 min. (b) 4.5 m (c) 1.5 min (d) 2 min. 20. Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60° and as seen from B are 60º and 45°. If AB is 30m, then the distance between the flagstaffs in metres is (a) 30 + 15 3 (b) 45 + 15 3 (c) 60 - 15 3 (d) 60 + 15 3 MATHEMATICS 74 24. Numeric Value Answer 21. Points D, E are taken on the side BC of an acute angled triangle ABC, such that BD = DE = EC. If ÐBAD = x, ÐDAE = y and ÐEAC = z then the value of 22. sin (x + y) sin (y +z) is _________. sin x sin z In triangle ABC, sin A sin B + sin B sin C + sin C sin A = 9/4 and a = 2, then the value of 3D, where D is the area of triangle, is _________. 23. Triangle ABC is right angled at A. The points P and Q are on hypotenuse BC such that BP = PQ = QC. If AP = 3 and AQ = 4, if BC = a b then AD, BE, CF are internal angular bisectors of A DABC and I is the incentre. If a (b + c) sec 2 B C ID + b(a + c) sec IE + c(a + b) sec 2 2 IF = kabc, then the value of k is 25. A man from the top of a 100 metres high tower see a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. If the distance (in metres) travelled by the car during this time is a b æ a ö , then ç - b ÷ = 3 è 20 ø | a - b |= 1 2 3 (c) (c) (b) 4 5 6 (d) (b) (a) 7 8 9 (c) (b) (b) 10 11 12 (c) (d) (b) ANSWER KEY 13 (a) 16 14 (a) 17 15 (c) 18 (a) (a) (b) 19 20 21 (c) (d) (4) 22 23 24 (3) (2) (2) 25 (7) Hints & Solutions CHAPTER Sets 1 1. B A B¢ – A¢ and A – B 2. 3. 4. 5. (b) Let A = {q : sin q = tan q} and B = {q : cos q = 1} sin q ü ì \ A = íq : sin q = ý cos qþ î = {q : sin q (cos q – 1) = 0} = {q = 0, p, 2p, 3p,.....} For B : cos q = 1 Þ q = p, 2p, 4p,...... This shows that A is not contained in B. i.e. A Ë B. but B Ì A. (c) n(P(S)) = 23 = 8 elements. n(P(P(S))) = 28 = 256 elements. (c) Let n be the number of newspapers which are read. Then 60 n = (300) (5) Þ n = 25 (a) From Venn-Euler’s Diagram. (A È B) ' A 7. Total 100 Chicken (60) Fish (50) c a b n U (A 'Ç B) 6. (a) bN = {bx : x ÎN}; cN = {cx : x ÎN} \ bN Ç cN = {x : x is multiple of b and c both} = { x: x is multiple of l.c.m. of b and c } = { x : x is multiple of b c} [given b and c are relatively prime \ l.c.m. of b and c = bc] \ bN Ç cN = {bc x : x ÎN} = dN (Given) \ d = bc. 9. (a) (A È B È C) È (A Ç B' Ç C' )' Ç C' = (A È B È C) Ç (A' È B È C ) Ç C' = [(A Ç A') È (B È C)] Ç C' = (f È B È C) Ç C' = (B È C) Ç C' = (B Ç C') È (C Ç C’) = (B Ç C') È f = B Ç C' 10. (a) Total number of persons = a + b + c + n = 100 8. (a) B \ (A È B) ' È (A 'Ç B) = A ' (b) Given set can be written as (A – B) È (B – A) = (A È B) – (A Ç B) (By definition of symmetric difference) Hence, (A \ B) È (B \ A) = (A È B) \ (A Ç B) (b) A = {x : |x| < 1} = (–1, 1) Since, |x| < 1 Þ –1< x < 1 B = {x : |x–1| ³ 1} = (– ¥, 0] È [2, ¥) Since, |x –1| ³ 1 Þ x–1£ –1 or x – 1³ 1 Þ x £ 0 or x ³ 2 \ A È B = (– ¥ , 0] È [2, ¥ ) È (–1, 1) = (– ¥, 1) È [2, ¥ ) = R – [1, 2) \ D = [1, 2) = { x : 1 £ x < 2} Do not prefer fish b + n = 50 60 prefer chicken hence b + c = 60 Do not like fish and chicken is n = 10 On solving these equations we will get a = 30, b = 40, c = 20 The number of persons who prefer both fish and chicken is = c = 20 11. (c) {(A – B) È (B – C) È (C – A)}¢ = (A – B)¢ Ç (B – C)¢ (C – A)¢ = [(U – (A – B)) Ç (U – (B – C) Ç (U – (C – A))] A B A– B C –A C By Venn - diagram = (A Ç B Ç C) MATHEMATICS 76 12. (d) n(A) = 1000, n(B) = 500, n(A Ç B) ³ 1 & n(A È B) = p Also, n(A È B) = n(A) + n(B) – n(A Ç B) Þ p = 1000 + 500 – n (A Ç B) Q 1 £ n(A Ç B) £ 500 Hence p £ 1499 and p ³ 1000 Þ 1000 £ p £ 1499 13. (a) Minimum value of n = 100 – (30 + 20 + 25 +15) = 100 – 90 = 10 14. (a) Let the number of students who take only Math be x and only Chemistry be y. C M x 30 y So, from the Venn diagram, we have total number of students who take Math = x + 30 and take Chemistry = y + 30. According to question, we have 10 ( x + 30) 30 = 100 12 ( 30 + y ) Þ x = 270 and 30 = 100 Þ y = 220 x + y + 30= 270 + 220 + 30 = 520. 15. (a) See the following Venn diagram M I 23 29 Hence a = 20k and b = 60k, then n = 100k – 20k – 60k = 20k The difference between those who opted for ‘Surf blue’ and those who were uncertain = 60k – 20k = 40k = 720 hence, k = 18, Hence total number of persons covered in survey = 100k = 1800 18. (c) From the given information (C) = 80, (F) = 40, and (C Ç F) = (n) Hence (U) = (C) + (F) – (C Ç F) + (n) Or 80 + 40 – x + x = 120 19. (d) If number of students who like chocolate =a+c Number of students who like Biscuit = b + c Number of students who like Both = c Number of students who like none = n = 10 From the given condition 100 = a + b + c + n Since 60 of them don’t like Chocolate, hence b + n = 60 or b = 50 And 50 of them don’t like Biscuit hence a + n = 50, a = 40 Hence 100 = 40 + 50 + c + 10 or c = 0 20. (b) The given information can be represented as follows. U = 100 F 4–x x W n (I) = 29 + 23 = 52; n (F) = 100 – 52 = 48 n (m È D ) = n ( m ) + n ( D ) - n ( m Ç D ) 24 = 23 + 4 - n (m Ç D) \ n (m Ç D) = 3 \ n( W Ç D) = 4 - 3 = 1 16. (b) C stands for set of students taking economics E b g a c C e d f M a + b + c + d + e + f + g = 40; a + b + d + g = 16 b + c + e + g = 22; d + e + f + g = 26 b + g = 5; e + g = 14; g = 2 Go by backward substitution e = 12, b = 3, d + f = 12, c + e = 17 Þ c = 5; a + d = 11 a +d + f = 18 Þ f = 7 \ d = 12 – 7 = 5 17. (c) Let those who opted for Nirma = a and those who opted Surf Blue = b and those who opted for none is n. 21. Now from the given information we can frame following equationsc = 18, f + c = 23, f + g = 8, c + f + g + e = 28, a + d + f + g = 48, d + g = 10, n = 24 People who read exactly two consecutive months is represented by d and e. f + c = 23 and c = 18 \ f= 5 f + g = 8 and f = 5 \ g=3 d + g = 10 and g = 3 \ d=7 c + f + g + e = 28, c = 18, f = 5 and g = 3 \ e = 2 or \ d + e = 9 (4) We have n (U) = 100, where U stands for universal set n (M Ç C Ç T) = 10; n (M Ç C) = 20; n (C Ç T) = 30; n (M Ç T) = 25; n (only M ) = 12; n (only C) = 5; n (only T) = 8 Filling all the entries we obtain the Venn diagram as shown below : Solutions 77 M C 12 15 10 10 8 = Total students – n(C È F È H) U n 28 = =7 4 4 23. (7) The given condition is as follows- 5 n(C¢ Ç F¢ Ç H¢)= 80 – 52 = 28, So 20 T \ n (M È C È T) = 12 + 10 + 5 + 15 + 10 + 20 + 8 = 80 \ n (M È C È T)' = 100 – 80 = 20 = n. n 20 =4 So, = 5 5 22. (7) Numbers which are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80 they are 16 in numbers. Now, Numbers which are divisible by 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 they are 11 in numbers. Also, total odd numbers = 40 Let C represents the students who opt. for cricket, F for football and H for hockey. \ we have n(C) = 40, n(F) = 16, n(H) = 11 Now, C Ç F = Odd numbers which are divisible by 5. CÇ H = Odd numbers which are divisible by 7. F Ç H = Numbers which are divisible by both 5 and 7. n(C Ç F), 8, n(C Ç H) = 6, n(FÇ H) = 2, n (C Ç F Ç H) = 1 We Know n(CÈFÈH) = n(C) + n(F) + n(H) – n(C Ç F) – n(C Ç H) – n(F Ç H) + n(C Ç H Ç F) n(CÈFÈH) = 67 – 16 + 1 = 52 \ n(C¢ Ç F¢ Ç H¢) CHAPTER 2 1. (b) Put x = y = 1, (f (1))2 = 3 f (1) – 2 Þ f (1) = 1 or 2 Let f (1) = 1, then put y = 1 f (x) . f (1) = f (x) + f (1) + f (x) – 2 Þ f (x) = 1 constant function \ f (1) ¹ 1, 2. hence f (1) = 2 (a) Given f ( x) = x14 - x11 + x 6 - x3 + x 2 + 1 for f (x) to be defined, x14 - x11 + x6 - x 3 + x 2 + 1 ³ 0 Case 1 : x ³ 1 x14 - x11 + x 6 - x3 + x 2 + 1 We know that {(a + d + e + g) + ( b + d + f + g) + (c + e + f + g)} – ( d + e + f) – 2g =a+b+c+d+e+f+g or 61x + 46x + 29x – 25x – 2g = 97x or 2g = 14x or g = 7x. So 7% of people watched all the three movies) 24. (4) n (A È B) is minimum when A Í B. In this case, (A È B) = B and hence minimum value of n (A È B) = n(B) = 7. n (A È B) is maximum when A and B are disjoint. \ Maximum value of n (A È B) = 4 + 7 = 11. So 11 – 7 = 4 25. (3) 2m - 2n = 112 Þ 2 n (2 m- n - 1) = 16.7 \ 2n (2m -n - 1) = 24 (23 - 1) Comparing, we get n = 4 and m – n = 3 Þ n = 4 and m = 7 So m – n = 7 – 4 = 3 Relations & Functions-1 = ( x14 - x11 ) + ( x 6 - x3 ) + ( x 2 + 1) > 0 Case 2 : 0 £ x £ 1 x14 - x11 + x 6 - x3 + x 2 + 1 = x14 - {( x11 - x6 ) + ( x3 - x 2 )} + 1 > 0 {Q x11 - x6 £ 0, x 3 - x 2 £ 0 } Case 3 : x < 0 x14 - x11 + x 6 - x3 + x 2 + 1 > 0 (Q x11 < 0, x3 < 0, x14 , x 6 , x 2 > 0) Thus, for all real x, x14 - x11 + x6 - x 3 + x 2 + 1 ³ 0 Hence, the domain of f (x) = R = (-¥, ¥) MATHEMATICS 78 3. (a) Given f (x) = cos (log x) \ f (xy) = cos (log xy) f (xy) = cos [log x + log y] æ xö æ è Replacing x + 1 by ....(1) 1 æ 1 ö af ç -1 ÷ + bf ( x + 1) = x +1 è x +1ø Eq. (1) × a – Eq. (2) × b ö yø And f ç ÷ = cos log x ç ÷ è yø æ xö f ç ÷ = cos (log x – log y) è yø Adding (1) and (2), we get ....(2) Þ ( a 2 - b 2 ) f ( x + 1) = a( x + 1) - a - 7. Then the value of f (x)f (y) ü 1 ì æ xö – í f ç ÷ + f ( xy)ý 2 î è yø þ 1 = f ( x) f ( y) - .2 { f ( x). f ( y)} = 0 2 (c) We must x 4 - 21x 2 ³ 0 and 8. 10 - x 4 - 21x 2 ³ 0 ( ) x 2 x 2 - 21 ³ 0 ( )( ) Þ (2x – 3 ) (2x +1) £ 0 Þ - x 2 - 25 x 2 + 4 £ 0 Þ x 2 - 25 £ 0 Þ -5 £ x £ 5 Domain is given by (as x2 + 4 > 0 always) æ x + 59 ö 3 f ( x) + 2 f ç ÷ = 10 x + 30 è x -1 ø For x = 7, 3f (7) + 2f (11) = 70 + 30 = 100 For x = 11, 3f (11) + 2f (7) = 140 f (7) f (11) -1 = = or f(7) = 4 -20 -220 9 - 4 (b) 6. (a) æ 1 ö af ( x + 1) + bf ç ÷ = ( x + 1) - 1 ...(1) è x +1 ø 9. 1 3 £x£ 2 2 1 3 Domain of g(x) = éê- , ùú ë 2 2û Hence, domain of (f + g) = intersection of their é 1 ö domains = ê - ,1÷ . ë 2 ø (b) We have f(x) = log4[log5{log3 (18x – x2 – 77)}] Since, loga x is defined for all x > 0, f(x) is defined if log5 {log3 (18x – x2 – 77)} > 0 and 18x – x2 – 77 > 0 or log3 (18x – x2 – 77) > 50 and x2 – 18x + 77 < 0 or log3 (18x – x2 – 77) > 1 and (x – 11)(x – 7) < 0 or 18x – x2 – 77 > 31 and 7 < x < 11 or 18x – x2 – 80 > 0 and 7 < x < 11 or x2 – 18x + 80 < 0 and 7 < x < 11 \ é -5, - 21 ù È é 21,5 ù È {0} ë û ë û 5. b 2a + b = 2 2 (d) f (x) = cos(sin x ) + log x {x} Domain cos(sin x) ³ 0 {x} > 0, x > 0, x ¹ 1, logx {x} ³ 0 (i) cos(sin x) ³ 0 for all x, x Î R [–1, 1] (ii) {x} > 0, x Ï Int. (iii) x > 0, x Î(0, ¥) (iv) x ¹ 1 (v) log x {x} ³ 0 Þ 1 > f ( x ) ³ 0 , so 1 > x ³ 0 logx f(x) is positive x Î[0, 1) Þ x Î(0, 1) (c) Given that, g ( x ) = 3 + 4 x - 4 x 2 = -(2 x - 3)(2 x + 1) 2 Eq. (2) Þ x - 21x - 100 £ 0 Þ b +b 2 1 æ px ö f ( x ) = - tan ç ÷ , -1 < x < 1 è 2ø 2 Here, domain of f(x) = (– 1, 1) and ...(1) 4 2 ...(2) and 100 ³ x - 21x Eq. (1) gives x = 0 or x £ - 21 or x ³ 21 4 b +b x +1 =a+ æ xö Þ f (xy) + f ç ÷ = 2 f (x). f (y) è yø Þ ...(2) 2 2 Putting x = 1, (a - b ) f (2) = 2a - a - æ xö f (xy) + f ç ÷ è yø = cos (log x + log y) + cos (log x – logy) = 2 cos (log x). cos (log y) 4. 1 , we get x +1 Solutions 79 or (x – 10) (x – 8) < 0 and 7 < x < 11 or 8 < x < 10 and 7 < x < 11 or 8 < x < 10 or x Î (8, 10) Hence, the domain of f(x) is (8, 10). 1 x -x 10. (a) Clearly, g(x) = (a + a ) and 2 1 h( x ) = (a x - a - x ) . Now g(x + y) + g(x – y) 2 1 x+ y 1 + a - ( x + y ) ) + (a x - y + a - x + y ) = (a 2 2 1 x y x -y -x y -x -y = (a a + a a + a a + a a ) 2 1 x y -y -x y -y = (a (a + a ) + a (a + a )) 2 æ1 x öæ1 ö = 2 ç ( a + a - x )÷ ç ( a y + a - y ) ÷ = 2 g ( x ) g ( y ) è2 øè2 ø 11. (b) Given f (l + x) = f (l – x) ...(1) f (2l + x) = – f (2l – x) ...(2) for l > 0 Replacing x by l – x in l – x in (1), we get f(2l – x) = f(x) ...(3) \ From (2) and (3), f(x) = –f(2l + x) Þ f(x) = –[–f (2l + 2l + x)] Þ f(x) = f(x + 4l) ...(4) Þ f(x) is periodic with period 4l. Further from (3), replacing x by –x, we get f (2l + x) = f (–x) ...(5) From (2), (3) and (5), we have f (–x) = f (2l + x) = –f (2l – x) = –f (x) i.e. f (–x) = –f (x) Þ f(x) is odd function Thus, f is odd and periodic function. 12. (d) f ( x) = ax3 - bx - (tan x )sgn x Since f (- x) = f ( x) Þ -ax3 + bx - tan x sgn x = ax 3 - bx - (tan x ) (sgn x ) ( ) Þ 2 ax 2 - b x = 0 " x Î R Þa = 0 and b = 0 \ [ a ]2 - 5 [ a ] + 4 = 0 and 6 {a}2 - 5{a} + 1 = 0 Þ ([a] – 1)([a] – 4) = 0 and (3{x} – 1) (2{x} – 1) = 0 1 1 Þ [a] = 1, 4 and {a} = , 3 2 \ 1 1 1 1 a = 1+ , 1+ , 4 + , 4 + 3 2 3 2 Sum of all possible values of a = 35 3 æ 5x ö 13. (c) f (x) = cos nx . sin ç ÷ ; è nø Period of cos nx = Period of sin 2p |n| 5x 2p 2 | n | p = = 5 n 5 n æ 2p 2 | n | p ö \ Period of f (x) = L.C.M. ç , = 2p è|n| 5 ÷ø (given) æ 1 | n |ö L.C.M. (1, | n |) , =1= =1 Þ L.C.M. ç è | n | 5 ÷ø H.C.F. (| n |, 5) |n| = 1 Þ H.C.F. (|n|, 5) = |n| H.C.F. (| n |, 5) If g.c.d. (|n|, 5) = 1 Þ |n| = 1 Þ n = 1 If g.c.d. (|n|, 5) ¹ 1 Þ |n| = 5m; m Î ¥ Þ g.c.d. (5m, 5) = 1 Þ |n| = 5 Þ n = ±5 \ n Î {±1, ±5} 14. (d) f : ¡ ® ¡ , g; ¡ ® ¡ We know that min. {f1(x), f2(x)} ( f1 ( x) + f 2 ( x )) - | f1 ( x) - f 2 ( x) | = 2 \ min {f(x) – g(x), 0} ( f ( x ) - g ( x ) + 0) - | f ( x) - g ( x) - 0 | = 2 ( f ( x ) - g ( x)) - | f ( x ) - g ( x) | = 2 15. (b) y = f(ex) + f(|ln| |x|) Domain f(x) = (0, 1) Þ 0 < ex < 1 Þ x < 0 ...(1) and 0 < ln |x| < 1 Þ 1 < |x| < e Þ x Î {–e, –1} È (1, e) ... (2) Taking intersection x Î (–e, –1) Þ 16. (b) 1ö æ f ç x + ÷ = f ( x ); f (2) =5; 2ø è æ9ö æ1ö f ç ÷ = 2, f (-3) - f ç ÷ = ? è4ø è4ø MATHEMATICS 80 Q f(x) is periodic with period 1 2 Þ nö æ Þ f(x) = f ç x + ÷ " x Î ¥ è 2ø 10 ö æ Þ f(–3) = f çè - 3 + ÷ø = f (2) = 5 2 Þ 20. (d) < 2- 1 4 x 1 1 -1 < <Þ x Î f (null set) 2 4 x y 1 æ1 æ 1ö æ 1ö ö æ 9ö and f ç ÷ = f ç + 4 ç ÷ ÷ = f ç ÷ = 2 è 4ø è 2ø ø è 4ø è4 0 æ 1ö \ f (-3) - f çè ÷ø = 5 - 2 = 3 4 17. 18. é 2x + 3 x £ 1 (c) f(x) = ê 2 êë a x + 1 x > 1 For x £ 1 ; f(x) £ 5 So for range of f(x) to be R. Þ a2 + 1 £ 5 and a ¹ 0 Þ a Î [–2, 2] Hence, a = {–2, – 1, 1, 2} (b) g ( 2 f ( x) + 3) = log 5 ( 2 (sin x - cos x) + 3) We know that - 2 £ (sin x - cos x) £ 2 " x Î ¡ éQ - a 2 + b2 £ a sin x + b cos x £ a 2 + b 2 ù ëê ûú Þ - 2 £ 2(sin x - cos x) £ 2 Þ 1 £ 2(sin x - cos x) + 3 £ 5 Þ 0 £ log 5 ( 2 (sin x - cos x ) + 3) £ 2 (Q logax is increasing for a > 1) 1 ö æ (d) –g(2, f(x)) = - log 2 ç 1 + ÷ è 4 xø Þ 1 ö æ –g(2, f(x)) – 1 = - log 2 ç1 + ÷ -1 è 4 xø æ1 ö \ g ç , - g (2, f ( x )) - 1÷ è2 ø æ 1 ö ö æ = log1/2 ç - log 2 ç 1 + ÷ - 1÷ 4 xø ø è è Þ 1 ö æ log 2 ç1 + ÷ +1< 0 è 4 xø 1 2 x 3 0 £ f (x) £ 1 Þ 0 £ 7 f(x) £ 7 Þ – 1 £ sin (7 f(x)) £ 1 21. (5.5) Here f ( - x ) = 1 1 + ex 1 1 + So, f ( x ) + f ( - x ) = -x 1+ e 1 + ex = ex + 1 ex + 1 1+ ex = ex +1 ex +1 =1 \ S = {f ( 5 ) + f ( -5 )} + ... + {f (1) + f ( -1)} + f ( 0 ) 1 11 = 2 2 1+ e (2) f(x + p) = 1 + {1 – 3f (x) + 3f 2(x) – f 3(x)}1/3 Þ f(x + p) = 1 + (1 – f(x)) = 2 – f(x) Þ f(x + p) = 2 – [2 – f(x – p)] Þ f(x + p) = f(x – p) Þ f(x) = f(x + 2p) Þ Period of f(x) = 2p = lp (given) Þ l=2 (9) Given f(x + 2) = f(x) + f(2) Put x = –1. Then f(1) = f(–1) + f(2) or f(1) = –f(1) + f(2) [as f(x) is an odd function] or f(2) = 2f(1) = 6 Now, put x = 1 We have f(3) = f(1) + f(2) = 3 + 6 = 9 (6) Given f(x, y) = f(2x + 2y, 2y – 2x) " x, y Î ¡ , f(x) = f(2x, 0) and f(x) is periodic with period k. Þ f(x) = f(2x, 0) = f(2.2x + 2(0), 2(0) – 2.2x) = f(2x + 1, 2x +1) = f(2.2x + 1 – 2.2x + 1, – 2.2x + 1 – 2.2x + 1) = f (0, – 2x + 3) = f(2.(–2x + 3), –2.2x + 3) = f(– 2x + 4, – 2x + 4) = f(–2x + 6, 0) = f(–2x +7, 2x + 7) = f(0, 2x + 9) = f(2x + 10, 20x + 10) = f(2x + 12, 0) = f(x + 12) Þ f(x) is periodic with period 12 Þ k = 12. = 1 + 1 + 1 + 1 + 1 + f(0) = 5 + 22. 23. Hence, range of g ( 2 f ( x) + 3) is [0, 2]. 19. 1 0 <1+ 24. 1 0 = 5+ Solutions 81 é x ù é 15 ù 25. (6) f(x) = ê ú ê - ú ë15 û ë x û 0 £ x < 15 15 £ x < 30 30 £ x < 45 45 £ x < 60 f(x) = – 3 60 £ x < 75 f(x) = – 4 75 £ x < 90 f(x) = – 5 Total integers in range f(x) = {0, –1, –2, –3, –4, –5} x Î (0, 90) f(x) = 0 f(x) = – 1 f(x) = – 2 CHAPTER Trigonometric Functions 3 1. (d) 2 + 1 2 cos a 1 + sin a 2 = 2. 1 2 + 2 4 + 4 a Þ sin 2 = 2 8 1 + sin a 1 + sin a 4. Þ b = p - a , g = 2p + a, d = 3p - a . So that the given expression is equal to 2p + a 3p - a a æ p-a ö + sin + 3 sinç ÷ + 2 sin 2 2 2 è 2 ø = 4sin a a a a + 3cos - 2sin - cos 2 2 2 2 aö a a æ a = 2çsin + cos ÷ = 2 1+ 2sin cos = 2 1+ k è 2 2ø 2 2 3. (c) 1 + cos a = 1 + = … (i) b b b b - cos 2 1 + 2 sin 2 - [1 - sin 2 ] 2 2= 2 2 2b 2b 1 + 2 sin 1 + 2 sin 2 2 1 + 2sin 2 5. 2 cos b - 1 2 - cos b + 2cos b - 1 = 2 - cos b 2 - cos b b cos 2 1 + cos b 2 a 2 = Þ cos = b 2 2 - cos b 1 + 2 sin 2 2 b cos 2 2a 2 =1Þ 1 - cos b 2 1 + 2sin 2 2 1 + 2 sin 2 6. …(ii) a 2 = 3 b tan 2 (b) Gives equations can be written as 2 cos a + 9 cos d = – 6 cos b – 7 cos g ..(i) 2 sin a – 9 sin d = 6 sin b – 7 sin g ...(ii) Square and add equations (i) and (ii), Þ 4 + 36 + 36 [cos a cos d – sin d sin a] = 36 + 49 + 84 [cos b cos g – sin b sin g] Þ 36 [cos(a + d)] = 84 [cos (b + g)] a b tan 2 = 3 tan 2 Þ 2 2 sin a = sin b = sin g = sin d = k 4 sin b 2 b 2 Divide equation (ii) by (i) 4 + + 1 - sin 4 a 1 + sin 4 a 1 + sin8 a 4 4 8 = + = 1 - sin 8 a 1 + sin 8 a 1 - sin16 a 40 8 1ö = 10 æ = = 16 çQ sin a = ÷ 1 4 5ø 1è 5 (b) a < b < g < d and 3 sin 2 tan cos (a + d) 84 7 m = = = ; m + n = 10 cos (b + g ) 36 3 n (a) sin q + sin 3q + sin 2q = sin a Þ 2 sin 2q cos q + sin 2q = sin a Þ sin 2q(2 cos q + 1) = sin a ...(i) Now cos q + cos 3q + cos 2q = cos a 2 cos 2q cos q + cos 2q = cos a cos 2q (2 cos q + 1) = cos a ...(ii) From (i) and (ii), tan 2q = tan a Þ 2q = a Þ q = a/2 (c) (sin 7a + sin 5a ) + 5(sin 5a + sin 3a) +12(sin 3a + sin a ) sin 6a + 5 sin 4a + 12 sin 2a 2sin 6a cos a + 5(2sin 4a cos a) + 12(2sin 2a cos a ) = sin 6a + 5sin 4a + 12sin 2a = 2 cos a = 5 +1 2 MATHEMATICS 82 7. (b) 2 é 2sin q cos q 2sin 3q cos3q 2sin9q cos9 q ù + + 1 êê cos q cos3q cos9q cos3q cos9q cos 27 q úú 2ê 2sin 27q cos 27q ú + ê cos 27q cos81q úû ë é sin(3q - q) sin(9q - 3q) sin(27q - 9q) ù + + ê ú 1 ê cos q cos3q cos3q cos9q cos9q cos 27 q ú = 2ê sin(81q - 27q) ú + ê cos 27q cos81q úû ë = 8. 1 1 é sin 80 q ù [tan 81q - tan q] = ê 2 2 ë cos q cos 81q úû 11. qö q æ = 4 ç 2 cos q× cos ÷ = 16 cos 2 q× cos 2 è ø 2 2 = 4 (1 + cos q) (1 + cos 2q) (a) Let ìï 22 p üï ìï 264 p üï ì 2p ü P = cos í cos ...cos ý í 64 ý í 64 ý î 264 - 1 þ îï 2 - 1 þï îï 2 - 1 þï Þ P= æ 22 p ö æ 22 p ö sin ç ÷ cos ç ÷ æ 2p ö çè 264 -1 ÷ø çè 264 -1 ÷ø 2sin ç ÷ è 264 -1 ø 1 p+q 1 p+q Þ = p-q sin q p - q Apply componendo and dividendo æ 264 p ö ...cos ç ÷ ç 264 - 1 ÷ è ø (b) cosec q = 1 + sin q p + q + p - q = 1 - sin q p + q - p + q Þ P= 2 2 q qü ì qü ì cos sin + 1 tan + ï ï 2 2ï = p 2ï = p Þ í Þí ý ý q q q q ï cos - sin ï ï1 - tan q ï î 2 2þ î 2þ qö q 2æp æp qö p Þ tan 2 ç + ÷ = Þ cot ç + ÷ = 4 2ø p è è 4 2ø q 9. (b) Given, sin 2q + sin 2f = 1/2 ...(i) and cos 2q + cos 2f = 3/2 ...(ii) Square and adding, \ (sin2 2q + cos2 2q) + (sin2 2f + cos2 2f) + 2[sin 2q sin 2f + cos 2q cos 2f] = 1/4 + 9/4 Þ cos 2q cos 2f + sin 2q sin 2f = 1/4 Þ cos (2q – 2f) = 1/4 Þ cos2 (q – f) = 5/8 10. (a) u = 1 + cos q + cos 2q + cos (q + 2q) = (1 + cos 3q) + (cos q + cos 2q) 3q 3q 3q + 2 cos + cos 2 2 2 3q é 3q qù cos + cos ú = 2 cos 2 êë 2 2û = 2 cos 2 Similarly v = 2 sin \ 3q é 3q qù cos + cos ú ê 2 ë 2 2û 3q qö æ u2 + v2 = ç cos + cos ÷ 2 2ø è 2 Þ P= æ 23 p ö æ 264 p ö sin ç ...cos ç ÷ ÷ ç 264 -1÷ æ 2p ö çè 264 -1÷ø è ø 2sin ç ÷ è 264 -1ø 1 æ 265 p ö 1 1 sin ç ÷ = 64 64 ç ÷ æ p ö 2 è 2 -1 ø 2 264 sin ç ÷ è 264 - 1 ø 12. (c) sin x sin y + 3 cos y + 4 sin y cos x = \ \ 26 3 cos y + (sin x + 4 cos x) sin y = 3 cos y + (sin x + 4 cos x) sin y 26 £ 9 + (sin x + 4 cos x) 2 £ 9 + 1 + 16 = 26 \ sin x sin y + 3 cos y + 4 sin y cos x = 26 cos y sin y cos x = Þ sin x sin y = 3 4 Þ 3 tan y = cosec x and tan x = 1/4 Þ 9 tan2 y = cosec2 x = (1 + cot2 x) = 17 1 9 + Þ tan2 x + cot2 y = 16 17 x x x x x × cos × × cos × ×.....cos × × cos × 13. (b) cos × 256 128 64 4 2 sin x = æ x ö 256 sin ç è 256 ÷ø 14. (a) – 5 £ 3 sin x – 4 cos x £ 5 10 £ 3 sin x – 4 cos x + 15 £ 20 log20 10 £ log20 (3 sin x – 4 cos x + 15) £ log20 20 Solutions 83 15. (c) Given expression sin 50° + sin 140° + sin 170° = 2 sin 25° sin 70° sin 85° 4 (Using sin 2A + sin 2B + sin 2C = 4 sin A 2 sin B sin C, where A + B + C = 180°) =2 = 2 2 16. (c) 32(1 + tan x) + 1 - 10 × 3tan x = 0 Þ 32 (3tan x )2 + 1 - 10.3tan x = 0 . 2 2 2 Put 3tan x = t ; 9t2 – 10t + 1 = 0 1 and t = 1 9 Þ 3tan 2 x = 3–2 and 3tan 2 x = 1 Þ tan2x = –2 gives no solution and tan2 x = 1 gives four solutions. 17. (c) (4cos2 2x + 4 cos2x + 1) Þ t= 2 Þ Þ Þ Þ Þ \ 18. (b) + (tan x – 2 3 tan x + 3) = 0 (2 cos 2x + 1)2 + (tan x – 3 )2 = 0 1 cos 2x = - and tan x = 3 2 1 2cos2 x – 1 = - and tan x = 3 2 1 cos x = ± - and tanx = 3 2 p 4p x= , 3 3 Number of solutions of equation = 2. 2 sin x - 3 2cos2 x - 3cos x + 1 =1 2 Case I: 2 cos x – 3 cos x + 1 = 0 1 p cos x = ,1 ; x = 0, 2 3 p But at x = L.H.S. = 0° 3 p \ x= (rejected) \ x = 0 is a 3 solution Case II: 2 sin x – 3 = 1, – 1 \ \ \ 3 +1 3 -1 3 -1 , sin x = 2 2 2 x has 2 values in [0, p]. Total number of solutions = 2 + 1 = 3. sin x = 19. (a) sin2 (sin x) – 3 sin (sin x) + 2 = 0 {sin (sin x) – 2} {sin(sin x) – 1} = 0 Equation has no solution. 20. (c) x2 + 12 + 3 sin (a + bx) + 6x = 0 Þ (x + 3)2 + 3 + 3 sin (a + bx) = 0 Þ (x + 3)2 + 3 = – 3 sin (a + bx) L.H.S. ³ 3 but R.H.S. £ 3 L.H.S. = R.H.S. = 3 \ x = – 3 and sin (a + bx) = – 1 Þ sin (a – 3b) = – 1 p or a – 3b = (4n – 1) n Î Z. 2 1 1 21. (50) 5 2 + 5 2 1 Þ Þ Þ + log5 (sin x) 1 + log5 cos x = 15 2 1 5 2 + 5 2 × 5log5 (sin x) = 151/ 2 ×15 log 15 cos x 51/2 + 51/2 × sin x = 151/2 × cos x 1 + sin x = 3 cos x 3 sin x 1 cos x = 2 2 2 pö p æ Þ cos ç x + ÷ = cos 6 3 è ø p p Þ x + = 2np ± , n Î Z 6 3 p p Þ x = 2np - , 2np + , n Î Z . 2 6 But we must have sin x, cos x > 0 \ x = 2np + p/6, n Î Z. sin x sin 3x sin 9x + + =0 22. (6) cos 3x cos 9x cos 27x 2sin x cos x 2sin 3x cos3x + or 2cos3x cos x 2cos9x cos3x 2sin 9x cos9x + =0 2cos 27x cos9x sin(3x - x) sin(9x - 3x) + or 2cos3x cos x 2cos9x cos3x sin(27x - 9x) + =0 2cos 27x cos9x or (tan3x – tan x) + (tan9x – tan3x) + (tan27x – tan9x) = 0 or tan 27x – tan x = 0 or tan x = tan 27 x np , nÎI Þ 27x = np + x, n Î I or x = 26 p 2p 3p 4p 5p 6p or , , , , , 26 26 26 26 26 26 Hence, there are six solutions. Þ MATHEMATICS 84 23. (4) Since – 2 £ sin x – 3 cos x £ 2, we have 2m - 3 4m - 6 £1 -2 £ £ 2 or -1 £ 4-m 4-m If 2m - 3 £ 1, we have 4-m (2m - 3) - (4 - m) 3m - 7 £ 0 or ³0. 4-m m-4 ...(i) m +1 2m - 3 £0 Þ Also, - 1 £ ...(ii) m-4 4-m x p æ x pö x = 2mp, mÎ Zor - ç + ÷ = np + , n ÎI 2 è 2 6ø 2 p . \ x = 0 or x = 3 p p 25. (2) Given < 3x - £ p 2 2 Þ p æ pö p ö -p æ < ç 3x - ÷ £ p or - p £ ç 3x - ÷ < 2 è 2ø 2ø 2 è 3p -p Þ p< 3x = or £ x < 0 2 6 é -p ö æ p p ù \ xÎê ,0 ÷ È ç , ú ë 6 ø è 3 2û Now, 1 + cos x + cos 2x + sin x + sin 2x + sin 3x = 0 Þ 2 cos2x + cos x + sin 2x + 2 sin 2x cos x = 0 Þ cos x (2 cos x + 1) + sin 2x (2 cos x + 1) = 0 Þ (cos x + sin 2x) (2 cos x + 1) = 0 Þ cos x (1 + 2 sin x) (2 cos x + 1) = 0 -1 Þ cos x = 0 or sin x = 2 (as for given interval, cos x > 0) p -p . Þ x = or x = 2 6 Hence, there are 2 solutions. Þ 7ù é From Eqs. (i) and (ii), we get m Î ê -1, ú . 3û ë Therefore, the possible integers are – 1, 0, 1, 2. 24. (2) We have, 2 æ pö æ æ p öö 2 ç cos ç x + ÷ - cos ç ÷ ÷ = sin x 6 6 è ø è ø è ø Þ æxö æ x pö æxö æxö 4sin2 ç ÷ × sin2 ç + ÷ = 4sin2 ç ÷ × cos2 ç ÷ è 2ø è 2 6ø è 2ø è2ø \ sin x x æ x pö = 0 or sin 2 ç + ÷ = cos 2 2 2 è2 6ø CHAPTER Principle of Mathematical Induction 4 1. 2. 3. (d) We note that P(1) = 2 and hence, P(n) = n (n + 1) + 2 is not true for n = 1. So the principle of mathematical induction is not applicable and nothing can be said about the validity of the statement P(n) = n (n + 1) + 2. (c) The product of r consecutive integers is divisible by r ! . Thus n (n + 1 ) (n + 2) (n + 3) is divisible by 4 ! = 24. 2 + 2 + 2 + ...... 2 = 2 + 2 + 2 + ...... 2 144 42444 3 ( k +1) terms æ p ö ÷÷ = 2 + 2 cosçç è 2 k +1 ø (a) Let f (n ) = 2 + 2 + ..... + 2 (number of roots is n) Then f(1) = 2 = 2 cos p or 2 sin p 4 4 \ f (1) may be true for (a) as well as for (b) Again f(2) = 2 + 2 = 2´ \ f (2) is true for (a). We check it for any integer. 2+ 2 p = 2 cos 2 8 æ p ö Let 2 + 2 + ...... 2 = 2 cosçç k +1 ÷÷ for some k - roots è2 ø ...(i) k ³1 Now , k terms [From (i)] p é p ù 2 p = 2ê1+ cos k+1 ú = 2.2 cos k+2 = 2 cos k+2 2 û 2 2 ë \ The result is true for n = k + 1. Hence, by the principle of mathematical induction 4. æ p ö 2 + 2 + ....+ 2 = 2 cosçç ÷÷ for all n Î N . è 2n +1 ø n -roots (a) It can be proved with the help of n mathematical induction that < a(n) £ n. 2 Solutions 85 200 < a(200) 2 Þ a(200) > 100 and a(100) £ 100. 1 1 1 1 + + .... + (a) Let Un = + n n +1 n + 2 2n - 1 1 1 1 1 Vn = 1 - + - + ..... + 2 3 4 2n - 1 Let T(n) be the statement Un = Vn Then T(1) is true. For U1 = 1 and V1 = 1, so that U1 = V1 Let T(k) be true for some positive integer K. Now, \ 5. 1 1 ù é 1 U k +1 - U k = ê + + .... + k + 1 k + 2 2 k + 1 úû ë 7. 1 1 1 1 = Þ = is true 1.2 1 + 1 2 2 thus P(n) is true for n = 1 Suppose that P(k) is true for some natural number ‘k’ 1 1 1 1 k + + + ..... + = ......(1) 1.2 2.3 3.4 k ( k + 1) k + 1 Now, 1 1 1 1 1 + + + ..... + + 1.2 2.3 3.4 k (k + 1) (k + 1)(k + 2) = k (k + 2) + 1 (k + 1)(k + 2) [From (1)] (k + 1)2 k +1 k +1 = = (k + 1)(k + 2) k + 2 (k + 1) + 1 Thus P(k + 1) is true whenever P(k) is true. Hence. by the principle of mathematical induction P(n) is true for all natural numbers. (a) Let us write the statement æ 2n +1ö 3öæ 5ö æ 7ö P(n) : çæ1+ ÷ç 1+ 1+ .......ç1+ = (n +1)2 2 ÷ø è 1øè 4÷ø çè 9÷ø è n We note that æ 3ö P : ç1 + ÷ = 4 = (1 + 1) 2 , which is true è 1ø Thus P (n) is true for n = 1 Suppose that P(k) is true for some natural number ‘k’ i.e., æ 3ö æ 5 ö æ 7 ö æ 2k + 1ö 2 çè1 + ÷ø çè1 + ÷ø çè1 + ÷ø ..... ç1 + 2 ÷ = ( k + 1) è 1 4 9 k ø 1 ù é 1 1 - ê1 - + - ..... + 2 3 2 k - 1 úû ë 6. k 1 + k + 1 (k + 1)(k + 2) = 1 1 ù é1 + .... + – ê + 2k - 1 úû ë k k +1 1 1 1 1 1 + - =+ = ...(i) 2k 2k + 1 k 2k 2k + 1 Also, 1 ù é 1 1 Vk +1 - Vk = ê1 - + - ...... + 2 3 2 k + 1úû ë 1 1 + =...(ii) 2k 2k + 1 From (i) and (ii), we find that U k +1 - U k = Vk +1 - Vk Since, Uk = Vk, therefore, it follows that Uk+1 = Vk+1 \ T (n ) is true for all n Î N . (a) Let us write the statement 1 1 1 1 n P(n) : 1.2 + 2.3 + 3.4 + ..... + n(n + 1) = n + 1 = ....(1) Now, æ 3ö æ 5 ö æ 7 ö æ 2k + 1ö çè1 + ÷ø çè1 + ÷ø çè1 + ÷ø ..... çè1 + 2 ÷ø 1 4 9 k ìï 2k + 3 üï í1 + 2ý îï ( k + 1) þï ì 2k + 3 ïü 2 ï = ( k +1) í1 + 2ý îï (k + 1) þï [Using (1)] 2 ï (k + 1) + 2k + 3üï 2 2ì = ( k +1) í ý = k + 2k + 1 + 2k + 3 (k +1)2 îï þï we note that P(1) : = k 2 + 4 k + 4 = ( k + 2) 2 = {( k + 1) + 1} 2 8. 9. Thus , P(k + 1) is true whenever P(k) is true. Hence by the principle of mathematical induction P(n) is true for all natural numbers. (d) Q 21 > 12. 22 = 22, 23 < 32, 24 = 42 But 25 > 52, 26 > 62 and so on. 4n (2n)! < (d) Let P(n) : n + 1 (n !)2 For n = 2, MATHEMATICS 86 16 24 42 4! < < Þ 2 3 4 2 + 1 (2) which is true. Let for n = m ³ 2, P(m) is true. P(2) : 4m (2m)! < m + 1 ( m !) 2 i.e. Now, = (2m)! 4(m + 1) 4m+1 4m 4(m + 1) · · < = (m !) 2 (m + 2) m +1 m + 2 m+2 (2m)!(2m + 1)(2m + 2)4(m + 1)(m + 1) 2 (2m + 1)(2m + 2)(m !) 2 ( m + 1) 2 ( m + 2) [2(m + 1)]! [2(m + 1)]! 2(m + 1) 2 · < 2 [(m + 1)!] (2m + 1)(m + 2) [(m + 1)!]2 Hence, for n ³ 2, P(n) is true. 10. (b) Given that, P(n) : 3n < n! Now, P(7) : 37 < 7! is true Let P(k) : 3k < k! Þ P(k + 1) : 3k+1 = 3.3k < 3.k! < (k + 1)! (Q k + 1 > 3) 11. (c) Let P(n) = 2 . 42n + 1 + 33n + 1 Then P(1) = 2 . 43 + 34 = 209, which is divisible by 11 but not divisible by 2, 7 or 27. Further, let P(k) = 2 . 42k + 1 + 33k + 1 is divisible by 11, that is, = 2.4 2 k +1 3k +1 +3 = 11q for some integer q. Now P (k + 1) = 2 . 42 k +3 + 33k + 4 2k +1 =2 . 4 2 3k +1 3 .4 + 3 2k +1 .3 =16 . 2.4 13. (a) n p - n is divisible by p for any natural number greater than 1. Trick : Let n = 4 and p = 2 Then (4)2 – 4 = 16 – 4 = 12, it is divisible by 2. So, it is true for any natural number greater than 1 14. (b) For n = 1, we have 49n + 16n + l = 49 + 16 + l = 65 + l = 64 + ( l + 1), which is divisible by 64 if l = – 1 For n = 2, we have 49n + 16n + l = 492 + 16 × 2 + l = 2433 + l = 64 × 38 + ( l + 1), which is divisible by 64 if l = – 1 Hence, l = – 1 15. (b) Let P(n) : 1 1 1 + + ..... + 1× 2 × 3 2 × 3 × 4 n ( n + 1) (n + 2 ) = n (n + 3) 4 (n + 1) (n + 2) For n = 1, + 27.3 = 16 . 2 . 42 k +1 + (16 + 11) . 33k +1 = 16 [2. 42 k +1 + 33k +1 ] + 11 . 33k +1 1(1 + 3) 1 = 4 (1 + 1) (1 + 2) 6 \ P(1) is true. Let P(k) is true, then P(k) : 1 1 1 + + ..... + 1× 2 × 3 2 × 3 × 4 k ( k + 1) (k + 2 ) k (k + 3) 4 (k + 1) (k + 2) For n = k + 1, 1 1 + + ..... P(k + 1) : 1× 2 × 3 2 × 3 × 4 = 16 . 11q + 11 . 33k +1 = 11(16q + 33k +1 ) = 11m where m = 16q + 33k + 1 is another integer. \ P(k +1) is divisible by 11. \ P(n ) = 2 . 42 n +1 + 33n +1 is divisible by 11 for all n Î N . + = (k + 1) ( k + 4) 4 ( k + 2 ) ( k + 3) + = ... (i) 1 1 + k (k + 1) (k + 2 ) (k + 1) (k + 2 ) (k + 3) L.H.S. = 2 12. (b) n(n - 1) = (n - 1)(n)( n + 1) It is product of three consecutive natural numbers, so according to Langrange’s theorem it is divisible by 3! i.e., 6 1 1 = and 1× 2 ×3 6 R.H.S.= = 3k +1 L.H.S.= 1 1 + + ..... 1× 2 × 3 2 × 3 × 4 1 1 + k (k + 1) (k + 2 ) (k + 1) (k + 2 ) (k + 3) k ( k + 3) 4 (k + 1) (k + 2 ) + 1 (k + 1) (k + 2) (k + 3) [from (i)] Solutions 87 (k + 1)2 (k + 4) (k +1)(k + 4) = = 4 (k + 1)(k + 2)(k + 3) 4 (k + 2)(k + 3) = R.H.S. Hence, P(k + 1) is true. Hence, by principle of mathematical induction for all n Î N, P(n) is true. 16. (b) 3.52n + 1 + 23n + 1 Put n = 1, we get (3 × 53) + 24 = 391, which is divisible by 17. Put n = 2, we get (3 × 55) + 27 = 9503, which is divisible by 17 only. 17. (b) Let the statement P(n) be defined as P(n) = 1.3 + 2.32 + 3.33 + ..... + n.3n = (2n – 1) 3n + 1 + 3 4 Step I : For n = 1, P(1) : 1.3 = (2.1 – 1) 31 + 1 + 3 = 32 + 3 4 4 9 + 3 12 = = 3 = 1.3, which is true. = 4 4 Step II : Let it is true for n = k, i.e. 1.3 + 2.32 + 3.33 + ..... + k.3k = (2k – 1) 3k + 1 + 3 ... (i) 4 Step III : For n = k + 1, (1.3 + 2.32 + 3.33 + ..... + k.3k) + (k + 1)3k+1 = (2k – 1) 3k + 1 + 3 4 [Using equation (i)] = (2k – 1) 3k + 1 + 3 + 4 (k + 1) 3k + 1 4 k +1 = + (k + 1)3 k+1 3 (2k – 1 + 4k + 4) + 3 4 [taking 3k + 1 common in first and last term of numerator part] = 3k + 1 (6k + 3) + 3 = 3k + 1 × 3 (2k + 1) + 3 4 4 [taking 3 common in first term of numerator part] k + 1) + 1 3( [ 2k + 2 – 1] + 3 = 4 k + 1) + 1 éë 2 ( k + 1) – 1ûù 3( +3 = 4 Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n. 18. (d) Let the statement P(n) be defined as 1 1 + + ..... P(n) : 1 + 1+ 2 1+ 2 + 3 + i.e. P(n) : 1+ 1 2n = 1 + 2 + 3 + ..... + n n + 1 1 1 2 2n + + ..... + = 1+ 2 1+ 2 + 3 n (n +1) n +1 é n (n + 1) ù êQ sum of natural numbers = ú 2 û ë Step I : For n = 1, 2 ´1 2 = = 1, which is true. P(1) : 1 = 1+1 2 Step II : Let it is true for n = k, 1 1 2 2k + + ..... + = i.e. 1 + 1+ 2 1+ 2 + 3 k (k + 1) k + 1 ...(i) Step III : For n = k + 1, æ 1 1 2 ö 2 ç1+ 1+ 2 + 1+ 2 + 3 + ..... + k (k +1) ÷ + (k +1)(k + 2) è ø = 2k 2 + [using equation (i)] k + 1 (k + 1) (k + 2) é 2 ù 2k (k + 2 ) + 2 2 ë k + 2k + 1û = = (k + 1) (k + 2) (k + 1) (k + 2) [taking 2 common in numerator part] 2 (k + 1) 2 (k + 1) 2 (k + 1) = = = (k + 1) (k + 2) k+2 ( k + 1) + 1 Therefore, P(k + 1) is true, when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n. 19. (c) 24 º 1 (mod 5) Þ (24)75 º (1)75 (mod 5) i.e. 2300 º 1 (mod 5) Þ 2300 × 2 º (1.2) (mod 5) Þ 2301 º 2 (mod 5) \ Least positive remainder is 2. 20. (d) Let P(n) be the statement given by P(n) : 32n when divided by 8, the remainder is 1. or P(n) : 32n = 8l + 1 for some l Î N For n = 1, P(1) : 32 = (8 × 1) + 1 = 8l + 1, where l = 1 2 MATHEMATICS 88 \ P(1) is true. Let P(k) be true. Then, 32k = 8l + 1 for some l Î N ... (i) We shall now show that P(k + 1) is true, for which we have to show that 32(k + 1) when divided by 8, the remainder is 1. Now, 32(k + 1) = 32k . 32 = (8l + 1) × 9 [Using (i)] = 72l + 9 = 72l + 8 + 1 = 8(9l + 1) + 1 = 8m + 1, where m = 9l + 1 Î N Þ P(k + 1) is true. Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true for all n Î N. 21. (133) Putting n = 1 in 11n + 2 + 122n + 1 We get, 111+2 + 122×1+1 = 113 + 123 = 3059, which is divisible by 133. 22. (27) Let P(n) be the statement given by P(n) : 41n – 14n is a multiple of 27 For n = 1, i.e. P(1) = 411 – 141 = 27 = 1 × 27, which is a multiple of 27. \ P(1) is true. Let P(k) be true, i.e. 41k – 14k = 27l ... (i) For n = k + 1, CHAPTER 2. 25. (8) 599 = (5)(52 )49 = 5(25) 49 = 5(26 - 1) 49 = 5 ´ (26) ´ (Positive terms) –5 , So when it is divided by 13 it gives the remainder –5 or (13 – 5) i.e., 8. Complex Numbers and Quadratic Equations 5 1. 41k + 1 – 14k + 1 = 41k 41 – 14k 14 = (27l + 14k) 41 – 14k 14 [using (i)] = (27l × 41) + (14k × 41) – (14k × 14) = (27l × 41) + 14k (41 – 14) = (27l × 41) + (14k × 27) k = 27(41l + 14 ), which is a multiple of 27. Therefore, P(k + 1) is true when P(k) is true. Hence,from the principle of mathematical induction, the statement is true for all natural numbers n. 23. (8) Let m = 2k + 1, n = 2k – 1 (k Î N) \ m2 – n 2 = 4k2 + 1 + 4k – 4k2 + 4k – 1 = 8k Hence, all the numbers of the form m2 – n 2 are always divisible by 8. 24. (16) 32n+ 2 – 8n – 9, " n Î N Putting n = 2 Þ 32 × 2 + 2 – 8 × 2 – 9 = 729 – 16 – 9 = 704 It is divisible by 16. (a) | x1z1 – y1z2 |2 + | y1z1 + x1z2 |2 = | x1z1 |2 + | y1z2|2 – 2Re(x1y1z1z2) + | y1z1 |2 + | x1z2 |2 + 2Re(x1y1z1z2) 2 2 = x1 | z1 | + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2 = x12 | z1 |2 + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2 = 2(x12 + y12) (42) = 32(x12 + y12) (d) Q | z | = | w| and arg z = p - arg w Let w = reiq , then z = rei (p-q) 4. 5. 2 + (1– i ) n2 z = reip . e-iq = (re -iq ) (cos p + i sin p) = w (-1) = -w (c) Q | z - 1| + | z + 3 | £ 8 \ z lies inside or on the ellipse whose foci are (1, 0) and (– 3, 0) and vertices are (– 5, 0) and (3, 0). Clearly the minimum and maximum values of | z - 4 | are 1 and 9 respectively, representing Þ 3. Observe that (1+ i ) n1 + (1– i )n1 = the distances PA and PA¢. Thus, 1 £ | z - 4 | £ 9 . A' (– 5, 0) S' (–3, 0) S A (4, 0) (1, 0) (3, 0) P i i2 (b) Let z = sin ln (ii )i + cos ln (i ) ì æ p ö –1 ü ì æ p ö –1 ü – ï ï ï ï = sin íln ç e 2 ÷ ý + cos íln ç e 2 ÷ ý ç ÷ ç ÷ ï è ø ï ï è ø ï î þ î þ p p = sin + cos = 1 2 2 (d) Let z = (1+ i ) n1 + (1– i )n1 + (1 + i )n2 6. np ( 2 ) ìíî2cos 14 üýþ , always real " n1 Î ¥ Similarly (1 + i )n2 + (1– i ) n2 is always real " n2 Î ¥ Þ n1, n2 may be any positive integers (cos a k + i sin a k )2 (d) wk = zk Solutions = 89 zk 2 rk2 zk = 1 zk2 = z We have zk zk zk k 1 1 1 + + z3 z1 z2 w1 + w2 + w3 = Þ æ1 1 1ö = ç + + ÷ =0 z ø èz z 1 7. 2 O 3 \ The origin O is the centroid of DA1A2A3. (c) If P(z1) is the reflection of Q(z2) through the line b z + b z = c in the argand plane. Then, 9. (d) z1 – z2 = cos a + i sin a z1 + z2 Þ cos a + i sin a + 1 2 z1 = cos a –1 + i sin a – 2 z2 P(z)1 æ z +z ö Rç 1 2÷ è 2 ø bz + bz =c = Q(z)2 C(z3) B(z2) 2 cos 2 a / 2 + 2i sin a / 2 cos a / 2 2i sin a / 2 cos a / 2 – 2sin 2 a / 2 2cos a / 2[cos a / 2 + i sin a / 2] = 2i sin a / 2[cos a / 2 + i sin a / 2] æ z1 + z2 ö R çè lies on the line. 2 ÷ø Þ z1 a z K = i cot Þ Given 1 = 3 1 z2 z2 ... (i) \ tan a/2 = – 1/K Since, PQ is perpendicular to the line b z + b z = c. Therefore Slope of PQ + Slope of the line = 0 Þ æ z1 + z2 ö æ z1 + z2 ö + =c b çè 2 ÷ø b çè 2 ÷ø Þ b z1 + b z1 + b z2 + b z2 = 2c æ z2 – z1 ö æ – b ö Þ ç z – z ÷ + çè ÷ =0 b ø è 2 1ø Þ b (z2 – z1) – b ( z2 – z1 ) = 0 8. æ 2 2 ö 2r 2 ( z1 – z2 ) 2 z2 – z1 ) = z3 r ç 2 2 ÷ z1 z2 è z1 z2 ø [\ |z1|2 = |z2|2 = r2] A(z1) 2z1 z2 z3 = r z1 + z2 Þ Þ b z2 – b z1 – b z2 + b z1 = 0 Adding Eqs. (i) and (ii), we get 2 ( b z1 + b z2) = 2c Þ b z1 + b z2 = c (b) As DOAC is a right triangle with right angle at A, |z1|2 + |z3 – z1|2 = |z3|2 Þ 2|z1|2 – z3 z1 – z1 z3 = 0 Þ 2z1 – z3 – z1 z =0 z1 3 z Similarly, 2z2 – z3 – 2 z3 = 0 z2 Subtracting (2) from (1) we get æz z ö 2 ( z2 – z1 ) = z3 ç 1 – 2 ÷ è z1 z2 ø ... (1) –2 / K 1–1/ K 2 Þ Þ tan a = 2 tan a / 2 1– tan 2 a / 2 –2 K K 2 –1 æ 2K ö ÷ Þ 2 tan– 1 (K) a = tan– 1 ç è 1– K 2 ø 10. (d) z2 + az + a2 = 0 Þ z = aw, aw2 (where 'w' is non-real root of cube unity) Þ Locus of z is a pair of straight lines and arg(z) = arg(a) + arg(w) or arg(a) + arg(w2) 2p Þ arg(z) = ± 3 Also, |z| = |a||w| or |a||w2| Þ |z| = |a|. 11. (c) Zp = r(cos q + i sin q) y Q ... (2) 2r p/4 O a q x r P(z) x MATHEMATICS 90 æ pö pöö æ æ 2 | z |2 ç cos ç q + ÷ + i sin ç q + ÷ ÷ è ø è è 4 4øø ZQ = æ -b ö a ´ ç ÷ + 2b b è a ø = = c b ac æ ö a 2 . + ab . ç ÷ + b2 a è a ø é æ pö pö ù æ 2r êcos ç q + ÷ + i sin ç q + ÷ ú è ø è 4 4ø û ë From the figure, = 2 2 2 2 2 2 2 p 2r + r – x 3r – x 3r – x = = 2 \1 = 2 4 2 × 2r × r 2r 2r2 Þ r2 = x2 Þ x = r Þ Triangle is right isosceles. cos 12. (d) = = aw + b + cw 2 aw2 + bw 2 + c aw 2 + bw + cw 3 w (aw 2 + bw + c) + + aw 2 + b + cw a + bw + cw 2 a w 3 + bw + c w 2 17. (a) Let t = x + x2 + b2 w (a + bw + cw 2 ) 1 (aw 2 + bw + c) (a + bw + cw 2 ) + w (aw 2 + bw + c) w (a + bw + cw 2 ) (Q w3 = 1) 2 = 1 1 2 2w + = = = 2w 2 w w w w3 13. (c) Q \ z= z+ 2 n–4 n–4 and 22 = 24.2 has last digit 6. =16 2 \ b=6–1=5 Hence, a2 + b2 = 12 + 52 = 26 14. (c) We have ||z1| – |z2|| £ | z1 – z2|| and equality holds only when arg z1 = arg z2 Þ ||z – w| – |z – w2|| £ |w2 – w| £ 3 and equality 1 . 2 15. (a) a and b are roots of ax2 + bx + c = 0 1 1 b c + Þ a + b = - and ab = = a a + b a b +b a a = ab + b + aa + b (aa + b) (ab + b) 2 Þ t - b = 2x & t + x2 + b2 - x b2 b2 = 2 x2 + b2 t æ b2 ö = ç 2a - t + ÷ (t ) = 2at - t 2 + b2 t ø è = – w, – w2 can hold only when |z| = 2 and not when |z| = 1 1 = = t x + x2 + b2 \ 2(a - x)( x + x2 + b2 ) [w is cube root of unity] and z2017 = (– w)2017 = – w, z2017 = (– w2)2017 = – w2 1 1ö æ \ a = z2017 + 2017 = – çè w + ÷ø w z = – (w + w2) = 1 n Þ t 1 = 1 Þ z2 – z + 1 = 0 z – (–1) ± (1– 4) 1 1 = (aa + b) (ab + b) ac \ Required quadratic equation x2 – (sum of roots) x + product of roots = 0 Þ acx2 – bx + 1 = 0 16. (c) c = – 4, – b = –3 So, x2 + bx + c = 0 becomes x2 + 3x – 4 = 0 or (x + 4) (x – 1) = 0 This gives 1 and – 4 as its roots. Þ a 2 + b2 - (a 2 - 2at + t 2 ) = a 2 + b 2 - (a - t )2 £ a 2 + b2 18. (b) a, b are the roots of x 2 + px + q = 0 Þ a + b = - p and ab = q ...(i) Again, a, b are roots of x 2 n + p n x n + q n = 0 Þ a 2 n + pn a n + q n = 0 and b 2 n + p nb n + q n = 0 Þ ( a 2 n - b 2 n ) + p n (a n - b n ) = 0 Þ a n + bn = - p n Now ...(ii) a is a root of ( x n + 1) + ( x + 1) n = 0 b æ an ö æ a ön Þ ç n + 1 ÷ + ç + 1÷ = 0 çb ÷ èb ø è ø Þ (a n + b n ) + (a + b ) n = 0 Þ - p n + (- p)n = 0, Which holds only if n is an even integer. Solutions 91 19. (a) Let f (x) = (x – sin b) (x – sin g) + (x – sin a) (x – sin g) + (x – sin a) (x – sin b) Now, f (sin a) = (sin a – sin b) (sin a – sin g) = (–) (–) = positive f (sin b) = (sin b – sin a) (sin b – sin g) = (+)(–) = negative f (sin g) = (sin g – sin a) (sin g – sin b) = (+)(+) = positive Þ Roots of f (x) = 0 are real and distinct. 20. (c) (a – 1) x2 – (a + 1)x + (a – 1) ³ 0 " x ³ 2 For a = 1, – 2x ³ 0 " x £ 2 Þ a ¹ 1 Þ a > 1 and f ' (x) ³ 0 " x ³ 2, f (2) ³ 0 Þ 2(a – 1)x – (a + 1) ³ 0 " x ³ 2, f (2) ³ 0 Þ 2(a – 1)x ³ (a + 1) " x ³ 2, 3a – 7 ³ 0 Þ Þ a +1 7 "x ³ 2, x ³ 2(a –1) 3 a +1 7 £ 2; a ³ 2(a - 1) 3 x³ 7 Þ a + 1 £ 4a – 4; a ³ ; a > 1 3 5 7 é7 ö Þ a ³ ; a ³ ; a > 1 Þ a Î ê , ¥÷ 3 3 ë3 ø 21. (1) Given that a, b, c are integers not all equal, w is cube root of unity ¹ 1, then | a + bw + cw2 | æ -1 + i 3 ö æ -1 - i 3 ö = a + bç ÷ + cç ÷ è 2 ø è 2 ø æ 2a - b - c ö æ b 3 - c 3 ö = ç ÷ ÷ +iç è 2 ø è 2 ø = 1 (2a - b - c)2 + 3(b - c) 2 2 = 1 4a2 + b2 + c2 - 4ab + 2bc - 4ac + 3b2 + 3c2 - 6bc 2 22. (45) If 2 – i is the root then 2 + i is also the root sum of roots = 4 Þ a = – 3 b Product of roots = = (2 – i)(2 + i) a Þ b = – 15 \ ab = 45 23. (0) As the coefficients of two equations are in reverse order, if the roots of ax2 + bx + c = 0 are a and b then the roots of second equation are 1 1 , . Given that one negative root is common, a b two possibilities may arise. 1 1 Either a = < 0 Þ a = -1 or a = < 0 b a c Þ ab = 1 Þ = 1 Þ c = a (not possible) a \ a = -1 is the common root. Put a = -1 in any of the equations, we get a - b + c = 0. 24. (7) x 2 - (3 + 2 log 2 3 –3 log 3 2 )x log 3 2 – 2(3 – 2log 2 3 ) = 0 Þ x2 – 3(x) – 2(2 – 3) = 0 Þ x2 – 3x + 2 = 0 Þ a = 2, b = 1 Þ a2 + ab + b2 = 4 + 2 + 1 = 7 25. (0) Let f(x) = (x – x1) (x – x2)(x – x3)(x – x4) Þ |f(i)| = 1 + x12 1 + x22 1 + x33 1 + x44 = 1 Þ x1 = x2 = x3 = x4 = 0 Þ All four roots are zero. Þ f (x) = x4 Þ a + b + c + d = 0 26. (2) The given relation can be rewritten as 2 1 1 1 + = + a+w b+w c+w w 1 1 1 2 + = 2 and 2 + 2 2 b+w a+w c+w w 2 Þ w and w are roots of 1 1 1 2 + + = a+ x b+ x c+ x x 2 3x 2 + 2(a + b + c) x + bc + ca + ab = x (a + x )(b + x)(c + x ) = a 2 + b2 + c 2 - ab - bc - ca Þ 1 [(a - b) 2 + (b - c ) 2 + (c - a) 2 ] 2 R.H.S. will be min. when a = b = c, but we cannot take a = b = c as per question. \ The min value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1. Thus b = c = 0, a = 1 gives us the minimum value 1. Þ x3 - (bc + ca + ab) x - 2abc = 0 = Two roots of the equation (i) are w and w the third root be a , then a + w + w2 = 0 Þ a = – w – w 2 = 1. \ a = 1 will satisfy equation (i) 1 1 1 + + =2 Þ a +1 b +1 c +1 ...(i) 2 . Let MATHEMATICS 92 k2 = Þ y x2 + k 2 - x k2 y \ 2x = x +k -x y 2 2 ( 29. (4) Q (1 + i )4 = [(1 + i )2 ]2 k2 27. (2) Let y = x + x 2 + k 2 = Let z = 2 ( k - x ) x + x 2 + k 2 = (1 + i 2 + 2i )2 = (1 - 1 + 2i) 2 = 4i2 = –4 ...(i) & ) æ k2 ö ÷y = çç 2k - y + y ÷ø è = k 2 - y 2 + 2ky = 2k 2 - ( k - y )2 £ 2k 2 \ z k 2 £2 Consider z = 2- + 0i+ and w = -1+ + 0 + i Y O or or = p - i - pi - p + p - pi - i - p p +1 = -2pi - 2i = -2i p +1 ...(ii) p 2 + r 2 + 2 pr - 4 pr ³ 0 4 Þ p 2 + r 2 - 14 pr > 0 X Þ p r r + > 14 Þ - 7 ³ 4 3 Þ k = 4 r p p Linear Inequalities 6 or (1- pi )( p - i ) ( p - i )(1- pi ) + p +1 1+ p Since D ³ 0 Þ CHAPTER (c) = æ | z | ö 2p = 4. ÷= Now, ç è amp( z ) ø p / 2 30. (4) If p, q, r are in A.P., p + r = 2q Þ| z - w |< 3 so k = 3 (as |z – w| < k and k is least upper bond). 1. p -i 1 + pi p = ( - 4)(- 2i ) = 2pi 4 [from Eqs. (i) and (ii)]: w z p +i + æ 1- pi p p -i ö Given, z = (1+ i )4 ç ç p + i + 1+ pi ÷÷ 4 è ø æ z-2ö p 28. (3) Arg ç gives semi-circle with ÷= è z+2ø 2 radius r = 2. æ w -1 ö p Arg ç ÷ = gives semi circle with radius è w +1 ø 2 r = 1. 1 - pi ...(i) 1 2x - 1 - 3 ³0 x + 1 x - x +1 x +1 2 2 - ( ) 2 ( x + 1) - x 2 - x + 1 - ( 2 x - 1) ( x + 1) ( x2 - x + 1) ³0 2. ( ) ³0 ( x + 1) ( x - x + 1) - x2 - x - 2 2 - ( x - 2 )( x + 1) ( x + 1) ( x 2 - x + 1) ³0 3. 2- x ³ 0 , where x ¹ – 1 x - x +1 or 2 - x ³ 0, x ¹ -1 (as x2 – x+ 1 > 0 for " x ÎR) Þ x £ 2, x ¹ - 1 Þ x Î (– ¥, – 1) È (–1, 2] (c) (2x + 1) (x – 3) (x + 7) < 0 Sign scheme of (2x + 1) (x – 3) (x + 7) is as follows : – + – + or 2 –7 –1/2 3 Hence, solution is (–¥, 7) È (–1/2, 3) (c) | x - 1| £ 3 Þ -3 £ x - 1 £ 3 Þ -2 £ x £ 4 and | x - 1 | ³ 1Þ x - 1 £ -1 Solutions 4. 93 or x - 1 ³ 1 Þ x £ 0 or x ³ 2 Taking the common values of x, we get x Î [-2, 0] È [2, 4] (b) The given equations are (iii) If x ³ 1 , then ì x + 3y = 5, x ³ 1 | x - 1| +3y = 4 Þ í î - x + 3y = 3, x < 1 and Q ( x + 2) - ( x - 1) < x - L (i) L (ii) L (iii) ì x - y = 1, y ³ 1 x - | y - 1| = 2 Þ í L (iv) î x + y = 3, y < 1 Solving (i) and (iii), we get x = 2, y = 1 Solving (i) and (iv), we get x = 2, y = 1 no solution (Q x ³ 1 and y < 1) Solving (ii) and (iii), we get x = 3, y = 2 no solution (Q x < 1, y ³ 1) 5. Þ x Îf 5 3 solving (ii) and (iv) we get, x = , y = 2 2 no solution (Q x < 1, y < 1) Hence solution is x = 2, y = 1(a unique solution) (c) 2- |1- | x || = 1 Þ 2- |1- | x ||= ±1 9 > 1. Þ common solution is 2 x> 9 æ9 ö Þ x Îç ,¥÷ 2 è2 ø ö æ9 \ Solution set is x Î ç , ¥ ÷ . 2 ø è 7. 12x (a) 4x 2 + 9 ³1Þ 12 | x | ³1 4x 2 + 9 Q 4x 2 + 9 > 0 Þ 4x 2 - 12 | x | +9 £ 0 Þ 4| x |2 -12| x | +9 £ 0 Þ (2 | x | -3) 2 = 0 Þ | x | = 8. 3 2 | x - 1| + a = 4 Þ | x - 1| + a = ±4 (b) Þ| x - 1| = - a ± 4 Þ|1- | x ||= 1 or 3 or 2 Þ x = 0 or ± 2 The above equation holds if -a + 4 ³ 0 or -a - 4 ³ 0 Þ a £ 4 or a £ -4 Þ a Î (-¥, 4] È (-¥, - 4] If 1- | x | = 3 Þ 1- | x |= ±3 Þ| x |= -2 or 4 Þ a Î (-¥, 4] If 1- | x | = 1 Þ 1- | x |= ±1 Þ| x |= 0 Þ| x |= 4 Þ x = ±4 [Q | x |¹ -2] \ Solution set is {– 4, – 2, 0, 2, 4}, hence 5 real roots in all. 6. 9 3 Þx> 2 2 3 (a) The inequality is | x + 2 | - | x - 2 |< x - . 2 Dividing the problem into three intervals : (i) If x < –2, then - ( x + 2) + ( x - 1) < x - 3 3 Þx > 2 2 3 > -2 , hence no common values 2 Þ x Îf (ii) If -2 £ x < 1, then But - ( x + 2) + ( x - 1) < x - But - 3 5 Þx<2 2 5 < -2 , hence no common values 2 9. (c) We know that |f(x)| = – f(x) if f ( x ) £ 0 \ Þ x 2 - 8x + 12 x 2 - 10x + 21 =- x 2 - 8x + 12 x 2 - 10x + 21 x 2 - 8 x + 12 £0 x 2 - 10 x + 21 ( x - 2)( x - 6) Þ £ 0, x ¹ 3, 7 ( x - 3)(x - 7) Þ ( x - 2)( x - 3)(x - 6)(x - 7) £ 0 , x ¹ 3, 7 + + 2 – 3 6 Þ 2 £ x < 3 or 6 £ x < 7 Þ x Î [2, 3) È [6, 7) 10. (a) Case (i) : When x ³ 0 \ |x| = x – + 7 MATHEMATICS 94 3- x (3 - x)(4 - x) ³0Þ ³0 4-x (4 - x) 2 Þ (x – 3) (x – 4) ³ 0 and x ¹ 4 Þ x £ 3 or x > 4 but x ³ 0 so x Î [0, 3] È (4, ¥) ...(i) Case (ii) : When x < 0 3+ x (x + 3)(x + 4) ³0Þ ³0 4+x (x + 4) 2 Þ x < – 4 or x ³ – 3 but x < 0 so x Î (– ¥, – 4) È [–3, 0) ...(ii) So union of (i) and (ii) gives [ -3,3] È ( -¥, - 4) È (4, ¥) \ 11. (b) |x| = –x \ x+y 2 x +y x+y 2 x +y 2 - 2 - 1 æ1 1ö ç + ÷ 2 èx yø æ x+y 1 1 ö = ( x + y) ç 2 ç x + y2 2xy ÷÷ 2xy è ø æ 2xy - (x 2 + y 2 ) ö = ( x + y ) çç 2 2 ÷÷ è (x + y ) (2 xy) ø æ - (x - y) 2 ö and ( x + y ) çç 2 ÷÷ £ 0 2 è (x + y ) (2 xy) ø Similarly, y+z y2 + z £ 2 1 æ1 1ö ç + ÷ and 2 èy zø x+z 1 æ 1 1ö £ ç + ÷ 2 2 2 èx zø x +z \ Adding; we get x+y y +z z+x 1 1 1 + 2 + 2 £ + + 2 2 2 2 x y z x +y y +z z +x and Hence A £ B 12. (d) xy + yz + zx < 0 and m = \ \ Þ Þ x 2 + y2 + z 2 xy + yz + zx (x + y + z)2 = x2 + y2 + z2 + 2 (xy + yz + zx) (x + y + z)2 ³ 0 x2 + y2 + z2 + 2 (xy + yz + zx) ³ 0 x2 + y2 + z2 ³ – 2 (xy + yz + zx) x 2 + y2 + z 2 £–2 xy + yz + zx 13. (a) We have (a1 + a2 + ... + an -1 + 2an ) (a a ...a 2a )1/n ³ 1 2 n–1 n n [Using A.M. ³ G.M.] 1/n Þ a1+ a2 + a3 + ..... + an– 1+ 2an ³ n(2c) Þ 14. (a) The inequality is log 0.2 x+2 £ 1 . The x x+2 >0 x Þ x( x + 2) > 0 Þ x < -2 or x > 0 . Solving the inequality, we get (note that base < 1) x+2 1 4 x + 10 x+2 1 - ³0Þ ³0 ³ 0. 2 = Þ x 5 5x x 5 L.H.S is valid if 5 or x ³ 0 . 2 Taking the intersection, we get 5 x£or x > 0 2 x (2x + 5) ³ 0 Þ x £ - æ 5ù Þ x Î çç - ¥, - ú È (0, ¥) 2û è 15. (a) The log functions are defined if (x + 3) 2 x 2 + 6x + 9 >0 > 0 and x + 1 > 0 Þ 2( x + 1) 2(x + 1) and x + 1 > 0 Þ x > –1 Now the inequality is log -1 2 x 2 + 6x + 9 < - log 2 ( x + 1) 2(x + 1) Þ - log 2 Þ log 2 Þ x 2 + 6x + 9 < - log 2 (x + 1) 2( x + 1) x 2 + 6x + 9 > log 2 ( x + 1) 2( x + 1) 2 x 2 + 6x + 9 > ( x + 1) Þ - x + 2x + 7 > 0 2( x + 1) 2( x + 1) Þ ( x + 1)(x 2 - 2x - 7) < 0 Þ x 2 - 2x - 7 < 0 [Q x + 1 > 0] Þ -1 - 2 2 < x < -1 + 2 2 , but x > -1 Þ -1 < x < -1 + 2 2 16. (b) For the validity of inequality ax2 + 4x + a > 0, which is possible if a > 0 and 16 – 4a2 < 0 Þ a >2 ...(i) Further, the inequality can be rewritten as log 5 5 + log 5 ( x 2 + 1) £ log 5 (ax 2 + 4x + a ) Þ 5( x 2 + 1) £ ax 2 + 4x + a Solutions Þ (a - 5) x 2 + 4 x + (a - 5) ³ 0 . It holds if a - 5 > 0 and 16 - 4(a - 5) 2 £ 0 ...(ii) Þ a > 5 and a £ 3 or a ³ 7 Þ a ³ 7 Combining the results of (i) and (ii) for common values, we get a Î [7, ¥) 17. (b) For x ³ -2 , x2 – x – 2 + x > 0 95 2sin 3x . cos 3x sin (2 . 3x ) 1 = £ 2x 2 2 x 3 +3 1 x –x Þ £ but (3 + 3 ) ³ 2 k 2 Þ Minimum value of k = 4 22. (2) As x, y Î R, and xy > 0, so x and y will be of same sign. = 2 Þ x > 2 Þ x Î (-¥, - 2 ) È ( 2 , ¥) Þ x Î [-2,- 2 ) È ( 2 , ¥) For x < – 2 x2 + x + 2 + x > 0 or x2 + 2x + 2 > 0 which is true for all x. Hence x Î (-¥, - 2 ) È ( 2 , ¥) 18. (c) log10 x = A x >0 log10 (x – 2) = B, x – 2 > 0 Þ x > 2 Þ A2 – 3AB + 2B2 < 0 Þ(A – 2B) (A – B) < 0 Þ (log x – 2 log (x – 2)(log x – log (x – 2)) < 0 Case - I : log x – 2 log (x – 2) < 0 and log x – log (x – 2) > 0 Case - II :log x – 2 log (x – 2) > 0 and log x – log (x – 2) < 0 From (1) & (2), x Î (4, ¥) 19. (a) x ¹ (2n + 1) p/2, npnÎ I. The given inequality can be written as log 2 (x 2 - 8x + 23) 3 > log 2 | sin x | log 2 |sin x | As log2 |sin x| < 0, we get; log (x2 – 8x + 23) < 3 Þ x2 – 8x + 23 < 23 = 8 Þ x2 – 8x + 15 < 0 Þ (x – 5) (x – 3) < 0 Þ 3 < x < 5 p 3p For x Î (3, 5), x ¹ p, , . Hence 2 2 æ 3p ö æ 3p ö x Î (3, p) È ç p, ÷ È ç , 5 ÷ . è 2 ø è 2 ø 3 x 2 + 9 x + 17 20. (b) y = 3 x2 + 9 x + 7 3x 2 ( y - 1) + 9 x ( y - 1) + 7 y - 17 = 0 D ³ 0 Q x is real 81( y - 1)2 - 4 ´ 3( y - 1)(7 y - 17) ³ 0 Þ ( y - 1)( y - 41) £ 0 Þ 1 £ y £ 41 \ Max value of y is 41 21. (4) Consider the L.H.S of the equation, (sin 3x) (cos 3x) Therefore, all the quantities 2x x 3 y 4y 2 , , y3 3 9x 4 are positive. Now, A.M. ³ G.M. 1/3 æ æ 2x ö æ x3 y ö æ 4y2 ö ö 2x x3 y 4y2 ÷÷ çç 4 ÷÷ ÷÷ Þ 3+ + 4 ³ 3 çç çç 3 ÷÷ çç 3 y 9x è è y ø è 3 ø è 9x ø ø =3× 2 =2 3 23. (1) x + 3 - x ³ 3 - x + 3 Þ x ³ 3 But 3 – x ³ 0 Þ x £ 3 Hence, x = 3 is the only integral solution. x+3 + x >1 24. (4) We have x+2 x+3 -2 x+3 + x -1 > 0 Þ >0 Þ x+2 x+2 Now two cases arise : Case I : When x + 3 ³ 0, i.e., x ³ – 3. Then x+3 -2 >0 Þ x+3- 2 >0 x+2 x+2 x +1 > 0 Þ {(x + 1) > 0 and x + 2 > 0} Þ x+2 or {x + 1 < 0 and x + 2 < 0} Þ {x > – 1 and x > – 2} or {x < – 1 and x < –2} Þ x > – 1 or x < – 2 Þ x Î (-1, ¥) or x Î ( - ¥, - 2) Þ x Î ( -3, -2) È ( -1, ¥) [Since x ³ -3 ]…(i) Case II : When x + 3 < 0, i.e., x < – 3 x+3 -2 -x - 3 - 2 >0 >0 Þ x+2 x+2 -( x + 5) x+5 >0 Þ <0 Þ x+2 x+2 Þ (x + 5 < 0 and x + 2 > 0 ) or (x + 5 > 0 and x + 2 < 0) Þ (x < – 5 and x > – 2) or (x > – 5 and x < – 2). It is not possible. ...(ii) Þ x Î (– 5, – 2) MATHEMATICS 96 Þ ab ³ 0 Combining (i) and (ii), the required solution is x Î (– 5, – 2) È (– 1, ¥ ), smallest integral member = – 4 \ Modulus of smallest integral member =|–4| = 4 25. (4) 4 - x2 4 - x2 = x+ x x As | a | + | b | = | a + b | |x|+ CHAPTER 7 1. 2. 3. 4. æ 4 - x2 ö \ x çç ÷÷ ³ 0, x ¹ 0 è x ø Þ x2 – 4 £ 0 Þ x Î [– 2, 2] – {0} Number of integral values = 4. Permutations and Combinations and 3, and a number cannot begin with 0) Similarly, the number of numbers with 5 in the middle = 5P4 – 4P3, etc. \ The required number of numbers = (4P4 – 3P3) + (5P4 – 4P3) + (6P4 – 5P3) + (a) Total number of words that can be formed using 5 letters out of 10 given different letters = 10 × 10 × 10 × 10 × 10 (as letters can repeat) = 1, 00, 000 Number of words that can be formed using 5 different letters out of 10 different letters = 10P5 (none can repeat) 10! = 30, 240 = 5! \ Number of words in which at least one letter is repeated = total words-words with none of the letters repeated = 1,00,000 – 30,240 = 69760 (c) X – X – X – X – X. The four digits 3, 3, 5, 5 can be arranged at (–) places in 4! = 6 ways. 2!2! The five digits 2, 2, 8, 8, 8 can be arranged at 5! = 10 ways. (X) places in 2!3! Total no. of arrangements = 6 × 10 = 60 ways. (c) The letter of word COCHIN in alphabetic order are C, C, H, I, N, O. Fixing first letter C and keeping C at second place, rest 4 can be arranged in 4! ways. Similarly the words starting with CH, CI, CN are 4! in each case. Then fixing first two letters as CO next four places when filled in alphabetic order give the word COCHIN. \ Numbers of words coming before COCHIN are 4 × 4! = 4 × 24 = 96 (d) The smallest number, which can occur in the middle is 4. The number of numbers with 4 in the middle = 4 P4 – 3 P3 (Q The other four places are to be filled by 0, 1, 2 9 ..................... + (9P4 – 8P3) = å ( n P4 - n - P3 ) 1 n=4 5. (b) C ...... = 4! = 24 D ...... = 4! = 24 M ...... = 4! = 24 S C ...... = 3! = 6 S D ...... = 3! = 6 S M C DW=1 6. S M C WD=1 (b) m = Number of permutations of letters of the word BHARAT, when B and H are never together 7. 5! 6! – × 2! (keeping (BH) together and 2! 2! permuting with A, R, A, T) = 360 – 120 = 240 and n = number of permutations of the letters of the word BHARAT when each word begins with B an d ends with 4! m 240 T= = 12 \ = = 20 2! n 12 (c) Let the square has dimensions n × n. If n is even, total no. of blue (diagonal) tiles = n + n = 2n If n is odd, then total no. of blue tiles = n + n – 1 = 2n – 1 (the middle tile is be counted twice). Hence, 2n – 1 = 121 Þ n = 61 So, number of red tiles = 61 × 61 – 121 = 3600 = Solutions 8. 97 (c) 1 9 8 . 2 = 2 = 256. 2 12. (a) Arrangement does not matter because of descending order there should be only one arrangement \ required number of ways = 15C3 × 24C2 1 ................ 24 25 26 ................ 40 = Number of digits Numbers ending with 0 Numbers ending with 5 Total 1 0 1 1 2 8 9 17 3 9.8 = 72 8.8 = 64 136 4 9.8.7 = 504 8.8.7 = 448 952 Total 9. 1106 (d) 10 identical boxes can contain 12 balls such that each box contains at least one ball in the following cases: Case (i) : 8 boxes contains single ball each + 2 boxes contains two balls each Single ball boxes Boxes with doubleball 6 474 8 2W,6B; 64 4744 8 (WW)(WW) 3W,5B; (WW)(WB) 4W, 4B; 5W,3B; (WW)(BB); (WB)(BB) (WB)(WB) 6W, 2B; (BB)(BB) Case (ii): 9 boxes single ball each + 1 box containing 3 balls Singleballboxes Triple ball boxes 6474 8 3W + 6 B; 4W + 5B; 6 474 8 (WWW ) (WWB ) 5W + 4 B; (WBB ) 6W + 3B; ( BBB ) \ Total 10 ways are there 10. (c) Required number of ways = (Total number of ways of arranging 7 persons) – (Number of ways where two American are together + Number of ways when two Britishian are together – Number of ways when both American and both Britishian are together) = 6! – [2 × 5! + 2 × 5! – 4! × 2! × 2!)] = 720 – [480 – 96]= 240 + 96 = 336 11. (b) The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers. \ The number of ways to be unsuccessful = 9 C9 + 9 C8 + 9 C7 + 9 C6 + 9 C5 = 9 C0 + 9 C1 + 9 C2 + 9 C3 + 9 C4 = 1 9 ( C0 + 9 C1 + ......... + 9 C9 ) 2 Selecting any 2 out of Selecting any 3 from 24 15 26 to 40 = C3 1 to 24 = C2 13. (b) Let the number of children in the class = n. Number of groups of 3 children = nC3 Each child would go to zoo n– 1C2 times. \ According the question: nC3 = 84 + n – 1C2 n(n - 1)(n - 2) (n - 1)(n - 2) = 84 + Þ 6 2 Þ n(n – 1) (n – 2) = 504 + 3(n – 1) (n – 2) Þ (n – 1) (n – 2) (n – 3) = 504 Þ (n – 1) (n – 2) (n – 3) = 7 × 8 × 9 Þ x = 10 14. (d) We will consider the following cases: Case Flags 4 alike and 2 others alike 4 alike and 2 others different 3alike and 3 others alike 3alike and 2 others alike and1 different 4 white and 2 red 4 white,1red and1blue 3 white,3red 3 white,1blue, 2 red or 3 red, 1 blue, 2 white Total No.of signals 6! =15 4!2! 6! = 30 4! 6! = 20 3!3! 2 C1 × 6! =120 3!2! 185 15. (c) Let the number of men = n Number of women = 2 \ Total participants = (n + 2) Now number of games played by men among the m selves = 2(nC2) Number of games men played with women = 4n According to question 2(nC2) = 4n + 66 Þ n(n – 1) = 4n + 66 Þ n2 – 5n – 66 = 0 Þ (n – 11)(n + 6) = 0 Þ n = 11 or n = – 6 \ Total number of participants = 11 + 2 = 13 16. (b) Suppose there n players in the beginning. The total number of games to be played was equal to nC2 and each player would have played n – 1 games. Let us assume that A and B fell ill. Now the total number of games of A and B is (n – 1) + (n – 1) – 1 = 2n – 3. But they have played 3 games each. Then their remaining number of games is MATHEMATICS 98 17. \ Number of ways é 1 1 1 1ù = 10C6 × 4! ê1– + – + ú ë 1! 2! 3! 4!û 2n – 3 – 6 = 2n – 9. Given, nC – (2n – 9) = 84 2 Þ n2 – 5n – 150 = 0 or n = 15 (b) Required ways = 8C5 – (Number of ways when R2 or R3 are unused) = R1 R2 R3 18. 23. = 8C5 – (6C5 + 6C5) = 8C5 – (12) = 56 – 12 = 44 (c) The number of triangles with vertices on different lines = p C1 × p C1 × p C1 = p3. The number of triangles with 2 vertices on one line and the third vertex on any one of the other two lines p( p - 1) 2 \ The required number of triangles = p3 + 3p2(p – 1). (c) Let A1, A2, ..., A10 be vertices of a regular polygon of 10 sides. The number of ways of selecting 3 vertices is 10C3. The number of ways of selecting 3 consecutive vertices is (i.e., A1 A2 A3, A2 A3 A4, ..., A10 A1 A2) = 10. The number of ways of selecting three vertices such that two vertices are consecutive = (First select 2 consecutive vertices, leave their neighbouring two vertices and select one more from the remaining 6 vertices) is 10 × 6C1 = 60 The total number of required selections is 10C – 10 – 60 = 120 – 70 = 50 3 (b) Division of objects can be (0, 1, 4) or (0, 2, 3) So total number of distribution ways = 3 C1 19. 20. 22. p 2 2p C1 } = 6p 5! 5! ´ 3!+ ´ 3! = 30 + 60 = 90 ways 0!.1!.4! 0!.2!.3! (8) a + b + c = 21 Þ 3b = 21 Þ b =7 [Q a + c = 2b] Þ a + b + c = 21 Þ a + c = 14 Þ l = 14 – 1C2 – 1 = 13C1 = 13 Hence l – 5 = 13 – 5 = 8 (9) If exactly 6 of his matching are correct, then the remaining 4 statements from column I must be incorrectly matched with 4 statements from column II = 21. { C ´ 24. 25. 10! é1 1 1 ù × 4! – + 6!4! êë 2 6 24 úû é12 – 4 + 1ù = (10 × 9 × 8 × 7) ê = 1890 ë 24 úû (6) Number of numbers beginning with 1 = 120 1 Number of numbers beginning with 2 = 120 2 Starting with 31 ................................. = 24 3 1 Starting with 3214 ................................. = 2 3 2 1 4 Finally = 1 3 2 1 5 4 6 Hence, unit place digit of 267th number is 6. (7) If x denotes the number of times he can take unit step and y denotes the number of times he can take 2 steps, then x + 2y = 7. We must have x = 1, 3, 5 If x = 1, the steps will be 1 2 2 2 4! Þ Number of ways = =4 3! If x = 3, the steps will be 1 1 1 2 2 5! Þ Number of ways = 2!.3! = 10 If x = 5, the steps will be 1 1 1 1 1 2 Þ Number of ways = 6C1 = 6 If x = 7, the steps will be 1 1 1 1 1 1 1 Þ 7C0 = 1 Hence total number of ways = N = 21 Þ N/3 = 7 (9) In this case number of participants from 10 different countries are 1, 2, 3… 10. number of ways that they can be arranged in a row such that all the participants from the same country were together is (10!)(1!)(2!)(3!)(4!)…. (10!) Hence K = (10!)(1!)(2!)(3!)(4!)…. (10!) highest power of 10 or that of 5 can be given by only (5!)(6!)(7!)(8!) (9!)(10!)(10!) which is 1+1+1+1+1+2+2=9 Solutions 99 CHAPTER Binomial Theorem 8 1. 2 60 = (1 + 7) 20 (a) 6. = 20 C0 . + 20 C1. 7 + 20 C2 . 72 + 20 2. n = 2160 = (3 – 1)160 = 3m + 1, m Î N \ (32) = (32) 3m+1 3. , remainder = 4 When 7 divides (32) n æ r –1 ö (d) å ç å n Cr .r C p .2 p ÷ ÷ r =1 çè p = 0 ø n – 1 r æ ö = å n C r ç å r C p .2 p ÷ ç p=0 ÷ r =1 è ø 7. 4. 5. 9 )log3 8. \ n n/2 (– 1) .(1/2) æ 1 ö = ç 5/3 ÷ è3 ø (a) 17 2009 + 112009 – 7 2009 = 2009 – 7 2009 is 0. Also, last digit in 112009 is 1. Last digit of 17 2009 + 112009 – 72009 is 0 + 1 = 1. … (i) 153 8. So, t5 = 10C4 (21/3)6 (–1/ 2 )4 = 10C4. = 10C10–4. (c) Let x = 2 2 \ a = x2 + 204 2x + 1 2 Let y = 2 \ b=y –y+1 Þ N = x16 – 1 Þ N = (x4 – 1) (x4 + 1)(x8 + 1) …(ii) log3 8. 5 - . 3log3 2 3 = 3-5 log 3 2 = 2 -5 = 3 ... (iv) 8 in equation (iv), from equation (iii) a Cleary last digit in 17 n (4 –1) – (3 –1) = 4 – 3 2 6 (d) Given expression is (1 + x - 2 x ) = (1/3. 3 n(n – 1) 2 a = 24 2 2009 r =1 (-2)6 = 1 - a1 + a2 – .... + a12 Adding (i) and (ii), 2 + 2a2 + 2a4 + .... + 2a12 = 64 64 - 2 = 31 Þ a2 + a4 + .... + a12 = 2 (a) The last term = nCn. (–1 / 2 )n ...(iii) C1 a = 8 Þ na = 8 (17 2008 + 17 2007.71 + ..... + 7 2008 ) + 11 n = 1 + a1 x + a2 x 2 + ..... + a12 x12 Put x = 1 both side; 0 = 1 + a1 + a2 + .... + a12 Put x = – 1 both sides; n (17 2009 – 72009 ) + 112009 = (17 – 7) = å n Cr { (1 + 2)r – 2r }= å n Cr 3r – n Cr 2r n ...(ii) éæ 8 ö 2 8 ù 2 Þ ê çè ÷ø – ú a = 48 Þ 64 – 8a = 48 a a ëê ûú Þ a=2 \ n=4 2–4 a–n 1 =– = \ 2+ 4 a+n 3 n r=1 = n C0 = 1 Þ 1 = 1 Put n = 3232 n C0 + n C1 ( ax )1 + n C2. (ax)2 + ........ = 1 + 8x + 24x2 + ... (i) and n C2 . a 2 = 24 Þ = 25(3m + 1) = 23(5m + 1) 2 2 = 4. 85m + 1 4(7 + 1)5 m + 1 = 4(7n + 1), n Î N = 28n + 4 \ n Compairing coefficients of x 0 , x1 , x 2 in equation (i), ............ + 20 C20 . 7 20 \ the remainder = C0 . = 1 (c) 32 = 25 Þ (32) 32 = (25 ) 32 3232 (c) 9. Þ N = (x4 – 1) (x2 + 2 x + 1 ) (x2 – 2 x + 1 )(x8 + 1) Þ N = y6 – 1 = (y3 – 1) (y3 + 1) Þ (y3 – 1)(y + 1)(y2 – y + 1) (c) N = 20C7 – 20C8 + 20C9 – 20C10 + ..... – 20C20 = (20C7 + 20C9 + 20C11 + ..... + 20C19) – (20C8 – 20C10 + ..... + 20C20) = (20C0 + 20C2 + 20C4 + 20C6) – (20C1 + 20C3 + 20C5) = (1 + 190 + 4845 + 38760) – (20 + 1140 + 15504) = 43796 – 16664 = 27132 = 3 × 4 × 7 × 19 × 17 MATHEMATICS 100 10. 11. (a) x = T7 = nC6(31/3)n – 6. (4–1/3)6 y = Tn – 5 = nCn – 6(31/3)6. (4–1/3)n – 6 Þ y = 12x n Cn – 6(31/3)6 (4–1/3)n – 6 = 12. nC 6(31/3)n – 6.(4–1/3)6 Þ 12 = (121/3)12 – n Þ n = 9 (d) Given expression = = ( x1/ 3 ) 3 + (1)3 x 2/3 1/ 3 –x +1 – n + 1ù é n So, from (iii) and (iv), x Î ê , n úû ën +1 x ( x 2 / 3 - x1/ 3 + 1) ( x1/ 2 + 1) ( x1/ 2 - 1) x1/ 2 ( x1/ 2 - 1) 1/ 3 æ x +1 x -1 ö Þ çè 2 / 3 1/ 3 ÷ x - x + 1 x - x1/ 2 ø 1/ 3 = (x -x 10 - x –1/ 2 )10 is Cr x æ 10 - r r ö - ÷ çè 3 2ø which is æ 10 - r r ö independent of x if çè 3 - 2 ÷ø =0 Þ r=4 4 Hence required coefficient = 10 C4 (-1) = 210. 12. (c) Middle term = 2n Cn . x n = t n+1 If tn + 1 is the greatest term also, then t n+1 ³ tn ....(i) t n+1 ³ t n+ 2 .....(ii) From (i), Þ 2n r \ The term independent of x = 10 C5 . sin 5 a . Cr ( x1/ 3 )10 - r . ( -1)r . ( x -1/ 2 )r r 10 r It is independent of x if r = 5 ) Tr +1 in ( x = ( -1) –1/ 2 –1/ 2 10 1/ 3 10 – x ( ) . As 2 and . 62 14. ( x1/ 3 + 1) ( x 2 / 3 - x1/ 3 + 1) = ( x1/ 3 + 1) – (1 + x -1/ 2 ) = x 100 – r 5 are coprime, t r +1 will be rational if 100 – r is a multiple of 8 and r is a multiple of 6. Also 0 £ r £ 100. \ r = 0, 6, 12, ............. , 96 Þ 100 – r = 4, 10, 16, ............, 100 But 100 – r is to be a multiple of 8. So, 100 – r = 0, 8, 16, 24, ... 96. The common terms in (1) and (2) are 16, 40, 64 and 88. \ r = 84, 60, 36, 12 give rational terms. \ The number of irrational terms = 101 – 4 = 97. r æ cos a ö (c) t r +1 = 10 Cr (x sin a )10 – r . çè x ÷ø . ( x1/ 2 - 1) – (8 5) (a) ( x - 1) 1/ 2 t r +1 = 100 C 13. Cn . x n ³ 2 n C n –1 . x n –1 (2n)! (2n)! Þ (n + 1) x ³ n ³ (n – 1)!(n + 1)! n! n! n Þ x³ n +1 From (ii), 2n Cn . x n ³ 2 n Cn +1 . x n +1 (2n)! (2n)! x Þ ³ (n –1)!(n + 1)! n!n ! n +1 Þ x£ n ...(iii) ...(iv) 15. 1 1 cos 5 a = 10 C5 . 5 (sin 2a )5 £ 10 C5 5 2 2 (10)! 1 = . (5!) 2 25 T (c) If Tr + 1 ³ Tr , then r +1 ³ 1 Tr 25 – r + 1 . 3x ³1 r 2 26 – r . Þ 3 ³1 r Þ (Q x = 2) 1 Þ 78 – 3r ³ r \ r £ 78 Þ r £ 19 2 4 Þ r = 19 Largest term = (r + 1)th term = 20th term 16. (c) Middle term in the expansion is æ 10 ö çè + 1÷ø 2 th i.e., 6th term. 7 1 63 Þ 10C5 5 . x5 sin5 x = 8 8 x 63 1 Þ 252.sin 5 x = Þ sin 5 x = 8 32 1 π Þ sin x = \ x = nπ + (-1)n 2 6 Thus T6 = 7 Solutions 101 17. (d) The cofficients of the integral powers of x are 40 Þ = 2[ n C0 2n + n C2 2n - 2.3 + n C4 .2n - 4.32 +.............] = 2 × Integer = Integer Q I + f + F is integer Þ f + F must be integer.. \ 0 £ f < 1 and 0 < F < 1 Þ 0 < f + F < 2 Þ f + F = 1 Þ F = 1 – f \ ( I + f ) (1 – f ) = (I + f ) F = (2 + 3)n (2 – 3) n = 1 C0 , 40 C2 . 22 , 40 C4 . 24 , ..... , 40 C40 . 2 40 (1 + 2) 40 = 40 C0 + 40 C1 2+ 40 C2 . 22 +...... + 40 C40 . 2 40 Þ (1 – 2) 40 = 40 C0 – 40 C1 . 2 4 + 40 C2 . 22 –......+ 40 C40 . 2 40 Adding, 3 40 + 1 = 2 × (required sum) 18. (b) 21. (1) (3 + x )4 4! 4! k =0 According to the question. n 1æ 1 öæ 2 ö æ n +1ö ÷ ÷........ç ÷ç = ç 1 è n + 1ø è n +1 ø è n -1ø n é C r +1 ù êQ n +1 r = ú C r +1 n + 1 úû êë = (n +1)! n+1 = n! n \ p (2002) (2002) 2001 = p (2001) (2001)! (n + 1) (n + 1) 19. (d) n n (n - 1)(n - 2) C1 C 3 C 5 + + + ..... = + 2 4! 2 4 6 + n (n - 1)(n - 2)(n - 3)(n - 4) + ...... 6! = 1 é (n + 1)n (n + 1)n (n - 1)(n - 2) + + ê n + 1 ë 2! 4! or and 0 £ f < 1. We note that (2 + 3 ) (2 – 3 ) n = 1. So let us assume that F = (2 – 3) . Clearly 0 < F < 1. Now, I + f + F = (2 + 3) n + (2 – 3)n = (3 + x)4 32 = 4! 3 4 (3 + x) = 256 or x + 3 = 4 or x=1 1 1 1 ù é 1 11! ê + + + ... + 11!1! úû ë1!10! 3!8! 5!6! 11 11 11 11 + + + ... + 1!10! 3!8! 5!6! 11! = 11C1 + 11C3 + 11C5 + 11C7 + 11C9 + 11C11 = sum of the even coefficient in the expansion of (1 + x)11 = 210 Therefore K = 10 (6) (1 – 2x + 5x2 – 10x3) [C0 + C1x + C2x2 + ...] = 1 + a1x + a2x2 + ... = = 20. (b) Given (2 + 3)n = I + f, where I is integer = å 22. (4) Expression = (1 – x)5 .(1 + x)4 (1 + x2)4 = (1 – x) (1 – x2)4 (1 + x2)4 = (1 – x) (1 – x4)4 \ so the coefficient of x13 = – 4C3 (–1)3 = 4 23. (0) We have 3C0 – 5C1 + 7C2 + ... + (–1)n (2n+3)Cn = 3C0 – 3C1 + 3C2 + ...+ (–1)n 3Cn – 2C1 + 4C2+... + (–1)n 2n Cn n = 3 [C0 – C1 + C2 + ....+ (–1) Cn] –2[C1 – 2C2 + .... (–1)nnCn] =3×0–2×0=0 24. (5) We have ù (n + 1)n (n - 1)(n - 2)(n - 3)(n - 4) + .....ú 6! û 1 n +1 [ C 2 + n +1C 4 + n +1C6 + .....] n +1 1 2n - 1 éë 2n +1-1 -n +1 C0 ùû = = n +1 n +1 øè ø k ö æ x ö 4! 3 = åç ÷ç ÷ ç (4 - k )! ÷ ç k ! ÷ 4! k =0 è øè ø 4-k 4 4 C .34- k x k k æ C n ö÷ 1 æç C 0 ö÷ æç C1 ö÷ .......ç = n +1 n +1 n +1 n +1 ç ÷ ç ç ÷ C0 è C1 ø è C2 ø C n +1 ÷ø è n ö æ xk ö k =0 è 4 æ n C 0 .n C1.n C 2 .....n C n p( n ) = p(n + 1) n +1 C 0 .n +1 C1.n +1 C 2 ......n +1 C n +1 n æ 34 - k å çç (4 - k )! ÷÷ çç k ! ÷÷ 25. Þ a1 = n – 2 and a2 = n(n - 1) – 2n + 5 2 a21 = 2a2. Therefore, we have (n – 2)2 = n (n – 1) – 4n + 10 Þ n2 – 4n + 4 = n2 – 5n + 10 \ n=6 MATHEMATICS 102 CHAPTER Sequences and Series 9 1. (d) We have å n2 - 4 å n + nùû = c2 é 4 ë Sn = 13 + 3. 23 + 33 + 3. 43 + 53 + ............ Let n = 2 m Then S2m é 4n(n + 1)(2n + 1) 4n(n + 1) ù = c2 ê + nú 6 2 ë û = ( 3 + 33 + 53 + ...... to m terms) 1 é 2(2n2 + 3n + 1) - 6(n + 1) + 3 ù = c2 n ê ú 3 ëê ûú + 3(23 + 43 + 63 + ...to m terms) = {13 + 23 + 33 + 43 + ...... + (2m – 1)3 + (2m)3} – {23 + 43 + ....+(2m)3} + 3{23 + 43 + 63 + ..... + (2m)3} 2 é 2m(2m + 1) ù = ê + 8 × 2{ 13 + 23 + 33 ú 2 ë û 4. é 4n2 - 1ù n(4n 2 - 1)c 2 = c2 n ê ú= 3 êë 3 úû (c) d = a2 – a1 = a3 – a2 = ... = an – an – 1 \ 3 + ... + m } m2 (m + 1)2 = m (2m + 1) + 16. 4 2 = 2 sin d [sec a1 sec a2 + sec a2 sec a3 + ... + sec an–1 sec an] sin(a 2 - a1 ) sin(a 3 - a 2 ) + + ... cos a1 cos a 2 cos a 2 cos a 3 + n2 (2n 2 + 6 n + 5) n [Put m = ] 4 2 (b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P. Þ 2 log9 (31– x+2) = 1 + log3 (4.3x – 1) Þ log3 (31 – x + 2) = log33 + log3 (4.3x – 1) Þ log3 (31– x + 2) = log3 [3(4 × 3x – 1)] Þ 31– x + 2 = 3 (4.3x – 1) = 2. Put 3x = t Þ sin a 3 cos a 2 - cos a 3 sin a 2 + ... cos a 2 cos a 3 = tan a2 – tan a1 + tan a3 – tan a2 + ... + tan an – tan an– 1 = tan an – tan a1 5. 3 (as 3 x ¹ - ve ) 4 æ 3ö è 4ø 3. Then, Tn = S n - S n -1 = cn 2 - c (n - 1)2 = (2n - 1)c \ Sum of squares of n terms of this A.P = åTn2 = å (2n - 1)2.c2 1 1 æ 1 ö (b) y – x = 3 çè 2 + 2 + 2 + ...÷ø 2 4 6 1 1 æ 1 ö x – z = çè 2 + 2 + 2 + ...÷ø 2 4 6 \ (y – x) = 3(x – z) Þ 4x = y + 3z Þ x = log3 ç ÷ or x = log3 3 – log3 4 Þ x = 1 – log3 4 (c) Given that for an A.P, Sn = cn2 sin a 2 cos a1 - cosa 2 sin a1 cosa1 cos a 2 + 3 + 2 = 12t - 3 or 12t2 – 5t – 3 = 0; t 1 3 Hence t = - , 3 4 Þ 3x = = sin(a n - a n -1 ) cos a n cos a n -1 x y z Þ2 = + 3 6 2 6. (b) Given Þb– a +b b+c , b, are in A.P.. 1– ab 1– bc a+b b+ c = –b 1– ab 1– bc Solutions 7. 8. 103 – a(b 2 + 1) c(b 2 + 1) = Þ 1– ab 1– bc Þ a + c = 2abc Now, given quadratic equation is 2ac x2 + 2abc x + 2abc = 0 (Substituing a + c = 2abc and then cancelling 2ac) Þ x2 + bx + b = 0 (d) Given mid terms tn = 1 and tn + 1 = 7 \ tn + tn + 1= 8 = t1 + t2n and tn + 1 – tn = 6 = d (common difference of A.P.) tn + tn + 1 = 8 \ a + (n – 1)d + a + nd = 8 \ a + 6n = 7 Now 4t1t2n = [(t1 + t2n)2 – (t2n – t1)2] = 64 – 36(2n – 1)2 [as t2n – t1 = (2n – 1) × 6] \ t1t2n = 16 – 9(2n – 1)2 \ 16 – 9(2n – 1)2 + 713 ³ 0 \–4£n£5 \ n=5 Hence, from a + 6n = 7, a = – 23 (d) Given, b 2 = ac, x = Now, = a c 2a 2c + = + x y a+b b+c 2( ab + ac + ac + bc) ab + ac + b 2 + bc Again, = 9. a+b b+c ,y= 2 2 =2 [Q b 2 = ac] b b 1 ù é 1 + = 2b ê + ú x y ëa +b b + cû 2b(b + c + a + b ) ab + ac + b2 + bc Þ (a – c)2 = 1 – 1 = 0 Þ a = c but a ¹ c as given that a < b < c \ We consider a + c = 1 and ac = – 1/4 Þ (a – c)2 = 1 + 1= 2 Þ a – c = ± 2 but a < c Þ a - c = - 2 Solving (i) and (ii) we get a = 1 1 2 2 x x = 5 Þ r = 11- r 5 10. (c) Since G.P. contains infinite terms \ – 1< r<1 x 5 x 5 Þ -1 < 1 - < 1 Þ - 2 < - < 0 11. Þ – 10 < x < 0. Þ 0 < x < 2 Þ 0 < x < 10. 5 (c) In the quadratic equation ax2 + bx + c = 0 b c D = b 2 - 4 ac and a + b = - , ab = a a a 2 + b2 = (a + b)2 - 2ab = = b2 a2 - 2c b2 - 2ac = a a2 3 3 and a + b =- b3 a 3 - 3c æ b ö ç- ÷ a è aø æ b - 3abc ö ÷ = -ç a3 è ø ATQ a + b, a2 + b2, a3 + b3 are in G.P. 3 2 3 Þ - b , - b - 2ac , - (b - 3 abc) are in G.P. a a2 a3 =2 æ a cö æ b bö \ ç + ÷ ç + ÷ = 4. è x yø è x yø (d) Given that a, b, c are in A.P. Þ 2b = a + c but given a + b + c = 3/2 Þ 3b = 3/2 Þ b = 1/2 and then a + c = 1 Again a2, b2, c2, are in G.P. Þ b4 = a2 c2 1 1 Þ b = ± ac Þ ac = or 4 4 and a + c = 1 Considering a + c = 1 and ac = 1/4 ....(ii) 2 æ b2 - 2 ac ö b æ b3 - 3 abc ö = ç ÷ ÷ Þ ç aè a3 è a2 ø ø 4 2 2 2 4 Þ b + 4a c – 4ab c = b – 3ab2c Þ 4a2c2 – ab2c = 0 Þ ac D = 0 Þ c D = 0 ( Q In quadratic a ¹ 0) 12. (c) Let first term = a, common ratio = r, where –1 < r < 1 Then, 2 ....(i) a3 a = 24 = 2 and 1- r 1 - r3 MATHEMATICS 104 \ 1- r3 (1 - r ) 3 (Q r > 1) 1 3 = \ i.e., 1 – 2r + r2 = 3 (1 + r + r2 ) or 2r 2 + 5r + 2 = 0 \ r = -2 or -1 2 16. As – 1 < r < 1 \ we have r = - 1 2 3 3 3 + - + ... 2 4 8 (b) Given b2 = ac (Q a, b, c are in G. P.) and 2(log 2b – log 3c) = log a – log 2b + log 3c – log a [Q given terms are in A. P.] 2 3c æ 3c ö æ 2b ö Þ log ç ÷ = log ç ÷ Þ b = è 2b ø è 3c ø 2 \ p2 - q2 = 14. 17. b2 + c2 – a 2 2bc 18. \ S = 100.2100 - 2100 + 1 = 99.2100 + 1 (a) Consider two real numbers (a + b) and (c + d). Using G. M. £ A. M. M = (a + b) (c + d) > 0. \ 0 < M £ 1 19. (a) Since r > 1 \ ar 2 is greatest side 2 Þ r2 – r – 1 < 0 1- 5 1+ 5 1+ 5 <r< Þ 1<r< 2 2 2 ( a + b) + ( c + d ) 2 2 M £ 2 Þ M£ 1 Also a, b, c, d are positive. Therefore yn = 2( 2)n –1 = ( 2)n + 1 \ a + ar > ar ( a + b)(c + d ) £ Þ 2. (c) Let sides of triangle be a, ar, ar 2 1(2100 - 1) - 100.2100 = 2100 - 1 - 100.2100 2 -1 We get 2, 2 2 , 2 4 , 2 8 , ................ is a G.P. where 15. (a - c )(a + c + 2b) = (a - c)b = p '2 - q '2 4 (d) Let S = 1 + 2.2 + 3. 22 + 4 .23 + 5.24 + .... + 100 . 299 \ 2S = 1.2 + 2.22 + 3.23 + ... + 99.299 + 100 . 2100 Subtracting, we get –S = 1 + 1.2 + 1.22 + .... + 1.299 – 100. 2100 = (1 + 2 + 22 + ... + 299) – 100 . 2100 = (c) y = 2 x , being in the first quadrant. The sequence of x-coordinate 1, 2, 4, 8, ................ \ the sequence of y-coordinate the common ratio is (a + b)2 - (b + c )2 4 = 3b 9c b2 = = 2 4 c \ a is the largest side. 9c 2 81 + c2 – c2 4 16 = = negative 3c 2´ ´c 2 \ A > 90° \ triangle is obtuse. a+b b+c ,q= 2 2 Again, p ' = ab and q ' = bc Now, a = Now, cos A = 1+ 5 <–r<–1 2 \ [– r] = – 2 [r] + [– r] = 1 – 2 = – 1. (c) We have 2b = a + c and a, p, b, q, c are in A.P. Þ p= \ The series is 3 - 13. [r] = 1. Also – = tn = n + 2 æ 1ö .ç ÷ n(n + 1) è 2 ø 2(n + 1) - n æ 1 ö .ç ÷ n(n + 1) è 2 ø n n Solutions 105 1 æ1ö = ç ÷ n è 2ø n-1 n 1 æ1ö – . . n + 1 çè 2 ÷ø ìï1 æ 1 ö° 1 æ 1 ö1 üï Sn = å t n = í ç ÷ - ç ÷ ý n=1 ïî1 è 2 ø 2 è 2 ø ïþ n ìï 1 æ 1 ö1 1 æ 1 ö 2 üï + í ç ÷ - ç ÷ ý + .......... ïî 2 è 2 ø 3 è 2 ø ïþ n ìï 1 æ 1 ö n -1 1 1 æ 1 ö üï + í ç ÷ ý =1– ç ÷ n è 2ø n + 1 è 2ø ï ( n + 1)2 n îï þ 20. (a) As A.M. ³ G.M., we get 1æ 2 b b ö æ 2 b b ö ç ax + + ÷ø ³ çè ax × × ÷ø 3è 2x 2 x 2 x 2x Þ 1 æ 2 b ö æ ab ö ç ax + ÷ø ³ çè ÷ø x 3è 4 1/ 3 But the least value of ax2 + æ ab2 ö 3ç ÷ è 4 ø 1/ 3 b is c, therefore x ³ c Þ 27ab2 ³ 4c3 21. (0.60) Let a = first term of G.P. and r = common ratio of G.P.; Then G.P. is a, ar, ar2 a = 20 1- r Þ a = 20(1 – r) Also a 2 + a 2 r 2 + Þ b 36 and c = br Þ b = 36 and a = r r r = 2, 3, 4, 6, 9, 12, 18 Let a = Þ æ 1ö Also 36 ç1– ÷ is a perfect cube. Þ r = 4 è rø Þ a + b + c = 9 + 36 + 144 = 189. 24. (3) We have 2 6 10 14 S = 1 + + 2 + 3 + 4 + .......¥ 3 3 3 3 Multiplying both sides by 1/ 3 Given S ¥ = 20 Þ From Eqs. (i) and (ii) , we get r2 – 2r + 1 = 16r – 64 Þ r2 – 18r + 65 = 0 (r – 5) (r – 13) = 0 r = 5 or 13 If r = 5, then a(80 – 64) = 64 and hence a = 4. In this case the numbers are 4,20,100 and their sum is 124. 23. (189) log6 (abc) = 6 Þ (abc) = 66 ... (i) a 2 r 4 + ... to ¥ =100 1 , we get 3 1 1 2 6 10 S = + 2 + 3 + 4 + ........¥ ....(ii) 3 3 3 3 3 Subtracting eqn. (ii) from eqn. (i), we get 2 1 4 4 4 S = 1 + + 2 + 3 + 4 + ........¥ 3 3 3 3 3 Þ 2 4 4 4 4 S = + 2 + 3 + 4 + ........¥ 3 3 3 3 3 2 Þ S= 3 a2 =100 1 - r2 ... (ii) Þ a2 = 100(1 – r)(1 + r) 2 2 From (i), a = 400(1 – r) ; From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)2 Þ 1 + r = 4 – 4r Þ 5r = 3 Þ r = 3/5 = 0.60 22. (124) Let b = ar and c = ar2. Given that a, ar, ar2 – 4 are in AP. Therefore a + (ar2 – 64) = 2(ar) ...(i) 2 Þ a(r – 2r + 1) = 64 Again a, ar – 8, ar2 – 64 are in GP. Therefore a(ar2 – 64) = (ar – 8)2 ...(ii) Þ a(16r – 64) = 54 ....(i) 4 3 = 4´3 Þ S =3 1 3 2 13 25. (990) Let S1= 2 + 3 + 5 +9 + 16 +............+ xn S1 = 2 + 3 + 5 + 9 + ..........x n -1 + x n 0 = 2 + [1 + 2 + 4 + 7 + ...... + to (n - 1) term] - xn \ x n = 2 + [1 + 2 + 4 + 7 + .....to (n - 1) terms] Again let S2 = 1 + 2 + 4 + 7 + ........+ tn–1 S2 = 1 + 2 + 4 + 7..... + t n - 2 + t n -1 0 = 1 + [1 + 2 + 3 + ....... + (n - 2) term] - t n -1 MATHEMATICS 106 (n - 2)(n - 1) n 2 - 3n + 4 t n -1 = 1 + = 2 2 n -1 \ S2 = 1 3 å t n -1 = 2 Sn 2 - 2 Sn + 2S1 n =1 = 1 ( n - 1) n ( 2n - 1) 3 n ( n - 1) + 2( n - 1) 2 6 2 2 28. é 2n 2 - n 3n ù = (n - 1) ê - + 2ú 4 ëê 12 ûú = 1 1– | cos x | = n -1 é 2 2n - n - 9n + 24 ù û 12 ë \ = n -1 é 2 n 3 - 6n 2 + 17n - 12 n - 5n + 12ù = û 6 ë 6 Þ |cos x| = \ x n = 2 + S2 = 2 + n 3 - 6n 2 + 17n - 12 6 n 3 - 6n 2 + 17n 6 So, x20 = 990 (5) Let k and k + 1 be removed. Then, 29. n (n + 1) - 2k - 1 105 2 = n-2 4 4m 2 + 103(1 - m) . Clearly 4 (1 – m) must be divisible by 4. Let m = 1 + 4t, then we get k = 16t2 – 95t + 1 and 1£ k < n 27. 1 1 Þ cos x = ± 2 2 p 2p 1 , Þ ( x1 + x2 ) = 1. p 3 3 (6) For the given A.P., we have 2(2a + b) = (5a – b) + (a + 2b) Þ b = 2a ...(1) Also for the given G.P., we have (ab + 1)2 = (a – 1)2 (b + 1)2 ...(2) Putting b = 2a from (1) in (2), we get a = 0, – 2, or 1 1 and b = 2a = 4 2 Hence, (a– 1 + b– 1) = 2 + 4 = 6. (2) Let the two quantities be a and b But a > 0, so a = 30. G = ab ; p = a + (kp – q) = ka + Þ 1 £ 16t 2 - 95t + 1 < 8t + 2 Þ t = 6 and so, n = 50 (8) Since a, b, c, d are in A.P., we have b – a = c – b = d – c = D (let common difference) or d = a + 3D Þ a – d = – 3d and d = b + 2D or b – d = – 2D Also c = a + 2D or c – a = 2D \ Given equation 2(a – b) + k(b – c)2 + 3 =6 1– | cos x | 1 4 Þ 2n 2 - 103n - 8k + 206 = 0 Since n and k are integers, so n must be even, say n = 2m, then k = 33/(1 – |cos x|) = 36 Þ Þ x= = 26. (c – a)3 = 2(a – d) + (b – d)2 + (c – d)3 becomes – 2D + kD2 + (2D)3 = – 6D + 4D2 – D3 Þ 9D2 + (k – 4)D + 4 = 0 Since D is real, we have (k – 4)2 – 4(4) (9) ³ 0 or k2 – 8k – 128 ³ 0 or (k – 16) (k + 8) ³ 0 \ k Î(– ¥, – 8] È [16, ¥) Hence, the smallest positive value of k = 16. (1) x Î (0, p), implies |cos x| < 1. Thus, S¥ = 1 + |cos x| + cos2 x + |cos3 x| + ... up to ¥ = 1 + |cos x| + |cos x|2 + |cos x|3 + ... up to ¥ –2 b–a 2(b – a ) ;q=a+ 3 3 k (b – a ) –a 3 (b – a ) b–a = a( k –1) + ( k – 2) 3 3 kq – p = ak + 2k (b – a) (b – a ) –a– 3 3 b–a (2k – 1) 3 \ (kp – q) (kq – p) = ab Þ a(k – 1) + \k=2 Solutions 107 CHAPTER Straight Lines 10 1. (a) Given P º (a, b) 13 - 3 (x – 4) 7-4 Þ 3y – 10x + 31 = 0. Now, AB is y – 3 = x y ...(i) + =1 a b If line (i) cuts x and y axes at A and B respectively, then A = (a, 0) and B = (0,b). Given line is Also the area of D OAB = S i.e. (7,13) Y 1 ab = S 2 Ry = x A (3,4) Þ ab = 2S Since line (i) passes through P( a , b ) P a b ab a + =1 Þ + =1 a 2 S b a A’ (4,3) X 2 Þ a b – (2S)a + 2 a S = 0 Since a is real, 4S 2 – 8abS ³ 0 Þ S ³ 2ab 2. Hence the least value of S = 2ab . (b) a, b, c are the roots of equation x 3 - 3 x 2 + 6x + 1 = 0. So, a + b + c = 3, ab + bc + ca = 6 abc = – 1 Now, the centroid of the triangle is 4. and æ aö b Since AO isperpendicular to BC, (– 1) ç - ÷ = – 1 è ø \ a =–b A (c) Consider a point A’ , the image of A in y = x \ Coordinates of A’ = (4, 3) [Notice that A and B lie to the same side with respect to y = x]. Then PA = PA’. Thus, PA + PB is minimum, if PA’+ PB is minimum, i.e. if P, A’, B are collinear. O =0 3. B + 0 -1 æ6 3 ö =ç , ÷ or (2, – 1) è 3 -3ø 2x 1= 2y æ ab + bc + ca a + b + c ö , è 3 3abc ÷ø i.e. ç 3y x+ 1 1 1ö æ + + ç ab + bc + ca ab bc ca ÷ , ç ÷ 3 3 ç ÷ è ø æ 31 31 ö It intersects y = x at ç , ÷ , which is the è 7 7ø required point P. (c) Equation of AO is 2x + 3y – 1 + l (x + 2y – 1) = 0 Where l = – 1, since the line passes through the origin. So, x + y = 0 ax + by - 1 = 0 C Similarly, (2x + 3y – 1) + m (ax – ay – 1) = 0 will be the equation of BO for m = – 1. BO is perpendicular to AC ì (2 - a) ü æ 1 ö Þ íý ç - ÷ = – 1. \ a = – 8, b = 8. î 3+ a þ è 2ø MATHEMATICS 108 5. (d) The coordinates of A are (a, 0) and of B are (0, b). If the coordinates of C are (x1, x1) then area of the D AOC = æ6 ö y = 6 intersects y = mx at Q ç , 6 ÷ èm ø 1 OA × x1 and the area of the 2 2 6 2 Thus PQ = æç - ö÷ + ( 6 - 2 )2 < 5 èm mø 1 OB × x1 . 2 According to the given condition, D BOC = 2 æ4ö Þ ç ÷ + 16 < 25 èmø 1 1 OA × x1 = 2 × OB × x1 Þ OA = 2OB 2 2 Þ a = 2b. = Ry B ( 0, b) 2 4 3 1 3 æ 4ö Þ ç ÷ < 9 Þ -3 < < 3 Þ - < < è mø m 4 m 4 \ x 8. C O Þ Þ Þ Þ Þ 2b a = 3 3 Hence the coordinates of C are Þ x1 = AP BP CP 1 = = = AQ BQ CQ 2 2AP = AQ Þ 4(AP)2 = AQ2 4[(x – 1)2 + y2] = (x + 1)2 + y2 4(x2 + 1 – 2x) + 4y2 = x2 + 1 + 2x + y2 3x2 + 3y2 – 8x – 2x + 4 – 1 = 0 3x2 + 3y2 – 10x + 3 = 0 10 x +1=0 ...(1) 3 \ A lies on the circle given by (1). As B and C also follow the same condition. \ Centre of circumcircle of DABC = centre of circle given by (1) Þ æ 2b 2b ö æ a aö çè 3 , 3 ÷ø or çè 3 , 3÷ø . (c) ry ry+3x+2=0 4y+x-14=0 7/2 9. 5/3 (a) Let P(1, 0) and Q(–1, 0), A(x, y) Given: A (a, 0) x y Equation of AB is therefore + =1 2b b or x + 2y = 2b Since C lies on it, x1 + 2 x1 = 2b 6. 4ö æ 4 ö æ m Îç - ¥, - ÷ È ç , ¥÷ è 3ø è 3 ø 3y-2x-5=0 x2 + y2 – æ5 ö = ç ,0÷ . è3 ø (b) Let C = (x1, y1) A(– 3, 2) 1 –5/2 –2 –1 –2/3 O 1 2 –1 –2 4 5 7 £ b £ . 3 2 æ2 ö (c) y = 2 intersects y = mx at P ç , 2 ÷ èm ø From diagram it is clear that 7. 3 X 2 E 1 B (– 2, 1) D æ x1 - 2 y1 + 1 ö , ç ÷ è 2 2 ø æ x - 5 y1 + 3 ö Centroid, E = ç 1 , ÷ 3 ø è 3 C (x1, y1) Solutions 109 Since centroid lies on the line 3x + 4y + 2 = 0 æ x -5ö æ y1 + 3 ö 3ç 1 ÷ + 4ç ÷+2 = 0 è 3 ø è 3 ø Þ 3x1 + 4y1 + 3 = 0 Hence vertex (x1, y1) lies on the line 3x + 4y + 3 = 0 10. (d) Circumcentre = (0, 0) \ æ (a + 1) (a - 1) Centroid = ç 2 , 2 ÷ è ø 2 2ö We know the circumcentre (O), Centroid (G) and orthocentre (H) of a triangle lie on the line joining the O and G. HG 2 Also, = Þ Coordinate of orthocentre GO 1 3(a + 1) 2 3(a - 1)2 , 2 2 Now, these coordinates satisfies eqn given in option (d) Hence, required eqn of line is (a – 1)2 x – (a + 1)2 y = 0 (a) The line passing through the intersection of lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0 is b , a since it is ^ to the line ax + by + c = 0 and it cuts the x-axis at (2,0). Hence, the required line passes 12. (d) Slope of the line in the new position is through (2, 0) and its slope is y-0 = b ( x - 2) a Þ ay = bx - 2b Þ ay - bx + 2b = 0 13. (d) Let L1 (x, y) = x – y – 1 and L 2 (x, y) = 2 (x – y ) + 5 Then, from figure P(a, 2) lies below L 2 , So 2a - 4 + 5 1 <0 Þ a > -2 2 = 11. ax + 2by + 3b + l (bx – 2ay – 3a) = 0 Þ (a + b l ) x + (2b – 2a l )y + 3b – 3 l a = 0 As this line is parallel to x-axis. \ a + b l = 0 Þ l = – a/b a Þ ax + 2by + 3b – (bx – 2ay – 3a) = 0 b 2a 2 3a 2 =0 y+ Þ ax + 2by + 3b – ax + b b æ 2a 2 ö 3a 2 =0 y ç 2b + ÷ + 3b + ç b ø÷ b è 2ö 2ö æ 2 æ 2 y ç 2b + 2a ÷ = - ç 3b + 3a ÷ ç ÷ ç ÷ b b è ø è ø 2 y= 2 -3(a + b ) 2(b2 + a 2 ) = -3 2 So it is 3/2 units below x-axis. b . Required eq. is a L 2 : 2(x – y) + 5 = 0 P (a, 2) L1 : x - y - 1 = 0 Also P(a, 2) lies above L1 , So a - 2 -1 >0 Þ a<3 -1 Taking intersection, we get - 1 <a<3 2 æ 1 ö Þ a Î ç - , 3÷ . è 2 ø 14. (b) A(5, – 1) H B (– 2, 3) (0, 0) C (a, b) MATHEMATICS 110 Let the third vertex of DABC be (a, b). Orthocentre = H(0, 0) Let A (5, – 1) and B (– 2, 3) be other two vertices of DABC. Now, (Slope of AH) × (Slope of BC) = – 1 Þ æ -1 - 0 ö æ b - 3 ö çè ÷ç ÷ = -1 5 - 0 ø è a + 2ø 2 a2 æ 2ö Þ a2 = ç ÷ + è 5ø 4 Þ a2 - Þ b – 3 = 5 (a + 2) Similarly, (Slope of BH) × (Slope of AC) = – 1 Þ Now, In DPCB, (PB)2 = (PC)2 + (CB)2 (By Pythagoras theoresm) ...(1) a2 = æ 3ö æ b + 1ö -ç ÷ ´ç = -1 è 2 ø è a - 5 ÷ø a4 4 3a 2 4 Þ = = 4 25 4 25 16 Þa= 75 \ Length of Equilateral triangle ( a) = Þ 3b + 3 = 2a – 10 Þ 3b – 2a + 13 = 0 On solving equations (1) and (2) we get a = – 4, b = – 7 Hence, third vertex is (– 4, – 7). 15. (b) ...(2) and L 2 (x, y) = 2x – 3y – 5 \ L1 (10, – 20) = 10 – 20 + 1 = – 9 Þ ‘–’ve and L 2 ( 10, – 20) = 20 + 60 – 5 = 75 Þ ‘+’ve Equation of the bisector will be x + y +1 2 C 3x + 4y = 9 Shortest distance of a point (x1, y1) from line ax1 + by1 - c a2 + b2 Now shortest distance of P (1, 2) from 3x + 4y = 9 is PC = d = 3(1) + 4(2) - 9 2 3 +4 2 = 2 5 Given that DAPB is an equilateral triangle Let 'a' be its side then PB = a, CB = a 2 æ 2 x - 3 y - 5ö ÷ è 13 ø = -ç Þ x ( 13 + 2 2 ) + y ( 13 – 3 2 ) + ( 13 B ax + by = c is d = 4 3 15 16. (a) Let, L1 (x, y) = x + y + 1 \ P (1, 2) A 16 4 3 4 3 = ´ = 75 5 3 15 3 –5 2 )=0 17. (c) If a point is equidistant from the two intersecting lines, then the locus of this point is the angle bisector of those lines. Now, let (h, k) be the point which is equidistant from the lines 4x – 3y + 7 = 0 and 3x – 4y + 14 = 0 Then 4h - 3k + 7 = ± 3h - 4k + 14 4 2 + ( -3) 2 32 + ( -4) 2 Þ 4h - 3k + 7 = ± (3h - 4k + 14) Þ h + k – 7 = 0 and 7h – 7k + 21 = 0 Hence locus of (h, k) is x + y – 7 = 0 and x – y + 3 = 0 Solutions 111 18. (7) The given lines are 2x + y = 9/2 … (i) and 2x + y = – 6 … (ii) Signs of constants on R.H.S. show that two lines lie on opp. sides of origin. Let any line through origin meets these lines in P and Q respectively then req. ratio is OP : OQ Y (0, 9/2) B P C (–3, 0) A X O x' (9/4, 0) Q Ð POA = Ð QOC (ver. opp. Ð' s) Ð PAO = Ð OCQ (alt. int. Ð' s) \ DOPA ~ DOQC (by AA similarity) OP OA 9 / 4 3 = = = OQ OC 3 4 Req. ratio is 3 : 4. = OP : OQ = 9 20 : -6 5 1 5 æQ Perpendicular distance of ö ç ÷ ç ax + by = c from (x1 , y1 )is ÷ ç ÷ ax + by1 + c ç ÷ p= 1 ç ÷ 2 2 a +b è ø Let AB = BC = AC = x So, in DABD , AD 2 + BD 2 = AB 2 m n Þ m + n = 7. 19. (15) Let ABC be an equilateral triangle with base BC. So, AD ^ BC A (2,–1) BC ö æ ÷ çQ BD = 2 ø è 4 2 2 Þx= = 15 K 15 K = 15 Þ x2 = Þ 20. (0) Since so = 3: 4 = 5 = 2 Now in DOPA and DOQC, \ 2 - 2 -1 æ 1 ö x2 ÷÷ + Þ çç = x2 4 è 5ø D (0, –6) y' \ = Þ 3 .1 – 4 + 1 < 0 , 3 sin q – cos q + 1 £ 0 1 3 1 sin q – cos q £ – 2 2 2 1 pö æ Þ sin ç q - ÷ £ – 2 6ø è Þ 11p p 4p 7p £ q £ 2p Þ £ q– £ 6 6 3 6 Þ maximum value of sin q is 0. 21. (4) Family of lines can be written as (3x + 7 y + 11)sec q + cosec q (5 x - 3 y - 11) = 0 B D C x+2y =1 Now, length AD = 2 + 2(-1) - 1 (1)2 + (2)2 Þ (5 x - 3 y - 11) + tan q (3 x + 7 y + 11) = 0 So, for any value of q this line passes through the point of intersection of lines 5 x - 3 y - 11 = 0and 3x + 7 y + 11 = 0 i.e from P(1, – 2). MATHEMATICS 112 Let A( x, y ) be a point on the line x- y+3=0 Þ k + h 5+ 4 = =9 k –h 5–4 Applying cosines rule in DPAB PB2 = PA2 + AB2 - 2PA. AB cos q ³ PA2 + AB2 A(1, – 2) 24. (5) 2x 3 ( PA - AB) 2 £ PB 2 y= +y x– Þ =0 -2PA. AB = ( PA - AB )2 So, maximum value of | PA - PB | is PB = 4 + 36 = 2 10. Þ k 2 = 40 B x + py = q æ q 2q ö çè 2 p 1 , 2 p 1÷ø C æ 3 p + q q 3ö çè p + 1 , p + 1÷ø 22. (1) Let the equation of L2 be L1 + lL = 0 Þ (1 + l)x + (2 + l)y + 3 + l = 0 Slopes of L2, L and L1 are – 1+ l , – 1, – 1/2 2+l Since L is the bisector of the angle between L1 and L2 1+ l +1 2+ l –1 + 1/ 2 Þl=–3 = 1+ l 1 + 1/ 2 1+ 2+l – \ So the equation of L2 is y + 2x = 0 Þ m = – 2 and 812m2 + 3 = 812 × 4 + 3 = 3251 23. (9) tana, tanb, tanl are the roots of equation t3 – 12t2 + 15t – 1 = 0 (given). So, tana, tanb, tanl = 12 and tana, tanb, tanl = 1. Also Stana tanb = 15 When A(tana, cota), B(tanb, cotb) and C(tanl, cotl) The centroid G(h, k) æ tan a + tan b + tan l cot a + cot b+ cot l ö , =ç ÷ 3 3 è ø æ 12 D tan a.tan b ö =ç , ÷ = (4, 5) 3 3(tan a tan b tan l) ø è P is the orthocenter. Therefore, AP ^ BC æ 1 ö æ 3 + 2ö 5 or ç – ÷ ç = – 1 or = 1 or p = 5 ÷ è p ø è 2 –1 ø p Since BP ^ AC, we have 27 – 2q =–1 18 + q or q = 27 + 18 or q = 45 \ p + q = 5 + 45 = 50 25. (4) Since y = x is the perpendicular bisector of the side AB and A = (1, 2), we have B = (2, 1). Since the image A'(x, y) of A in the angular bisector x – 2y + 1 = 0 lies on the line BC, we have x –1 y – 2 – 2(1– 2(2) + 1) 4 = = = 1 –2 5 12 + 2 2 æ 9 2ö Therefore, A' = ç , ÷ è 5 5ø Since equation of BC is the equation of BA', we have the equation of BC as y–1= 1– (2 / 5) ( x – 2) 2 – (9 / 5) Þ y – 1 = 3(x – 2) Þ 3x – y – 5 = 0 So that a = 3, b = – 1. Hence, a – b = 4. Solutions 113 CHAPTER Conic Sections 11 1. 2. (a) Since 4x + 3y – 4 = 0 is dividing the circumference in the ratio 1 : 2, angle subtended at the centre = 2 p / 3. Also the perpendicular distance from the centre of the given line is 5 Þ Radius = 10 Þ Equation of the circle is Þ y = x + (5 2 – 4) For no solution c > 5 2 – 4 5. \ c Î (5 2 – 4, ¥ ). (b) Equation of the given circle is x 2 + y 2 + 2gx + 2fy + c = 0 x 2 + y 2 - 10 x - 6 y - 66 = 0. (b) y C(4 cos q , 4 sin q ) ......(i) Y given circle P R A B O (– 4, 0) x Q O 3. 1 A = . 8 . 4 sinq = |16 sin q| 2 1 2 15 Now, sinq can be equal to , ,... 16 16 16 i.e. there are 15 points in each quadrant. (c) If (a, 0) is the centre C and P is (2, – 2), then Ð COP = 45°. Since the equation of OP is x + y = 0. \ Equation of the chord of contact PQ, drawn from the origin (0, 0) to the given circle will be gx + fy + c = 0 .....(ii) Eq. of any circle passing through the intersection points of the given circle and the chord PQ can be written as 2 2 ( x + y + 2gx + 2fy + c ) + l (gx + fy + c) = 0 ....(iii) If this circle passes through the origin, then we have, c + l c = 0 gives l = – 1 Putting the above value of l in equation (iii) gives the equation of the required circle as OP = 2 2 = CP. Hence OC = 4 Y O C A B circumcircle X of DOPQ X x 2 + y 2 + gx + fy = 0 P , (2 –2 6. (b) ) The point on the circle with the greatest x coordinate is B. a = OB = OC + CB = 4 + 2 2 . 4. 2 (d) Since y = | x | + c and x + y 2 – 8 | x | – 9 = 0 both are symmetrical about y-axis we consider the case x > 0, when the equations become 2 y = x + c and x + y 2 – 8x – 9 = 0 . Equation of 2 P (–2, –1) (0, –1) Any line through ( – 2, – 1) is y + 1 = m ( x + 2 ) It touches the circle if tangent to circle x 2 + y 2 – 8x – 9 = 0 parallel to y = x + c is y = (x – 4) + 5 1 + 1 2 x + y =1 B Þ m = 0, 4 3 2m - 1 1 + m2 =1 MATHEMATICS 114 4 \ Equation of PB is y + 1 = ( x + 2) 3 Þ 4 x– 3y+ 5=0 A point on PB is (– 5, – 5), (we can choose some other point as well) Its image by the line y = – 1 is P¢ ( – 5 , 3). Hence equation of incident ray PP¢ is 3 +1 ( x + 5) Þ 4 x + 3 y + 11 = 0 -5 + 2 (b) Circle : x2 + y2 + 3x = 0, y -3 = 7. æ 3 ö Centre, B = ç – , 0÷ è 2 ø 3 Radius = units. 2 B æ 3 ö çè – , 0÷ø 2 8. X y = mx + 1 y-intercept of the line = 1 \ A = (0, 1) OA Slope of line, m = tan q = OB 1 2 Þ m= = Þ 3m – 2 = 0 3 3 2 (b) Let the variable circle is x 2 + y 2 + 2 gx + 2 fy + c = 0 It passes through (a, b) \ a 2 + b2 + 2 ga + 2 fb + c = 0 2 2 (i) cuts x + y = 4 orthogonally \ 2( g ´ 0 + f ´ 0) = c - 4 Þ c = 4 \ from (ii) a 2 + b2 + 2 ga + 2 fb + 4 = 0 \ Locus of centre (–g,–f ) is a 2 + b2 - 2ax - 2by + 4 = 0 or 2ax + 2by = a 2 + b 2 + 4 ...(ii) 50 - l - 2 < 3 2 or 3 2 < 50 - l + 2 50 - l < 4 2 or 2 2 < 50 - l Þ 50 - l < 32 or 8 < 50 - l Þ l > 18 or l < 42 Required interval is (18, 42) 10. (b) Equation of required circle : S : (x – 1)2 + (y – 1)2 + l (x – y) = 0 S' : x2 + y2 – 2y – 3 = 0 Common chord of S = 0 and S¢ = 0 is S – S ' = 0 (l – 2) x – (l + 4) y + 5 = 0 Centre of S ' : (0, – 1) lies on common chord Þ l =–9 9 S : (x – 1)2 + (y – 1)2 – 9 (x – y) = 0 Þ r = 2 11. (c) 2 y = 8x Y O ...(i) r1 = 50 - l , r2 = 2 For exactly two common tangents we have r1 - r2 < C1C2 < r1 + r2 Þ 50 - l - 2 < 3 2 < 50 - l + 2 Þ Y¢ Line : (b) The equations of the circles are x 2 + y 2 - 10 x - 10 y + l = 0 and x 2 + y 2 - 4 x - 4 y + 6 = 0 C1 = centre of (i) = (5, 5) C2 = centre of (ii) = (2, 2) d = distance between centres = C1C2 = 9 + 9 = 18 Þ A(0, 1) X¢ 9. A 2 2 2 x +y =3 B 2 ...(i) ...(ii) We have : x + (8x) = 9 2 Þ x + 9x – x – 9 = 0 Þ x (x + 9) – 1 (x + 9) = 0 Þ (x + 9) (x – 1) = 0 Þ x = –9, 1 for x = 1, y = ± 2 2 x = ± 2 2 L1 = Length of AB (2 2 + 2 2) 2 + (1 - 1) 2 = 4 2 L2 = Length of latus rectum = 4a = 4 × 2 = 8 L1 < L2 12. (c) Equation of the tangent at P (x1, y1) to y2 = 4ax is yy1 – 2ax – 2ax1 = 0 ...(i) Equation of the chord of y2 = 4a(x + b) whose mid-point is (x', y') is = Solutions 115 yy' – 2ax – 2ax' – 4ab = y'2 – 4a x' – 4ab or yy' – 2ax – (y'2 – 2a x') = 0 ...(ii) Equation (i) and (ii) represent the same line. y1 2 ax 2a = =– \ 2 y' 2a y ' - 2ax ' This gives y' = y1 and then 2ax1 = y'2 – 2ax' = y12 – 2ax' = 4ax1 – 2ax' \ x' = x1 \ mid-point (x', y') = (x1, y1). 13. (d) The given curve is y = x2 + 6 Equation of tangent at (1, 7) is 1 ...(i) ( y + 7) = x .1 + 6 Þ 2x - y + 5 = 0 2 As given this tangent (1) touches the circle x2 + y2 +16x + 12y + c = 0 at Q Centre of circle = (– 8, – 6). =0 +5 y – 2x P(1,7) Q C (–8,–6) 9 4 2 \ a = b 2 (1 - e 2 ) = 9 and centre of the ellipse is (2, 1) \ Equation of the required ellipse is ( x - 2) 2 ( y - 1)2 =1 + 9 25 16. (a) For ellipse a = 4, b = 3 x2 2 4 + y2 32 =1 2 Then equation of CQ which is perpendicular to (1) and passes through (– 8, – 6) is 1 y + 6 = – ( x + 8) Þ x + 2y + 20 = 0 ...(ii) 2 Now Q is pt. of intersection of (i) and (ii) \ Solving eqs (i) & (ii) we get ; x = – 6, y = – 7 \ Req. pt. is (– 6, – 7). 14. (a) Since S = (a, 0) = (1, 0), the circle is of the form (x – 1)2 + y2 = r2 Suppose AB is a common chord. Since this is equidistant from the focus and the vertex. M(1/2,0) lies on AB and AB is double ordinate of the parabola, let A = (1/2, y) so that æ1ö y2 = 4 ç ÷ Þ y = ± 2 è2ø æ1 ö æ1 ö Þ A = ç , 2 ÷ and B = ç , - 2 ÷ 2 2 è ø è ø Since DAMS is right-angled triangle, we have 1 9 SA2 = SM2 + MA2 = + 2 = = (Radius)2 4 4 Hence, the equation of the circle is (x – 1)2 + y2 = 15. (d) As per the definition, the locus must be an ellipse, with given points as foci and 10 as its major axis. Since the line segment joining (2, –3) and (2, 5) is parallel to y-axis, therefore, ellipse is vertical. 4 \ 2 be = 8 and 2b = 10 Þ b = 5 and e = 5 Þ 7 æ3ö e = 1–ç ÷ = \ Foci are è4ø 4 ( 7, 0) and ( – 7,0) Centre of circle is at (0, 3) and it passes through ( ± 7, 0) , therefore radius of circle = ( 7 ) + ( 3)2 = 4 2 17. (d) x2 = 8y ...(i) 2 x + y2 = 1 3 ...(ii) 8y 1 + y 2 = 1 Þ y = – 3, 3 3 When y = – 3, then x2 = – 24, which is not possible. From (i) and (ii), 2 6 1 , then x = ± 3 3 Point of intersection are æ 2 6 1ö æ 2 6 1ö , ÷ and ç , ÷ ç 3 3ø è 3 3ø è When y = Required equation of the line, y Þ 3y – 1 = 0 1 =0 3 y2 x2 + =1 ... (i) 4 16 Equation of a circle centered at (1, 0) can be written as (x –1)2 + y2 = r2 ... (ii) The abscissae of the intersection points of the circle and the ellipse is given by the equation 18. (c) Given ellipse is MATHEMATICS 116 ( x - 1)2 + See figure B (4, 2) is one end of the minor axis of 16 - x 2 2 =r 4 (4, 0) (1, 0) 2 2 ( x - 4)2 y 2 + = 1 and (– 1, 2) is one 25 4 end of the major axis of the second ellipse. Therefore, AB = 5, OB = 16 + 4 = 20 , OA = 1 + 4 = 5 We have (OA)2 + (OB)2 = 25 = (AB)2 p Therefore Ð AOB = 2 21. (1) C the ellipse i.e. 4 ( x – 2x + 1) + 16 – x = 4 r 2 2 2 i.e. 3 x – 8x + 20 – 4 r = 0 If the circle lies inside the ellipse, then the roots of the above equation must be imaginary or equal Q i.e. D £ 0 i.e. 64 + 12 (4 r 2 – 20) £ 0 Þ r= £ S 11 3 B R Hence, greatest value of r = equation of required circle is 11 ( x - 1)2 + y 2 = 3 11 and the 3 2 i.e. 3 ( x + y 2 ) – 6x – 8 = 0. x2 y 2 + =1 16 3 Now, equation of normal at (2, 3/2) is 16 x 3 y = 16 - 3 2 3/ 2 13 Þ 8x – 2y = 13 Þ y = 4 x 2 13 Let y = 4 x - touches a parabola 2 y2 = 4ax. We know, a straight line y = mx + c touches a parabola y2 = 4ax if a – mc = 0 æ 13ö \ a - ( 4) ç - ÷ = 0 Þ a = – 26 è 2ø Hence, required equation of parabola is y2 = 4 (– 26)x = – 104 x y 20. (a) 19. (a) Ellipse is A B O A P Point A ( 33 + 3, 0) lies on the given circle, x2 + y2 – 6x – 8y – 24 = 0 PQ and AB intersect inside the circle. Let PR = a, RS = b, QS = c Since PR × RQ = AR × RB Þ a(b + c) = 3 × 7 Also, QS × SP = 3 × 7 Þ c(a + b) = 3 × 7 Þ a = c \ PR/QS = 1 22. (11) The centre C of the circle = ( 5,7 ) and the radius = 52 + 7 2 + 51 = 5 5 PC = 122 + 52 = 13 Þ q = PA = 13 - 5 5 and p = PB = 13 + 5 3 \ G.M. of p and q = pq = (13 - 5 5(13 + 5 5)) = 169 - 125 = 2 11 = 2 k Þ k = 11. A C (5, 7) B P (–7, 2) (0, –1) 2 23. (4) Tangent to y = 8 (x + 2) is x y = m (x + 2) + c = 2m + 2 m 1ö 2 c æ = çè m + ÷ø Þ m m 2 Solutions 117 1 c ³2 Þ ³ 2 Þ c³4 m 2 Þ The minimum value of c = 4. 24. (3) The locus of the point of intersection of tangents to the parabola y2 = 4 ax inclined at an angle a to each other is tan2a (x + a)2 = y2 – 4ax Given equation of Parabola y2 = 4x {a = 1} Point of intersection (–2, –1) tan2a (–2 + 1)2 = (–1)2 – 4 × 1 × (–2) Þ tan2a = 9 Þ tan a = ± 3 Þ |tan a| = 3 5 4 x2 y2 2 25. (27) + =1 Þ e =1 - = 9 9 9 5 2 Þ e= 3 æ 5ö One end of latusrectum is ç 2, ÷ è 3ø y 2x æ 5ö =1 Equation of tangent at ç 2, ÷ is + 3 9 è 3ø Q m+ Q 2ö æ ç - ae, b ÷ ç a ÷ø è R F¢ 2ö æ ç- ae, - b ÷ ç a ÷ø è 2ö æ ç ae, b ÷ ç a ÷ø è C F O P 1 9 27 Area of D CPQ = × ×3= 2 2 4 \ Area of quadrilateral PQRS = 4 × 27 = 27. 4 26. (4) Focus of the parabola y 2 = 4x is (1, 0) So diagonals are focal chord. AS = 1 + t 2 = c (say) 1 1 =1 Q + 25 c -c 4 For c = 5, 1 + t 2 = 5 Þ t = ± 2 æ1 ö A º ç , 1÷ , B º (4 , 4), C º (4, – 4) and è4 ø ö æ1 D º ç , -1÷ ø è4 AD = 2 and BC = 8, distance between AD and BC 15 = 4 \ Area of trapezium ABCD 1 15 75 = ( 2 + 8) × = sq. units. 2 4 4 27. (2) Due to symmetry the desired area 1 = 4 ´ area of DS1OS3 = 4 ´ ae ´ be1 2 S 3 2ö æ ç ae, - b ÷ ç ÷ a è ø S 25 25 c – c 2 Þ 4c 2 – 25 c + 25 = 0 = 4 4 5 Þ c= ,5 4 5 5 1 1 Þ t2 = For c = , 1 + t 2 = Þt =± 4 4 4 2 1 1ù é 1 êëQ AS + CS = a úû S2 S4 Where e1 is eccentricity of conjugate hyperbola = 2 ´ 2e ´ 3e1 = 12ee1 Now b 2 = a 2 (e 2 - 1) Þ e 2 = 13/ 4 1 1 13 and 2 + 2 = 1 Þ e12 = 9 e e1 \ Required area = 12 ´ 28. (9) P A 2 O D C 13 13 ´ = 26 2 3 C y B S S1 L A Ö12 30° N(4, 0) B M(1, 0) D x=1 30° Q(4 + 4Ö3, 0) x MATHEMATICS 118 Common chord of both the circles is x = 1. Now, we have to find the ratio of areas of equilateral triangles ANB and CQD. Now in triangle OPN, ON = OP cosec 30° = 2 × 2 = 4. Area of triangle NAB 1 MN . AB = MN . AM = MN. MN tan 30° 2 = (ON – OM)2 tan 30° 1 9 sq. units. = (4 - 1) 2 = 3 3 Now in triangle NLQ, NQ = NL cosec 30° = 4 3. 1 Since area of tr iangle CQD = QM .CD 2 = QM .CM For maximum length of the common chord, it must pass through the centre of the smaller circle ( r2 , r2 ) , so r 4r2 = r1 + r2 Þ 1 = 3 r2 Þ Þ 30. 1 57 + 24 3 = sq. units. 3 3 57 + 24 3 So, ratio of area of trianlges = . 9 (5) Let r be the radius of the circle. Its equation is x 2 + y 2 - 2r ( x + y ) + r 2 = 0. Since it passes through P(a , b) a 2 + b 2 - 2r ( a + b ) + r 2 = 0 r1 = a + b + 2ab ...(1) r2 = a + b - 2ab Now, the equations of two circles are x 2 + y 2 - 2r1 ( x + y ) + r12 = 0 and x 2 + y 2 - 2r2 ( x + y ) + r2 2 = 0 Þ 2 ( r2 - r1 ) ( x + y ) + r12 - r2 2 = 0 CHAPTER 6b ± 36b2 - 4b2 = (3 ± 2 2 ) b 2 a = 3± 2 2 b (5) The tangent at any point A(2sec q, tan q) x sec q y tan q = 1. is given by 2 1 It meets the line x - 2 y = 0 2 x sec q x tan q Þ =1 Þ x = sec q - tan q 2 2 2 1 æ ö Þ Q ºç , ...(1) ÷ è sec q - tan q sec q - tan q ø Also, the tangent meets the line x + 2 y = 0 at R, so x x sec q + tan q = 1 Þ 2 2 2 Þ x= sec q + tan q 2 -1 æ ö ...(2) Þ Rºç , ÷ è sec q + tan q sec q + tan q ø Now, 2 2 + 12 2 2 + 12 (sec q - tan q) 2 (sec q + tan q) 2 Limits & Derivatives 12 lim f ( x) = lim {- h}cot{-h} x ®0 - \ lim f ( x ) does not exist. h ®0 x ®0 = lim (1 - h) cot(1 - h) = cot 1 h ®0 lim f ( x ) = lim x ®0+ a 2 - 6 ab + b 2 = 0 = 22 + 12 Þ CQ.CR=5 Þ 2 ( x + y ) = r1 + r2 (d) a= CQ.CR = The common chord is S1 - S2 = 0 1. = 3 Þ 2 ( a + b ) = 4 2 ab Þ = (3 + 4 3) 2 Solving a + b - 2 ab Þ ( a + b ) 2 = 8ab Þ QM. QM tan 30° = (MN + NQ) 2 tan 30° 29. a + b + 2ab tan 2 {h} h ®0 h 2 - [ h ]2 = lim h®0 tan 2 h h2 2. =1 (a) x æ 4x +1 ö æ x 2 + 5x + 3 ö lim ç 2 ÷ = xlim çè1+ 2 ÷ ®¥ x + x + 2ø x ®¥ çè x + x + 2 ÷ø x Solutions 119 ( 4 x +1) x é x2 + x + 2 ù x2 + x + 2 êæ 4 x + 1 ö 4 x+1 ú ú = lim êç1 + 2 ÷ x ®¥ ê è x + x + 2ø ú êë úû 4+ lim 2 4x + x lim 3. 1 x 6. x ®¥ 1+ 1 + 2 x x2 2 4 1 e x - sin x [using L¢ Hospital's rule] = 2 lim x® 0 ( x - 2) x n - 3 For this limit to be finite, n – 3 = 0 Þ n=3 (c) lim [(sin x)1/x + (1/x)sin x] x® 0 1/ x =e = e x®¥ x + x + 2 = e (b) According to the question æ æ x2 ö x3 ö lim ç 3 – ÷ £ lim f ( x) £ lim ç 3 + ÷ 12 ÷ø x®0 9 ÷ø x ®0 ç x ®0 ç è è Þ (3 – 0) £ lim f ( x) £ (3 + 0) = lim (sin x ) x®0 Hence lim f ( x) = 3 (from Sandwitch Theorem) =e x®0 Let y 1 k k k 1/ n = lim k ((n + 1) (n + 2) ...(n + n) ) n®¥ n 1 Þ ln y = lim n®¥ n k k æ æ n + 1 ök æn+2ö æn+nö ö ç ln ç ÷ + ln ç + ... + ln ç ÷ ÷ ÷ ç è n ø è n ø è n ø ÷ø è k æ æ n + 1ö æ n + 2ö æ n + nö ö = lim ç ln ç ÷ + ln çè ÷ + ... + ln çè ÷ n ø n ø ÷ø n ®¥ n è è n ø (c) n å = lim k . n®¥ r =1 n r =1 7. xn number = lim n x (1 + cos x ) x®0 æ sin x ö æ e x - cos x ö æ 1 ö = lim ç 2 ÷ . ç ÷ . çè ÷ n 2 x®0 è x ø è x ø 1 + cos x ø 2 2 e = lim 1 . x ®0 x - cos x 1 . 2 xn - 2 ì æ aö ü = lim í1 + ç1 - ÷ ý è xø þ x®a î tan px 2a p æ h ö tan ( a + h) ç ÷ = lim h® 0 è a + h ø 2a e = lim h®0 e h æ p ph ö tan ç + ÷ a + h è 2 2a ø = lim h®0 h æ æ ph ö ö - cot ç ÷ ÷ ç è 2a ø ø a+hè e = lim h® 0 e = finite non-zero (1 - cos x ) (1 + cos x )(e x - cos x ) px 2a aö px px æ x - aö = lim ç tan ÷ x 2a ex®a è x ø 2a e Let x – a = h we get the limit x x®0 tan æ ln (n + r ) - ln n ö ÷ø n (cos x - 1) (cos x - e ) (c) Limit is of the form 1¥ , so = lim ç1 - ÷ tan x®a è ø k æ æ 1ö ö æ 4ö æ 4ö = k . ç 2 ç ln 2 - ÷ ÷ = ln ç ÷ . Þ y = ç ÷ . è eø è è è eø 2ø ø (c) Given that, lim =e -1/ x lim x ® 0 - cosec x cot x [Using L' Hospital rule] aö æ lim ç 2 - ÷ xø x®a è k 5. [Q | sin x | < 1 when x ® 0] - log x lim x ® 0 cosec x sin x lim . tan x x x ® 0 = e0 = 1 =e æ ln (n + r ) - ln n ö çè ÷ø n æ çè å n®¥ = k . lim sin x æ 1ö lim sin x log ç ÷ è xø = 0 + e x®0 x ®0 4. æ 1ö + lim ç ÷ x®0 è x ø -h p 1 . . æ ph ö 2 a p / 2 a (a + h) tan ç ÷ è 2a ø 2a 2 -2a ph / 2a . = e pa = e p = lim eh®0 p(a + h) tan(ph / 2a) 8. (a) Q ax 2 + bx + c = 0 has roots a and b a b then 2 + + c = 0 x x 1 1 i.e., cx 2 + bx + a = 0 has roots and . a b MATHEMATICS 120 1ö æ 1ö æ Þ cx 2 + bx + a = c ç x - ÷ ç x - ÷ è aø è bø cos2 (sin2(sin2).........(sin2(x)).....) ( x + 4 + 2) . p x®0 ìï æ (x + 4) - 2ö üï sin ípç ý ÷ x ø þï îï è æ (x + 4) - 2ö çp ÷ x è ø = lim æ 1 - cos(cx 2 + bx + a ) ö ç ÷ 1 è 2(1 - ax ) 2 ø x® Now lim a 1 ì 2 æ cx 2 + bx + a ö ü 2 ï sin ç ÷ï 2 è øï ï = lim í ý 2 x ®1/ a ï (1 - ax ) ï ï ï î þ = 10. æ cx + bx + a ö sin ç ÷ 2 è ø = lim (1 - ax) x®1/ a 2 = lim x®1/ a c æ 1 ö æ 1ö çx- ÷ x2 è aø èç bø÷ t t cos2 x t 1/ t é ù 1 = t ³ 1ú ê Putting 2 cos x ë û 1/ t c æ 1ö x2 èç bø÷ . lim -a x®1/ a t 1/ t = lim ( n ) t ®¥ 1/ t 11. (b) ì log e n lim í n®¥ î log e (n - 1) ´ (b) Let P cos2 (1 - cos 2 (1 - cos 2 (......cos 2 ( x))...... x ®0 ïì æ ( x + 4) - 2 ö ïü sin í p ç ÷ø ý è x ïî ïþ 2 2 2 2 cos sin (1 - cos (......cos ( x))...... = lim x®0 ïì æ ( x + 4) - 2 ö ïü sin í p ç ÷ý x ø þï ïî è 2 2 2 cos (sin (sin (......(sin 2 ( x))......) = lim x®0 ïì æ x + 4 - 2 ö ïü sin íp ç ÷ýì x ø ïþ ï æ x + 4 - 2 ö üï ïî è íp ÷ý x ìï æ x + 4 - 2 ö üï îï çè ø ïþ íp ç ý ÷ x ø þï ïî è t éæ 1 öt æ 2 öt ænö ù êç ÷ + ç ÷ + ...... + ç ÷ ú è n ø úû êëè n ø è n ø t éæ 1 öt æ 2 öt ænö ù = n lim êç ÷ + ç ÷ + ...... + ç ÷ ú t ®¥ êè n ø ènø è n ø úû ë = n (0 + 0 + ...... + 1)0 = n c æ 1 1ö 2 çè a bø÷ c æ 1 1ö = 1. = 2a èç a bø÷ -a 9. 2 2 2 (b) lim éê11/ cos x + 12 / cos x + ..... + n1/ cos x ùú û x ®p / 2 ë (1 + 2 + .......n ) = tlim ®¥ ìc æ 1ö æ 1ö ü sin í ç x - ÷ ç x - ÷ ý è ø 2 a è bø î þï = lim 1ö x®1/ a æ -a ç x - ÷ è aø æ c æ 1 ö æ 1ö ö sin ç ç x - ÷ ç x - ÷ ÷ è 2 è aø è bø ø cos 2 0 (2 + 2) 4 . = 1 p p = lim 12. ´ log e ( n + 1) log e n log e (n + 2) log e n k üï ´ .... ´ ý log e ( n + 1) log e (n k - 1) ïþ ìï log e n k üï log e n lim í ý = k lim log ( n 1) log n®¥ î n ®¥ e ( n - 1) ï e þï 1 Let n = , then h æ 1ö log e ç ÷ è hø - log e ( h) = k lim = k lim æ 1ö h® 0 h® 0 log e (1 - h) - log e ( h ) log e ç 1 - ÷ è hø 1 1 = k lim =k× =k (1 - 0) h®0 ì log e (1 - h) ü í1 ý loge ( h) þ î (b) Q lim 1 {[12 x] + [22 x] + [32 x] + ... + [ n2 x]} n®¥ n3 ì n 2 ü æ n 2 ö 2 ï å [r x ] ï ç å r x - {r x}÷ í r =1 ý ç r =1 ÷ = lim ï ï = lim ç ÷ø n ®¥ î n 3 þ n®¥ è n3 Solutions 121 æ n (n + 1)(2 n + 1) ö x. n {r 2 x}÷ ç 6 -å 3 ÷ = lim ç n®¥ n3 r =1 n ÷ çè ø (1)(1)(2) x = x. -0= 6 3 2 æxö 2 x ç ÷ tan a 2ø è 2 x 2 = a lim ´ = lim ´ tan 2 x®0 g ( x ) x ®0 g ( x ) 2 æxö ç ÷ è2ø x2 x2 4b \ lim = x ®0 g ( x ) x ®0 4 g ( x ) a lim f ( x) = lim f ( x ) 13. (c) x ®3- Þ lim = a lim x ®3+ ( x +3) x (27) 27 - 9 = lim l 3x - 27 x ®3+ 2 æ x + 3 x -18 ö ç ÷ 9 - 1÷ 32 ç 3 ç ÷ l è ø= Þ lim 2 33 (3x -3 - 1) x ®3x ®3- Þ lim 1 - cos( x - 3) ( x - 3) 2 2 1 x + 3x -18 l 1 2 l = Þ ×9 = Þ l = . 9( x - 3) 2 27 2 3 x®3 3 æ p ö sec 2 ç ÷ öö è 2-bx ø æ æ p 14. (a) Here, lim ç sin 2 ç ÷÷ x ®0 è è 2 - ax ø ø æ p ö sec 2 ö ü çè 2 -bx ÷ø ì æ p = lim í1 - cos 2 ç ÷ý x ®0 î è 2 - ax ø þ =e ì ü 1 ïï 2 æ p ö ïï lim - ícos ç ý ÷× è 2 - ax ø cos 2 æ p ö ï x®0 ï ç ÷ è 2 -bx ø þï îï ì pa ü æ p ö æ p ö ï 2sin ç 2 - ax ÷ cos ç 2- ax ÷´ ï è ø è ø (2- ax )2 ï ï lim - í ý pb ï æ p ö æ p ö x ®0 ï 2sin ç ÷ cosç ÷´ 2 ï è 2 -bx ø è 2 -bx ø (2 -bx ) ïþ î =e [using L'Hospital's rule] æ 2p ö sin ç ÷ 2 è 2 - ax ø × a × (2 - bx ) - lim p 2 ö b (2 - ax )2 x ®0 æ sin ç ÷ 2 - bx ø è =e =e 15. (c) a (2 -bx )3 - lim × x ®0 b (2 - ax )3 lim =e f (1 - cos x) - a b. x ®0 g ( x ) sin 2 x xö æ æ 2 xö f ç 2 sin 2 ÷ ç 2sin ÷ 2ø 2ø è è = lim ´ x®0 x x xö æ ö æ öæ g ( x ) ç 2 sin 2 ÷ 4 ç sin 2 ÷ç cos 2 ÷ 2ø 2 øè 2ø è è Now, lim g (1 - cos 2 x) x ®0 x 4 = lim g (2sin 2 x) x®0 x4 g (2sin 2 x) (2sin 2 x) 2 a a ´ = ´4 = . x ® 0 (2sin 2 x ) 2 x4 4b b = lim 16. (a) Let f ( x ) = lx 2 + mx + n Þ f '( x) = 2lx + m Now , f (1) = f ( -1) Þ l + m + n = l - m + n Þ m=0 \ f '(x) = 2lx \ f '(a1) = 2la, f '(a2 ) = 2la2 , f '(a3 ) = 2la3 As a1, a2 , a3 are in A.P.. \ f '(a1 ), f '(a2 ), f '(a3 ) are in A. P.. 17. (b) Since, f(x) is a polynomial function satisfying æ1ö æ1ö f ( x) × f ç ÷ = f ( x ) + f ç ÷ , è xø è xø \ f ( x) = x n + 1 or f ( x ) = - x n + 1 If f ( x) = - x n + 1, then f (4) = -4n + 1 ¹ 65 So, f ( x ) = x n + 1 Since, f(4) = 65 \ 4 n + 1 = 65 Þ n = 3 \ f ( x) = x3 + 1 Þ f '( x) = 3 x 2 \ f '(l1 ) = 3l12 , f '(l2 ) = 3l22 , f '(l3 ) = 3l32 Since, l1, l2, l3 are in GP. \ f '(l1 ), f '(l2 ), f '(l3 ) are also in GP.. g ( x) f (a) - g (a) f ( x) x-a x®a g (a + h) f (a) - g (a) f (a + h) = lim h h®0 [For x = a + h] 18. (5) = lim h®0 lim g (a + h) f (a) - g (a) f (a) + g (a) f (a) - g (a) f (a + h) h MATHEMATICS 122 é g (a + h ) - g ( a ) ù é f (a + h) - f (a) ù lim f (a ) ê g (a ) ê ú - hlim ú = h ®0 h h ë û ®0 ë û 1- x ïì - ax + sin ( x - 1) + a ïü1- x 1 lim í = ý x ®1 ï 4 î x + sin ( x - 1) - 1 ïþ 22. (2) = f (a) g' (a) – g (a) f ' (a) = 2 × 2 – (– 1) × 1 = 5 lim (sin x) 19. (1) x® = e tan x p 2 2 lim = 20. (0.50) ì sin( x - 1) ü ïï - a + ï x -1 ï Þ lim í ý sin ( x - 1) ï x ®1 ï 1+ ïî x - 1 ïþ æ0 ö çè form÷ø 0 2sin x cos x -cos x - sin x 2 = e0 =1 Required limit, (x + x + x ) - x l = lim x ®¥ x+ x+ x + x Dividing numerator & denominator by get x ®¥ 1+ x+ x x2 1+ x ®¥ x , we x®0 +1 \y= 1 x = x (1 + a cos x ) - b sin x { f ( x )}3 1 1 = = 0.50 1+1 2 =1 ìï x 2 x 4 üï ìï üï x3 x5 x + ax í1 + - ......ý - b í x - + - ......ý 3! 5! ïî 2! 4! þï îï þï = 1 Þ lim 3 x ®0 { f ( x)} Þ lim æ a bö 1+ a - b æ a b ö + ç - + ÷ + x 2 ç - ÷ + ....... è 2! 3!ø è 4! 5!ø x2 x ®0 3 ì f ( x) ü í ý î x þ a b Þ 1 + a - b = 0 and - + = 1 2! 3! Þa=- x 5/ 3 x = y x2 /3 or y 2 + ( x5/ 3 ) y - x 5/ 3 = 0 1 1 1+ + 3/2 +1 x x 21. (6) lim 1 æ -a + 1 ö Þç ÷ = Þ a = 0 or 2 4 è 2 ø \ Largest value of a is 2. 23. (1) Let x x = y= 3 1 x x x + 2/3 x+ 3 3 x x x ....¥ + x + x ....¥ = x 1+ x l = lim = lim 1+ x ìï a (1 - x ) + sin ( x - 1) üï ý Þ lim í x ®1 ï î ( x - 1) + sin ( x - 1) ïþ 1+ x sin 2 x - sin x lim cos x x®p x® p e 2 lim ( tan x (sin x -1)) x® p =e 2 5 3 and b = - . Thus b – 3a = 6 2 2 =1 x 5/ 3 +y x+ - x5 / 3 ± x10 / 3 + 4 x5 / 3 2 - x 5 / 3 + x10 / 3 + 4 x5 / 3 2 4 x5 /3 = 2( ( x10/ 3 + 4 x 5/ 3 ) + x5 / 3 ) 2 = 4 ö æ ç1 + 5/ 3 ÷ + 1 è x ø 2 2 \ lim y = = = 1. x ®¥ 1+ 0 + 1 2 = (Q y > 0) a( x3 - 1) + ( x - 1) = 0 24. (1) or ( x - 1)(ax 2 + ax + a + 1) = 0 a, b ¹ 1 so, a, b are roots of ax 2 + ax + a + 1 = 0 a +1 a + b = -1, ab = a lim x® (1 + a) x3 - x2 - a 1 (e1-ax - 1)( x -1) a = lim x® 1 a ( x3 - x2 ) + a( x3 -1) (e1-ax -1)( x -1) Solutions = lim 123 [x2 + a(x2 + x +1)] (e1-ax -1) 1 x® a (1 + a)x2 + ax + a 1-ax 1 -1 ö x® æ e aç ç 1 - ax ÷÷ (1 - ax) è ø = lim 25. (36) Let 3x = t2 27 t 2 + 2 - 12 t t 4 - 12t 2 + 27 lim = lim 1 3 t ®3 t -3 t ®3 t t2 éæ1+ a ö 2 ù êç a ÷ x + (1)x +1ú 2 è ø û = lim a (abx -(a+b)x +1) = lim a ë 1 1 (1- ax) (1-ax) x® x® a = lim a 1 x® a (1 - (a ) x )(1 - (b) x) a (a - b) = . a (1 - ax ) CHAPTER Mathematical Reasoning 13 1. 2. 3. 4. 5. 6. 7. 8. (t 2 - 3)(t + 3)(t - 3) t -3 t ®3 = (32 – 3) (3 + 3) = 36. = lim a (a) Inclusive “or”. 17 is a real number or a positive integer or both. (c) p ® (~ p Ú q) has truth value F. It means p ® (~ p Ú q) is false. It means p is true and ~ p Ú q is false. Þ p is true and both ~ p and q are false. Þ p is true and q is false. (b) ~p : Ashok does not work hard Use '®' symbol for then (~p ® q) mean = If Ashok does not work hard then he gets good grade. (d) When p is false and q is true, then p Ù q is false, pÚ ~ q is false. (Q both p and ~q are false) and q Þ p is also false, only p Þ q is true. (d) ~ p Ù q = ~ (q ® p) (a) p Ù q p Ú q ~(p Ú q) (p Ù q) Ù ~ (p Ú q) p q T T T T F F T F F T F F F T F T F F F F F F T F \ (p Ù q) Ù (~ (p Ú q)) is a contradiction. (b) (p Ù ~ q) Ù (~ p Ù q) = (p Ù ~ q) Ù (~ q Ù q) = f Ùf = f (By using associative laws and commutative laws) \ (pÙ ~ q) Ù (~ p Ù q) is a contradiction. (b) p Þ q is logically equivalent to ~ q Þ ~ p \ (p Þ q) Û (~ q Þ ~ p) is a tautology but not a contradiction. 9. (a) p q T T p Ù q (p Ù q)Þ p T T T F F F F F F T F T T T \ ( p Ù q) Þ p is a tautology.. 10. (b) 11. pÞq ~q Þ ~p p Þ q Û ~q Þ ~p T T T F F T T T T T T T (b) The truth value of ~(~p) « p as follow ~(~p ) ~(~p) ®p p ®~(~p ) ~(~p )«p p ~p T F T T T T F T F T T T Since last column of above truth table contains only T. Hence ~ (~p) ® p is a tautology. 12. (c) p T T F F q p®q ~p ~pÚq (p®q)«~(pÚq) T T F T T F F F F T T T T T T F T T T T MATHEMATICS 124 13. 14. (d) p Þ (~ p Ú q) is false means p is true and ~ p Ú q is false. Þ p is true and both ~p and q are false. Þ p is true and q is false (c) Let p : 2 + 3 = 5, q : 8 < 10 Given proposition is : p Ù q . 18. (c) The inverse of the proposition (p Ù ~ q) ® r is ~ (p Ù ~ q) ® ~ r º ~ p Ú ~ (~q) ® ~ r º ~ p Ú q ® ~ r 19. (a) ~ ((~ p) Ù q) º ~ (~ p) Ú ~ q º p Ú (~ q) 20. (c) ~ ( p Þ q) º p Ù ~ q \ ~ (~ p Þ ~ q) º ~ p Ù ~ (~ q) º ~ p Ù q . Its negation is ~ ( p Ù q) = ~ p Ú ~ q 15. 16. 17. \ we have 2 + 3 ¹ 5 or 8 </ 10. (b) We know that p ® q is false only when p is true and q is false. So p ® (~ p Ú q) is false only when p is true and (~ p Ú q) is false. But (~ p Ú q) is false if q is false because ~ p is false. Hence p ® (~ p Ú q) is false when truth value of p and q are T and F respectively. (a) We know that the contrapositive of p ® q is ~ q ® ~ p. So contrapositive of p ® (~q ® ~r) is ~ (~q ® ~r) ® ~p º ~ q Ù [~ (~r)] ~p Q ~ (p ® q) º p Ù ~q º ~ q Ù r ® ~p (a) ~ [ p Ú (~ p Ú q)] º ~ p Ù ~ (~ p Ú q) 21. 22. 23. 24. 25. º ~ p Ù (~ (~ p)Ù ~ q) º ~ p Ù ( p Ù ~ q ) . Thus ~ (~ p Þ ~ q) º ~ p Ù q (c) ~ [ (p Ú q) Ù (q Ú ~ r)] º ~ ( p Ú q) Ú ~ (q Ú ~ r) º (~ p Ù ~ q) Ú (~ q Ù r) (c) Statement given in option (c) is correct. ~ [p Ú (~ q) ] = (~ p) Ù ~ (~ q) = (~ p) Ù q. (d) Since ~ (p Ú q) º ~ pÙ ~ q (By De-Morgans’ law) \ ~ (p Ú q) ¹ ~ p Ú ~ q \ (d) is the false statement. (d) We know that ~ (p ® q) º p Ù ~q \ ~((p Ù r) ® (r Ú q)) º (p Ù r) Ù [~(r Ú q)] º (p Ù r) Ù (~r Ù ~q) (d) Let P = A Í B, Q = B Í D, R = A Í C Contrapositive of (P Ù Q) ® R is (® ~ R ® ~ (P Ù Q)). ~R®~ PÚ~Q CHAPTER Statistics 14 1. (a) X = n n å r. nCr X = r =n1 å Cr n r =1 2. C1 + nC2 + .... + n Cn n = n å n -1 Cr -1 r =1 n å Cr n n.2n -1 = n . 2 -1 r =1 (a) Let the mean of the remaining 4 observations be x1 . Then, M = 3. \ 1. n C1 + 2. n C2 + 3. n C3 + .... + n. nCn a + 4 x1 nM - a Þ x1 = . (n - 4) + 4 4 n1 x1 + n 2 x 2 n1 + n 2 (a) x= Q x1 = 400, x 2 = 480, x = 430 5. n1 (400) + n 2 (480) Þ 30n1= 50n2 n1 + n 2 n1 5 = n2 3 (c) Sum of 6 numbers = 30 × 6 = 180 Sum of remaining 5 numbers = 29 × 5 = 145 \ Excluded number = 180 – 145 = 35. (c) Let the items be a1, a2, ........, an. a + a + ........ + a n then X = 1 2 . n Now, according to the given condition: (a + 1) + (a 2 + 2) + ........ + (a n + n) X= 1 n 1 + 2 + 3 + ........ + n n(n + 1) =X+ =X+ n 2n (using sum of n natural nos.) n +1 =X+ . 2 Þ 4. 430 = Solutions 6. 7. 125 (a) Median is given as N -F M = l+ 2 ´C f where l = lower limit of the median - class f = frequency of the median class N = total frequency F = cumulative frequency of the class just before the median class C = length of median class. Now, given, M = 25, N = 100, F = 45, C = 20 – 30 = 10, l = 20. \ By using formula, we have 50 – 45 25 = 20 + ´ 10 f 50 50 25 – 20 = Þ5= Þ f = 10. f f 101 + d (1 + 2 + 3 + ......+100) (b) Mean = 101 d × 100 × 101 =1+ = 1 + 50 d 101 × 2 Q Mean deviation from the mean = 255 1 [|1 - (1 + 50d ) | + | (1 + d ) - (1 + 50d ) | + Þ 101 | (1 + 2d )| -(1 + 50d ) | +....+ | (1 + 100d ) - (1 + 50 d ) |] = 255 Þ 2d [1 + 2 + 3 + ... + 50] = 101´ 255 Þ 2d ´ Þ 8. 50 ´ 51 = 101´ 255 2 101´ 255 25755 d= = = 10.1 50 ´ 51 2550 (c) Clearly mean A = 0. Standard deviation s = 2= 2 å ( x - A)2 2n 2 ( a - 0) + ( a - 0) + ...(0 - a) 2 + ... 2n a 2 .2n = | a | . Hence | a | = 2. 2n (c) Using, = 9. s= 22 n1 (s12 + d12 ) + n 2 (s22 + d 22 ) = . n1 + n 2 3 10. (a) We know that Q.D 11. 5 5 = ´ M .D. = ´12 = 10 6 6 3 3 \ S.D = ´ Q.D. = ´ 10 Þ S .D. = 15. 2 2 (d) Since S.D. £ Range = b – a Var ( x ) £ (b – a )2 or (b - a)2 ³ Var ( x) . 12. (d) Since 0 < y < x < 2y x x \ y> Þ x- y < 2 2 \ x – y < y < x < 2x + y. y+x = 10 Hence median = 2 Þ x + y = 20. ...(i) And range = (2x + y) – (x – y) = x + 2y. But range = 28 \ x + 2y = 28 ...(ii) From equations (i) and (ii), x = 12, y = 8 ( x - y ) + y + x + (2 x + y ) 4 x + y = \ Mean = 4 4 y 8 = x + = 12 + = 14. 4 4 13. (b) Let xi be n observations, i = 1, 2, ...n Let X be the mean and M.D be the mean \ deviation about X . If each observation is increased by 5 then new mean will be X + 5 and new M.D. about new n æ xi ö = Q Mean ç å ÷ mean will be M.D. nø è i =1 14. (a) The first n natural numbers are 1, 2, 3, ................n 1 + 2 + 3 + 4 + ... + n Their mean, x = n n(n + 1) n + 1 = . = 2n 2 [Q The sum of Ist n natural numbers is n(n + 1) ] 2 2 ( x x ) å i Now, Variance = s 2 = n 1é 2 2 ù = å ( xi - 2 xx i + x ) û në = å xi2 - 2 x å xi + x 2 .n n n n MATHEMATICS 126 = å xi2 - 2 x 2 + x 2 = å xi2 - æ å xi ö 2 ç ÷ n n è n ø [Since frequency of each variate is one] n(n + 1)(2n + 1) Q å xi2 = . 6 \ Variance = n(n + 1)(2n + 1) æ (n + 1) ö -ç è 2 ÷ø 6n 2 (n + 1)(2n + 1) (n + 1)2 = 6 4 æ 2n + 1 n + 1ö (n + 1)(n - 1) = (n + 1) ç ÷ = è 6 4 ø 12 n2 - 1 . 12 15. (b) Let the observations be x1, x2, ...., x20 and x be their mean. Given that, variance = 5 and n = 20. We know that, = 20 ( x i - x )2 ( ) n1 å i=1 Variance s 2 = i.e. 5 = or 20 1 20 ( x i - x )2 å 20 i =1 å ( x i - x ) = 100 2 ...(i) i =1 If each observation is multiplied by 2 and the new resulting observations are yi, then 1 yi = 2xi i.e., xi = yi . 2 Therefore, y = = 2. 1 20 å xi 20 i =1 1 20 1 20 yi = å å 2x i n i =1 20 i=1 1 y. 2 On substituting the values of xi and x in eq. (i), we get i.e., y = 2x or x = 2 20 1 ö æ1 å çè 2 yi - 2 y ÷ø = 100 i =1 20 i.e., å ( yi - y )2 = 400 . i =1 Thus, the variance of new observations 1 ´ 400 = 20 = 22 ´ 5 . 20 (d) If initially all marks were xi then = 16. å ( xi - x )2 si2 = i . N Now each is increased by 10 2 å ëé( xi +10)-( x +10)ûù i å ( xi - x )2 i = = si2 N N Hence, variance will not change even after the grace marks were given. 17. (a) C.V. (1st distribution) = 60, s1 = 21 C.V. (2nd distribution) = 70, s2 = 16 Let x1 and x2 be the means of 1st and 2nd si2 = distribution, respectively, Then s C.V. (1st distribution) = 1 ´ 100 x1 21 21 ´ 100 = 35 ´100 or x1 = \ 60 = 60 x1 s2 ´100 and C.V. (2nd distribution) = x2 16 16 ´ 100 or, x = ´ 100 = 22.85 i.e., 70 = 2 x2 70 18. (93.32) When each observation is multiplied by 2, then variance is also multiplied by 2. We are given, 2, 4, 5, 6, 8, 17. When each observation multiplied by 2, we get 4, 8, 10, 12, 16, 34. \ Variance of new series = 22 × Variance of given data = 4 × 23.33 = 93.32. 19. (0) We know that, s Coefficient of variation = ´ 100 x s1 ´ 100 \ CV of 1st distribution = 30 s1 Þ 50 = ´ 100 [CV of 1st distribution = 50 30 (given)] Þ s1 = 15 s2 ´ 100 Also, CV of 2nd distribution = 25 s2 ´100 Þ 60 = 25 60 ´ 25 Þ s2 = Þ s2 = 15 100 Thus, s1 – s2 = 15 – 15 = 0. Solutions 127 20. (78) å x = 170, å x 2 = 2830 increase in 1 i.e., 8.24 = [(3.4)2 + (2.4)2 + (1.6)2 + x2 + y2 5 – 2 × 4.4 (x + y) + 2 × (4.4)2 ] or 41.20 = 11.56 + 5.76 + 2.56 + x2 + y2 – 8.8 ×13 + 38.72. Therefore, x2 + y2 = 97 … (ii) But from eq. (i), we have x2 + y2 + 2xy = 169 … (iii) From eqs. (ii) and (iii), we have 2xy = 7 2 … (iv) On subtracting eq, (iv) from eq. (ii), we get x2 + y2 – 2xy = 97 – 72 i.e. (x – y)2 = 25 or x – y = ± 5 … (v) So, from eqs. (i) and (v), we get x = 9, y = 4 when x – y = 5 or x = 4, y = 9 when x – y = –5. Thus, the remaining observations are 4 and 9. Required difference = 5. 24. (18) Var (1, 2, ....., n) = 10 å x = 10 , then å x¢ = 170 + 10 = 180 Increase in å x 2 = 900 - 400 = 500 , then å x¢2 = 2830 + 500 = 3330 Variance = 1 æ1 ö x '2 - ç å x ' ÷ å n èn ø 2 2 = 1 æ1 ö ´ 3330 - ç ´ 180 ÷ = 222 - 144 = 78. 15 è 15 ø 21. (8.25) 2 11 ×12.23 æ 11.12 ö -1 ç -1 ÷ 2 6 s = -ç 2 ÷ = 8.25 10 è 10 ø 22. (5) We have, CV of 1st distribution (CV1) = 50. CV of 2nd distribution (CV2) = 60 s1 = 10 and s2 = 15. s We know that, CV = ´ 100 x s 10 \ CV1 = 1 ´ 100 Þ 50 = ´ 100 x1 x1 2 Þ 2 (n + 1)(2n + 1) æ n + 1 ö -ç ÷ = 10 6 è 2 ø Þ n2 – 1 = 120 Þ n = 11 Var (2, 4, 6, ....., 2m) = 16 Þ Var (1, 2, ....., m) = 4 Þ m2 – 1 = 48 Þ m = 7 Þ m + n =18 25. (52) Mean Þ 10 ´ 100 = 20 . 50 s Also, CV2 = 2 ´ 100 x2 Þ x1 = Þ 60 = 15 ´100 x2 15 ´100 Þ x 2 = 25. 60 Thus, x 2 - x1 = 25 - 20 = 5. 23. (5) Let the other two observations be x and y. Therefore, the series is 1, 2, 6, x, y. Also, variance (s2 ) = 8.24 = 1 n 5 å ( xi - x )2 i =1 Þ Variance = s2 = s2 = … (i) 3 + 7 + 9 + 12 + 13 + 20 + x + y = 10 8 x + y = 16 =x = Þ x2 = 1+ 2 + 6 + x + y Now, mean ( x ) = 4.4 = 5 or 22 = 9 + x + y. Therefore, x + y = 13 12 + 2 2 + ..... + n2 æ 1 + 2 + ..... + n ö -ç ÷ = 10 n n è ø ...(i) S ( xi )2 - ( x )2 = 25 8 9 + 49 + 81+144 +169 + 400 + x 2 + y 2 - 100 = 25 8 Þ x2 + y2 = 148 ...(ii) From eqn. (i), (x + y)2 = (16)2 Þ x2 + y2 + 2xy = 256 Using eqn. (ii), 148 + 2xy = 256 Þ xy = 52 MATHEMATICS 128 CHAPTER Probability-1 15 1. 2. (d) Total number of outcomes S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)} n(S) = 16 Number of favourable outcomes E = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (4, 4)} n(E) = 7 n(E) 7 = Required probability = n(S) 16 (b) Given equation 100 x+ > 50 Þ x 2 - 50x + 100 > 0 x So, for each ai Î A , there are four possibilities. \ Total no. of cases = 4 × 4 × ............ × 4 (n times) = 4n Further out of above four possibilities, first three satisfy ai Î P È Q So, the number of cases, when exactly r element of A belong to P È Q = n Cr (3)r \ Required probability = 5. Þ ( x - 25) 2 > 525 Þ x - 25 < - (525) or x - 25 > (525) Þ x < 25 - (525) or x > 25 + (525) As x is positive integer and (525) = 22.91 , we must have x £ 2 or x ³ 48 Let E be the event for favourable cases and S be the sample space. \ E = {1, 2, 48, 49, ......100} 3. \ n(E ) = 55 and n(S) = 100 Hence the required probability n( E) 55 11 P(E) = = = . n( S ) 100 20 (b) Total number of combinations of numbers on the cube and the tetrahedron = 6 × 4 = 24 Favourable number of ways of getting a sum not less than 5 = sum of coefficients of x6, x7, .... x10 in the product 2 3 4 5 6 = (x + x + x + x + x + x ) ( x + x 2 + x3 + x 4 ) = ( x 2 + 2 x 3 + 3 x 4 + 4 x5 + 4 x 6 4. + 4 x7 + 3x8 + 2 x9 + x10 ) = 4 + 4 + 4 + 3 + 2 + 1 = 18 18 3 \ Required probability = = 6´ 4 4 (b) Let A = {a1, a2, a3, ........... an}. For any ai Î A , 1 £ i £ n , we have following choices (1) ai Î P and ai Î Q (2) ai Î P and ai Ï Q (3) ai Ï P and ai Î Q (4) ai Ï P and ai Ï Q n Cr (3)r 4n (a) Since the chairs are numbered, so for counting of total number of cases it is equivalent to linear permutation. Hence, total number of cases = 10! If two particular persons A and B sit together then the total number of linear arrangements =2! 9!. Consider one of such arrangements in which the arrangement started at chair 1 (C1) and ends at chair 10 (C10). C1 – C2 –C3 – ......... –C9 –C10 If two persons sit at C1 and C10 then it will lead to 2! 8! new arrangements. So the favourable number of cases = 2! 9! + 2! 8! = 2!8! (10) 2!8!(10) 2 = 10! 9 (b) Total ways of distribution = 45 Þ n(S) = 45 Total ways of distribution so that each child get \ Probability = 6. atleast one game = 45 - 4 C1 35 + 4C2 25 - 4 C3 7. = 1024 - 4 ´ 243 + 6 ´ 32 - 4 = 240 n(E) = 240 n( E ) 240 15 = = . required probablity = 64 n(S ) 45 (d) A chess board is a square divided into 64 equal squares. In 1st diagonal we have only 1 square In 2nd diagonal we have only 2 squares In 3rd diagonal we have 3 squares so selection can be done in 3C3 ways In 4 diagonal we have 4 squares so selection can be done in C34 ways And so on Hence the total number of ways in which 3 squares can be chosen 2(3C3 + 4C3 + 5C3 + 6C3 + 6C3 + 7C3) + 8C3 [Note that we do not have 2 × 8C3] Hence the total number of favourable ways m Solutions 129 = 4(3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 2 × 8C3 = 392. 64 × 63 × 62 64 C3 And total number of ways = 1× 2 × 3 8. 9. = 32 × 21 × 64 Hence the required probability m 392 7 = = = n 32 × 21 × 62 744 (c) The total numbers of arrangements is 11! 11! = 2!2!2! 8 The number of arrangements in which C, E, H, 6! 11! I, S appear in that order 11 C5 = 2!2!2! 8 × 5! 11! 1 \ Required Probability = 8 × 5! = 11! 120 8 (c) Player A can win if A throws (1, 6) or (6, 1) and B throws ((1, 1), (2, 2), (3, 3), (4, 4), (5, 5) or (6, 6)). Thus the number of ways is 12. Similarly the number of ways in which B can win is 12. Total number of ways in which either A wins or B wins = 24. Thus the number of ways in which none of the two wins = 64 – 24. \ The required probability = 6 4 - 24 6 4 = 53 . 54 10. (c) Value 50p. 25p. 10p. No. 2 5 15 5 coins out of 22 can be chosen in 22C2 ways. Let x be the no. of 50 P coins selected. y be the no. of 25p. coins selected and z be the no. of 10 p. coins selected. We desire that 50x + 25y + 10z < 150 or 10x + 5y + 2z < 30 where 0 £ x £ 2; or 0 £ y £ 5, 0 £ z £ 14 and x + y + z = 5. Thus following cases are impossible. Case (i) : x = 2, y = 2, z = 1 No. of ways of such selections = 2C2 ´ 5C2 ´ 15C1 = 1´10 ´15 = 150. Case (ii) : x = 1, y = 4, z = 0 No. of ways of such selections = 2C1 ´ 5C4 = 10. Case (iii) : x = 2, y = 3, z = 0. No. of ways of such selection = 2C2 ´ 5C3 ´ 15C0 = 1´10 = 10. \ Total = 150 + 10 + 10 = 170 ways are to be rejected. 170 \ Required probability = 1 - 22 . C5 11. log b log a Let b = 2m and a = 2n where m and n denote the exponents on the base 2 in the given set then (b) log a b = m n Therefore, logab is an integer only if n divides m. Now, total no. of ways m and n can be chosen = 25 × 24 = 600. For favourable cases Let n = 1, So m can take values 1, 2, 3, 4, 5, 6, ..., 24 = 24 if n = 2 m = 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 = 11 n = 3m = 6, 9, 12, 15, 18, 21, 24 =7 n = 4m = 8, 12, 16, 20, 24 =5 n = 5m = 10, 15, 20, 25 =4 n = 6m = 12, 18, 24 =3 n = 7m = 14, 21 =2 n = 8m = 16, 24 =2 n = 9, 10, 11, 12 m = 1 for each =4 62 log a b = Hence, required probability = 62 = 31 . 600 300 12. (d) n(S) = Number of total ways 14! = 14 P12 = = 7 ´ 13! 2! The girls can be seated together in the back seats leaving a corner seat in 4 × 3! = 24 ways and the boys can be seated in the remaining 11 seats in 11! 1 11 = ´ 11! ways P9 = 2! 2 \ n(E) = Number of favourable ways 1 = 24 ´ ´11! = 12! 2 n( E ) 12! 1 = = The required probability = n(S ) 7 ´13! 91 13. (a) We know that P (exactly one of A or B occurs) = P (A) + P (B) – 2P (A Ç B). Therefore, P (A) + P (B) – 2P (A Ç B) = p … (i) Similarly, P (B) + P (C) – 2P (B Ç C) = p … (ii) and P (C) + P (A) – 2P (C Ç A) = p …(iii) MATHEMATICS 130 Adding (i), (ii) and (iii) we get 2 [P (A) + P (B) + P (C) – P (A Ç B) – P (B Ç C) – P (C Ç A)] = 3p Þ P (A) + P (B) + P (C) – P (A Ç B) – P (B Ç C) – P (C Ç A) = 3p/2 … (iv) We are also given that, P (A Ç B Ç C) = p2 … (v) Now, P (at least one of A, B and C) = P (A) + P (B) + P (C) – P (A Ç B) – P (B Ç C) – P (C Ç A) + P (A Ç B Ç C) 3p 3 p + 2 p2 + p2 = [using (iv) and (v)] 2 2 14. (d) Since, one and only one of the three events E1, E2 and E3 can happen, therefore P (E1) + P (E2) + P (E3) = 1 .....(i) Q Odds against E1 are 7 : 4 4 4 Þ P(E1 ) = = .....(ii) 4 + 7 11 Q Odds against E2 are 5 : 3 3 3 Þ P (E 2 ) = = ......(iii) 3+5 8 From (i), (ii) and (iii), we have, 4 3 + + P(E 3 ) = 1. 11 8 4 3 88 - 32 - 33 Þ P(E3 ) = 1 - - = 11 8 88 23 23 = = 88 23 + 65 Hence odds against E3 are 65 : 23. 15. (a) Let p1, p2 be the chances of happening of the first and second event, respectively; then, according to the given conditions, we have = p1 = p22 and Þ 1 - p22 p22 1 - p1 æ 1 - p2 ö =ç ÷ p1 è p2 ø P ( A) 52 3 C4 C2 ´ 12 C2 52 C4 = = 3 ´ 12 ´ 36 1296 = 270725 270725 3 ´ 66 198 = 270725 270725 Now, E = A È B and A Ç B = f. This implies 1494 . 270725 (0.80) Total no. of arrangements of the letters 10! of the word UNIVERSITY is . 2! No. of arrangements when both I's are together = 9! So. the no. of ways in which 2 I’s do not together 10! - 9! = 2! 10! - 9! 10!- 9! 2! 2! = \ Required probability = 10! 10! 2! 10 ´ 9!- 9!.2! 9![10 - 2] 8 4 = = = = = 0.80 10! 10 ´ 9! 10 5 (0.50) Total number of numbers = 4! = 24 For odd nos. 1 or 3 has to be at unit's place If 1 is at unit place, then total number of numbers = 3! = 6 And if 3 is at unit place, then total number of numbers = 3! = 6 \ Total number of odd number = 6 + 6 = 12 12 1 = = 0.50 \ Required probability = 24 2 P ( E ) = P ( A È B) = P ( A) + P ( B ) = 18. 3 3 1 1 and so p1 = . 3 9 (c) Let P(A) = probability that Ashmit will pass, and P(B) = probability that Bishmit will pass, then from the given information P(A) = 0.5, 0.1 £ P(A Ç B) £ 0.3. We have to find the value of P(B). Since P(A È B) = P(A) + P(B) – P(A Ç B) Or P(B) = P(A È B) + P(A Ç B) – P(A) 1( 3C1 )( 12 C1 )( 36C1 ) P( B) = æ 1 - p2 ö 2 =ç ÷ Þ p2 (1 + p2 ) = (1 - p2 ) p è 2 ø Þ 3 p2 = 1 Þ p2 = 16. For maximum value of P(B) both P(A È B) and P(A Ç B) should be maximum, maximum value of P(A È B) is 1 and maximum value of P(A Ç B) = is 0.3. Hence P(B) = 1 + 0.3 – 0.5 = 0.8. Similarly For minimum value of P(B) both P(A È B) and P(A ÇB) should be minimum, minimum value of P(AÈB) = P(A) = 0.5 and minimum value of P(A Ç B) = is 0.1. Hence P(B) = 0.5 + 0.1 – 0.5 = 0.1. So required range is (0.1, 0.8). 17. (a) Total number of outcomes = 52C4 = 270725 E = Event of exactly two spade cards and exists two aces. A = Event of 1 spade ace, 1 non-spade ace and 1 spade card and 1 non-spade card B = Event of 2 non-spade aces and 2 spade cards. 19. Solutions 20. (0.33) = 131 Total number of cases = 3 n C 2 3n(3n - 1) 2 Now x3 + y3 = ( x + y ) ( x 2 - xy + y 2 ) x3 + y3 is divisible by 3 if 3 divdes x + y or divides x2 – xy + y2 we arrange the 3n numbers in 3 sequences. A : {1, 4, 7, ............., 3n – 2} B : {2, 5, 8, .............., 3n – 1} C : {3, 6, 9, ............., 3n} Clearly we must choose either one number from the first sequence and other number from the second sequence or both numbers from the third sequence only. \ Number of favourable cases n (n - 1) n = (3n - 1) = n × n + n C2 = n 2 + 2 2 n (3n - 1) 1 2 = = 0.33 \ Required probability = 3n(3n - 1) 3 2 21. (3) Total no. of arrangements = 15! Extreme chairs are occupied by girls, thus there are four gaps among 5 girls where boys can be seated. Let the number of boys in these four gaps be 2x + 1, 2y + 1, 2z + 1 and 2t + 1, then 2x + 1 + 2y + 1 + 2z + 1 + 2t + 1 = 10 Þ x+y+z+t=3 Where x, y, z, t are integers and 0 £ x £ 3, 0 £ y £ 3, 0 £ z £ 3, 0 £ t £ 3 \ The number of ways of selecting positions for boys = coefficient of x3 in (1 + x + x2 + x3)4 4 æ 1 - x4 ö = coefficient of x3 in ç ÷ è 1- x ø = coefficent of x3 in (1 – x4)4 (1 – x)–4= 6C3 = 20 \ Number of arrangements of boys and girls with given condition = 20 × 10 ! × 5 ! \ Required probability = 20 ´ 10!´ 5! = 20 15! 3003 n 3003 Þ = =3 1001 1001 22. (3) Let a and b be the roots of the quadratic equation. According to question, a + b = a 2 + b2 and ab = a 2b2 Þ ab(ab - 1) = 0 Þ ab = 1 or ab = 0 Þ a = 1, b = 1; a = w, b = w2 [Cube roots and unity] a = 1, b = 0, a = 0, b = 0 \ n( S ) = Number of quadratic equations which are unchanged by squaring their roots = 4 and n(E) = Number of quadratic equations have equal roots = 2. n( E ) 2 1 p = = = . \ Required probability = n(S ) 4 2 q Þ p + q = 3. 23. (4) The number of ways to answer a question = 25 – 1 = 31. i.e. In 31 ways only one correct. Let number of choices = n n 1 > Now, according to the question 31 8 31 Þn> Þ n > 3.8 \ Least value of n = 4. 8 24. (5) Let S be the sample space and E be the event of getting a large number than the previous number. \ n( S ) = 6 ´ 6 ´ 6 = 216 Now, we count the number of favourable ways. Obviously, the second number has to be greater than 1. If the second number is i(i > 1), then the number of favourable ways = (i – 1) × (6 – i) \ n(E) = Total number of favourable ways 6 = å (i - 1) ´ (6 - i ) i =1 =0+1× 4+2×3+3×2+4×1+5×0 = 4 + 6 + 6 + 4 = 20. Therefore, the required probability, n( E ) 20 5 P( E ) = = = =p (given) n(S ) 216 54 \ 54 p = 5 25. (5) Let A be the event of a boy and B the event of having cat eyes. So 20 1 20 1 P ( A) = = and P ( B ) = = 40 2 40 2 10 1 = Now, P ( A Ç B) = 40 4 \ P ( A È B ) = P( A) + P ( B ) - P ( A Ç B ) 1 1 1 3 a = + - = = 2 2 4 4 b = a 2 + b 2 = 32 + 4 2 = 5 MATHEMATICS 132 CHAPTER Relations & Functions-2 16 1. (d) P = {( a, b) : sec 2 a - tan 2 b = 1} Þ x = w1 y and y = w2z for some w1, w2 ÎQ For reflexive : Þ x = w1 ( w2 z ) = ( w1 × w2 ) z sec 2 a - tan 2 a = 1 (true " a) For symmetric : sec2 b – tan2 a = 1 Þ xRz as w1 × w2 ÎQ Now let L.H.S. 1 + tan 2 b - (sec 2 a - 1) 2. = 1 + tan 2 b - sec 2 a + 1 = – (sec2 a – tan2b) + 2 = – 1+2 = 1 So, Relation is symmetric. For transitive : if sec2 a – tan2 b = 1 and sec2 b – tan2 c = 1 sec2 a – tan2 c = (1 + tan2 b) – (sec2 b – 1) = –sec2b + tan2b + 2 = – 1 + 2 = 1 So, relation is transitive. Hence, relation P is an equivalence relation. (b) Reflexivity : xRx as x = 1 . x and 1 is a rational number. 3. ( 5, 1) Î R because 5 - 1 + 5 = 2 5 - 1 which is an irrational number. But (1, 5) Ï R \ R is not symmetric. We have, ( 5, 1), (1, 2 5) Î R because 5 -1+ 5 = 2 5 -1 If 1 - 2 5 + 5 = 1 - 5 are irrational numbers. Also, ( 5, 2 5) Î R and 5-2 5+ 5 = 0 which is not an irrational numbers. \ ( 5, 2 5) Ï R So, R is not transitive. 1 x w Þ yRx provided w ¹ 0, but if x = 0 and y is any non-zero integer, then x = 0y and there exist no rational number w for which y = wx. Þ y R x thus R is not symmetric. Now, let m p S n q Þ mq = pn Þ pn = mq Þ (a) x Î R Þ x - x + 5 = 5 is an irrational number. \ ( x, x) Î R So, R is reflexive. m m S iff m . n = m . n, which is true. n n Thus, R and S are reflexive relations. Symmetricity : Let xRy Þ x = wy for some rational number w Þy= m p p r S and S ; where n, q, t ¹ 0 . n q q t p m S q n Þ S is a symmetric relation. Transitivity : Let xRy and yRz. 4. (a) f ( x) is onto \ S = range of f (x). Now f (x) = sin x - 3 cos x + 1 pö æ = 2sin ç x - ÷ + 1 è 3ø pö æ Q -1 £ sin ç x - ÷ £ 1 è 3ø pö æ -1 £ 2sin ç x - ÷ + 1 £ 3 è 3ø \ f ( x) Î[ -1, 3] = S . Solutions 5. 6. 7. 133 (b) (c) Let f (x) ¹ 2 be true and f (y) = 2, f (z) ¹ 1 are false Þ f (x) ¹ 2, f (y) ¹ 2, f (z) = 1 Þ f (x) = 3, f (y) = 3, f (z) = 1 but then function is many one, similarly two other cases. (c) Total number of functions from A ® B = 34 = 81 and number of onto mappings = coefficient of x4 in 4! (ex – 1)3. = coefficient of x4 in 4! (e3x – 3e2x + 3e2x + 3ex – 1) æ 34 3.24 3.14 ö = 4! çç 4! – 4! + 4! – 0 ÷÷ è ø = 81 – 48 + 3 = 81 – 45 = 36 \ Number of functions from A ® B, which are not onto is 81 – 36 = 45. 8. 2 (b) Let f ( x) = ax + bx + c as it touches X-axis at x = 3 -b =3 2a Þ Þ Also, 9. (b ) ...(1) ...(2) ...(3) 3 18 27 ,b=- ,c= 25 25 25 Þ f ( x) = \ Range of f = (0, ¥) Ì R. 11. (b) f (x) = sin x + cos x, g (x) = x2 – 1 Þ g (f (x)) = (sin x + cos x)2 – 1 = sin 2x p p £ 2x £ 2 2 [Q sin q is invertible when – p/2 £ q £ p/2] Clearly g (f (x)) is invertible in – p p £x£ . 4 4 12. (b) Here Þ – g 2 ( x ) = gog ( x) = g{g ( x)} = g (3 + 4 x) = 15 + 42 x = (42 - 1) + 42 x 7 4 8 \ x - x + x - x -1 < 0 For x Î (1, ¥ ); x < x 4 , x 7 < x8 \ x - x 4 + x 7 - x8 - 1 < 0 7 g n ( x) = (4n - 1) + 4n x = y (say) \ g - n ( x) = ( x + 1)4- n - 1 For x Î (0, 1); x - 1 < 0, x - x < 0 4 = 63 + 43 x = (43 - 1) + 43 x Generalizing, we get Þ g - n ( y ) = ( y + 1)4- n - 1 3 2 ( x - 6 x + 9) 25 7 Also, f is not onto as f ( x) ³ 0, "x Î R. then x = ( y + 1 - 4n )4 - n f ( x ) = sgn( x - x 4 + x 7 - x8 - 1) 4 f (-1) = f (-2) = 0 but -1 ¹ -2 g3 ( x) = gogog(x) = g(15 + 42 x) = 3 + 4(15 + 42 x) b = -6 a 9a + 3b + c = 0 4a – 2b + c = 3 Þa= 10. (d) We have, f ( x) = x + x2 = x+ | x | Clearly, f is not onto as 8 Also for x = 1; x - x + x - x - 1 = -1 Thus x - x 4 + x7 - x8 - 1 < 0 for all x Î R + Hence sgn( x - x 4 + x 7 - x8 - 1) = -1"x Î R + . Therefore f (x) is many-one and into. 13. (d) f ( x) = 3 | x | - x - 2 and g (x) = sin x 3 | sin x | - sin x - 2 which is defined if 3 |sinx | – sin x – 2 ³ 0 If sin x > 0 then 2 sin x – 2 ³ 0 Þ sin x ³ 1 for fog (x) = p 2 If sin x < 0 then – 4 sin x – 2 ³ 0 Þ sin x = 1 Þ x = 2np + Þ – 1 £ sin x £ – 1 2 7p 11p ù Þ x Î é2np + , 2np + êë 6 6 úû MATHEMATICS 134 7p 11p ù ì pü é x Î ê 2np + , 2np + È í 2m p + ý , n , m Î I . 6 6 ûú î 2þ ë æ 3 ö 1 f ( x) = sin 2 x + ç cos x + sin x ÷ ç 2 ÷ 2 è ø 14. (b) 2 æ1 ö 3 + cos x ç cos x sin x ÷ ç2 ÷ 2 è ø 3 1 3 = sin 2 x + cos 2 x + sin 2 x + cos x sin x 4 4 2 1 3 + cos 2 x cos x sin x 2 2 = 5 5 (sin 2 x + cos 2 x ) = 4 4 æ5ö Now, y = g (f (x)) = g ç ÷ = 1. è4ø Clearly y = 1 is a straight line. 2 15. (a) Given that f (x) = x and g(x) = sin x, " x Î R. Then (gof) (x) = sin x2 Þ (gogof) (x) = sin (sin x2) Þ (fogogof) (x) = sin 2 (sin x2) As given that (fogogof) (x) = (gogof) (x) Þ sin2 (sin x2) = sin (sin x2) Þ sin (sin x2) = 0, 1 p Þ sin x = np or ( (4n + 1) where n Î Z 2 2 2 2 Þ sin x = 0 Q sin x Î[ -1,1] 2 Þ x = np Þ x = ± np where n Î W 16. (b) Let h(x) = g( f (x)) where, g(x) is injective and h(x) is surjective. \ Codomain of h(x) = Range of h(x). Range of h(x) = [0, 2] Þ codomain of g is [0, 2]. So, g(x) must be surjective. Now, domain of g = [–1, 1] which must be range of f. But codomain of f = [–1, 1] Þ f must be surjective. 17. (d) We have, –1< f (0) < f (1) <1 g = [–1,1] ®[0,1] and gof ( x) = x, " xÎ[0,1] Only condition that g(x) should satisfy for gof ( x) = x, " xÎ[0,1] is that g(x) should attain all values in [0, 1] when rangte of f (x) a subset of (–1, 1) is used as image for g(x). Thus, there can be infinite such functions g(x) with domain [–1, 1] and range [0, 1] 18. (14) If set A has m elements and set B has n elements then number of onto functions from A to B is n å (-1)n-r nCr r m where 1 £ n £ m r =1 Here E = {1, 2,3, 4}, F = {1, 2} m = 4, n = 2 \ No. of onto functions from E to F 2 = å (-1)2-r 2Cr (r )4 r =1 = (-1) 2C1 + 2C2 (2) 4 = – 2 +16 = 14. 19. (0) f ( x) = (a - xn )1/n , x > 0 Þ f ( f (x)) = [a -{(a - xn )1/ n}n ]1/ n = x Also g (x) = x 2 + px + q \ g (x) – x = 0 is quadratic equation x 2 + ( p - 1) x + q = 0 Given that this equation has imaginary roots \ x 2 + ( p - 1) x + q > 0 for all real x [\ coefficient of x2 = 1 > 0] \ g ( x) - x > 0 for all real x Þ g (g(x)) – g (x) > 0 " x Î R Now g (g (x)) – f (f (x)) = g (g (x)) – x = [g (g(x))– g (x)] + [g(x) – x] > 0 " x Î R \ The equation g (g (x)) – f (f (x)) = 0 has no real root. 20. (1) g (x) = 1 + x – [x]; ì –1, x < 0 ï f (x) = í 0, x = 0 ï 1, x > 0 î Solutions 135 For integral values of x; g (x) = 1. For x < 0; (but not integral value) x – [x] > 0 Þ g (x) > 1. For x > 0;(but not integral value) x – [x] > 0 Þ g (x) >1 \ g (x) ³ 1, " x \ f (g (x)) = 1, " x. 21. (6) Let y = g ( x) = e x - e- x 2 Þ e2 x - 1 = 2 ye x Substituting ex = t, we have t2 – 2yt – 1 = 0 Þt= 2 y ± 4 y2 + 4 = y ± y2 +1 2 (Q e x > 0) Þ e x = y + y2 + 1 ù æp ö é æp ö Þ f (1) = cos2 1 + cos ç + 1÷ × êcos ç + 1÷ - cos1ú è3 ø ë è3 ø û æp ö é æ p ö pù Þ f (1) = cos2 1+ cos ç +1÷ × ê-2sin ç +1÷ sin ú è3 ø ë è 6 ø 6û æp ö æp ö Þ f (1) = cos 2 1 - cos ç + 1 ÷ sin ç + 1 ÷ è3 ø è6 ø Þ f (1) = cos 2 1 - 1é 1ù Þ f (1) = cos 2 1 - êcos 2 - ú 2ë 2û 1 1 3 = cos 2 1 - (2 cos 2 1 - 1) + = 2 4 4 2 Þ x = ln( y + y + 1) \ gof (1) = f ( f (1)) = f (3/ 4) = 2. Þ g -1 ( y ) = ln( y + y 2 + 1) -1 24. (2) Since, f (x) is symmetric about y = x. 2 \ g ( x ) = f ( x) = ln( x + x + 1) (Q g ( f ( x)) = x Þ g -1 ( x) = f ( x)) æ 22 æ e 22 - 1 ö ç e -1 + \fç = ln ÷ ç 2e11 ÷ ç 2e11 ç è ø è æ 22 e -1 = ln ç + ç 2e11 ç è 22. (4) 2 ö æ e22 - 1 ö ÷ + 1 çç ÷ 11 ÷ ÷ 2 e ÷ è ø ø 2 ö æ e 22 + 1 ö ÷ = ln[e11 ] = 11. çç 11 ÷ ÷ ÷ è 2e ø ÷ ø ìï - x + 1, f -1 ( x ) = í ïî1 + - x , x ³1 x£0 the solution of f ( x) - f -1 ( x) = 0 are x = –1, 1, 0, 2. 23. (2) 1é æp ö æ p öù sin ç + 2 ÷ + sin ç - ÷ ú 2 êë è 2 ø è 6 øû æp ö f ( x) = cos 2 x + cos 2 ç + x ÷ è3 ø æp ö æ 3ö - cos x × cos ç + x ÷ ; g ç ÷ = 2; è3 ø è 4ø Þ f -1 ( x) = f ( x ) \ f 2 ( x) = ( f -1(x))2 - px × f (x) f -1(x) + 2x2 f (x). Þ f 2 ( x) = f 2 ( x ) - px × f ( x ) × f ( x) + 2 x 2 f ( x ) Þ f ( x ) × {2 x 2 - px × f ( x )} = 0 Þ f ( x) = 2x2 px Þ f ( x) = 2x Þ p = 2. p 25. (7) Here, f(1) = 1 f(2) = f [f (1)] + f [2 – f (1)], using f(1) = 1 f(2) = f(1) + f (1) = 2. f(3) = f [f(3)] + f[3 – f(2)] = f (2) + f(1) = 2 + 1 = 3. Thus, f (n) = n \ 1 20 1 å f (r ) = 30 [1 + 2 + 3 + .... + 20] 30 r =1 = 1 20(20 + 1) ´ = 7. 30 2 gof(x) = ? æp ö æp ö f (1) = cos 2 1 + cos 2 ç + 1÷ - cos1cos ç + 1÷ 3 è ø è3 ø [as f ( x ) ¹ 0 ] MATHEMATICS 136 CHAPTER Inverse Trigonometric Functions 17 1. (c) Let tan–1 x = a and tan –1y = b Þ tan a = x, tan b = y The given system of equations is : a tan a + b sec a = c and a tan b + b sec b = c \ a and b are roots of a tan q + b sec q = c 2 Þ (b sec q ) = (c - a tan q) 3. cos -1 b1 + cos-1 b2 + .... + cos-1 b k = k We know that cos 2 Þ (a 2 - b 2 ) tan 2 q - 2ac tan q + c 2 - b 2 = 0 -1 \ cos br £ 2ac k \ tan a + tan b = \ x+ y = Þ 1 - xy = 2 a -b 2 and Þ c 2 - b2 tan a tan b = a 2 - b2 2ac 2 a -b 2 and xy = 2 c -b p "x ³ 0 2 p "r = 1, 2,3,...., k 2 kp å cos-1b r £ 2 r =1 k Thus A= r =1 (1 + x 2 )1/ 3 - (1 - 2 x)1/ 4 x ®0 1ù 1ù é é f ( x) = sin-1 ê x2 + ú + cos-1 ê x2 - ú 2û 2û ë ë 1ù 1 ù é é = sin -1 ê x 2 + ú + cos-1 ê x 2 + - 1ú 2 2 û ë û ë 1ù é sin -1 ê x 2 + ú is defined only for these two 2û ë values. 1ù é Hence ê x2 + ú = 0 2û ë é 2 1ù Þ f ( x) = sin -1 0 + cos -1 (-1) = p ê x + ú = 1 2û ë -1 ( ) -1 Þ f ( x) = sin 1 + cos 0 = p . å ( br ) r = 0 . Required limit = lim (c) æé 1ù 1ù ö é = sin -1 ê x2 + ú + cos-1 ç ê x2 + ú - 1÷ 2 2û ø ë û èë 1 1 é 1ù Since x 2 + ³ , ê x2 + ú = 0 or 1 as 2 2 ë 2û p 2 Þ b1 = b2 = .... = b k = 0 . a2 - b2 2 Therefore range of f ( x) = {p} x£ cos -1 b1 = cos -1 b2 = ..... = cos -1 bk = 2 a 2 - c2 2 -1 p 2 So the given equality holds only if a -b x+ y 2ac \ = 2 1 - xy a - c 2 2. (a) Given 4. x + x2 1 1 (1 + x 2 ) -2 / 3 (2 x ) - (1 - 2 x) -3 / 4 ( -2) 4 = lim 3 1+ 2x x®0 (L’ Hospital Rule) 1 = 2 (c) sin–1 (log[x]) is defined if -1 £ log[ x ] £ 1 and [ x] > 0 Þ 1 £ [ x ] £ e Þ [x] = 1, 2 Þ x Î [1, 3) e Again, log(sin -1[ x ]) is defined if sin -1[ x ] > 0 and -1 £[ x ] £ 1 Þ [ x ] > 0 and - 1 £ [ x] £ 1 Þ 0 < [ x] £ 1 Þ x Î[1, 2) \ Domain of f ( x) = [1, 2) For 1 £ x < 2, [x] = 1 \ f ( x) = sin -1 0 + log p p = log , "x Î [1, 2) 2 2 Solutions 5. 137 pü \ Range of f ( x) = ì ílog ý . î 2þ (c) We have S1 = S x1 = sin 2b S2 = S x1x2 = cos 2b S3 = S x1x2x3 = cos b S4 = x1x2x3x4 = – sin b 4 So that å tan -1 xi = tan -1 i =1 S1 - S3 1 - S2 + S4 sin 2b - cos b 1 - cos 2b - sin b cos b ( 2sin b - 1) = tan -1 sin b ( 2sin b - 1) = tan -1 = tan cot b = tan -1 ( tan ( p / 2 - b ) ) = p / 2 - b 6. (d) f ( x ) = tan æ -1 ç ( 12 - 2) x 2 ö÷ ç x4 + 2 x 2 + 3 ÷ è ø æ ö ç 2 3 -1 ÷ = tan ç ÷ ç x2 + 3 + 2 ÷ ç ÷ x2 è ø 3 2 x + ³ 2 3 As [using AM > GM] x2 3 Þ x2 + 2 + 2 ³ 2 + 2 3 x æ 2 3 -1 ö -1 ÷= p \ ( f ( x ) ) max = tan çç 2 3 + 1 ÷ 12 è ø ( -1 ) ( ( 7. (c) ) ) ( x + 1) - 4 x = ( x - 1) = | x - 1 | 2 2 2 2 2 2 Þ |tan–1 | x || = | x2 – 1 | Draw the graphs of y = |tan –1 |x|| and y = |x2 – 1| y 2 y = |x – 1| p/2 –1 y = |tan – |x|| –3 8. –2 –1 0 1 1 1 + tan -1 + ...... + 1 + (n - 1)n 1 + n(n + 1) 1 n + 19 tan -1 = tan -1 1 + (n + 19) (n + 20) n + 21 1 -1 n - 1 + tan -1 Þ tan n +1 1 + n ( n + 1) 1 + tan -1 1 + (n + 1) (n + 2) 1 n + 19 + ...... + = tan -1 1 + (n + 19) (n + 20) n + 21 1 1 + tan-1 + ...... + Þ tan-1 1+ n(n +1) 1+ (n +1)(n + 2) 1 n -1 -1 n + 19 - tan -1 = tan 1 + ( n + 19) ( n + 20) n + 21 n +1 1 ö 1 æ ö -1 æ Þ tan-1 ç 2 ÷ + tan ç 2 ÷ +...... + è n + n +1 ø è n + 3n + 3 ø 1 -1 tan 1 + ( n + 19) ( n + 20) æ n + 19 n - 1 ö ç ÷ = tan -1 ç n + 21 n + 1 ÷ çç 1 + n + 19 ´ n - 1 ÷÷ n + 21 n + 1 ø è 20 -1 = tan =S 2 n + 20n + 1 20 \ tan S = 2 n + 20n + 1 1 1 -1 -1 9. (c) sin x - -1 = cos x sin x cos -1 x -1 (sin x . cos-1 x + 1) Þ (sin-1 x - cos-1 x) =0 sin -1 x . cos-1 x + tan 2 3 From the graph, it is clear that equation has four roots. (c) We know that, 1 1 1 tan-1 + tan-1 + tan -1 + ...... 1+ 2 1+ 2 ´ 3 1+ 3´ 4 -1 Þ sin-1 x = cos-1 x or sin -1 x cos-1 x +1 = 0 1 æp ö Þ x= or sin -1 x ç - sin -1 x ÷ + 1 = 0 2 2 è ø 2 æ ö p p ± ç + 4÷ ç 4 ÷ 2 1 è ø or sin -1 x = Þ x= 2 2 æ p2 ö p - ç1 + ÷ ç 16 ÷ 4 2 è ø 10. (b) (cot–1 a)x2 – (tan –1 a)3/2 x + 2(cot–1 a)2 = 0 Equation has real roots Þ x= 1 or sin -1 x = MATHEMATICS 138 ( \ D ³ 0 Þ tan -1 a 13. (b) The given relation is possible when ) - 8 ( cot -1 a ) ³ 0 3 3 a 2 a3 + + ... = 1 + b + b 2 + ... 3 9 Also a- \ tan -1 a ³ 2cot -1 a = p - 2 tan -1 a p Þa³ 3 3 Þ tan -1 a ³ tan -1 a ) ( Sum of roots > 0 Þ 3/ 2 2 cot -1 a >0 tan–1 a > 0 Þ a > 0 Product of roots > 0 Þ 2 cot–1 a > 0 Þ aÎR Þ a ³ 3 11. (b) Tn = sec Þ sec 2 Tn = Þ sec 2 Tn = Þ sec2 Tn = Þ tan Tn = Þ tan Tn = Þ \ \ 12. (c) ( n2 + 1)( n2 - 2n + 2) 2 ( n2 - n + 1) -1 (n 2 + 1)2 (n2 - 2n + 2) (n2 - n + 1)2 (n + 1)2 + (n 2 + 1) - 2 n( n 2 + 1) 2 (n2 - n + 1)2 1 + (n2 + 1 - n)2 (n2 - n + 1)2 1 1ö 1ö æ æ = tan –1 ç r + ÷ – tan –1 ç r – ÷ è è 2ø 2ø n2 - n + 1 n - ( n - 1) 1 + n ( n - 1) tan–1 n – tan–1 (n – 1) S = T1 + T2 + T3 +....+ Tn S = tan –1 n (sin–1 x)3 – (cos–1 x)3 + (sin –1 x) (cos–1 x) p3 (sin–1 x – cos–1 x) = 16 (sin–1 x – cos–1 x) {(sin –1 x)2 + (cos–1 x)2 + (2 cos–1 x sin –1 x)} = (sin–1 x – cos–1 x) (sin –1 x + cos–1 x)2 = (sin–1 x – cos–1 x) p2 p3 = 4 16 p p 3p = ; 2 sin –1 x = 2 4 4 3 p 3 x x or cos Þ sin–1 x = Þ x = sin 8 8 8 2sin -1 x - a2 a3 + + ... £ 1& -1 £ 1+ b + b2 + ... £ 1 3 9 a 1 Þ | b |< 1 Þ| a |< 3 and = a 1- b 1+ 3 3a 1 Þ = , there are infinitely many a + 3 1- b solution. But in given options it is satisfied only 1 when a = 1 and b = - . 3 2 æ ö 1 –1 4 r + 3 –1 ÷ = tan 14. (d) Tr = cot ç 1 è 4 ø 1+ r2 – 4 1ö æ 1ö æ çè r + ÷ø – çè r – ÷ø 2 2 = tan –1 1ö æ 1ö æ 1+ ç r + ÷ ç r – ÷ è 2ø è 2ø -1 £ a - p3 16 p3 16 1ö 1 æ Sn = å Tr = tan -1 ç n + ÷ – tan –1 2 2 è ø æ 4n ö = tan –1 ç ÷ è 2n + 5 ø 1 S ¥ = lim S n = tan –1 2 = cot –1 . 2 n®¥ æ (n + 1) - n ö 1 æ ö 15. (c) Tn = tan -1 ç 2 = tan-1 ç è n + n + 1÷ø è 1 + (n + 1) n ÷ø –1 = tan (n + 1) – tan –1 n –1 Þ T1 = tan (1/3) = tan –1 2 – tan–1 1 T2 = tan–1 (1/7) = tan –1 3 – tan–1 2 ..... .......................................... Tn = tan–1 (n + 1) – tan–1 n On adding T1 + T2 + T3 +.....+Tn = tan–1 (n + 1) – tan–1 1 æ n ö = tan -1 ç è n + 2 ÷ø \ lim (T1 + T2 + T3 + ..... + Tn ) n®¥ æ 1 ö = lim tan -1 ç ÷ n®¥ è 1+ 2 / n ø Solutions 139 p 4 16. (b) x3 + bx2 + cx + 1 = 0 ; f(–1) = b – c < 0 f(0) = 1 > 0 Þ –1 < a < 0 a = –B B Î (0, 1) æ 1 ö ì -1 æ 1 ö -1 æ 1 ö ü = sin -1 ç ÷ + ísin ç ÷ - sin ç ÷ý è 3 øþ è 2ø î è 2ø = tan–1 1 = ì æ 1 ö -1 æ 1 ö ü + ísin -1 ç ÷ - sin ç ÷ý è 4 øþ è 3ø î æ 2 sin B ö y = -2 tan -1 ( cosec B ) - tan -1 ç ÷ è cos 2 B ø 2cosec B ö 2sin B æ = - ç p + tan -1 - tan -1 = -p 2 ÷ è 1 - cosec B ø cos 2 B p 17. (a) cos -1 x 3 + cos -1 x = is given equation 2 ...(i) Now cos -1 x 3, cos-1 x ³ 0 and their sum is Þ p 2 cos-1 x 3, cos-1 x both must belong to é pù ê 0, 2 ú ë û Þ x 3, x Î [0,1] Now from (i) cos -1 x 3 + cos -1 x = p - cos-1 x 2 Þ cos -1 x 3 = Þ cos -1 x 3 = sin -1 x Þ cos -1 x 3 = cos -1 1 - x 2 Þ x 3 = 1- x function 2 p 2 \ S = sin æ1 a3 a ö b3 bö æ1 cosec 2 ç tan -1 ÷ + sec 2 ç tan -1 ÷ bø 2 aø 2 è2 è2 1 1 + b3 æ -1 æ a ö ö æ -1 b ö 1 + cos ç tan 1 - cos ç tan ç ÷ ÷ ÷ aø è è b øø è 1 = a3 æ æ öö b 1 - cos ç cos -1 ç ÷÷ çè a 2 + b2 ÷ø ÷ çè ø 1 +b3 æ ö a ÷ 1 + cos ç cos -1 çè a 2 + b2 ÷ø = a3 1 1- 1 2 1 But x Î[ 0, 1] Þ x = is the only possible 2 solution Þ (a) is the correct option. é n - n -1 ù -1 ú 18. (a) Let Tn = sin ê ê n ( n + 1) ú ë û é 1 1 1 1ù = sin -1 ê 11- ú n + 1 nû n +1 ë n æ 1 ö 1 ö -1 æ Tn = sin -1 ç ÷ - sin ç ÷ n n +1 ø è ø è -1 æ 19. (c) = a3 as cos –1 x is one-one 3x 2 = 1 - x 2 Þ 4 x2 = 1 Þ x = ± ì æ 1 ö -1 æ 1 ö ü + ísin -1 ç ÷ ý + ... ÷ - sin ç è 4ø è 5 øþ î p p æ 1 ö = 2 sin -1 ç ÷ =2. = 4 2 è 2ø 1 ö ç ÷ + T2 + T3 + T4 + ... è 2ø 1 + b3 b 2 a +b a 1+ 2 2 a + b2 æ ö a3 b3 ÷ = a 2 + b2 ç + çè a 2 + b2 - b a 2 + b2 + a ÷ø æ 2 2ç 3 = a +b ç a ç è ( ) ( a2 +b2 +b a2 3 a2 +b2 -a +b b2 ) ö ÷ ÷ ÷ ø é ù = a 2 + b2 êa æç a 2 + b2 + b ö÷ ú øû ë è +b æç a 2 + b2 - a ö÷ è ø = a 2 + b2 ( a + b ) a 2 + b2 = (a + b)(a 2 + b 2 ) 20. (a) cot–1(2.12) + cot–1(2.22) + cot–1(2.32) +... ¥ ¥ æ 1 ö = å cot -1 2.r 2 = å tan -1 ç 2 ÷ è 2r ø r =1 r =1 ( ) MATHEMATICS 140 ¥ é (1 + 2r ) + (1 - 2r ) ù = å tan -1 ê ú ëê1 - (1 + 2r )(1 - 2r ) ûú r=1 ¥ = å é tan -1 (1 + 2r ) + tan -1 (1 - 2r ) ù ë û r =1 = tan–1 3 – tan–1 1 + tan–1 5 – tan–1 3 + tan–1 7 – tan–1 5 +....+ tan –1 ¥ p p p =- + = 4 2 4 21. æp -1 ö (1) Let y = lim ç - tan x ÷ ø x®¥ è 2 –1 1/x = lim (cot x) x®¥ 1/ x ö log cot -1 x æ ¥ çè form÷ø ¥ x®¥ x \ log y = lim –1 log y = lim x®¥ (1 + x 2 ) cot -1 x log y = – lim x®¥ -2 x cot æ0 ö çè form÷ø 0 x (1 + x 2 )2 -1 x®¥ = – lim 1 + x2 x lim = – 2 x®¥ 1 + x2 \ y = e0 =1 1 =0 x® ¥ 2 x = 2 lim 22. (0) We know that cot A > 1 if 0 < A < and cot A < 1 if p p <A< 4 2 p 4 tan -1 (cot A) + tan -1 (cot 3 A) = p + tan -1 cot A + cot 3 A 1 - cot 4 A , p 4 and tan–1(cotA) + tan–1 (cot3A) p p cot A + cot 3 A if < A < = tan -1 4 4 2 1 - cot A If 0 < A < Also, cot A + cot 3 A 4 1 - cot A = 2 (sin A + cos 2 A)(sin 2 A - cos 2 A) sin 2A 1 == - tan 2A 2 cos 2 A 2 Hence, æ1 ö tan -1 ç tan 2 A÷ + tan -1 (cot A) + tan -1 (cot 3 A) è2 ø p ì if 0 < A < ïï p 4 [Q tan -1 (- x) = - tan -1 x] =í p p ï0 if < A < ïî 4 2 é p pù 23. (2) (sin–1x) Î ê - , ú ë 2 2û ( \ sin -1 x ( 0. ¥ form) (1 + x 2 )-1 -1 sin A cos A = cot A cosec2 A.sin 4 A sin 4 A - cos 4 A 2 ) £ p4 Þ (sec y) , (tan z) > 0 2 –1 2 –1 2 p2 p2 \ (sin–1 x)2 = 4 4 \ (sec–1 y)2 + (tan–1 z)2 = 0 or sec–1 y = tan –1 z = 0 p -1 \ sin x = ± , y = 1, z = 0 2 \ x = ±1, y = 1, z = 0 24. (3) cos (2 sin–1 (cot (tan –1 (sec (6 cosec–1x))))) = –1 p sin–1 (cot (tan–1 (sec (6 cosec–1 x)))) = ± 2 cot (tan–1 (sec (6 cosec–1x))) = ± 1 p tan–1 (sec (6 cosec–1x)) = ± 4 sec (6 cosec–1 x) = ±1 –1 6 cosec x = ± 3p, ± 2p, ± p 2 p p p ,2 cosec–1x = ± , ± , ± Þ x = 1, 3 2 3 6 \ RHS ³ (Q x > 0) 25. (7) f ( x ) = 3cos -1 ( 4 x ) - p is defined 1 p 1 Þ 4 x £ or x £ ...(i) 3 2 8 -1 1 £x£ Also, -1 £ 4 x £ 1 or ...(ii) 4 4 Therefore, from eqs. (i) and (ii), we have domain: é -1 1 ù x Î ê , ú Þ 4a + 64b = 7 ë 4 8û If cos -1 4x ³ Solutions 141 CHAPTER Matrices 18 1. é 2 - 1ù é3 2ù é1 1ù \ (B -1 )(C -1 ) = ê úê ú=ê ú ë- 3 2 û ë5 3û ë1 0û é cos 2 q cos q sin q ù (a) AB = ê ú êëcos q sin q sin 2 q úû é1 1 ù \A = ê ú ë1 0û é cos2 f cos f sin f ù ê ú êë cos f sin f sin 2 f úû écos q cos f + cos q cos f sin q sin f =ê 2 2 êë cos f cos q sin q + sin q sin f sin f 2 2 3. cos 2 q cos f sin f + sin 2 f sin q cos qù ú cos q cos f sin q sin f + sin 2 q sin 2 f úû écos q - sin q ù \ (I – P) ê sin q cos q ú ë û écos q cos f(cos q cos f + sin q sin f) =ê ë sin q cos f(cos q cos f + sin f sin f cos q sin f(cos q cos f + sin q sin f ù sin q cos f(cos q cos f + sin q sin f)úû écos q cos f cos( q - f) cos q sin f cos(q - f)ù =ê ú ë sin q cos f cos( q - f) sin q sin f cos(q - f) û Clearly AB is the zero matrix if cos( q - f) = 0 i.e. q - f is an odd multiple of 2. p . 2 1 tan (q / 2) ù écos q - sin q ù é = ê - tan q / 2 ú . ê sin q cos q ú 1 ë û ë û é cos q + tan( q / 2) sin q - sin q + tan(q / 2) cos q ù = ê - tan(q / 2) cos q + sin q tan(q / 2) sin q + cos q ú ë û é ù -2sin(q / 2)cos(q / 2) 1- 2sin2 (q / 2) ê 2 + tan(q / 2)(2cos2 (q / 2) -1) ú + q 2sin ( / 2) ú =ê ê- tan (q / 2)(2cos2 q / 2 -1) tan(q / 2)(2sin(q / 2) cos(q / 2))ú ê ú +(1- 2sin2 (q / 2)) ë+2sin (q / 2) cos(q / 2) û - tan (q / 2) ù é 1 = ê ú = I+P 1 ë tan q / 2 û é2 1ù é- 3 2 ù and C = ê (a) Let B = ê ú ú ë 3 2û ë 5 - 3û -1 -1 Given BAC = I Þ B (BAC) = B I Þ I(AC) = B -1 Þ AC = B-1 Þ ACC -1 = B -1C -1 Þ AI = B -1C -1 0 - tan (q / 2) ù é1 0 ù é (c) I– P = ê –ê ú ú 0 û ë0 1 û ë tan (q / 2) 1 tan (q / 2) ù é = ê - tan (q / 2) ú 1 ë û 4. é 1 – xù é 1 – y ù (c) (1 – x)–1 (1 – y)–1 ê úê ú ë – x 1 û ë– y 1 û –( x + y ) ù é 1 + xy = (1 + xy – (x + y))–1 ê –( x + y ) 1 + xy ú ë û \ A = (B -1 )(C -1 ) 1 é 2 - 1ù é 2 - 1ù -1 Now B = ê ú=ê ú 4 - 3 ë- 3 2 û ë- 3 2 û 1 é - 3 - 2 ù é3 2 ù C -1 = ê ú=ê ú 9 - 10 ë- 5 - 3 û ë5 3 û é æ x+ y ö = ç1 – ÷ + xy ø 1 è –1 ê 1 ê ê x+ y ê – 1 + xy ë Þ A( x) A( y) = A( z) – x+ yù 1 + xy ú ú = A( z) ú 1 ú û MATHEMATICS 142 5. (d) We have 0 ù é1 0 ù é1 0 ù é1 0 ù é1 A2 = ê ú ê1/ 2 1ú = ê 2(1/ 2) 1ú = ê1 1ú 1/ 2 1 ë ûë û ë û ë û 8. é0 (c) Let A = ê a ê ëê a é1 0 ù é1 0 ù é1 0 ù A4 = ê úê ú=ê ú; ë1 1û ë1 1û ë 2 1û b -c aù -b ú ú c úû a ù é1 0 0 ù é 0 2b c ù é 0 a ê ú ê Þ a b - c 2b b -b ú = ê 0 1 0 ú ê úê ú ê ú ëê a -b c ûú êë c -c c ûú ëê 0 0 1 ûú é 4b2 + c 2 ê Þ ê 2b 2 - c 2 ê ê -2b2 + c 2 ë é 3 1 ù ê ú 2 2 ú (a) Given that, P = ê ê 1 3ú ê– ú ë 2 2 û é1 1ù T T 2005 A=ê P ú and Q = P A P and X = P Q ë0 1û We observe that Q = P A PT Þ Q2 = (P A PT) (P A PT) = P A (PT P) A PT = PA (I A) PT \ P A2 PT Proceeding in the same way, we get Q 2005 = P A2005 PT é1 2005 ù = IA2005 I = A2005 = ê 1 úû ë0 a \ AAT = I é1 0ù A50 = ê ú ë 25 1û (d) As aij = (i2 + j2 – ij) (j – i) aji = (j2 + i2 – ji) (i – j) = – aij Þ A is skew symmetric Þ Tr(A) = 0. Also |A| = 0. é1 2005 ù A2005 = ê 1 úû ë0 Now, X = PTQ2005 P = PT(PA2005 PT)P = (PT P) A2005 (PT P) cù -c ú . ú c ûú Q A is orthogonal. Þ é1 1ù é1 2 ù Þ A2 = ê Also A = ê ú ú ë0 1û ë0 1 û and proceeding in the same way -b é0 Now A = ê 2b ê êë c 0ù é1 In general, by induction An = ê ú ë n / 2 1û 7. b T é1 0 ù é1 0 ù é1 0 ù A8 = ê 2 1ú ê 2 1ú = ê4 1ú ë ûë û ë û 6. 2b -2b 2 + c 2 ù ú a 2 - b2 - c 2 ú ú a 2 + b2 + c 2 ú û 2b 2 - c 2 a 2 + b2 + c 2 a 2 - b2 - c 2 é1 0 0ù = ê 0 1 0ú ê ú ëê 0 0 1ûú Equating the corresponding elements, we get 4b2 + c2 = 1 ....(i) 2b2 – c2 = 0 ....(ii) a2 + b2 + c2 = 1 ....(iii) a2 – b2 – c2 = 0 ....(iv) From solving (i), (ii) and (iii) we get, a=± 9. (a) 1 2 ;b=± 1 6 ; c=± 1 3 ; é 0 1 0 ù é 0 1 0ù A2 = AA = ê 0 0 1 ú ´ ê 0 0 1 ú ê ú ê ú êë p q r úû êë p q r úû é0 ê =ê p ê ë pr ù ú q r ú ú p + qr q + r 2 û 0 1 Solutions 143 Again é0 0 1 ù é 0 1 0ù ê ú A = A A=ê p q r ú ´ êê 0 0 1 úú ê 2 ú ê p q r úû ë pr p + qr q + r û ë p q r é ù ê ú 2 p + qr q+r = ê pr ú ê 2 2 2 3ú p + 2qr + r û ë pq + r p pr + q + qr 3 ép = êê 0 êë 0 2 0 p 0 0ù é 0 q ê ú 0ú + ê 0 0 ê 2 ú pû ë pq q 0ù ú qú ú qr û é 0 ê + ê pr ê 2 ë pr 0 X = A1 + 3 A 33 + ... ( 2 n – 1 ) ( A 2n -1 )2n -1 ù ú r ú 3ú pr + r û r 2 qr pr + qr 2 é1 0 0 ù é 0 1 0ù = p êê 0 1 0 úú + q êê 0 0 1úú êë 0 0 1 úû êë p q r úû é0 ê +r ê p ê ë pr = pI + qA + rA2 2 ù ú q r ú ú p + qr q + r 2 û XT = - [A1 + 3 A33 + ... ( 2 n – 1) ( A2n-1 ) 2n -1 ] = –X, so skew-symmetric éa b c ù ê ú 14. (c) A = ê d e f ú A2 = A × A = A Aq ëê g h i ûú 1 \ A3 - rA2 - qA = pI . 10. (a) Given A = 2A – I Þ A A = (2A – I) A Þ A3 = 2A2 – A = 2 (2A – I) – A = 4A – 2I – A = 3A – 2I Again A3 A = (3A –2 I) A = 3 A2 – 2A Þ A4 = 3 (2A – I) – 2A = 4A – 3I ........ ........ In general An = nA – (n –1) I 11. (d) PQ = I Þ P–1 = Q Now the system in matrix notation is PX = B \ X = P–1 B = QB 2 1 ù é1 ù é2 éxù úê ú ê ú 1ê \ ê yú = ê 13 - 5 m ú ê1 ú 9 êë- 8 1 5 úû êë5úû êëz úû We know that if A is a skew-symmetric matrix then AT = –A [Q Given that A = Aq] 0 2 1 (13 - 5 + 5m) 9 Þ – 27 = 8 + 5m (Given y = – 3) \m=–7 12. (b) We have, AB = A and BA=B. Now, AB = A Þ (AB) A = A . A Þ A (BA) = A2 Þ AB = A2 (Q BA = B) Þ A = A2 (Q AB = A) Again, BA = B Þ (BA)B = B2 Þ B(AB) = B2 Þ BA=B2 (Q AB = A) Þ B = B2 ( Q BA = B) \ A and B are idempotent matrices. 13. (b) Given that \ y= Þ éa b c ù éa d g ù ú ê ú ê A = êd e f ú êd e h ú êë g h i úû ê c f i ú ë û 2 é| a |2 + | b |2 + | c |2 ad + be + cf ù ag + bh + ci ê ú 2 2 2 ú Þ ê da + eb + fc | d | + | e | + | f | dg + eh + fi ê ú 2 2 2 ê ga + hb + ic gd + he + if | g | +| h | +| i | ú ë û é0 0 0ù ê ú = ê 0 0 0 ú (Q Given A2 = O) êë 0 0 0 úû Þ |a|2 + |b|2 + |c|2 = 0 Þ Re(a) = Re(b) = Re(c) = Im(a) = Im(b) = Im(c) = 0 Similarly Re(d) = Re(e) = Re(f) = Im(d) = Im(e) = Im(f) = 0 and Re(g) = Re(h) = Re(i) = Im(g) = Im(h) = Im(i) = 0 MATHEMATICS 144 é 0 + 0i 0 + 0i 0 + 0i ù ê ú \ A = ê 0 + 0i 0 + 0i 0 + 0i ú êë 0 + 0i 0 + 0i 0 + 0i úû = Null matrix of order 3 × 3 15. (d) (XY)T = YTXT Y T = (ABC – CBA)T = CTBTAT – ATBTCT = – CBA + ABC = Y XT = (ABC + CBA)T = CTBTAT + ATBTCT = – CBA – ABC = – X = BN+1 + ( N + 1) BNC = BN[ B + (N + 1) C] Thus K = N Þ K/N = 1. 19. (1023) (AB)×(AB) = A(BA)B = A3B2 (AB)(AB)(AB) = A7B3 n s o (AB)n = A2 - 1 Bn s o k = 210 – 1 = 1023 20. (27) Observe A1A2 é1 1 ù é1 2 ù é1 3 ù é1 (2 + 1) ù =ê úê ú=ê ú=ê 1 úû ë0 1û ë0 1û ë0 1û ë0 é cos q sin q ù 16. (d) A = ê ú ësin q – cos q û \ AAT = I ...(i) Now, C = ABAT Þ ATC = BAT ...(ii) Now ATCnA = ATCCn – 1 A = BATCn – 1 A (from (ii)) = BATCCn – 2 A = B2ATCn – 2 A = ... é 1 0ù = Bn – 1ATCA = Bn – 1BATA = Bn = ê ú ë – n 1û é x1 ù ê ú 17. (c) Let X = ê x2 ú , (x1 x2 x3) êë x3 úû æ a11 a12 a13 ö æ x1 ö ça a a ÷ çx ÷ =0 ç 21 22 23 ÷ ç 2 ÷ çè a ÷ç ÷ 31 a32 a33 ø è x3 ø a11 x12 + a22 x22 + a33 x32 + (a 12 + a 21 )x 1 x 2 + (a13 + a31)x1x3 + (a23 + a32)x2x3 = 0 It is true for every x1, x2, x3, then a11 = a22 = a33 = 0, a12 + a21 = 0, a13 + a31 = 0, a23 + a32 = 0 \ A is a skew symmetric matrix a21 = – 2008; a31 = – 2010 a32 = 2012 18. (1) We have, BC = CB, and AN + 1 = (B + C) N + 1 = N+1C0 B N+1 + N+1C1 BNC + N+1C2BN–1C2 + . . . + N+1CrB N+1–rC r + . . . . But given that C2 = 0 Þ C3 = C4 = .... = Cr = 0 Hence, AN+1 = N+1CNBN+1 + N+1C1BNC é1 3 ù é1 3 ù é1 (3 + 2 + 1) ù and A1A2A3 = ê úê ú=ê ú 1 ë 0 1û ë 0 1û ë0 û So, in general A1A2A3 ... An é1 n + ( n –1) + ( n – 2) + ... + 3 + 2 + 1ù =ê ú 1 ë0 û n(n + 1) = 378 Þ n = 27 2 (4) Given that AT A = I So 21. éa b cù where A = êê b c a úú êë c a b úû éa b c ù \A = êb c a ú ê ú êë c a b úû T \ AT A = I éa Þ êb ê êë c b c a c ù éa a ú êb úê b úû êë c b c a c ù é1 a ú = ê0 ú ê b úû êë 0 0 1 0 0ù 0ú ú 1 úû é a2 + b2 + c2 ab + bc + ca ab + bc + ca ù ê ú Þ ê ab + bc + ca a2 + b2 + c2 ab + bc + ca ú ê ú ê ab + bc + ca ab + bc + ca a 2 + b2 + c 2 ú ë û é1 0 0 ù = êê0 1 0 úú êë0 0 1 úû Solutions 145 Þ a2 + b2 + c2 = 1 ...(i) and ab + bc + ca = 0 ...(ii) abc = 1 (given) ...(iii) Now a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) Þ a3 + b3 + c3 = (a + b + c) [1 – 0] + 3 × 1 [Using (i), (ii) and (iii)] 3 3 3 Þa +b +c =a+b+c+3 Now (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca) Þ (a + b + c)2 = 1 Þ (a + b + c) = 1 [Q (a + b + c) ¹ – 1 as a, b, c all are + ve numbers.] \ We get a3 + b3 + c3 = 4 22. (1) z = -1 + i 3 Þ z3 = 1 and 1 + z + z2 = 0 2 é (- z) r P = ê 2s êë z 2 z 2s ù é( - z) r úê z r úû êë z 2s é z 2r + z 4s ê =ê 2s r r êë z (-z) + z ( ) z 2s ù ú z r úû ( ) z 2s (- z)5 + z r ù ú ú 4s 2r z +z úû For P2 = – I we should have z2r + z4s = 1 and z2s ((–z)r + zr) = 0 Þ z2r + zs + 1 = 0 Þ r is odd and s = r but not a multiple of 3. Which is possible when s = r = 1 \ only one pair is there. 23. (1) é 1 0 0ù é 0 0 0ù ê ú P = ê 4 1 0ú = I + êê 4 0 0úú = I + A êë16 4 0úû êë16 4 1úû é 0 0 0ù é0 0 0ù ê ú 3 ê A = ê 0 0 0ú and A = ê 0 0 0 úú ëê16 0 0úû ëê 0 0 0 ûú 2 \ An = O, n ³ 3 Now P50 = (I + A)50 = 50C0 I50 + 50C1 I49 A + 50C2 + I48 A2 + O = I + 50A + 25 × 49 A2. 50 \ Q = P – I = 50A + 25 × 49 A2. Þ q21 = 50 × 4 = 200 Þ q31 = 50 × 16 + 25 × 49 × 16 = 20400 Þ q32 = 50 × 4 = 200 \ q31 + q32 20600 q + q 32 =1 = = 103 Þ 31 q 21 200 103 q 21 –1 ù é 2 –1 ù é1 0 ù é2 24. (2) A2 = ê ú ê ú= ê ú = I; ë 3 – 2 û ë 3 – 2 û ë0 1 û x = 2, 4, 6, 8.... Þ S(cosx q + sinx q) = (cos2 q + sin2 q) + (cos4 q + sin4 q) + (cos6 q + sin6 q) + ... + ... = cos 2 q + sin 2 q 1 – cos 2 q 1– sin 2 q = cot2 q + tan2 q Using AM ³ GM, 1 (cot 2 q + tan 2 q) ³ cot 2 q tan 2 q 2 Þ (cot2 q + tan2 q) ³ 2 Þ So, minimum value = 2 25. (2) As A2 = O, Ak = O k Thus, (A + I)50 = I + 50A Þ (A + I)50 – 50A = I \ a = 1, b = 0, c = 0, d = 1 2. MATHEMATICS 146 CHAPTER Determinants 19 1. (a) Given determinant 2 1 a a cos(n - 1) x cos nx cos(n + 1) x sin (n - 1) x sin (n + 1) x sin nx =0 sin x = 0 or cos x = 1 + a2 2a æ 1 + a2 ö ÷ >1 è 2a ø Þ cos x > 1 It is not possible. \ sin x = 0 As a ¹ 1 \ç a 2. 2 c (b + 1) ( a - 1) 2 ( b - 1) 2 ( c - 1) 2 a2 b2 c2 a b c (a - 1)2 (b - 1)2 (c - 1)2 Þ4 2 2 2 ( a + 1) (d) b 2 ( c + 1) 2 = 0 Þ b2 c =0 1 (using R 3 ® R3 + 2 R2 ) æ C ® C2 - C1 ö c-a = 0 ç 2 ÷ è C3 ® C3 - C1 ø 0 b-a 0 a2 (d) b sin A c sin A D = b sin A c sin A Using, 1 cos A cos A 1 sin A sin B sin C = = =k a b c Þ sin A = ak , sin B = bk , sin C = ck So, a2 abk ack D = abk 1 cos A . cos A 1 ack Take a common from C1 and R1 both, we get 1 D = a bk bk 1 ck cos A ck cos A 1 1 = a 2 sin B sin B 1 sin C cos A 2 =0 (taking R2 ® R2 - R3 ) a2 b 1 Þ a = b or b = c or c = a Þ D ABC is an isosceles. 3. Þ (1 + a - 2a cos x)sin x = 0 Þ a 1 Þ (b2 - a 2 )(c - a) - (b - a)(c 2 - a 2 ) = 0 Þ (b - a)(c - a )(b - c ) = 0 By applying C1 ® C1 + C3 – 2 cos x C2 By expanding (1 + a2 – 2a cos x) [cos nx sin (n + 1) x – sin nx cos (n + 1) x]= 0 Now, (1 + a2 – 2a cos x) sin (n + 1 – n) x = 0 2 c2 Þ 1 1 sin (n + 1) x sin nx b2 a2 b2 - a 2 c 2 - a 2 1 + a 2 - 2a cos x a a2 0 cos nx cos(n + 1) x = 0 Þ 0 a2 c2 a b c =0 1 - 2a 1 - 2b 1 - 2c (u sin g R3 ® R3 - R1 ) sin C cos A 1 Operate C2 ® C2 - C1 sin B and C3 ® C3 - C1 sin C Solutions 147 1 2 D = a sin B 0 0 2 1 - sin B Where 0 < a , b, q < cos A - sin B sin C sin C cos A - sin B sin C 1 - sin 2 C 1 1 1 D = sin( b a ) cos( b a ) cos( b a + q) \ - cos b sin b sin(b - q) = a 2 [(1 - sin 2 C - sin 2 B + sin 2 B sin 2 C ) - (cos A - sin B sin C ) 2 ] 2 2 2 Operating C3 - C1 sin q - C2 cos q , we get 2 = a [sin A - sin B - sin C + 2sin B sin C cos A] The above expression does not represent area or perimeter of the triangle. 4. 1 1 1 - sin q - cos q D = - sin(b - a ) cos(b - a ) 0 - cos b sin b 0 1 1 + sin P sin P (1 + sin P) (d) D = 1 1 + sin Q sin Q (1 + sin Q) 1 1 + sin R sin R (1 + sin R) = (1 - sin q - cos q)[cos b cos(b - a) - sin b sin(b - a)] Þ D = [1 - (sin q + cos q)] cos(2b - a) Q 0 < a , b, q < 1 sin P sin P + sin 2 P n -1 1 sin P sin 2 P 6. = 1 sin Q sin 2 Q (C3 ® C3 – C2) å r = 1 + 2 + 3 + ... + (n – 1) = n(n2- 1) (d) r =1 n -1 å (2r - 1) = 1 + 3 + 5 + ... + [2 (n – 1) – 2] 1 sin R sin 2 R r =1 = sin 2 P 0 sin Q - sin P sin 2 Q - sin 2 P 2 2 0 sin R - sin P sin R - sin P (R2 ® R2– R1, R3 ® R3 – R1) = (sinQ – sinP) (sinR – sinP) 1 sin P sin 2 P 0 1 0 1 sin Q + sin P sin R + sin P = (sinQ – sinP) (sinR – sinP) (sinR – sinQ) Now 5. \ sin q + cos q ¹ 1 Also 2b - a < 1 sin R sin R + sin 2 R sin P p 4 p Þ cos(2b - a ) ¹ 0 4 \ D ¹ 0 Þ the three points are non-collinear.. = 1 sin Q sin Q + sin 2 Q (C2 ® C2 – C1) 1 p 4 D > 0 if P < Q < R D < 0 if P > Q > R Hence the sign of D cannot be determined. (d) The given points are P ( - sin(b - a), – cos b ) , Q (cos(b - a ),sin b) R (cos(b - a + q),sin(b - q)) = (n – 1)2 n -1 å (3r - 2) = 1 + 4 + 7 + .. + (3n – 3 – 2) r =1 = (n - 1)(3n - 4) 2 n -1 \ å Dr r =1 Sr n = 2 n(n - 1) 2 S (2r - 1) S (3r - 2) n -1 a (n - 1)2 (n - 1)(3n - 4) 2 n -1 å D r consists of (n – 1) determinants in r =1 L.H.S. and in R.H.S every constituent of first MATHEMATICS 148 row consists of (n – 1) elements and hence it can be splitted into sum of (n – 1) determinants. f "( x ) g "( x ) h "( x ) a n -1 \ å Dr (n - 1)2 (n - 1)(3n - 4) 2 n -1 a (n - 1)2 (n - 1)(3n - 4) 2 a b =0 a (a + b + 2) a +b 1 (b + c + 2) c2 + a 2 1 2 b +c = 1 4 0 = 65 8. 1 1 = (b) |A | = |A| 5 | (AB)T | = |AB| = |A×(adj A)| = |A|×|adj (A)| = 5 × 52 =53 1 1 ||A– 1|(AB)T| = | (AB)T| = 3 |AB| = 1 5 5 (a) Differentiating given equation w.r.t.x, we get 9. c p = 2(3n – r) (b) We have q r 1 1 a b c = (a – b)(b – c)(c – a)(a + b + c) 3 3 c3 b a b c p q r ...(1) Again differentiating w.r.t.x, we get ...(1) 1 1 1 b c 1 1 a 3 b 3 c3 a 2 b2 c2 1 Also, a b 1 a c = abc 1 (taking a, b, c common from R1, R2, R3) bc ac ab = 1 1 = 4mx3 + 3nx2 + 2rx + 5 1 1 a f '( x) g '( x ) h '( x) ...(5) b –1 \ c a a 0 1 4 b f "'(0) – f "(0) g "'(0) – g "(0) h "'(0) – h "(0) 10. 4 0 1 ...(4) p q r From (5) and (4), we get 2 (c + a + 2)2 a 6n = 1 2 c f "'(0) g "'(0) h "'(0) both 'a' and 'n'. (a) We have a2 + b2 + c2 + ab + bc + ca £ 0 \ (a + b)2 + (b + c)2 + (c + a)2 £ 0 \ a + b = 0, b + c = 0 and c + a = 0 \ a=b=c=0 2 b p q r Putting x = 0 in (3), we get å Dr is independent of 2 ... (3) f "(0) g "(0) h "(0) 2r = r= 1 2 c = 24mx + 6n p q r putting x = 0 in (2), we get n -1 Hence, value of Þ = 12mx2 + 6nx + 2r ...(2) f "'( x ) g "'( x ) h "'( x ) (Q R1 and R3 are identical) 7. c p q r Again differentiating w.r.t.x, we get r =1 n(n - 1) 2 n = 2 n(n - 1) 2 b 2 1 b 1 1 2 c (Multiplying R1 by abc) 2 1 = a 2 b2 c 2 then, bc ac ab Solutions 149 1 D= 1 1 a+a a a ( x – a)2 ( x – b) 2 ( x – c) 2 ( x – b)( x – c) ( x – c)( x – a) ( x – a)( x – b) = (a – b)(b – c)(c – a)(3x – a – b – c) Now given that a, b, c are all different, then 1 D = 0. Therefore, x = (a + b + c) 3 11. Operate R2 ® R2 - R1 ; R3 ® R3 - R1 , then l= 2 t + 3t + 4 Þ (l - 1)t 2 + 3(l + 1)t + 4(l - 1) = 0 Since, t is real Þ (3l + 3 - 4l + 4)(3l + 3 + 4l - 4) ³ 0 Now, D = 1 6 1 £l£7 7 a 1 1 1 b 1 =0 14. (c) 1 1 c -3 5 l [Determinant of coefficients of equations] é 1 ù êQ 7 £ l £ 7 ú ë û Hence the given system of equations has a unique solution. (c) Determinant of coefficients = 7 (l + 5) ¹ 0 12. t t +1 t -1 1 -1 = t +1 t t + 2 = t + 1 -1 1 t -1 1 t -1 t + 2 t t 3 1 -1 t = 2t + 1 0 0 2t - 1 4 0 = -4(2t + 1) 1 . 2 13. (a) The given system of equations will have a non-trivial solution if, For non-trivial solution t = - a+a a a a+b a a a a+c =0 0 0 b – 1 1 – c = 0, 1–c 1 1 c –1 0 +1 0 –1 1 1 1– a 1 1– b c 1– c = 0, æ 1 ö æ 1 ö C1 ® ç ÷ C1 , C 2 ® ç ÷ C2 , è1 – a ø è1 – b ø Þ (1 – a) (1 – b) (1 – c) æ 1 ö C3 ® ç ÷ C3 è1 – c ø ì 1 ü 1 c (0 + 1) – (–1 + 0) + (1– 0)ý = 0 í (1– a) (1– b) 1– c î þ 1 1 c + + = 0, 1– a 1– b 1– c as a ¹ 1, b ¹ 1, c ¹ 1 (given) Þ 1 1 c – 1+ 1 + + =0 1– a 1– b 1– c 1 1 1 Þ + + =1 1– a 1– b 1– c Þ a a –1 Þ R1 ® R1 – R3, R2 ® R2 – R3 Þ (1 – a) (1 – b) (1 – c) 4 2 c 1 æ 1 1 1ö = -ç + + ÷ è a b cø a Þ 9(l + 1) 2 - 16(l - 1) 2 ³ 0 3 -1 0 =0 0 Þ aab + c(ab + ab + aa ) = 0 since a, b, c ¹ 0 \ Þ ( 7 - l )(7l - 1) ³ 0 Þ b -a Þ a (bc + ca + ab) + abc = 0 t 2 - 3t + 4 (a) -a MATHEMATICS 150 15. (a) The given system of equations will have a non-trivial solution if a+a a a a a+b a 18. a+a a a 16. D= 17. b 0 =0 –a 0 c or aab + c(ab + ab + aa) = 0 or a(bc + ca + ab) + abc = 0 1 æ 1 1 1ö = –ç + + ÷ (Q a, b, c ¹ 0) or è a b cø a (c) The given system is consistent. Therefore, 1 1 2 –1 –1 –c =0 or – b 3b – c c + bc – 6b + b + 2c + 3bc = 0 or 3c + 4bc – 5b = 0 or c= 5b 4b + 3 Now, c < 1 5b 5b < 1 or –1 <0 Þ 4b + 3 4b + 3 b–3 æ 3 ö or < 0 Þ b Î ç – ,3÷ è 4 ø 4b + 3 (c) We have, dn dx np ö (-1)n n ! æ 0 cos ç x + ÷ è 2 ø ( x + 3)n +1 np 2 f ( x )] = 0 n[ cos a a3 19. 20. c1 + ra1 D¢ = a2 + pb2 b2 + qc2 c2 + ra2 a3 + pb3 b3 + qc3 c3 + ra3 21. (-1)n n ! c1 + ra1 = a2 b2 + qc2 c2 + ra2 a3 b3 + qc3 c3 + ra3 pb1 b1 + qc1 c1 + ra1 + pb2 b2 + qc2 c2 + ra2 pb3 b3 + qc3 c3 + ra3 In the first determinant, apply C3 ® C3 – rC1 and then C2 ® C2 – qC3. In second determinant take p common from C1 and then apply C2 ® C2 – C1. Then take q common from C2 and apply C3 ® C3 – C2. Finally taking r common from C3, we have ultimately D¢ = (1 + pqr)D. (d) det{Adj(Adj A)} = (14)4 = |A|(n – 1)2 Þ |A| = 14 = 3x + 11 Þ x = 1 (c) A(Adj A) = |A|In; |A| = xyz – (8x + 4y + 3z) + 28 (0) Apply C3 ® C3 – C1 22. sin(2x + 5p)sin(2p) Þ sin çè x + ÷ø 2 3p ö æ 5pö sin ç x + ÷ è 2ø æ 5pö 2cos ç x + ÷ sin(p) = 0 è 2ø æ 3pö sin ç x - ÷ è 2ø æ 5pö sin ç x - ÷ è 2ø æ 5pö 2cos ç x - ÷ sin(-p) è 2ø æ a (Q R1 and R2 are identical) b1 + qc1 æ 3pö æ 5p ö sin2 ç x + ÷ sin2 ç x + ÷ è è 2ø 2ø 3n +1 5 np ( -1)n n ! 2 3n +1 n d np ( -1)n n ! f x ( ) = 0 cos =0 [ ] x=0 2 dx n 3n +1 a a3 a5 a1 é 68 0 0 ù ê ú = 60 – 20 + 28 = 68 Þ A(Adj A) = ê 0 68 0 ú êë 0 0 68úû 0 cos \ b1 + qc1 =0 a a a+c Operating R2 ® R2 – R1 and R3 ® R3 – R1, we get –a (d) a1 + pb1 Q All elements of C3 are zero. (20) The matrices in the form é a11 a12 ù êë a21 a22 úû , aij Î {0, 1, 2}, a11 = a12 are é 0 0/1/ 2ù é 1 0/1/ 2ù é 2 0/1/ 2ù êë0/1/ 2 0 úû , êë0/1/ 2 1 úû , êë0/1/ 2 2 úû Solutions 151 At any place, 0/1/2 means 0, 1 or 2 will be the element at that place. Hence there are total 27 = 3 × 3 + 3 × 3 + 3 × 3) matrices of the above form. Out of which the matrices which are singular are é 0 0 / 1/ 2 ù é 0 0 ù é1 1ù é 2 2ù , , , êë 0 0 úû êë1/ 2 0 úû êë1 1úû êë 2 2úû Hence there are total 7(= 3 + 2 + 1 + 1) singular matrices. Therefore number of all non-singular matrices in the given form = 27 – 7 = 20 é3 0 0ù ê ú 23. (0.20) A = ê0 3 0ú êë0 0 3úû det (adj(adj(A))) = |A|4 = 274 cos x x sin x cos x 2 2 2 - tan x 2cos 2 x - x3 5 Operate 1 1 1 R2 , R3 ® R3 , R2 ® R2 , x x x respectively on three determinants R2 ® cos x - sin x f '( x) tan x = x x x 2 x sin 2 x - x2 1 5x sin x cos x tan x + 2 x - sec 2 x -3x 2 + x - x2 x sin 2 x 2 2cos 2 x 5 2 5 x 24. (0) log a + (n –1)log r log a + (n +1)log r log a + (n + 3)log r log a + (n + 9)log r log a + (n +11)log r log a + (n +13)log r log a + (n +15log r D = C3 ® C3 – C2 sin x + 2 x - sec x -3x + x 2 x sin 2 x 5x 2 x ìï 27 4 üï 1 í ý = = 0.20 îï 5 þï 5 log a + (n + 5)log r log a + (n + 7)log x C2 ® C2 – C1 sin x cos x x 1 1 0 0 0 1 f '( x ) = 0 -1 0 + 0 -1 0 Lt x ®0 x 0 0 0 2 2 5 0 log a + (n –1) log r 2 log r 2log r log a + (n + 5) log r 2log r 2 log r = 0 Þ 2 log a + (n + 11) log r 2 log r 2log r (1 + ap)2 (1 + bp)2 (1+ cp)2 25. (16) (1 + aq)2 (1 + bq)2 (1+ cq)2 (1 + ar )2 (1 + br )2 (1 + cr ) 2 1 2a a 2 1 p p2 = 1 2b b2 ´ 1 q q 2 = 2 × 2D1 × 2D2 1 2c c 2 1 r r2 = 8D1D2 = 8 × 1 × 4 = 16 2 26. (4) f '( x) = x - tan x 2 x sin 2 x - x3 5x 2 5 27. (0) The two roots of the equation are 1 ± i 3 , so that we can take æ1 3ö a = 1 + i 3 = 2 çç 2 + i 2 ÷÷ = 2eip/3 and è ø æ1 3ö b = 1 – i 3 = 2 çç 2 - i 2 ÷÷ = 2e–ip/3 è ø ( - sin x 2 1 = 0 + 2(0 + 1) + 2(0 + 1) = 4 i2 p / 3 + e-i2p / 3 a + b = 2, a2 + b2 = 22 e cos x 1 0 + 0 -1 0 = 22 . 2 cos ( a4 + b4 = 24 ( e 2p = – 22 , 3 ) ip -ip a3 + b3 = 23 e + e = – 24 , i4 p / 3 ) + e-i4p / 3 = – 24 ) MATHEMATICS 152 ( 29. (1) We have, éa b ù A= ê ú ëc d û ) i5 p / 3 + e-i5p / 3 = 25 and a5 + b5 = 25 e 2 -22 -2 4 Determinant = -22 -24 -2 4 -2 4 -24 25 1 2 -2 -2 2 = – 28 1 2 -1 -1 28. éa b ù éa b ù Þ A2 = ê úê ú ëc d û ëc d û 1 -2 -22 = – 28 0 6 0 -3 2 1 6 -3 (1 + x) 2 (1 + x )4 (1 + x)6 3 (1+ x)6 (1 + x) 9 (1 + x) 4 (1 + x )8 (1 + x)12 (0) Let f (x) = (1 + x ) Coefficient of 'x' is f ' (0) = 0. é a 2 + bc ab + bd ù ú =ê êë ac + cd bc + d 2 úû Given, A2 = A and ad – bc = 0 é a 2 + bc ab + bd ù é a b ù \ ê ú=ê ú êë ac + cd bc + d 2 úû ë c d û Þ ab + bd = b Þ b(a + d) = b Þ a+d=1 x x2 2 30. (2) –3x4 + 1 x 1 x3 x4 1 x6 =0 2(1 + x )2 4(1 + x )3 6(1 + x) 5 3 f ' (x) = (1+ x ) (1 + x)6 (1 + x)9 (1 + x )4 (1 + x)8 (1 + x )12 Þ –3x4 + x3 (1 + x )2 (1 + x )4 (1 + x )6 Now, apply R1 ® R1 – R2, R2 ® R2 – R3 + 3(1 + x)3 6(1 + x )5 9(1 + x)8 (1 + x )4 (1 + x )8 (1 + x )12 (1 + x )2 (1 + x)2 (1+ x) 6 + (1 + x )3 (1 + x)6 (1 + x )9 4(1+ x)3 8(1 + x)7 12(1 + x)11 Put x = 0, f ¢(0) = 0 1 1 1 1 x x2 1 x2 x4 =0 1- x 1 - x2 2 Þ –3x4 + x3 0 x - x 1 x2 x 2 - x4 0 x4 =0 Þ –3x4 + x3[x2(1 – x2)(1 – x) – x(1 – x)(1 – x2)] = 0 Þ x4[– 3 + (1 – x) (1 – x2) (x – 1)] = 0 x = 0, x = 2 Solutions 153 CHAPTER Continuity and Differentiability 20 1. (b) f is continuous at x = p / 4 , if lim f ( x ) = f (p / 4) . 3. (d) If f (a) = a, then obviously x = a is the solution. Let f (a) > a and g(x) = f (x) – x then g (a) > 0 and g( f (a)) = f ( f (a)) – f (a) = a – f (a) < 0 Since g(x) is continuous, so at least for one c Î(a, f (a)) , g(c) = 0. Similarly we can argue for f (a) < a. 4. (b) x ®p / 4 2 Now, L = lim (sin 2 x ) tan 2 x x ®p / 4 Þ log L = lim tan 2 2x log sin 2x x ®p / 4 log sin 2 x æ ¥ ö = lim ç ÷ x ®p /4 cot 2 2 x è ¥ ø 1 = lim =2 x ®p / 4 - 2 cot 2x cos ec 2x.2 2 or L = e -1 / 2 \ f (p / 4) = e -1 / 2 = 1 / e =e =e æ sin x ö tan x - lim ç è x ÷ø + x®0 =e 1x x ® 0+ - cosec x cot x = e -1´ 0 = 1 sin x ù é x > 0, f '( x ) = esin x ln(| x|) êcos x ln(| x |) + x úû ë f '(1) = 1(0 + sin1) = sin1 f (-1) = - a + 1 Þ a = 1 - sin1 ,x < 0 ì1 ï f (2x) = í1 - [ 2 x ] , x > 0 ï1 + [ 2 x ] î 5. g ( x) = (1 - sin1) x + 1. (b) We apply check for continuity at x = 0 LHL = lim f ( x) = lim f (0 - h) x ® 0- h® 0 ¥ = lim (cos h + sin h) -cosec h ( 1 from ) h ®0 = exp{ lim (cos h + sin h - 1) ´ - cosec h} h® 0 1 2 1 and x = 1 2 f (x) be a discontinuous function. continuous function and for x = lim + Let g (x) = ax + b Þ b = 1 Þ g ( x ) = ax + 1 For ,x < 0 ì1 ï f (x) = í1 - x ,x ³ 0 îï1 + x Þ f (x), for all values of x where x < ln x x ® 0+ cosec x lim sin x ln x = e x®0 \ f (0- ) = g (0) = 1 ì1 - [ x] ï , x ¹ -1 (d) f (x) = í 1 + x and ïî1 , x = -1 ,x< 0 ì1 ï 1 ,0 £ x < ï1 2 ï Þ f ( 2x) = í 1 , £ x £1 ï0 2 ï 1 3 ï,1 £ x < 2 î 3 sin x x ®0 + lim 2 cot 2x 2. f (0+ ) = lim x ì h hö æ 2h = exp í lim ç -2sin 2 + 2sin 2 cos 2 ÷ h ® 0 è ø î ´- 1 ü ï h hý 2sin cos ï 2 2þ MATHEMATICS 154 h höü ì æ sin - cos ï ç ïï 2 2 ÷ ï = e-1 = exp í lim ç ÷ý h ïh®0 ç ÷ï cos è ø ïþ 2 îï 7. (a) f (x) = min {x + 1, | x | + 1} Þ f (x) = x + 1 " x Î R Y h ®0 x ®0 = lim 1/ h e = lim h® 0 e (e -2 / h 3/ h e (0, 1) + e 2/ h + e3/ h ae 2/ h + be3/ h h ®0 3/ h {ae +e -1/ h -1/ h X' + 1) + b} = 1 b é ù e-1/ h = 0ú êëQhlim ®0 û 8. \ For continuity at x = 0 1 e -1 = a = b -1 Þ a = , b = e e 6. y=x+1 y=–x+1 RHL = lim+ f ( x ) = lim f (0 + h) x® k according as k Î (2n, 2n + 1) or k Î (2n + 1, 2n + 2) ( x - 1) n Where g (x) = log cos m ( x - 1) m, n are integers, m ¹ 0, n > 0 \ we get, lim ( x - 1)n x ®1+ log cos Þ lim m ( x - 1) hn h® 0 log cos m h , 0 < x < 2, = -1 = -1 hn = -1 h®0 m(log cos h) [Using L' Hospital's rule] 1 ì 1ï t ï lim (1- sin px) , 2n < x < 2n +1 1 ït®¥ 1+ ï (1+ sin px)t f (x) = í ï (1+ sin px)t +1 2n +1 < x < 2n + 2 ï lim , ï t®¥ (1+ sin px)t +1 ï x = 0,1,2,..... 0, î If k Î I, f (k) = 0, but lim f ( x) = 1 or – 1 Y' Hence, f (x) is differentiable everywhere for all x Î R. (c) As per question, p = left hand derivative of |x –1| at x = 1 Þ p = –1 Also lim g ( x ) = p x ®1+ ìï (1 + sin px )t - 1 üï (b) f (x) = lim í ý t ®¥ ï (1 + sin px )t - 1 ï î þ Q sin p x > 0 (in I and II quadrants) \ 2n p < p x < (2n + 1) p Þ 2n < x < 2n + 1, n Î I and sin px < 0 (in III and IV quadrrants) \ (2n + 1) < p x < (2n + 2) p Þ 2n + 1 < x < 2n + 2, n Î I and sin p x = 0 if x = 0, 1, 2,......... ì 1, 2n < x < 2n + 1 ï = í -1 2n + 1 < x < 2n + 2 ï 0, x = 0,1, 2.......... î X (– 1, 0) Þ lim n h n-1 cos h = -1 h®0 m( - sin h) [Using L' Hospital's rule] Þ lim n hn - 2 cos h = 1 Þ n = 2 and m = 2 sin h ö h® 0 m æç ÷ è h ø (b) Let | f (x) | £ x2, "x Î R Now, at x = 0, | f (0) | £ 0 Þ f (0) = 0 Þ lim 9. f (h ) - f (0) f (h) = lim ...(1) h-0 h® 0 h® 0 h \ f ¢ (0) = lim Solutions 155 f (h) £|h| Now, h 2 (Q | f ( x ) | £ x ) f (h) f (h) ® 0 ...(2) £ | h | Þ lim h® 0 h h (using sandwich Theorem) \ from (1) and (2), we get f ¢ (0) = 0, i.e. - f (x) is differentiable, at x = 0 Since, differentiability Þ Continutity \ | f (x) | £ x2, for all x Î R is continuous as well as differentiable at x = 0. Þ - | h |£ 10. (a) æ xx – x– x ö æ ( x) 2 x – 1 ö f ( x) = cot –1 ç ÷ = cot –1 ç ÷ ç ÷ ç 2xx ÷ 2 è ø è ø æ 2x x ö –1 x = tan –1 çç x 2 ÷ ÷ = –2 tan x + –1 ( ) x è ø \ 11. f ¢( x ) = – 2 1 + ( x x )2 ´ x x (1 + log e x ) –2 f ¢(1) = ´ 1 = –1 \ 2 (c) We have, g = inverse of f = f –1 Þ g(x) = f –1 (x) Þ f [g(x)] = x Differentiating w.r.t. x, we get f ' [g (x)]. g ' (x) = 1 \ g ' (x) = 1 = 1 + [g (x)]3 f '[ g (x)] é ù 1 1 \ f ¢[ g ( x)] = êQ f ¢ ( x) = 3 3ú 1+ x 1 + [ g ( x)] úû êë 12. (b) Since, 2 < x < 3, \ [x] = 2 æ 2p ö f ( x) = sin ç - x 2 ÷ \ è 3 ø \ f ¢( x) = cos çæ 2p - x 2 ÷ö (-2 x) è 3 ø æ 5p ö 5p \ f ¢ç =2 ç 3 ÷÷ 3 è ø 13. (c) Put x n = cos a , y n = cos b Þ Þ Þ a -b = -a 2 cos–1(xn) – cos–1( yn) = 2 tan–1(– a) tan dy x n -1 x n -1 = = n -1 y 1 - y 2 n dx 1 - x2n y n -1 14. (d) Let x3 = cos p and y3 = cos q 3 Given Þ (1 - cos 2 p ) + (1 - cos 2 q ) Þ = a(cos p – cos q) sin p + sin q = a(cos p–cos q) Þ 3 (1 - x 6 ) + (1 - y 6 ) = a (x – y ) æ p + qö æ p - qö 2sin ç cos ç è 2 ÷ø è 2 ÷ø = –2a sin æç è p - qö æ p + qö ÷ sin çè ÷ 2 ø 2 ø Þ 1 æ p - qö tan ç =- Þ è 2 ÷ø a Þ cos–1x3 – cos–1 y3 = tan–1 çè - a ÷ø æ 1ö a p – q = tan–1 çè - ÷ø æ 1ö Differentiate w.r.t. x, we have - 3x2 1 - x6 + dy x 2 1 - y 6 dy = =0 Þ . dx y 2 1 - x 6 1 - y 6 dx 3 y2 Hence, f (x, y) = x2 / y2 15. (b) f ¢ (x) = n (1 + x)n –1, f ¢¢ (x) = n (n – 1) (1 + x)n – 2 \ f n (x) = n!, f n (0) = n! Þ n (n –1) n! + ... + 2! n! = nC0 + nC1 + nC2 + ... + nCn = 2n 1+ n + 16. (c) Given ea cos a = 1 ....(i) b and e cos b = 1 .....(ii) –x Let f(x) = e – cos x, then f(x) is continuous and differentiable. Also, f(a) = f(b) = 0 (from (i) and (ii)) Therefore by Rolle's MVT, f '(x) = 0 has at least one root in (a, b). Þ - e - x + sin x = 0 for at least one x Î ( a, b) 17. (b) We have, f(x) = x|sin x|, x Î R x Î(2 np,(2 n + 1) p) ì x sin x, f ( x) = í î – x sin x, x Î(2n + 1) p , 2pn) f '(np ) = lim x ®nx f ( x ) – f ( np ) x – np x | sin x | x – np Clearly, f(x) is differentiable for all x except x = np, n = ±1, ±2, ±3,.... = lim x ® nx MATHEMATICS 156 18. (c) We have, ì æ pö f ( x ) = í x 2 cos ç ÷ è xø î 0 For x = 0, f (0) = 0 20. (d) for x ¹ 0 R.H.D. sin h 2 -0 f (0 + h) - 0 = lim = lim h =1 h h® 0 h h®0 æ pö h 2 cos ç ÷ – 0 p è hø = lim = lim – h cos = 0 . h x®0 –h x ®0 Simillarly, R f ¢ (0) = 0 Þ differentiable at x = 0. sin h2 -0 f (0 - h) - 0 L.H.D. = lim = lim -h =1 -h h® 0 -h h® 0 R.H.D. = L.H.D. p =0 x p p p cos = 0 Þ = (2 n +1) x x 2 2 2 x +1 f (x) is not differentiable at x= 2 x Î[–1,1]: x = , n ÎZ 2n + 1 19. (d) Let f : R ® R be a continuous function such that f (x2) = f (x3) ...(i) for all x Î R Put x = –x, in eqn. (i), then f (x2) = f (–x3) from (i) f (x3) = f (–x3) Let x3 = t, then f (t) = f(–t) Þ f (x) is an even function Now take x3 = t, then from (i) f (t2/3) = f (t) Put t = t2/3 2 2/3 Þ f (t (2/3) ) = f (t ) (2/3) 2 ) Þ f (t) = f (t 2/3 ) = f (t 3 n = f (t (2/3) ) ... = f (t (2/3) ) This is true for all t Î R and any n Î I. n x2 Now, to check the differentiability of f (x) f (0 – h) – 0 Now Lf ¢ (0) = lim –h x ®0 Q x ®0 x®0 Þ f (x) is continuous p =0. n x®0 Þ f (x) is continuous at x = 0 2 and lim x cos Þ sin x 2 æ sin x 2 ö ç ÷ x = (1)(0) = 0 = f (0) = xlim ®0 è x 2 ø for x = 0 Now, f (x) is not differentiable at cos lim f ( x ) = lim æ 2ö When n ® ¥, ç ÷ ® 0 è 3ø 0 Then f (t) = f (t ) = 1 Hence, f (x) is a constant function and therefore it is differentiable everywhere Þ f (x) is differentiable Now, differentiate f (x) w.r.t. x, we get f ¢(x) = x cos( x 2 )2 x - sin( x 2 ) x2 2 f ¢(x) = 2cos x - sin x 2 x2 2 ì 2 sin x , x¹ 0 ï 2cos x - 2 f ¢(x) = í x ï 1 , x =0 î Now, to check the continuity of f ¢(x) 2 lim f ¢ ( x ) = lim 2cos x 2 - sin x = 2 – 1 = 1 x®0 x ®0 x2 Þ f ¢(x) is continuous 21. (0) f '(x) = lim h®0 | f '( x) | = lim h®0 f ( x + h ) - f ( x) h f ( x + h) - f ( x) (h)2 £ lim h h®0 h Þ | f '( x) | £ 0 Þ f '( x) = 0 Þ f (x) = constant As f (0) = 0 Þ f (1) = 0. Solutions 157 22. (4) Q x = cosec q - sin q Þ 2 x + 4 = (cosec q - sin q) + 4 = (cosec q + sin q)2 ....... (i) and y 2 + 4 = (cosec n q - sin n q) 2 + 4 = (cosec n q + sin n q)2 é æ xö æ xö æ x ö æ x öù F ¢( x) = ê f ç ÷ . f ¢ ç ÷ + g ç ÷ g ¢ ç ÷ ú 2 2 è 2 ø è 2 øû ë è ø è ø Here, g (x) = f ' (x) and g' (x) = f '' (x) = – f (x) 24. (5) 2 ........(ii) æ dy ö ç ÷ dy è dq ø = Now , dx æ dx ö ç ÷ è dq ø x x x x So, F ¢( x) = f çæ ÷ö g çæ ÷ö - f çæ ÷ö g çæ ÷ö = 0 2 2 2 2 è ø è ø è ø è ø Þ F (x) is constant function So, F (10) = 5 25. (2) Given, | f (x) – f (y) | = | x – y | for all in [0, 1] Þ n(cosecn-1q)(-cosec q cot q) - n sin n-1 q cos q = -cosec q cot q - cos q n -1 n Þ | f '( x) – f (y) | =1 | x– y | lim x® y | f (x) – f (y) | = 1Þ | f '(x) | =1 x– y Þ f '( x) = ±1Þ f ( x ) = ± x xÎ[0,1] \ f (x) has exactly 2 function. 26. (3) Taking logarithm of both sides, we get = n(cosec q cot q + sin q cos q) (cosec q cot q + cos q) = n cot q (cosecn q + sin n q) cot q (cosec q + sin q) é æ 1 öù log y = x êlog ç1 + ÷ ú x øû è ë = 2 n(cosecn q + sin n q) n y + 4 = (cosec q + sin q) x2 + 4 Þ [From (i) and (ii)] Squaring both side, we get 2 n 2 ( y 2 + 4) æ dy ö çè ÷ø = dx ( x 2 + 4) 2 æ dy ö ( x 2 + 4) ç ÷ - n 2 y 2 = 4n2 = kn 2 è dx ø Þ k = 4. 23. (1) We have, f (x) = cos x cos 2x cos22 x cos 23 x ..... cos 2n–1x or Þ f ( x) = Þ f '( x) = Þ 1 æ 1ö + log ç 1 + ÷ x +1 xø è Since y (2) = (1 + 1/2)2 = 9/4 =- æ 1 æ 1 öö y2 ( x ) = y1 ( x ) ç + log ç 1 + ÷ ÷ x øø è è x +1 æ 1 x æ 1 öö + y ( x ). ç + ç- ÷÷ è ( x + 1)2 x + 1 è x 2 ø ø sin 2 x 2n cos 2n x sin x - sin 2n x cos x 2n sin 2 x n æ p ö 2 cos 2 f 'ç ÷ = è 2ø 2n n-1 p = cos 2n–1 p = (–1) 2 n – 1 = 1 ......... (1) 3ö æ 1 So, y1 (2) = (9/4) çè - + log ÷ø 3 2 Multiplying (1) by y and then differentiating, we get n 2n sin x 1 x2 æ 1 ö 1ö æ y1 ( x ) = ç÷ + log ç 1 + ÷ y x + 1 è x2 ø xø è So, 3ö æ 1 æ1 1ö y2 (2) = y1 (2) ç - + log ÷ + y(2) ç - ÷ 2ø è 3 è9 6ø 2 3ö 1 æ 9ö æ 1 = ç ÷ ç - + log ÷ è 4ø è 3 2ø 8 MATHEMATICS 158 27. (3) Here, y= So, y = ( x - sin x ) + ( x - sin x) + ...¥ \ y2 = x – sin x + y ( x - sin x) + y Differentiating, we get : 2 y At x = ) dy dy = 1 - cos x + dx dx p 2 p 1 p 3 , y - y = -1 Þ y2 - y + = 2 2 4 2 4 2 1ö 2p - 3 æ Þ (2 y - 1) = ± 2p - 3 Þ çy- ÷ = è ø 2 4 From Eq. (i), we get (2 y - 1) \ dy = dx x= p 2 2p - 3 Þ 2 28. (2) Þ lim f ( x) = x ®0 dy = 1 - cos x dx dx - 2p = 3 dy x= p 2 1 3 x × a cot x + b 1 = x®0 3 x2 Þ lim xa + b tan x 1 lim = x ® 0 x 2 tan x 3 æ ö x3 ax + b ç x + + ...¥ ÷ 3 è ø 1 = Þ lim x®0 3 3 æ tan x ö x ç è x ÷ø æb ö (a + b) x + ç ÷ x3 + ...¥ è3ø 1 = Þ lim 3 x®0 3 x So a + b = 0 Also, b = 1 Þ a = –1 Þ a2 + b2 = 2 ì 1 1 , | sin x |< ï 3 - cos x 2 2 ï 29. (4) f`(x) = í ï 2 + cos x + 1 , | sin x |³ 1 ïî 2 2 ì 1 1 , | cos x |> ï 3 - cos x 2 2 ï =í ï 2 + cos x + 1 , | cos x |£ 1 ïî 2 2 1 1 ì ïï 3 - cos x + 2 , | cos x |> 2 =í ï 2 + cos x + 1 , | cos x |£ 1 ïî 2 2 Thus, f (x) is discontinuous at p 3p 5 p 7 p 1 or x = , , , | cos x |= 4 4 4 4 2 30. (3) Put cos f = 2 3 3 ;sin f = ; tan f = 2 13 13 -1 -1 y = cos {cos( x + f)} + sin {cos( x - f)} -1 = cos {cos( x + f)} + p - cos -1{cos(f - x )} 2 p -f+ x 2 p y = 2 x + ; z = 1 + x2 2 dy dy dx 2 1 + x 2 Now, = = dz dz x dx 10 æ dy ö \ç ÷ = 3 è dz ø x= 3 = x+f+ 4 Solutions 159 CHAPTER Application of Derivatives 21 1. 2. (d) Given f (x) = tan –1 (sin x + cos x) 1 .(cos x - sin x ) f '(x) = 1 + (sin x + cos x) 2 1 æ 1 ö 2. ç cos x sin x÷ è 2 ø 2 = 2 1 + (sin x + cos x ) p p æ ö çè cos .cos x - sin .sin x÷ø 4 4 = 1 + (sin x + cos x) 2 pö æ 2 cos ç x + ÷ è 4ø \ f '(x) = 1 + (sin x + cos x ) 2 if f ' (x) > 0 then f (x) is increasing function. p p p Hence f (x) is increasing, if - < x + < 2 4 2 3p p Þ <x< 4 4 æ p pö Hence, f (x) is increasing when n Î ç - , ÷ è 2 4ø (b) If f ( x ) = x1/ x , then 1 f ' ( x ) = 2 é x1/ x (1 - ln x ) ù û x ë f is decreasing if x > e and f is increasing if x < e. As e < 3 <4 < 5 < 6 < 7 \ i.e. 3( x 2 - 6 x + 8)e < 1 i.e. x2 – 6x + (8 – 1/3e) < 0 i.e. ( x - (3 + 1 + 1/ 3e ))( x - (3 - 1 + 1/ 3e ) < 0 Û 3 - 1 + 1/ 3e < x < (3 + 1 + 1/ 3e ) Hence x Î (4, 3 + 1 + 1/ 3e ) Similarly if x < 2, then f ' (x) < 0, If log 3(x2 – 6x + 8) e > 0 i.e., x < 3 - 1 + 1/ 3e or x > 3 + 1 + 1/ 3e 4. -(4a2 + 8a - 14) > 0"x Î R But 21 / 2 = 41 / 4 < 31 / 3 3. 5. + (6 x - 18) log( x 2 - 6 x + 8) + (3 x 2 - 18 x + 24)(2 x - 6) x2 - 6x + 8 = 6 ( x - 3 ) éê log 3 + log x 2 - 6 x + 8 + 1ùú ë û ( ( = 6 ( x - 3) log 3e x 2 - 6 x + 8 ) ) For f (x) to be defined x 2 - 6 x + 8 > 0 Þ x < 2 or x > 4 If x > 4 then f '( x) < 0 if log 3( x 2 - 6 x + 8)e < 0 4cos2 x - 8(a + 1)cos x - (4a2 + 8a -12) > 0 Þ cos 2 x - 2(a + 1) cos x - ( a 2 + 2a - 3) > 0 (cos x - a )(cos x - b) > 0 where a = (a + 1) + 2a 2 + 4a - 2 and b = (a + 1) - 2a 2 + 4a - 2 Þ cos x - b < 0 since cos x - a < 0 "a > 0 Þ (a + 1) - 2a 2 + 4a - 2 > cos x "x Î R Þ (a + 1) - 2a 2 + 4a - 2 > 1 Þ 2 2 a 2 + 4a - 2 < a Þ a + 4 a - 2 < 0 a Î (-2 - 6, 6 - 2) Þ \ Max {1, 21 / 2 } = 21 / 2 \ the greatest number is 31 / 3 (a) f '( x) = 2 x log 27 - 6 log 27 Þ Þ Max {31/ 3 , 41/ 4 ,51/ 5 , 61/ 6 , 71/ 7 } = 31/ 3 Also 1 < 2 < e Hence x Î (3 - 1 + 1/ 3e , 2) (a) Here f ¢ ( x) > 0"x Î R Þ 2cos 2 x - 8(a + 1)cos x Hence a Î (0, 6 - 2) since a > 0. x ,0 < x £ 1 (c) We have f (x) = sin x sin x - x cos x Þ f ¢ ( x) = sin 2 x where sin2 x is always +ve, when 0 < x £ 1 . But to check Nr., we again let h (x) = sin x – x cos x Þ h ¢( x ) = x sin x > 0 for 0 < x £ 1 Þ h (x) is increasing Þ h (0) < h (x), when 0 < x £ 1 Þ 0 < sin x – x cos x, when 0 < x £ 1 Þ sin x – x cos x > 0, when 0 < x £ 1 Þ f ' (x) > 0, x Î (0,1] Þ f (x) is increasing on (0, 1] x Again g ( x) = tan x MATHEMATICS 160 Þ g ¢( x) = tan x - x sec 2 x , when 0 < x £ 1 tan 2 x Here tan2 x > 0 But to check Nr. we consider p (x) = tan x – x sec2 x p ¢ ( x ) = sec 2 x - sec2 x - x.2 sec x.sec x tan x 2 6. 7. Þ p ¢( x ) = -2 x sec x tan x < 0 for 0 < x £ 1 Þ p (x) is decreasing, when 0 < x £ 1 Þ p (0) > p (x) Þ 0 > tan x – x sec2 x \ g ¢( x) < 0 Hence g (x) is decreasing when 0 < x £ 1 . p -1 u (c) Given that g(u) = 2 tan (e ) 2 p -1 \ g(–u) = 2 tan ( e - u ) 2 1 p æ ö -1 = 2 tan ç u ÷ èe ø 2 p ép ù p -1 u = 2cot ( e ) - = 2 ê - tan -1 ( eu ) ú 2 ë2 û 2 p p -1 u -1 u = p - 2 tan (e ) - = - 2 tan (e ) = – g(u) 2 2 \ g is an odd function. 2eu Also g '(u) = > 0 , " u Î (– ¥, ¥) 1 + e 2u \ g is strictly increasing on (–¥, ¥) . x2 in the (b) Consider the function f (x) = 3 x + 200 interval [1, ¥). Since the derivative f ¢(x) = x(400 – x3 ) 3 2 is ( x + 200) positive at 0 < x 3 400 and negative at x > 3 400 , the 8. function f(x) increases at 0 < x < 3 400 < 8 it follows that the largest term in the sequence can be either a7 or a8. Since a7 = 49/534 > a8 = 8/89, the largest 49 term in the given sequence is a7 = . 543 (c) We have, x2 + 1 f ( x) = , for all x Î [1, 4) [ x] ì ï x 2 + 1, for all x Î[1, 2) ï 2 ï x +1 , for all x Î[2, 3) Þ f ( x) = í ï 2 ï x2 +1 , for all x Î[3, 4) ï î 3 9. ì ï2 x, for all x Î[1, 2) ï for all x Î[2,3) Þ f ¢ ( x ) = í x, ï 2x ï , for all x Î[3, 4) î3 Clearly, f ¢ (x) > 0 in each of the intervals (1, 2), (2, 3) and (3, 4). So, f (x) is increasing in each of these intervals. It is to note here that f (x) is not increasing on [1, 4] because values of f (x) in (1.5, 2) are greater than the values of f (x) in (2, 3). p (b) We have, A + B = 3 p 3 - tan A - A Þ tan B = 3 1 + 3 tan A Let Z = tan A. tan B. Then, \B = Z = tan A. Þ Z= Þ 3 - tan A 1 + 3 tan A 3x - x 3 tan A - tan 2 A = 1 + 3 tan A 2 1 + 3x , where x = tan A dZ ( x + 3 ) ( 3x - 1) =dx (1 + 3x ) 2 For max Z, dZ 1 =0Þ x = , - 3. dx 3 x ¹ - 3 because A+B = p/3 which implies that x = tan A > 0. It can be easily checked that d 2Z dx 2 < 0 for x = for x = 1 3 1 3 . Hence, Z is maximum i.e. tan A = 1 3 or A = p / 6. 1 . 3 f '( x) = (4a - 3) + ( a - 7) cos x For this value of x, Z = 10. (a) For non-existence of critical point f '( x) ¹ 0 for any x Î R . Þ cos x ¹ 3 - 4 a for any x Þ 3 - 4a > 1 a-7 a-7 Þ | 4a - 3 | - | a - 7 |> 0 (i) a ³ 7 Þ 4a - 3 - a + 7 > 0 Þ a > - Hence a ³ 7 4 . 3 Solutions 161 3 £ a < 7 Þ 4a - 3 + a - 7 > 0 4 Þ 5a > 10 Þ a > 2 Þ 2 < a < 7 3 (iii) If a < Þ 3 - 4a + a - 7 > 0 4 4 Þ a<3 4ö æ \ a Î ç -¥, - ÷ È (2, ¥) 3ø è 11. (a) Let H be the height of the cone and a be its semi vertical angle. Suppose that x is the radius of the inscribed cylinder and h be height then h = QL = OL – OQ = H – x cot a V= volume of cylinder = px 2 ( H - x cot a) 1 Also p = p(H tan a) 2 H ...(i) 3 dV = p(2 Hx - 3 x 2 cot a ) dx O Y (ii) If Q L So, a x P M 2 dV = 0 Þ x = 0 , x = H tan a; 3 dx d 2V dx 2 x = 2 H tan a = -2pH < 0 3 So, V is maximum when x = 2 H tan a and 3 4 1 q = Vmax = p H 2 tan 2 a H 9 3 4 p3 p tan 2 a 4 = = p [ from (i)] 27 p tan 2 a 9 Hence p : q = 9 : 4 2 12. (a) The graph of y = 2 - x + 5 x + 6 is drawn in the adjacent figure. Clearly f (x) will have maxima at x = –2 only if a 2 + 1 ³ 2 Þ | a | ³ 1 2 –3 X O –2 13. (b) f '( x ) = sin x cos x(3sin x + 2l) For maximum or minimum 2l 3 æ 2l ö p p \ Critical pointsin æç - , ö÷ arex = 0, sin -1 ç ÷ 2 2 è 3 ø ø è One of these is point of minima and other is point 2l <1 of maxima, provided - 1 < 3 3 3 Þ – <l< 2 2 But if l = 0 , then sinx = 0, which gives only one critical points f '(x) = 0 Þ sin x = 0 or cos x = 0 or sin x = - æ 3 3ö \ l Î ç - , ÷ - {0} è 2 2ø p 14. (a) For 0 < x £ ; [cos x] = 0 2 æ pù Hence, f (x) = 1 for all çè 0, ú 2û æ pö Trivially f (x) is continuous on çè 0, ÷ø 2 This function is neither strictly increasing nor strictly decreasing and its global maximum is 1. 15. (c) Let R and S be the positions of men P and Q at any time t. Since velocities are same \ OR = OS = x (say) and given dx = v and let SR = y dt S x Q 45° O P y x R Now in triangle ORS, applying cosine rule, we get y2 = x2 + x2 – 2x . x cos 45º = 2 x 2 - x 2 2 \ y = x (2 - 2) MATHEMATICS 162 dy dx = { (2 - 2)} = v (2 - 2) dt dt Hence the required rate at which they are Dividing (1) by (2) being separated is v 2 - 2. 3( cos q + sin q ) (cos q - sin q) dy = dx eq ( sin q + cos q ) \ 16. (a) Here y = 2 dy 3(cos 2 q - sin 2 q) 3cos 2q = q = q dx e (sin q + cos q) e (sin q + cos q) 2 c dy c Þ =x+a dx + ( x a)2 Slope of normal is Þ ( x + a) 2 c2 > 0 (for all x) \ x cos a + y sin a = p is normal if, cos a > 0 or cot a < 0 sin a i.e., a lies in II or IV quadrant. So, p 3p æ ö æ ö a Î ç 2 np + , (2n + 1)p ÷ È ç 2np + , (2n + 2)p ÷ 2 2 è ø è ø 17. (c) Any point on x2 – y2 = a2 is (a sec q, a tan q) This point is nearest to y = 2x if the tangent at this point is parallel to y = 2x dy x Now, x 2 - y 2 = a 2 Þ = dx y æ dy ö = cosec q ç ÷ è dx ø(a sec q ,a tan q ) cosecq = 2 Þ q = p 6 p pö æ Hence the point is ç a sec , a tan ÷ i.e. 6 6ø è 18. æ 2a a ö , ç ÷ , Clearly they lie on the line 2y = x è 3 3ø (c) Given, y = 3 sin q.cos q dy = 3[sin q(- sin q) + cos q(cos q)] dq dy = 3[cos 2 q - sin 2 q] = 3 cos 2q ...(1) dq and x = eq sin q dx = eq cos q + sin q eq dq dx = eq (sin q + cos q) dq Given tangent is parallel to x-axis then 0= - Þ dy 3(cos q - sin q) = dx eq ...(2) dy =0 dx 3(cos q - sin q) eq or cos q – sin q = 0 Þ cos q = sin q p tan p Þ q= Þ tan q = 1 Þ tan q = 4 4 19. (c) We have, y3 – 3xy + 2 = 0 dy æ dy ö Þ 3y2 - 3 ç x + y÷ = 0 è dx ø dx dy y = 2 dx y - x If the tangent is parallel to x-axis, then dy y =0Þ 2 = 0 Þ y = 0. dx y -x But, y = 0 does not satisfy equation (i). So, there is no point on the curve where tangent is parallel to x-axis. Therefore, H = f. 20. (a) We have, 1 p(x) > x2, p(0) = 0, p”(0) = 2 Let g(x) = p(x) –x2 g(x) > 0, " x ¹ 0 and g(0) = p(0) – 0 = 0 Þ x = 0 should be minima. \ g¢¢(x) should be ³ 0 at x = 0 Now, g¢(x) = p¢(x) – 2x g¢¢(x) = p¢¢(x) – 2 1 3 g¢¢(0) = p¢¢(0) – 2 = – 2 = – 2 2 But g¢¢(0) ³ 0 Þ No polynimial exists. 21. (40) Let the speed of the train be v and distance to be covered be s so that total time taken is s/v hours. Cost of fuel per hour = kv2 (k is constant) 3 Also 48 = k. 162 by given condition \ k = 16 Þ Solutions 163 3 2 v . 16 Other charges per hour are 300. Total running cost, 3s 300 s ös æ 3 C = ç v 2 + 300 ÷ = v + 16 v øv è 16 \ Cost to fuel per hour 24. (9.22) Equation of parabola y = x2 – 4 and 2y = 4 – x2 dC 3s 300 s = - 2 = 0 Þ v = 40 dv 16 v 2 d C dy x + y + 1 = dx x + y - 1 a+b dt = = ¥ when a + b = 1. dx (a,b ) a + b - 1 23. (260) 120° O B Let OA = x km, OB = y km, AB = R (AB)2 = (OA)2 + (OB)2 – 2 (OA) (OB) cos 120° æ 1ö R2 = x2 + y2 – 2 xy ç - ÷ = x 2 + y 2 + xy è 2ø ...(1) R at x = 6 km, and y = 8 km R = 62 + 82 + 6 ´ 8 = 2 37 Differentiating equation (1) with respect to t dx dy æ dy dx ö dR = 2 + 2y + ç x + y ÷ dt dt è dt dt ø dt 1 [2 ´ 8 ´ 20 + 2 ´ 6 ´ 30 + (8 ´ 30 + 6 ´ 20)] 2R dR 1 260 k [1040] = = = dt 2 ´ 2 37 37 37 = A B y= 4 - x2 2 æ 4 – h2 ö Let point B = ç h, ÷ è 2 ø æ 4 – h2 ö A = ç – h, C = (h, h2 – 4) ÷ è 2 ø D = (–h, h2 – 4) \ Area of rectangle ABCD æ 4 – h2 ö = AB × BC = 2h × ç – h 2 + 4÷ è 2 ø Þ A 2R C 600 s = > 0 \ v = 40 results in minimum dv 2 v3 running cost 22. (1) Given x + y – ln (x + y) = 2 x +5 dy 1 æ dy ö Þ 1+ ç1 + ÷ = 2 dx x + y è dx ø Þ y = x2 – 4 D A = 12h – 3h3 Þ dA = 12 – 9h2 dh For maxima of minima dA = 0 Þ 12 – 9h2 = 0 dh 2 2 h= ± \ maximum at h= 3 3 3 24 8 æ 2 ö æ 2 ö – \ A = 12 ç ÷ – 3 ç ÷ = è 3ø è 3ø 3 3 16 ´173 16 16 3 = 9.22 = ÞA= 3 3 3 25. (2) Q P(x) is non -zero polynomial and P(1 + x) = P(1 – x) for all x Now, differentiate w.r.t. x, we get P¢(1 + x) = –P¢(1 – x) Put x = 0, P¢(1) = –P¢(1) Þ P¢(1) = 0 and P(1) = 0 Þ P(x) touch x-axis at x = 1 \ P(x) = (x – 1)2 Q(x) Since, m is the lergest integer such that (x – 1)m divides polynomial P(x). Þ m = 2 such that (x – 1)m divides P(x) for all such P(x) = MATHEMATICS 164 26. (2) f '( x) = (b 2 - 3b + 2)(-2 sin 2 x) + b - 1 ¹ 0 for any x Î R Þ (b - 1){1 - (b - 2)(2sin 2 x)} ¹ 0 Þ b ¹ 1 and 1 æ 3 ö æ 5ö > 1 Þ b Î ç , 2 ÷ È ç 2, ÷ 2(b - 2) è2 ø è 2ø When b = 2, f (x) = x + sin 3 Þ f '( x) = 1 ¹ 0 æ 3 5ö \ b Î ç , ÷ Þ integral value of b = 2 è 2 2ø 27. (1) Equation of given curves are 2 2 ax + by = 1 2 ...(i) 2 from (i) and (ii), ax 2 + by 2 = a1x 2 + b1 y 2 or (a - a1 ) x 2 = (b1 - b) y 2 b -b = 1 2 a - a1 y ...(vi) x2 y2 x22 + x12 + x1 x2 = 3 x12 Þ 2 x12 - x1 x2 - x22 = 0 Þ x2 = -2 x1 Similarly x3 = -2 x2 = 4 x1 x4 = -2 x3 = -8 x1 .................................... x2n = -22n -1 x1 = ( x13 + x23 + x33 + ......... + x2 n3 ) + 2n x3 (( -8) 2 n - 1) + 2 = - 1 (82 n - 1) + 2n -8 - 1 9 i.e. y1 + y2 + y3 + .... + y2n = x13 3 æx ö æ 2n ö = - ç 1 ÷ (26 n - 1) + ç ÷ , è 3 ø è 3ø Now put n = 30 and x1 = 2 then y1 + y2 + y3 + ...... + y60 = - æç 2 ö÷ è3ø dy = x 2 Þ Equation of tangent at dx P1 ( x1 , y1 ) is y - y1 = x12 ( x2 - x1 ) 3 ( 2180 - 1) + 20 æ 2183 - 8 ö 2183 - 8 + 20 or S + ç ÷ = 20 è 27 ø 27 Þ 5k = 20 Þ k = 4 29. (9) Sectorial area AOB is removed and the remaining part be folded into a cone of height h and radius r. ArcAB ArcAB = = Arc AB Q q = Angle = radius 1 Þ S =- A I O I q q I in (v), we get aa1 æ b1 - b ö ç ÷ = -1 bb1 è a - a1 ø a - a1 b - b b - b1 1 1 1 1 or or - = =- 1 = a1 a b1 b aa1 bb1 bb1 28. (4) Þ A x2 Putting the value of x23 - x13 = 3x12 ( x2 - x12 ) 3( y1 + y2 + y 3 ........ + y2n ) and a1 x + b1 y = 1 ...(ii) dy dy a x From (i) 2ax + 2by =...(iii) =0\ dx b y dx dy From (ii) 2a1 x + 2b1 y = 0, dx a x dy \ =- 1 ...(iv) dx b1 y Curve (i) and (ii) will cut each other at right dy angles if the product of the values of for dx the two curves is –1 æ a x ö æ a1 x ö aa1 x 2 i.e., ç = -1 ÷ = -1 or ÷ç bb1 y 2 è b y øè b1 y ø ...(v) \ i.e. O q I B circular sheet r B after removed part 2 2 2pr = 2p - q and r + h = 1 \ volume of cone 1 1 V = pr 2 h = pr 2 1 - r 2 3 3 p2 r 4 (1 - r 2 ) 9 2 4 p (r - r 6 ) Let, y = V 2 = 9 Þ V2 = I h conical vessel Solutions 165 The point P (-(a - 2), 6) lies on it. dy p2 = (4r 3 - 6r 5 ) and dr 9 \ \ d2y p2 (12r 2 - 30r 4 ) = 9 dr 2 For max or min of y, æ2ö è3ø dy =0 dr or \ r= ç ÷ Then, 2 d y dr 2 r = 2 / 3 = or 2 p æ .2 4ö 16 2 çè 12 - 30. ÷ø = - p 9 3 9 27 b 2 = 1 2 .(1) .q 2 A (–a, 0) or b2 = \ 2 3 = 3+ 6 = m + n = 2 2 113 3 1+ 1 x2 a2 + ( a - 2) 2 = a2 4(a - 1) a2 ( a - 1) a2 C (0,b) y2 b2 D O X B (a, 0) 9a2 1 ü ì = 9 ía + 1 + ý (a - 1) - 1þ a î d (b) 2 1 ïü d 2 (b 2 ) 18 ïì = 9 í1 = and ý 2 2 da da (a - 1)3 îï ( a - 1) þï For extremum value of b, db 2 =0 da 9(2)2 Þ b = 6m (2 - 1) Hence greatest height of the arch is 6 m. Þ we get a = 2 \ b2 = m + n = 3 + 6 = 9. 30. (6) Equation of the ellipse is =1 6 Also, A = p =1 CHAPTER Indefinite Integration 22 sin( x - a + a) sin x ò sin( x - a) dx = ò sin( x - a) dx 1. (b) 2. sin( x - a ) cos a + cos( x - a)sin a = ò dx sin( x - a ) = ò {cos a + sin a cot( x - a )}dx = (cos a) x + (sin a) logsin( x - a) + C \ A = cos a, B = sin a (d) Let gn ( x ) 2 9 = 1- b2 2 æ 2ö = p(1 - r ) = p çç1 ÷ [from (1)] 3 ÷ø è Þ b 2 36 + (– (a–2),6) P 1 1 .q = (2p - 2pr ) 2 2 A \ = A1 36 ....(1) \ Removed sectorial area = A1 = a2 Y \ y is maximum and hence V is also maximum æ 2ö at r = ç ÷ è 3ø ( a - 2) 2 4 = 1 + x + x + .... + x 2n = = hn ( x ) = g 'n ( x ) = ( x 2 - 1)2 Now f ( x) = lim hn ( x ) = n®¥ Thus ò f ( x )dx = ò 2x ( x - 1) 2 2x 2 ( x - 1) 2 2 as 0 < x <1 dx 1 1 = +c = - 2 x - 1 1 - x2 x2n+ 2 - 1 x2 - 1 So, 2 x + 4 x 3 + ..... + 2nx 2 n-1 2 x (nx 2 n + 2 - (n + 1) x 2 n + 1) 3. (c) æ f ( x) g '( x) - f '( x) g ( x ) ö æ g ( x ) ö ÷ ln ç ÷ dx f ( x ) g ( x) ø è f ( x) ø ò çè MATHEMATICS 166 Let Þ 4. g ( x) =t f ( x) f ( x ) g '( x) - g ( x) f '( x ) 1 + xn x Here, f ( x) = f '( x ) = 8. 1 + xn , then 1 - xn nx n -1 (1 - x n ) 1 - x 2 n and ò e ( f ( x) + f '( x ))dx = e x f ( x) + C x (d) lim n 2 ( x1/ n - x1/(n +1) ) n®¥ = lim 1 h® 0 h 2 ( x h h h - x +1 ) æ æ h ö ö h 1 è èç h - h +1ø÷ ø h +1 = lim -1 x x h® 0 h 2 2 æ h ö ç x h +1 - 1÷ æ 1 ö h = lim ç ÷ .ç ÷ x h +1 = ln x.1.1 h®0 ç h2 ÷ è h + 1ø çè ÷ h +1 ø \ f ( x) = ln x So, I = ò xf ( x)dx = ò x ln xdx 6. 7. x2 x2 1 x2 x2 ln x = ln x - ò . dx = +C 2 2 x 2 4 x 2n - 1 (a) f ( x ) = lim 2n = -1 (Q 0 < x < 1) , n® ¥ x +1 so ò sin -1 x( f ( x))dx = - ò sin -1 x dx = - [x sin -1 x + 1 - x 2 ] + C x æ ln aa x /2 ln bb ö (b) I = ò x ç 5 x /2 3 x + 2 x 4 x ÷dx ç 3a b 2a b ÷ø è dx Then t ln a 2b3dx = dt . Therefore, 1 ln t I =ò dt 2 3 2 6 ln a b t 1 -1 ö æ - ln t = - ò 2 dt ÷ 2 3ç 6ln a b è t t ø 1 æ ln et ö =ç ÷+k 6 ln a 2 b3 è t ø æ ln a 2 x b3 x e ö 1 =÷+k 2 3ç 6ln a b çè a 2 x b3 x ÷ø +C 1 - xn ln a2 x b3 x 6a 2 x b 3 x Let a 2 x b3 x = t . dx = dt ; ( f ( x )) 2 ln t (ln t )2 dt = +C ò t 2 1 + nx n -1 - x 2 n (c) We have ò e x dx (1 - x n ) 1 - x 2 n é 1 + xn ù nx n -1 x údx + = òe ê ê 1 - x n (1 – x n ) 1 - x 2 n ú ë û =e 5. =ò I n = ò (sin x + cos x)n dx (c) (sin x + cos x) n -1 2( n - 1) (sin x - cos x) + I n -2 n n t4 t4 8 Þ I5 = (sin x - cos x) + I3 = (sin x - cos x ) 5 5 5 2 æ 8 t 4 ö + ç (sin x - cos x) + I1 ÷ 5 çè 3 3 ÷ø t4 8 = (sin x - cos x ) + t 2 (sin x - cos x ) 5 15 32 + (sin x - cos x) + C 15 (sin x - cos x ) 4 = (3t + 8t 2 + 32) + C ; 15 where t = (sin x + cos x). 9. (b) Let I be the given integral. We can see that = cos q + sin q ö the derivative of loge æç ÷ is 2 sec 2q, è cos q - sin q ø so that using integration by parts we have I= sin 2q æ cos q + sin q ö loge ç ÷ 2 è cos q - sin q ø -ò = sin 2q × (2 sec 2q)d q 2 sin 2q æ cos q + sin q ö log e ç ÷ - ò tan 2q d q 2 è cos q - sin q ø = sin 2q æ cos q + sin q ö 1 log e ç ÷ - log e (sec 2q) + c 2 è cos q - sin q ø 2 Therefore f ( x) = log e (sec 2q). Solutions 167 10. (d) We have, 1 ò x [log ex e × loge2 x × loge3 x e] dx 1 dx =ò x loge ex log ee 2 x.loge e 3 x =ò Put tan x = t =- 1 x(loge e + loge x)(log e e2 + loge x) dx (loge e3 + loge x) d (loge x) =ò (1 + loge x)(2 + log e x)(3 + log e x) 1 =ò dt , where t = log x e (1 + t )(2 + t )(3 + t ) 1 1 1 ö æ1 1 = òç × + × ÷ dt è 2 1+ t 2 + t 2 3 + t ø 1 = log | 1 + loge x | - log | 2 + log e x | 2 1 + log | 3 + loge x | +C 2 1 = log{log e ex} - log{loge e 2 x} 2 1 + log{log e e 3 x} + C 2 11. (a) I = ò =- t 2 -1 ò (t 2 + 2t + 1) t 3 + t 2 + t dt òæ 1ö 1 çt + 2 + ÷ t +1+ t t è ø 2du ò 1 + u 2 = -2 tan-1 u + c, 1 tan x b sin q a 2ax dx = 2b sin q cos qd q b 2b sin q cos q dq = cos qd q Þ dx = a 2ax b cos qd q b 2 ö a æ çè a + b. sin q÷ø . b cos q a ò I= = a = a ( x - b) ( x - a)(b - x ) dq ò (a 2 + b 2 sin 2 q) sec 2 qd q ò (a 2 sec2 q + b2 tan 2 q) Putting tan q= z Þ sec 2 qd q = dz then dz I= a 2 a (1 + z 2 ) + b 2 z 2 ò = a ( x - b) = ( a - b) sin 2 q 2(a - b) sin q cos q dq (a - b) sin 2 q (b - a ) sin q cos q = 2 dq 2 = cos ec 2 q dq ò 2 b - a sin q b - a ò 2 2 (- cot q) + c = cot q + C = b-a a -b Now x = a sin 2 q + b cos 2 q = = Þ x(1 + cot 2 q) = a + b cot 2 q dz ò (a 2 + b 2 ) z 2 + a 2 a (a + b ) ò 2 2 2 a a 2 + b2 \I= Þ x cos ec 2 q = a + b cot 2 q 12. (a) I = - ò 1 t 2 dt , put 1 + t + = u 13. (c) Let ax 2 = b sin 2 q Þ x = dx x-a 2 ; \I= a -b b-x 1 t2 where u = 1 + tan x + Also, ( x - a) = (b - a ) cos 2 q \cot q = 1- \ I=- Put x = a sin 2 q + b cos 2 q dx = 2(a - b) sin q cos q dq \I = ò (t - 1) ò (t + 1) t 3 + t 2 + t dt I =- z + dz a2 2 (a + b2 ) æ z a2 + b2 ö a 2 + b2 ÷+c tan -1 ç a ç ÷ a . è ø æ z a 2 + b2 ö tan -1 ç ÷ + c, 2 2 çè a ÷ø a(a + b ) 1 where z = tan q . x-a +C b- x (tan x - 1) sec 2 xdx (tan x + 1) tan 3 x + tan 2 x + tan x dx 14. (b) Given ò f ( x ) sin x cos xdx 1 = ln f ( x) + c 2 2( b - a 2 ) MATHEMATICS 168 Differentiating both sides w.r.t.x then 1 f '( x ) f ( x) sin x cos x = . 2 2 2(b - a ) f ( x ) f '( x) Þ 2(b 2 - a 2 ) sin x cos x = { f ( x )}2 Þ 2b 2 sin x cos x - 2 a 2 sin x cos x = +(n - 1) ò (sin x + cos x)n- 2 [2 - (sin x + cos x) 2 ] dx f '( x) { f ( x )}2 Integrating both side w.r.t. x we get 1 -b cos x - a sin x = f ( x) 1 or f ( x) = 2 2 ( a sin x + b 2 cos2 x ) sec6α sec18α sec54α 15. (a) We have + + cosec2α cosec6α cosec18α sin 2a sin 6a sin18a = + + cos 6a cos18a cos 54a sin 2a 1 sin 4a = Now, cos 6a 2 cos 2a cos 6a 1 é sin 6a cos 2a - cos 6a sin 2a ù = ê ú 2ë cos 2a cos 6a û 1 = (tan 6a - tan 2a ) 2 sin 6a 1 Similarly, = (tan18a - tan 6a ) cos18a 2 and sin18a = 1 (tan 54a - tan18a ) cos 54a 2 1 Thus integral = ò (tan 54a - tan 2a ) dx 2 1 é l n | sec54a | l n | sec 2a | ù = ê úû + c 2ë 54 2 16. (b) Let I n = ò (sin x + cos x ) n -1 (sin x + cos x ) dx 2 2 2 2 Take, u = (sin x + cos x )n -1 and dv = (sin x + cos x )dx so that, v = sin x - cos x Therefore, using integration by parts we have I n = (sin x + cos x)n -1 (sin x - cos x) -ò (n - 1)(sin x + cos x) n- 2 ´ (cos x - sin x )(sin x - cos x ) dx = (sin x + cos x )n -1 (sin x - cos x ) +(n - 1) ò (sin x + cos x)n -2 (cos x - sin x) 2 dx = (sin x + cos x )n -1 (sin x - cos x ) = (sin x + cos x )n -1 (sin x - cos x ) +2(n - 1) I n -2 - (n - 1) I n Therefore, I n + (n - 1) I n - 2(n - 1) I n- 2 = (sin x + cos x )n -1 (sin x - cos x ) Þ nI n - 2( n - 1) I n- 2 = (sin x + cos x )n -1 (sin x - cos x ). 17. (d) We have, f ' (x) = ln(x2 + 1) f (x) = ò ln ( x 2 –1) dx 2x2 f (x) = xln(x2 – 1) – ò 2 dx x -1 æ x 2 –1 1 ö ÷ dx è x –1 x –1ø f (x) = xln(x2 – 1) – 2ò ç 2 + 2 æ x –1ö f (x) = xln(x2 – 1) – 2x – ln ç +C è x + 1÷ø æ 1ö f (2) = 2ln(3) – 4 –ln çè ÷ø + C = 0 [Q f (2) = 0] 3 Þ C = 4 – 3ln3 ( x –1) + 4 – 3ln 3 \ f (x) = xln(x2– 1) – 2x – ln ( x +1) \ the above function is defined for S infinite C values possible in set S such that f ¢ (x) = ln(x2 – 1) 1/ x f ( x) ö æ 18. (1) We have lim ç1 + x + ÷ x ø x ®0 è = e3 1/ x æ f ( x) ö ö æ Þ lim ç1 + x ç1 + ÷÷ x ®0 è è x2 ø ø = e3 x é ù ê æ x 2 + f ( x) ö x 2 + f ( x ) ú Þ lim êç 1 + ÷ ú ÷ x ®0 êç x ú è ø êë úû 2 x + f ( x) Þ = 3 Þ f ( x) = 2 x2 x2 Therefore 2 ò f ( x)log e x dx = 2ò x loge x dx x2 + f ( x) 1 x x = e3 é x3 x3 1 ù = 2 ê loge x - ò × dx ú (By Parts) 3 x úû êë 3 2 2 2 æ 1ö = x3 log e x - x3 + c = x 3 ç log e x - ÷ + c 3 9 3 è 3ø Þ a+b=1 Solutions 169 19. (0.60) Let I = ò =ò (cos 2 x + sin 2 x ) (2 cos x - sin x) 2 dx dx (2 cos x - sin x )2 Integrating by part , taking cos x as the first (cos x + 2 sin x ) as the second function, and (2 cos x - sin x ) 2 we have 1 - sin xdx ì ü = cos x í ý-ò (2cos x - sin x) î 2 cos x - sin x þ 1 - sin x dx ì ü = cos x í ý+ò (2cos x - sin x) î 2 cos x - sin x þ cos x = (2 cos x - sin x) 1 2 - (2 cos x - sin x ) - ( -2 sin x - cos x) 5 5 dx +ò (2 cos x - sin x ) d é g ù êë N = l Dr + m dx Dr úû cos x 1 = (2 cos x - sin x) 5 2 ( -2 sin x - cos x) ò dx - 5 ò 2 cos x - sin x dx cos x 1 = - x (2cos x - sin x ) 5 2 - ln | 2 cos x - sin x | +c 5 Þ |a + b| = 0.60. 1 - (cot x)2010 20. (2012) Let I = ò dx tan x + (cot x) 2011 =ò (sin x )2010 - (cos x ) 2010 (sin x ) =ò 2010 × (sin x ) 2011 cos x (sin x ) 2012 + (cos x ) 2012 [(sin x)2010 - (cos x)2010 ]sin x cos x (sin x )2012 + (cos x )2012 dx dx Put t = (sin x) 2012 + (cos x )2012 . Then dt = [2012(sin x )2011 cos x +2012(cos x )2011 (- sin x )]dx = 2012[(sin x)2010 - (cos x )2010 ](sin x cos x) dx Therefore 1æ 1 ö 1 I =ò ç log e | t | + c ÷ dt = t è 2012 ø 2012 1 = log e | (sin x) 2012 + (cos x) 2012 | + c 2012 q 2 dq q cos3 q + cos2 q + cos q 2 q qö æ 2q çè 2sin .cos ÷ø .sin 2 2 2 dq =ò 2q 3 2 2 cos cos q + cos q + cos q 2 q 2 sin 2 sin q d q 2 =ò 2q 3 2 cos cos q + cos2 q + cos q 2 Put cos q = t Þ - sin q d q = dt q q Also cos q = 2cos 2 - 1 = 1 - 2sin 2 = t 2 2 1- t ( - dt ) 2 \ I= ò (1 + t ) t 3 + t 2 + t 1 (t 2 - 1)dt = ò 2 (t + 1)2 t 3 + t 2 + t æ 1ö çè1 - 2 ÷ø (dt ) 1 t = ò 2 æ 1 ö 1 çè t + + 2÷ø t + + 1 t t 1 Put t + + 1 = u 2 Þ æç1 - 1 ö÷ dt = 2u du t è t2 ø 1 2udu I= ò 2 1+ u2 u 21. (3) (cos x + 2sin x ) cos x I =ò sin 3 ( cos ) 1 u = tan -1 1 + + 1 + c t -1 = tan (cos q + sec q + 1)1/ 2 + c So, f (q) = cosq + secq + 1 > 2 + 1 =3 f ( x + h) - f ( x ) 22. (2) f '( x ) = lim h h® 0 æ æ h öö f ç x ç 1 + ÷ ÷ - f ( x) è x øø = lim è h h ®0 æ hö f ( x) × f ç 1 + ÷ - f ( x) è xø = lim h h ®0 (Q f ( xy ) = f ( x ) × f ( y ) " x, y Î R ) = tan -1 æ hö f ç1 + ÷ - 1 xø = f ( x ) lim è h h ®0 MATHEMATICS 170 ì hæ æ h öö ü 1 + ç1 + g ç ÷ ÷ - 1 ï ï x ï è x øø ï è = f ( x ) í lim ý h ï h ®0 ï ïî ïþ ì 1æ æ h ö öü = f ( x ) × í lim ç1 + g ç ÷ ÷ ý è x ø øþ îh ®0 x è h As h ® 0; ® 0 x æhö Þ g ç ÷ ® 0 as lim g ( x ) = 0 (given) x®0 èxø f ( x) 1 \ f '( x ) = f ( x) × Þ =x x f '( x) Þò 23. (1) f ( x) x2 dx = + C Þ k = 2. f '( x ) 2 I = ò {sin(100 x + x) × (sin x)99 } dx = ò {sin(100 x ) cos x + cos100 x sin x}(sin x )99 dx = ò sin(100 x ) cos x × (sin x )99 dx 1424 3 1442443 I II + ò cos(100 x ) × (sin x)100 dx = sin(100 x)(sin x )100 100 100 cos(100 x)(sin x )100 dx 100 ò + ò cos(100 x )(sin x )100 dx sin(100 x)(sin x )100 +C 100 l 100 Þ l = 100, m = 100 Þ = = 1. m 100 dx 24. (8) ò (cos x - sin x )(1 + sin x cos x) (cos x - sin x )dx = 2ò (cos x - sin x) 2 (2 + (sin x + cos x) 2 - 1) (cos x - sin x )dx = 2ò 2 ((sin x + cos2 x ) - 2sin x cos x) (1 + (sin x + cos x ) 2 ) (cos x - sin x)dx = 2ò (2 - (sin x + cos x ) 2 )(1 + (sin x + cos x) 2 ) dt = 2ò where t = sin x + cos x 2 (2 - t )(1 + t 2 ) dt 2é 1 dt ù = 2ò = êò dt + ò ú 2 2 2 3 (2 - t )(1 + t ) 2 - t2 û ë 1+ t = æ 2 + t öù 2 é -1 1 ln çç ê tan (t ) + ÷ú + C 3 ëê 2 2 è 2 - t ÷ø ûú 2 1 \A= , B = 3 3 2 2 1 \12 A + 9 2 B - 3 = 12 × + 9 2 - 3 = 8. 3 3 2 sin8 x sin8 x × cos x dx = ò dx 25. (4) I = ò cos x cos2 x = =ò sin8 x (1 - sin 2 x ) × cos x dx Put sin x = t , cos x dx = dt t8 t8 -1 + 1 \I = ò dt = ò dt 1- t2 1- t2 (t 2 - 1)(t 2 + 1)(t 4 + 1) 1 =ò dt + ò dt 2 1- t 1 - t2 1 = ò (t 2 + 1)(-1 - t 4 ) dt + ò dt 1- t2 1 = ò (-t 6 - t 4 - t 2 - 1) dt + ò dt 1- t2 -t 7 t 5 t 3 1 1+ t = - - - t + ln +C 7 5 3 2 1- t - sin 7 x sin 5 x sin 3 x = 7 5 3 1 1 + sin x - sin x + ln +C 2 1 - sin x 2 x xö æ ç sin + cos ÷ 2 2ø Now, ln è 2 ...(1) x xö æ ç cos - sin ÷ 2 2ø è 1 + tan x / 2 æp xö = 2 ln = 2 ln tan ç + ÷ ...(2) 1 - tan x / 2 è4 2ø \ From (1) and (2) - sin 7 x sin5 x sin 3 x \I = - sin x 7 5 3 æp xö + ln tan ç + ÷ + C è 4 2ø 7 5 3 - sin x sin x sin x sin x = + + a b c d æp xö + ln tan ç + ÷ + C (given) è 4 2ø Þ a = 7, b = -5, c = 3, d = -1 Þ (a + b + c + d ) = (7 - 5 + 3 - 1) = 4. Solutions 171 CHAPTER Definite Integration 23 1. (a) Since the required function is a polynomial, the absciassae of the points of inflection can only be among the roots of the second derivative. Consequently p ''( x ) = ax( x - 1)( x + 1) = a ( x3 - x ) Since at the point x = 0, p '(0) = tan 60° = 3, x æ x 4 x2 ö P '( x ) = P ''( x ) dx + 3 = a ç - ÷+ 3 4 2ø è 0 hen, since P(1) =1, we get P (x) = x æ x 5 x3 7 ö - + ÷ + 3( x - 1) + 1 P '( x )dx + 1 =a ç è 20 6 60 ø ò ò 1 Since, P (-1) = -1 so , a = Thus 1 60( 3 - 1) 7 1 é 3 -1 5 ù 3 ò P( x)dx = ò êë 7 (3x - 10 x ) + x 3 úûdx 0 0 3 - 1 æ x6 5 4 ö x2 3 ç - x ÷+ 7 è 2 2 ø 2 = 0 3 -1 æ 1 5 ö 3 3 3 2 = + ç - ÷+ 7 è 2 2ø 2 14 7 = p 2. 1 I= ò (d) (cos2 px + sin2 qx – 2 cos px sin qx) dx -p Q sin 2 qx, cos2 px are even functions of x and cos px . sin qx is an odd function. p p 2 2 ò cos px dx = 2ò cos px dx \ -p p p 0 p p 2 2 I = 2 ò cos px dx + 2 ò sin qx dx = 0 \ p 0 p p 0 0 I= (c) òe cos q cos(sin q)d q = Real part of 0 2p òe cos q {cos(sin q) + i sin(sin q) d q 0 2p = Real part of ò ecos q ei sin q d q = Real part of 0 2p òe cos q+i sin q 2p d q = Real part of 0 ei q ò e dq 0 = Real part of 2p é 2iq ù e3iq iq e 1 + e + + + ............ ú d q ê òê 2! 3! ûú 0 ë = Real part of 2p é ù 1 ò êë1 + (cos q + i sin q) + 2! (cos 2q + i sin 2q) + ...úû d q 0 = 2p é ù 1 ò êë1 + cos q + 2! cos 2q + .......úû dq 0 2p sin 2 q é ù = ê q + sin q + + .......ú = 2p 2.2! ë û0 4. (a) Let q = p + d , r = p + 2d , s = p + 3d p + sin x p + d + sin x -2d + sin x \ f (x) = p + d + sin x p + 2d + sin x -1+ sin x p + 2d + sin x p + 3d + sin x 2d + sin x = ò (1 + cos 2 px ) dx + ò (1 - cos 2 qx) dx 0 2 p + d + sin x p + 2d + sin x -1 + sin x p + 2d + sin x p + 3d + sin x 2d + sin x = 2[( p + d + sin x)( p + 3d + sin x) -( p + 2d + sin x )2 = -2d 2 0 æ 1 + cos 2 px ö æ 1 - cos 2qx ö = 2ò ç ÷ dx + 2 ò ç ÷ dx 2 2 è ø è ø 0 0 p 3. f ( x) = ò cos px sin qx dx = 0 -p 2p 0 0 p and p Applying R ® R1 + R3 - 2 R2 , we get 2 2 ò sin qx dx = 2ò sin qx dx -p p sin 2qx ù é sin 2 px ù é = êx + ú + ê x - 2q ú = 2p p 2 ë û0 ë û0 2 Given ò f ( x )dx = -4 Þ 2 0 2 ò ( -2d )dx = -4 d 2 = 1 Þ d = ±1 0 MATHEMATICS 172 p /2 5. dx ò a2 cos2 x + b2 sin2 x (a) 0 p /2 sec2 x dx ò b2 tan 2 x + a 2 = = 0 = Þ 1 b2 ¥ dt ò 2 æ aö 2 0t +ç ÷ è bø 1 p/2 ò b 2 0 Þ b-a £ 2 sec x dx æ aö tan 2 x + ç ÷ è bø 2 (Putting tan x = t ) = 8. 1 p 24 ò ò 7. p/4 3 1 f ( x ) f '( x ) Þ ³1 f ( x) 1 + f 4 ( x) Integrating on the interval (a, b), we get b b f ( x ) f '( x) dx ³ ò 1 + f 4 ( x) ò dx a a (d) f '( x ) ³ f ( x ) + 8 2 2 + 2 2 ò1 (1 + cos8 x)(ax 2 + bx + c) dx ò 1 (1 + cos x)(ax + bx + c) dx = 0 Þ - a 2 + b2 , which is least if a = b = 2 2 6. 1 ò 0 (1 + cos x )( ax + bx + c) dx ò 0 (1 + cos8 x )(ax 2 + bx + c) dx 1 = ò (1 + cos8 x )(ax 2 + bx + c ) dx 0 ¥ Þ - a 2 + b 2 ³ - 8 + 8 = -4 sin x (a) Let f ( x) = x x cos x - sin x \ f '( x ) = x2 ( x - tan x) cos x ép p ù < 0" x Îê , ú = 2 ë4 3 û x sin x \ f ( x) = is decreases on the interval x ép p ù ê4, 3 ú ë û Þ The least value of the function æ p ö sin( p / 3) 3 3 = m= f ç ÷ = è 3ø ( p / 3) 2p and the greastest value of the function æ p ö sin(p / 4) 2 2 = M = fç ÷= è 4ø ( p / 4) p therefore p/3 sin x æ p pö 3 3 æ p pö 2 2 < dx < ç - ÷ èç 3 4 ÷ø 2p è 3 4ø p x p/4 [Mean Value Theorem of Integral Calculus] p/3 3 sin x 2 Hence, < dx < x 8 6 ù 1é -1 2 -1 2 ê lim- tan f ( x) - lim + tan f ( x) ú 2 ëê x ® b x®a ûú (b) Given = b bt 1 æ pö p ´ tan -1 = ç ÷= 2 a è ø a ab 2 2 ab b 0 p p = Thus Þ ab = 8 2ab 16 Now minimum value of a cos x + b sin x is = 1 b tan -1 f 2 ( x ) a ³ b - a 2 8 2 Now we know that if b òa f ( x)dx = 0 then it means that f (x) is + ve on some part of (a, b) and – ve on other part of (a, b). But here 1 + cos8 x is always + ve, \ ax2 + bx + c is + ve on some part of [1 , 2] and – ve on other part [1, 2] \ ax 2 + bx + c = 0 has at least one root in (1, 2). Þ ax2 + bx + c = 0 has at least one root in (0,2). x2 9. f ( x) = (c) ò (t - 1)dt x 2 Þ f '(x) = (x -1)2x - (x -1) = (x -1)(2x2 + 2x -1) \ f '(w ) = (w - 1)(2w 2 + 2w - 1) = (w - 1){2( -1) - 1} = 3(1 - w ) | f '(w) |= 3 | 1 - w | \ Now, | 1 - w |2 = (1 - w )(1 - w) = (1 - w )(1 - w 2 ) 10. = 1 - w - w 2 + w3 = 3 \ | 1 - w |= 3 Þ | f '(w) |= 3 3 (a) We have d é h ( x) ù f (t ) dt ú = f ( h( x)) h '( x) - f ( g ( x)).g '( x) dx êë g ( x ) û ò = f (2) ´ 2 ´ 2 ´ 1 8 = f (2) p p 2´ 4 2 sec x Let L = lim p x® 4 ò2 x2 - f (t )dt 2 p 16 é0 ù êë 0 form úû Solutions 173 On applying L' Hospital's rule, we get 2 ù d é sec x f (t )dt ú ê dx êë 2 úû L = lim 2 p d æ 2 p ö x® x - ÷ 4 dx çè 16 ø If is possible when f (x) = 0. Hence, " p ÎR. There is no non-zero continous ò function. Hence, S ÎR. 13. (a) We have, f (sec 2 x ).2sec 2 x tan x p 2x x® Q L = lim 1 1 < (1+ x 2 )n 1+ nx 2 1 1 dx dx Þ ò <ò n 2 0 (1 + x ) 0 1 + nx 2 1 1 dx Þ ò [tan –1 nx ]10 < 0 (1 + x 2 ) n n 1 1 dx tan –1 n Þ ò < 0 (1 + x 2 ) n n Þ L n 1 1 (tan –1 n ) = lim n ò dx < lim n 2 0 x®¥ n x ®¥ (1+ x ) Þ 4 pö p p æ f ç sec 2 ÷ .2.sec 2 . tan 8 4 4 4 è ø = = . f (2) p p 2. 4 p 11. (c) Let I = ò (1– | sin 8 x) |) dx p 0 p Þ I = ò dx – ò | sin 8 x ) | dx Þ I= p– 0 p 8´ 8 0 ò | sin 8 x | dx 0 Þ p 8 I = p –8ò sin8x dx 0 [Q sin 8x is periodic with period p /8 p ] 8 é – cos8 x ù I = p –8 ê ë 8 úû 0 Þ I = p + (cos p – cos0) = p – 2 12. (d) Given, Þ 0 x ò0 f (t )dt = 0, " x Î R L < tan –1 ¥ = \ 1 <L<2 2 14. (a) f (x) = 2ò t f (t )dt + 1 ...(i) 0 ...(1) Differentiating, we get, Þ f (x) = p f ¢(x) ...(2) f '( x ) 1 = Þ f ( x) p Integrating, we get, x Þ log f (x) = p + C Þ f (x) = Aex/p ...(3) Putting x = 0 in eqn. (i) we get, 0 = p f (x) pf (0) = 0 Case I f (0) = 0, p ¹ 0 Þ A=0 (from 2) Þ f (x) = 0 pf (0) = 0 Þ f ¢ (x) = 0 (from (2)) Þ there is no non-zero continuous to f (x). Case II p = 0 \ p 2 é p ù êëQ 2 < 2 úû Þ x x ò f ( x)dt = pf ( x) 1 1 dx 0 x ®¥ (1+ x 2 )n (1 + x2)n > 1 + nx2 nò L = lim \ f (0) = 1 Now, differentiate (i) w.r.t. x f ¢(x) = 2x f (x) f ¢ ( x) = 2x f ( x) Integrate both sides, we get ln | f (x) | = x2 + c Q f (0) = 1 Þ c = 0 Then from eqn. (ii), ln | f (x) | = x2 f (x) = e Þ 1 15. (a) J = ò x2 x 0 1+ x 8 Þ f (1) = e dx Q 0 < x8 < 1 1 1 x 1 é x2 ù 1 Þ J > ò 2 dx Þ J > 2 ê 2 ú Þ J > 4 ëê ûú 0 1 2 ù1 0 éx x 1 J < ò 1 + 0 dx Þ J < ê 2 ú Þ J < 2 ê ú ë û0 0 Hence, (I) is true and (II) is false. ...(ii) MATHEMATICS 174 16. (a) It is given that f (x) = | sin x |, ( f ( x ))2 - 42 x -1 x ®1 2 f ( x). f ¢( x ) = lim (L-Hospital rule) 1 x ®1 = 2f (1) . f ' (1) = 2 × 4 × 2 = 16 x = lim g(x) = ò f (t ). dt 0 2 x p and p(x) = g(x) – 2 (x + p) p Now, p(x + p) = g(x + p) – p+ x ò = p 19. (0) 2 f (t )dt - x - 2 p 0 2 x-2 p p 0 Q f (x) is periodic function with period p p+ x ò p p 1 1 0 1 0 20. (0.30) We have lim 2 x-2 p p 0 1 2 1 2 f ( x) ò 1- x 1 2 dx = 2 1 - x2 ò (1 - x 2 ) dx 1 2 = ésin ë 3e - x - 3 + 3x 2 p p p xù = - = û1/2 4 6 12 x ®1 2 3 e- x - 1 + x 2 lim 5 x® 0 x® 0 5x4 x4 0ö æ ç form ÷ 0ø è Again using L' Hospital’s rule, we get = 2 3 e - x .( -2 x ) - 0 + 2 x lim 5 x® 0 4 x3 2 0ö æ 3 2 - e- x + 1 ç form ÷ = . lim 2 0ø è 5 4 x®0 x Again using L' Hospital’s rule we get 2 3 -e - x .(-2 x ) 3 3 = lim 1 = = 0.30 10 x® 0 (2 x) 10. 10 ò q.sin nq dq . 0 p/2 f ( x) 18. (16) lim 0ö æ ç form ÷ 0ø è p/2 1/ 2 ò x5 2 lim 1 - x2 -1 0 Using L’ Hospital’s rule we get 21. ( 2) I (n) = dx 1 2 2 x®0 1 ò f ( x)dx £ 4 Þ f (x) = 1 - x2 ò x2 + 1 0 1 = x tan x x éx ù 2 1 -1 ò f ( x)dx £ êë 2 1 - x + 2 sin ( x) úû0 0 1 2 - tan x 3ò e-t dt - 3x + x3 2 ò f ( x)dx £ ò 1 - x dx Þ cos x -p /3 f (x) £ 1 - x 2 Þ x sin x p /3 x 0 Þ f (- x ) = -cosec x Þ ò f ( x) dx = 0. 2 = 2 + g(x) – x – 2 p Þ p(x + p) = p(x) for all x 17. (a) If is given that, x2 + (f (x))2 £ 1 Þ x2 -1 f (t )dt = ò f (t )dt Þ p(x + p) = ò sin x dx + g ( x ) - Þ sec x = - f ( x ) Þ f ( x) is an odd function p+ x = ò f (t )dt + ò f (t )dt \ - sin x x -1 I ( n) = = lim x ®1 (t 2 ) |4f ( x ) x -1 ò q.sin n -2q(1 - cos2 q)d q 0 p/2 = I (n - 2) – 2t dt 4 Þ ò (q.cos q).cos q.sin 0 = I (n - 2) - q.cos q. sin n -1 q p / 2 n -1 0 n -2 qd q Solutions 175 p/2 + n -1 ò [q.(- sin q) + cos q]. sin q d q n -1 0 1 = I (n - 2) (n - 1) p/2 ò q.sin q dq n 0 p/2 1 + n -1 ò cos q. sin n -1 =- q dq 0 x ò ò 22. (0) We have x (1 - t ) f (t )dt = t f (t )dt 0 0 Differentating both sides with respect to x we get, x 2 2q sin q cos p + cos qt.cos ptdt p p ò 1 ù q sin pt sin qtdt ú p ú 0 û 2 2 q æ sin p cos q q I ö = - sin q cos p + + p p çè p p 2 ÷ø é q2 ù 2 2q or, I ê1 ú = - sin q cos p + 2 sin p cos q 2 p p ëê p ûú + ò x 2 f ( x ) = (1 - t ) f (t )dt 0 Differentiating again with respect to x on both sides, we get x 2 f ¢( x ) + 2 xf ( x ) = (1 - x ) f ( x) f ¢ ( x) 1 - 3 x = 2 Þ f ( x) x Integrating both the sides, we get 1 ln f ( x ) = - + 3lnx + l x 1 3 Þ ln éë x f ( x) ùû + = l and f (1) = 1 x Þ l =1 Þ f ( x) = e x3 Thus lim f ( x) = 0 x ®¥ . ò é 2q tan p 2 ù = cos p cos q ê - tan q ú 2 p êë p úû As p and q are different roots of equation, tanx = x Þ tan p = p and tan q = q \ I = 0. 3 24. (2) 1 3 0 1 ò [ x ] dx = ò [ x ] dx + ò [ x ] dx 0 é êQ ë 0 x æ 1ö çè 1- ÷ø x 0 0 ò 1 ò 1 2 2q éæ sin pt ö êç cos qt = - sin q cos p + p p êè p ÷ø 0 ë x(1 - x ) f ( x) + (1 - t ) f (t )dt = xf ( x ) Þ 1 1 æ cos pt ö cos pt I = ç -2 sin qt + 2q cos qt. dt ÷ è p ø0 p 1 p/2 1 1 = I (n - 2) .I (n) + .sin n q 0 (n - 1) ( n - 1)( n) n 1 Þ I (n) = I (n - 2) + , n -1 (n - 1)(n) n -1 1 Þ I (n) - I (n - 2). = 2. n n 1 Þ n I ( n ) - ( n - 1) I ( n - 2 ) = n Put n = 2010 1 Þ 2010 I ( 2010 ) - 2009 I ( 2008) = 2010 -1 2009 I (2008)] [2010 I (2010) = 2010 Þ x 23. (0) Using integration by parts 1 3 0 1 ì0, if éë x ùû = í î1, if 0 £ x < 1ù ú 1 £ x < 3û = ò 0 dx + ò 1dx = [ x ]1 = 2 3 x 25. (7) ò f (t )dt F ( x) 1 lim = Þ lim x –1 x ®1 G ( x) 14 x®1 ò t f ( f (t )) dt –1 1 1 Q ò-1 f (t )dt = 0 and ò-1t f ( f (t )) dt = 0 f(t) being odd function \ Using L Hospital’s rule, we get f ( x) 1 lim = 14 x ®1 x f ( f ( x )) f (1) 1 1/ 2 1 = Þ = Þ f ( f (1)) 14 æ 1 ö 14 fç ÷ è 2ø Þ æ 1ö æ 1ö f ç ÷ =7 Þ fç ÷ =7 è 2ø è 2ø MATHEMATICS 176 CHAPTER Application of Integrals 24 b 1. (d) Given ò f (x)dx = b + 1 - 2 2 = 2[ln | sec x ||0p / 4 +ln | sin x ||pp // 24 ] 1 ù é = êln 2 + ln 1 - ln ú = 2[2ln 2] = ln 4. 2û ë y 1 2. Differentiate with respect to b x b Þ f ( x) = f (b) = 2 x2 +1 b +1 (b) If 1 £ x < 2 Þ [x] = 1 Þ [ y] = ±1 Þ y Î[-1, 0) È [1, 2) 5. p/4 O p/2 x (d) Eliminating y from two equations, We get 2mx + 4 = x 2 Þ x 2 - 2m - 4 = 0 x1 and x2 are roots of this quadratic equation 3. y = mx + 2 If 2 £ x < 3 Þ [x] = 2 Þ y = ±2 Þ y Î[-2, -1) È [2, 3) If 3 £ x < 4 Þ [x] = 3 Þ y = ±3 Þ y Î[ -3, -2) È [3, 4) If 4 £ x < 5 Þ [x] = 4 Þ y = ±4 Þ y Î[-4, -3) È [4, 5) Clearly, the required area consists of eight squares, each of area unity (see figure) \ Required area = 8 sq. units (a) As point lies on the line. Locus of the point is straight line perpendicular to given line passing through (2 3, -1) i.e. y + x 3 2 x1 =1 = y = 3x - 7 P 6. 4. (c) Here f ( x ) = min{| tan x |,|cot x |} Required area = 2 p/4 p/2 0 p/4 ò tan xdx + 2 ò cot xdx x2 \ x1 + x2 = 2m and x1x2 = -4 x2 æ x2 ö 2 mx + ç ÷ dx Now, ò ç 2 ÷ø x1 è 1 3 m 2 2 3 = ( x2 - x1 ) + 2( x2 - x1 ) - ( x2 - x1 ) 2 6 1 2 ém 2 ù = ( x2 - x1 ) ê ( x2 + x1 ) + 2 - ( x1 + x1x2 + x2 )ú 6 ë2 û 3 ´1 . Þ required area = 2 O O ( x2 + x1 ) 2 - 4 x1 x2 1 ém ù 2 ê 2 ( x2 + x1 ) + 2 - 6 {( x2 + x1 ) - x2 x1}ú ë û é m2 + 4 ù 2 = 4m + 16 ê ú êë 3 úû Clearly above is minumum if m = 0 (d) Required area = 4A , where p p 0 0 A = ò ( x + sin x)dx - ò xdx Solutions 177 Both the curves pass through origin. 1 y=x ò ( xe x - xe- x )dx \ Required area = A = –1 f (x) p 7. 2a ò 2a ydx = 0 1 x3 / 2 ò 2a - x dx 0 p/2 é3 1 pù ò 8a sin q dq = 8a 2 êë 4 . 2 . 2 úû 2 ò 4 0 9. 1ö æ 2 -x 1 x 1 = = ç e + ÷ - e +e è 0 eø 0 e (a) If both x + y - 1 and 2 x + y - 1 are positive, then x + y – 1 + 2x + y – 1 = 1 Þ 3x + 2y = 3 x y Þ + =1 1 3/ 2 If x + y - 1 is positive and 2 x + y - 1 is negative, then (x + y – 1) – (2x + y – 1) = 1 Þ – x – 1 = 0 Þ x = –1 If first negative and second positive, then –( x + y – 1) + (2x + y – 1) = 1 Þ x–1=0 Þ x=1 If both negative, then – (x + y – 1) – (2x + y – 1) = 1 Þ– 3x – 2y + 1 = 0 x y + =1 Þ – 3x – 2y + 1 = 0 Þ 1/ 3 1/ 2 Y 0 3 ò 2 3/2 1 –1 3 X' 1/2 O 1/3 = 2y y= +2 + 3x m +1 n +1ö p/2 G G æ 2 2 ÷ ç using sin m x cos n x dx = m+n+2 ÷ çè 2G ÷ 0 ø 2 2 3pa = sq. unit 2 Y (a) y = x ex y = xe–x 3x 8. I= 1 x -x x -x = x(e + e ) |0 - (e + e )dx 2p Put x = 2a sin2q and dx = 4a sin q cosq dq \ ò x(e x - e- x )dx 0 p2 p2 A = - cos p + cos0 =2 sq. units. 2 2 (b) Let the equation of curve y2(2a – x) = x3 ...(i) and equation of line x = 2a ...(ii) The given curve is symmetrical about x-axis and passes through origin. x3 From (i) we have, y2 = 2a - x x3 But < 0 for x > 2a and x < 0 2a - x So, curve does not lie in the portion x > 2a and x < 0, therefore curve lies in 0 £ x £ 2a . \ Area bounded by the curve and line = 0 1 p y = sin x + y = f(x) 1 X –1 1,e A( O Y' ) B(1, 1 e) x=1 \ X Given curves are y = xe x and y = xe - x æ 1ö Line x = 1 meets the curves at A(1, e) and B ç 1, ÷ . è eø Required area 2 ö 1 æ1 ö 1æ4 = ç ´ 2 ´ 3÷ - ç ´ 2÷ + ´ 1 ´ è2 ø 2è3 ø 2 3 = 2 sq units 10. (c) y = 4 - x 2 is the parabola EMA y = |x – 2| is the pair of straight lines DCA and ABF, y = (x – 2)1/3 is the curve IAHBG MATHEMATICS 178 Y D (–2, 4) G C(–1, 3) y = ( x–2) H E (–2, 0) Also when x ® 0, t ® 0 and when x ® y = x –2 M (0, 4) O 1/3 B (3, 1) p 8 tan X A (2, 0) t ® tan \ A= ò p 8 2 (1 + t ) 1 - t 0 I (0, –2) Thus f (x) = max. {4 – x2, |x – 2|, (x – 2)1/3} is 2 -1 4t -1 ò 2 (2 - x )dx + ò y = x3 y = x2 (p ,0) (0.0)(1,0) x = p 0 4 1 ò 0 x 2 1 - tan 2 1 x3 x4 = 6 3 4 Þ 1 0 + p 0 ( ) 1 1 æ 1 1 ö æ p 4 p3 1 1 ö = ç - ÷ +ç - + ÷ 6 è 3 4ø è 4 3 4 3ø 1 1 1 1 1 3 p 4 – 4 p3 – + + – = 6 3 4 4 3 12 3 p ( 3 p - 4) Þ = 0 Þ p3 (3p – 4) = 0 12 4 Þ p = 0 or 3 Since, it is given that p > 1 \ p can not be zero. 4 Hence, p = 3 13. (d) Let I be the smaller portion and II be the greater portion of the given figure then, Þ Y x 2 1 dx x 1 x = t Þ sec2 dx = dt 2 2 2 2 dt Þ dx = 1+ t2 Let tan x 4 x3 4 3 p 2 æ x xö 1 - tan ÷ ç 1 + tan 2 2 A= ò ç ÷ dx x x 0 ç 1 - tan 1 + tan ÷ è 2 2ø p/4 ) x+ 7 3 3 59 sq. units. +9+ + = 2 4 2 4 11. (b) The given curves are x 1 + tan 1 + sin x 2 y= = ...(1) x cos x 1 - tan 2 x 1 - tan 1 - sin x 2 = and y = ...(2) cos x x 1 + tan 2 \ The area bounded by the above curves, by p the lines x = 0 and x = is given by 4 = ( 2 3 3 2 Required area = ò x - x dx + ò x - x dx 3 = dt ò ò 2 tan (1 + t ) 1 - t 2 3 + ( x - 2)dx p/4 0 4t 2 (4 - x) dx + ( x - 2)1/3 dx -1 2 ò p , 4 12. (d) Given curves are y = x2 and y = x3 Also, x = 0 and x = p, p > 1 Now, intersecting point is (1, 1) ì 2 - x - 2 £ x < -1 ï 2 ï4 - x , - 1 £ x < 2 f ( x) = í 1/ 3 ï( x - 2) , 2 £ x < 3 ï x - 2, 3 £ x £ 4 î Now, area bounded by the curve and x-axis. = 2 dt = y= y=2–x (0, 2) I II X¢ (– 2, 0) Y¢ X Solutions 179 0 The graph of given region is as follows- Area of I = ò éê 4 - x 2 - ( x + 2 ) ùú dx û -2 ë 0 0 é x2 ù éx 4 æ xö ù 4 - x 2 + sin -1 ç ÷ ú - ê + 2 x ú =ê è 2ø û 2 ë2 -2 ëê 2 ûú -2 4 p é ù 1 = ëé 2sin ( -1) ûù - ê - + 4ú = 2 ´ - 2 = p - 2 2 ë 2 û Now, area of II = Area of circle – area of I. = 4p – (p – 2) = 3p + 2 area of I p-2 Hence, required ratio = = area of II 3p + 2 14. (a) The given curve is y = tanx ...(1) p when x = , y = 1 4 Equation of tangent at P is pö æ 2 pö æ y – 1 = ç sec ÷ ç x - ÷ è 4ø è 4ø Y y = 2x + 1 – p 2 æp ö P ç ,1÷ è4 ø O X¢ L X M y = tan x Y¢ p or y = 2x + 1 – 2 Area of shaded region = area of OPMO – ar (DPLM) ...(2) S î 1é 1ù = êlog 2 - ú sq. unit 2ë 2û 15. (c) y³ B -1 ,10 2 0 A O –1/2 = 10 – 2 ò ( x 2 + x +10)dx ....(i) –1 2 parabola C(10, 10) þ x+9 and x £ 6 ....(ii) 5 Solving (i) and (ii) we get intersection points as (1, 2), (6, 3), (– 4, 1), (–39, –6) Also y £ 0 –1/2 x + 3 Þ y2 = x + 3 ìï ( x + 3 ) if x < -3 Þ y2 = í ïî ( x + 3 ) if x ³ -3 R T (– 3, 0) Area of rectangle OABC – 2 ò p 1 4 tan x dx - (OM - OL)PM 0 2 p 1 ì p p - 2ü log sec x 04 - í ý ´1 2 4 4 ] Q Required area = Area (trap PQRS) – Area (PST + TQR) 1 1 é -3 ù = ´ (1 + 2 ) ´ 5 - ê ò-4 -x - 3 dx + ò-3 x + 3 dx ú 2 ë û éæ 3/2 ö-3 æ 3/2 ö1 ù 2 ( x + 3) 15 ê 2 ( -x - 3) ÷ +ç ÷ ú - êç = ç ÷ ç ÷ ú 2 3 -3 ø-4 è ø-3 úû êëè 15 é 2 16 ù 15 3 - + - 6 = sq. units = = 2 êë 3 3 úû 2 2 16. (b) Since, y = x 2 + x +10 1 1 Þ x2 + x + = y –10 + 4 4 2 æ 1 ö æ 39 ö Þ ç x + ÷ =ç y - ÷ è 2ø è 4ø Þ Latusrectum of the above parabola is 1. \ length of chord is also 1. Area is maximum when chord is latusrectum. Area of shaded reion = = ò = [ (1, 2) (– 4, 1) P 0 é x3 x2 ù = 10 – 2 ê + +10 ú ëê 3 2 ûú –1 2 é –1 1 10 ù 1 = 10 – 2 ê + – ú = ë 24 8 2 û 6 MATHEMATICS 180 sin x x sin x x 2 + e and y = + x x 2 Required area of the region æ sin x x 2 ö üï 2ì ï sin x x + e + ÷ ý dx í = ò1 ç 2 øï è x îï x þ 2 2 3 é x x ù 2æ x x ö = ò1 ç e - 2 ÷ dx = ê e - 6 ú è ø êë úû1 1ö 7 2 8 æ 2 = e - -çe- ÷ = e -e6 è 6ø 6 18. (1) The graph of f (x) = min{x – [x], – x – [–x]} is shown as in figure. Therefore Y 17. (b) y = O 1 21. (4) The desired region consists of four disjoint squares. So the area = 1 × 4 = 4 square unit. 6 5 4 3 -2 -1 2 -2 1ö æ1 = Area of shaded region = 4ç ´ 1 ´ ÷ = 1 2 2ø è Y 1 2 –2 –3/2 –1 –1/2 0 1/2 1 A=4 log e n æ -x ò çè e 0 3/2 2 X 22. (1) The curves y = [a] x2 and 1 y = [ a ] x 2 represent parabolas which are 2 symmetric about y-axis. The equation y2 – 3y + 2 = 0 gives a pair of straight lines y = 1, y = 2 which are parallel to x-axis. Thus the area bounded is shown as; Y y=2 y=1 1 n O – logen logen 1 –1+ n Now, n - 2 ³ 2loge n Þ 2n - 2 - 2log e n ³ n Þ n - 1 - log e n 1 ³ n 2 1 =2 2 20. (3.33) Here both functions f (x) and g(x) are periodic. Thus required area \ From (1), A ³ 4 ´ 1 æ1 ö é x 3/2 x 3 ù 2 ç ÷ = 10 ( x - x )dx = 10 ê - ú ç ÷ êë 3 / 2 3 úû 0 è0 ø 10 = = 3.33. 3 ò 2 2 y = [a]x y =1/2 [a]x 1ö æ n - 1 - loge n ö ...(1) - ÷ dx = 4 ç ÷ø è nø n 1– l 1 2 3 45 -2 -3 ò min {x - [ x], - x - [- x]}dx 19. (2) The area of the region is X 3 2 X From above figure the required area y =2 2 æ 2y = 2 ò ( x2 - x1 ) dy = 2 ò ç [ a] y =1 1 è y ö dy [ a ] ÷ø 2 = 2( 2 - 1) 2( 2 - 1) 2 ò y dy = [a] - 3 .( y3/2 )12 [a] 1 = 4 ( 2 - 1) 3/2 . (2 - 1) 3 [a] 4 (5 - 2 - 2 2 ) 4 (5 - 3 2 ) = 3 [a ] 3 [a] \ Area is greatest, when [a] is least, i.e., 1. \ Area is greatest, when [a] = 1 = Þ a Î éë1, 2 ) Solutions 181 f ( x + h) - f ( x) h h® 0 2x 2 ( x - 4) . f ( x) meet the x-axis 3 at (0, 0) (–2, 0) and (2, 0). Since f ( - x) = - f ( x ) , the curve y = f ( x ) is symmetrical about the origin. Thus f ( x) = 23. (2) f ¢(x)= lim f ( x(1 + h / x )) - f ( x ) h h® 0 = lim = f ( x) f (1 + h / x ) - f (1) lim x h® 0 h/x = f ( x) · f ' (1) x Also, as a = 2, f '( x) > 0 for x < -2 / 3, x > 2 / 3 and f '( x) < 0 for f '( x) 2 2 f ( x) or = x f ( x) x Integrating both sides, we get, f (x) = cx2 Since f (1) = 1, \ c = 1 Y f (x) = x2 \ f ¢ (x) = -2 / 3 < x < 2 / 3 . Thus, shape of y = f ( x ) is as shown in figure Y E B(2, 0) O A y= C(2, 0) X D 2 x2 + y 2 = 4 1+ x2 O (–1,0) (1,0) X So, f (x) = x 2 2 = x2 Þ x 4 + x2 - 2 = 0 Now, 2 1+ x Þ x2 = 1 Þ x = ± 1 é1 æ 2 ù 2ö x ú dx Required area = 2 ê ò ç ÷ ø ú êë 0 è 1 + x 2 û 1 3ù é ép 1 ù = 2 ê 2 tan -1 x - x ú = 2 ê - ú ë2 3 û ë 3 û0 F Area of the region ACBDOEA = 2p - ò But ò 0 -2 0 2 -2 f ( x ) dx + ò f ( x ) dx 0 | f ( x) | dx = - ò 0 -2 0 2 2 0 f ( x )dx = - ò f ( -t )( -1) dt = ò f (t )dt \ Area (ACBDOEA) = 2p Thus, the area of the other region AEODBFA = 2p. 25. (7) Y 2ö æ = çè p - ÷ø sq. units. 3 24. (0) Since y = f ( x ) has relative extremes at 4 3 x = ± 2 / 3, these points are critical points, and hence they must be roots of f '( x) = 0 (clearly is differentiable everywhere). Thus 2 f '( x) = a ( x - 2 / 3)( x + 2 / 3) 1 2 = a ( x - 4 / 3) Þ f ( x ) = a ( x 3 / 3 - 4 x / 3) + b . This passes through (0, 0) and (1, –2). So b = 0 and a (1/ 3 - 4 / 3) = -2 Þ a = 2 . –2 –Ö 2 –1 –Ö3 O 1 Ö2 2 Ö3 X MATHEMATICS 182 \ Area of region bounded by y = {x} 2 between x-axis for the x Î [0, 2] is As we know that fractional part of any thing must lie between 0 and 1 thus Y {x}2 1 2 3 0 1 2 = ò x 2 dx + ò ( x 2 - 1) dx + ò ( x 2 - 2) dx 2 x 2 –1 x 2 – 2 x 2 – 3 + ò ( x 2 - 3) dx 3 x2 A0 = O 1 2 3 CHAPTER 7ö æ Required area = 2A0 = 2 ç 2 + 3 - ÷ è 3ø \ X (c) Let P (x, y) be the point on the curve passing through the origin O (0, 0) and let PN and PM be the lines parallel to the x-and y-axis, respectively. If the equation of the curve is y = y (x), the area POM x x 0 0 ³1 1 + f 4 ( x) Integrating with respect to x, from x = a to x = b b 1 tan –1 ( f 2 ( x)) ³ b – a a 2 ü 1ì or (b – a) £ í lim (tan –1 ( f 2 ( x ))ý 2 î x ®b þ ( Assuming that 2(POM) =PON, we therefore have x x x 0 0 0 f ( x ). f '( x) or equals ò ydx and the area PON equals xy - ò ydx . ) x ®a + y O 3. x Differentiating both sides of this gives dy dy dx dy Þ 3y = x + y Þ 2 y = x =2 y x dx dx 2. Þ log y = 2log x + C Þ y = Cx 2 With C being a constant. This solution represents a parabola. We will get a similar result if we have started instead with 2 (PON) = POM 1 3 (c) f '( x ) ³ f ( x ) + f ( x) 4 or f '( x ). f ( x) ³ 1 + f ( x ) ) p 24 (c) Differentiating w.r.t. x the given equation, Þ b–a £ P(x, y) M ( – lim tan –1 ( f 2 ( x)) 2ò y dx = xy - ò y dx Þ 3ò y dx = xy N 7 3 Differential Equation and its Applications 25 1. 2 2+ 3- we have 1 - ( f '( x)) 2 = f ( x) If y = f(x), then we have, 2 dy æ dy ö 1 - ç ÷ = y2 = ± 1- y2 dx è dx ø dy Þ = ±dx 1 - y2 Þ sin -1 y = C ± x Þ y = sin(C ± x) Since f(0) = 0 so 0 = sin C Þ C = 0 Thus f ( x ) = ± sin x As f ( x ) ³ 0 for x Î [0, 1] so we have f ( x ) = sin x. Since sin x < x " x > 0, we get f ( x) < x for 0 < x £ 1. Thus f (1/ 2) < 1/ 2 and f (1/ 3) < 1/ 3. Solutions 4. 183 (a) Put y = x sec q dy dq \ = x sec q tan q + sec q dx dx From the given equation, dq æ ö x 3 ç x sec q tan q + sec q÷ dx è ø = x3 sec3 q+ x3 sec2 q tan q Dividing both sides by x 3 sec q , we get dq x tan q + 1 = sec 2 q+ sec q tan q dx dq x = (tan q+ sec q) dx dx or (sec q- tan q)d q = x Integrating we get ln(sec q+ tan q) - ln sec q = ln x + ln c 7. where p = y 2 - x2 = cx Þ y + y 2 - x 2 = cxy y where c is an arbitrary constant. (a) Put x2y2 = z dy Given, x 2 × 2 y + y 2 × 2 x = tan( x 2 y 2 ) dx d 2 2 ( x y ) = tan( x 2 y 2 ) put x2y2 = z dx Now, given expression transform to dz = tan z \ ò dx = ò cot z dz dx x = ln (sin z) + C p p Þ C=1 Þ z= 2 2 x = ln sin (x2y2) + 1 \ ln sin (x2y2) = x – 1 sin (x2y2) = ex – 1 dy y / x + 1 Putting y / x = v , the last = dx y / x - 1 equation reduces to dv v + 1 v 2 - 2v - 1 x = -v= dx v - 1 v -1 dx 2(v - 1) Þ 2 =dv x v 2 - 2v - 1 Þ Þ 2ln x = - ln (v 2 - 2v - 1) + ln C Þ ln x 2 ( y 2 / x 2 - 2 y / x - 1) = ln C when x = 1, y = \ 6. (b) yö æ x ç cos ÷ ( y dx + xdy ) xø è æ yö = y sin ç ÷ ( x dy - y dx ) èxø 8. Þ y 2 - 2 xy - x 2 = C , which does not represent a pair of straight lines if C ¹ 0 . (a) The given equation can be written as æ dx dy ö ( x 2 dy – y 2 dx) =0 ç - ÷+ y ø ( x – y) 2 è x Þ 2 Dividing both sides by x dx we get y ö æ y dy ö y æ æ y ö æ dy y ö Þ ç cos ÷ ç + ÷ = sin ç ÷ ç - ÷ è ø è ø è x ø è dx x ø x x dx x Which is homogeneous equation dy dv Putting y = vx we get = v+x dx dx dv ö æ æ dv ö or cos v ç v + v + x ÷ = v sin v ç x ÷ è è dx ø dx ø dy . dx Replacing p by p + tan p / 4 = p + 1 , 1 - p tan p / 4 1 - p p +1 we have y + x=0 1- p or 1 + 5. Þ 2v cos v = x dv (v sin v - cos v) dx Separating the variables, we get 2dx æ v sin v - cos v ö Þ =ç dv è v cos v ÷ø x Integrating 2 ln x + ln c = ln sec v - ln v sec v Þ cx 2 = Þ sec v = cx 2 v v y Þ sec = cxy x where ‘c’ is an arbitrary constant. (c) Differentiating xy = C , we get y + xp = 0 æ dy dx ö – ÷ ç æ dx dy ö è y 2 x 2 ø =0 ç – ÷+ y ø æ 1 1 ö2 è x ç – ÷ è y xø dy dx – y 2 x2 æ dx dy ö =0 ç - ÷+ y ø æ 1 1 ö2 è x ç – ÷ èx yø Integrating, we get: ln | x | – ln | y | – Þ 1 æ1 1ö ç x – y÷ è ø =c MATHEMATICS 184 x xy x xy + =c – = c Þ ln y x- y y y-x a 2 dx x y (b) Here, × = + -2 xy dy y x Þ 9. ln dy 2 ( x + y 2 - 2 xy ) dx dy Þ ( x - y )2 × = a2 , dx dy dv = put x - y = v \1 dx dx æ dv ö Þ v 2 ç1 - ÷ = a 2 è dx ø Þ a2 = dv v 2 dv Þ 2 = dx dx v - a2 æ a2 ö Þ ç1 + 2 dv = dx ç v - a 2 ÷÷ è ø Integrating both the sides, we get Þ v 2 - a 2 = v2 a v-a v + log = x+C 2 v+a Þ (x - y) + æ x- y-aö a log ç ÷ = x+C 2 è x- y+aø æ x- y-aö a log ç ÷ 2 è x- y+aø It passes through (3a, a) Þ y+C = Þ a+C = a æ1ö log ç ÷ 2 è 3ø a æ1ö æ 2 + log 3 ö Þ C = -a + log ç ÷ Þ C = - a ç ÷ 2 2 è3ø è ø æ x- y-aö a \ y = (2 + log 3) + log ç ÷ 2 è x- y+aø Þ \ x- y-a = e y - k , where k = a (2 + log 3) x- y+a 2 y -k x - y 1+ e = a 1 - e y-k æ 1 + e y -k ö , Þ x = y + aç ç 1 - e y -k ÷÷ è ø a where k = (2 + log 3) 2 Which is required equation of curve. Hence, (b) is the correct answer. 10. (b) The given equation can be written as ( ydx - xdy) + x xy (x + y)dx + y xy ( x + y)dy = 0 Þ ( y dx - x dy ) + ( x + y ) xy ( x dx + y dy ) = 0 y dx - x dy æ x ö x æ x 2 + y 2 ö Þ + ç + 1÷ × =0 dç ç ÷÷ y2 èy ø y è 2 ø æ x2 + y 2 ö æ x ö x æxö Þ dç ÷+d ç +1 =0 ç 2 ÷÷ ç y ÷ y è yø ø è øè æxö dç ÷ æ x2 + y2 ö è yø Þ d çç =0 ÷÷ + è 2 ø æ x + 1ö × x ç ÷ èy ø y æ -1 æ x ö ö æ x2 + y 2 ö or d çç ÷÷ + 2d çç tan çç ÷÷ ÷÷ = 0 è 2 ø è y øø è Integrating both the sides, we get x2 + y2 x + 2 tan -1 =C 2 y Hence, (b) is the correct answer. dy y3 11. (a) = 2x dx e + y 2 Þ Dividing by e2x : \ dy y 3e -2 x = dx 1 + y 2 e-2 x \ dy + y 2 e -2 x dy = y 3e -2 x dx dy \ ò 2 + ò 2( ye -2 x dy - y 2 e -2 x dx ) = 0 y dy \ ò 2 + ò d (e -2 x y 2 ) + c = 0 y \ 2ln | y | + e-2 x y 2 = c ...(i) 12. (b) Given curve is a n -1 y = x n Differentiating equation (i) w.r.t. x, we get dy a n -1 = nx n-1 ...(ii) dx Eliminating a from equations using (i) & (ii), we get x n dy = nx n-1 y dx Putting - ...(iii) dy dx , in equation (iii) at the place of dx dy we get - xdx x n dx = n, - . = nx n -1 , Þ ydy y dy Þ – xdy = ny dy, x2 ny 2 Þ +c = , Þ ny2 + x2 = constant. 2 2 Solutions 185 13. (c) Given, equation of normal at P (1,1) is ay + x = a + 1 æ dy ö =a \ Slope of tangent at P = a Þ ç ÷ è dx ø(1,1) Given æ dy ö dy dy =k=a µy Þ = ky Þ ç ÷ è dx ø (1,1) dx dx Y Q P(1, 1) Normal curve O P X dy dy = ay Þ = adx dx y (variable being separated) Þ ln y = ax + c It is passing through (1, 1) then c = – a Þ equation of the curve is y = e a ( x -1) dy ö æ 14. (a) Let P be (x, y). C is ç x + y , 0 ÷ and B is dx ø è dy ö æ çè 0, y - x ÷ø . Centre of the circle through O, dx C, P and B has its centre at the mid-point of BC. Let it be (a, b) then dy dy 2a = x + y and 2b = y - x dx dx B P O C A D Now, (a, b) lies on y = x so, y - x dy = x + y dy Þ dy = y - x dx dx dx x + y dy - y log e 2 = 2sin x (cos x - 1) log e 2 dx This is linear differential equation 15. (a) I.F. = e - loge 2 ò dx = e - x loge 2 = 2- x (i) Solution is y 2 - x = ò 2 - x 2sin x (cos x - 1) log e 2 dx (ii) put sin x - x = t Þ (cos x - 1)dx = dt \ y 2 - x = log e 2 ò 2t dt \ y 2- x = 2t + c \ y = 2 x + t + c2 x \ y = 2sin x + c2 x 16. (c) Rearranging the terms of equation, we get dt g '( x ) t2 -t =dx g ( x) g ( x) 1 dt 1 g '( x) 1 Þ- 2 + = ...(i) t dx t g ( x ) g ( x) 1 1 dt dz = Let z = Þ - 2 t t dx dx Thus, from (i) we obtain dz + g '( x ) z = 1 dx g ( x ) g ( x) dz which is clearly linear in z and with dx g '( x ) dx ò I.F. = e g ( x ) = elog[ g ( x )] = g ( x) Þ Thus, complete solution is 1 z × g ( x) = ò g ( x ) × dx + C g ( x) 1 g ( x) Þ g ( x) = x + c Þ =t t x+c 2 2 17. (a) 1 = lim t f ( x) - x f (t ) æç 0 form ö÷ t®x t-x è0 ø 2 2tf ( x) - x f '(t ) = lim t ®x 1 Þ 2 xf ( x) - x2 f '( x) = 1 dy Þ x2 - 2 xy = -1, y = f ( x ) dx dy 2 1 Þ - y=- 2 dx x x which is linear equation whose I.F. 1 = e -2log x = 2 . x Multiplying both sides of the last equation by I.F., we get d æ y ö 1 y 1 ç ÷ = - 4 Þ 2 = 3 + K. dx è x 2 ø x x 3x Since y = f(1) = 1. 2 1 2 2 + x so K = . Thus y = 3 3x 3 x dy 18. (2) The equation is y = e dx Þ x dy = ln y dx MATHEMATICS 186 Again differentiating both sides w.r.t. x y ( x) = x 2 y '( x) + y ( x )2 x + xy ( x ) 19. (0) (2ny + xy logex) dx = x logex dy ö dy æ 2n Þ =ç + 1÷ dx y è x log e x ø or (1 - 3x ) y ( x ) = x 2 y '( x) y '( x ) æ 1 3 ö or = y ( x ) çè x 2 x ÷ø Integrating, we get 1 l n y ( x ) = - - 3l n x + l n c x æ x3 y ( x ) ö 1 x3 y ( x) or l n ç = e -1/ x ÷ = - or x c è c ø Þ log( y ) = 2n log | log x | + x + c and c = 0 (Q curve passes through (e, ee)) \ y = e x + log(log x) 2n = e x (log x )2n Þ f ( x) = e x (log x)2 n Now, ì ï® ¥ if ï ï g ( x) = lim f ( x) = í 0 if n ®¥ ï ï ® ¥ if ï î x< 1 e 1 <x<e e x>e or y ( x) = ò g ( x) dx = 0 2 1/ e 20. (2) The given differential equation can be dx written as = xy[ x 2 sin y 2 + 1] dy 1 dx 1 y = y sin y 2 3 dy 2 x x This equation is reducible to linear equation, so 1 putting - 2 = u, the last equation can be written as x du + 2uy = 2 y sin y 2 dy Þ 2 The integrating factor of this equation is e y . So required solution is 2 2 ue y = ò 2 y sin y 2 .e y dy + C 2 2 Þ 2 = x 2 [cos y 2 - sin y 2 - 2Ce - y ] 21. (8) Differentiating both sides w.r.t., x of the given equation x x. y ( x ) + ò y (t )dt .1 = ( x + 1) x. y ( x ) + ò ty (t )dt x 0 x or ò y (t )dt = x 2 y ( x ) + ò ty (t )dt 0 0 2 x -y = r 2 y x then differentiating (1) we get, 2 xdx - 2 ydy = 2rdr So, ... (1) and sin q = ...(2) or xdx - ydy = rdr and differentiaing (2) we get, ... (3) xdy - ydx x2 = cos qd q or xdy - ydx = x 2 cos qd q = r 2 sec 2 q cos qd q = r 2 sec qd q ...(4) Substituting values from (3) and (4) in the given differetial equation, we get æ 1+ r2 ö 1+ r2 = = ç ÷ ç r2 ÷ r r 2 sec 2 q d q è ø dr or = sec q d q (1 + r 2 ) Intergrating both sides, rdr (t = y ) = ò (sin t )et dt + C 2 2 1 y 2 2 Þ ue y = e (sin y - cos y ) + C 2 2 Þ 2u = (sin y 2 - cos y 2 ) + Ce - y x x3 æ 1ö So, y(1) = e Þ c = e2 \ y çè ÷ø = 8 2 22. (5) Put x = r sec q and y = r tan q e \ ce -1/ x 0 ln(r + (1 + r 2 )) = ln(sec q + tan q) + ln c Where c is an arbitrary constant. or (r + (1 + r 2 )) = c(sec q+ tan q) æ ö x+ y or ( ( x2 - y2 ) + (1+ x2 - y2 ) = c ç ÷ çè (x2 - y2 ) ÷ø 23. (1) Area of curvilinear trapezoid Solutions 187 x OABCO = ò f ( x )dx according to question 0 x n +1 ò f ( x)dx µ { f ( x)} 0 x or ò f ( x ) dx = k{ f ( x )}n +1 0 Y y x) = f( B A x æ1ö ln | f ( x) - 1 | + ln f ç ÷ - 1 = c èxø æ æ1ö ö Þ f ( x ) - 1) ç f ç ÷ - 1 ÷ = ec = k è èxø ø æ1ö æ1ö Þ f ( x) f ç ÷ - f ( x) - f ç ÷ + 1 = k è xø è xø Þ ( f (1))2 - f (1) - f (1) + 1 = k (substituting x = 1) (Q f (1) = 2, given) Þ k =1 C O Integrating, we get æ1ö æ1ö Þ f ( x) f ç ÷ = f ( x) + f ç ÷ x è ø è xø X Where k is constant of proportionality. Differentiating both sides w.r.t. x, f ( x ) = k (n + 1)( f ( x )) n f '( x ) 1 or { f ( x)n-1} f '( x) = k (n + 1)' Integrating both sides w.r.t. x, x { f ( x)}n = +c n k (n + 1) { f (0)}n Putting x = 0 = 0+ c Þ 0= 0+c n (Q f (0) = 0) n { f ( x )} x = n k (n + 1) nx n Þ { f ( x)} = k (n + 1) ....... (1) (From functional equation) Þ f ( x) = 1 ± x n Since f(1) = 2, we must have f(x) = 1 + xn Since f(5) = 26, 26 = 1 + 5n Þ n = 2 \ f ( x) = 1 + x 2 25. (1) Here, \ f (6) = 1 + (6)2 = 37 2 2 4 f ' ( x ) = 1 + 2 x 2 + × 4 x 4 + × × 6 x 6 + ....¥ 3 3 5 æd ö = 1 + x ç ( xf ( x)) ÷ dx è ø Þ f '( x) = 1 + x | xf '( x) + f ( x) | Þ (1 - x2 ) f '( x) = 1 + xf ( x) dy x 1 ×y= 2, dx 1 - x 2 x \ Þ Again putting x = 1 then { f ( x )}n = dx log|1- x 2 | ò I.F. = e 1- x2 = e 2 = 1- x2 1 \ y × 1 - x2 = ò × 1- x2 + C 2 1- x -x n =1 k (n + 1) (Q f (1) = 1) From (1), ( f ( x ))n = x or f ( x ) = x1/ n Þ (f(10))n = 10 Þ k =1 24. (7) Given æ1ö æ1ö æ1ö x2 f '( x) f ç ÷ - f ( x) f ' ç ÷ = x 2 f '( x) - f ' ç ÷ èxø è xø èxø é ù 1 1 æ ö æ ö Þ x 2 f '( x ) ê f ç ÷ - 1ú = f ' ç ÷ [ f ( x) - 1] èxø ë èxø û ì æ1ö ü 1 æ1ö Þ f '( x) í f ç ÷ - 1ý = 2 f ' ç ÷ { f ( x ) - 1} è xø î èxø þ x æ 1 öæ 1 ö f ' ç ÷ç - 2 ÷ f '( x) x x ø Þ + è øè =0 f ( x) - 1 æ1ö f ç ÷ -1 èxø 1 Þ y 1 - x 2 = sin -1 x + C , as f (0) = 0 Þ C = 0 Þy= sin -1 x 1- x2 Þ A=ò 3 / 2 sin 1/ 2 -1 x 1 - x2 p dx æ t 2 ö 2 1 é p2 p2 ù = ò t dt = ç ÷ = ê - ú ç 2 ÷ p 2 4 36 p/ 6 è ø ë û p/ 3 6 \[4 A] = 1. MATHEMATICS 188 CHAPTER Vector Algebra 26 1. (b) r r a ( x ) and b ( x) are collinear if and only if a b g + cosec2 + cosec2 = 2 2 2 2 which is not possible as Þ cosec 2 cos x = x . Now let f ( x ) = x - cos x , then f '( x ) = 1 + sin x ³ 0 Þ f ( x) is increasing and hence f (x) = 0 for a unique value of x. p p For x ³ , f ( x ) > 0 and x < , f ( x) < 0 . 6 3 Thus cos x = x , for a unique value of cosec2 3. a a c applying R2 ¾¾ ® R1 ® R2 - R1 and R3 ¾¾ or 1 1 1 - cos a cos b - 1 1 - cos a 0 1 =0 cos g - 1 or cos a(cos b - 1)(cos g - 1) 4. cos a 1 1 + + =0 1 - cos a 1 - cos b 1 - cos g -(1 - cos a) + 1 1 1 Þ + + =0 (1 - cos a) (1 - cos b) (1 - cos g ) 1 1 1 + + =0 1 - cos a 1 - cos b 1 - cos g 1 1 1 Þ + + =1 1 - cos a 1 - cos b 1 - cos g Þ -1 + 0 c c b c c b–c Operating C3 ® C3 – C1 Expanding along R2, we get a c–a – c b – c = c (c – a) – a (b – c) = 0 Þ c2 – ac – ab + ac = 0 Þ c2 = ab Þ a, c, b are in G.P. \ c is the G.M. of a and b. (c) In DABC, let AD is angle bisector of angle A. A -(1 - cos a)(cos g - 1) - (1 - cos a)(cos b - 1) = 0 Dividing through out by (1 - cos a )(1 - cos ß)(1 - cos g ) ; we get a a c–a \ 1 0 1 =0Þ 1 0 r r r (d) Suppose that a , b , c are coplanar.. cos a 1 1 Þ 1 cos b 1 =0 1 1 cos g cos a So the vectors cannot be coplanar. (b) a, b, c are distinct non-negative numbers and the vectors ai$ + a $j + ck$ , $i + k$ and r ci$ + c $j + bk are coplanar.. æp p ö x, x Î ç , ÷ . è6 3 ø 2. a g ß ³ 1, cosec2 ³ 1, cosec2 ³ 1. 2 2 2 a q q 2 2 b B ak D bk C \ BD = ak , DC = bk \ BC = (a + b)k Applying cosine formula, we have cos q = = ( AB)2 + ( AC )2 - ( BC ) 2 2( AB)( AC ) a 2 + b 2 - ( a + b) 2 k 2 2ab ...(1) Solutions 189 Also in DADC and DABD uur uur uur uur 1 1 Þ a . b - 2. = l Þ a . b = l + . 2 4 Again, uur uur uur uur uur uur uur uur uur b - 2 c = l a Þ b.b - 2b.c = l b.a q b 2 + ( AD)2 - b 2 k 2 a 2 + ( AD)2 - a 2 k 2 cos = = 2 2b AD 2a AD Þ ( AD) 2 = ab(1 - k 2 ) 2 2 ïì a + b - 2ab cos q ïü = ab í1 ý (a + b)2 îï þï 5. 4a 2 b2 cos 2 q / 2 2ab cos q / 2 ( a + b) ( a + b) r r r r uuur (ab + b a) ab æ a b ö \ AD = ± =± ç + ÷ (a + b) (a + b) è a b ø ab =± ( aˆ + bˆ) (a + b) uuur uuur AD (aˆ + bˆ) \ AD = =± AD 2cos q / 2 uur uur uur uur uur uur (a) Let A B = a and AD = b and AC = c = 2 uur uur uur uur uuuur uur uur uur uur uur uur uur Now, DB. AB = ( a - b ).( a ) = a . a - b . a c 2 - a 2 - b 2 3a 2 + b 2 - c 2 = 2 2 é a 2 + b2 - c 2 ù ê\ In DABC , cos(p - q ) = ú 2ab êë úû D C ® b 6. 2 Þ 8 - l - l - 2(1) = l æç 1 ö÷ è 4ø 2 4 Þ l 2 + l - 12 = 0 Þ l = -4, 3. 7. ur r r (b) Let the triangle be PQR with sides p, q, r ur r r r r Let p = 3(aˆ ´ b ) and q = b - (aˆ. b) aˆ ur r r r r \ p.q = 3(aˆ ´ b).{b - (aˆ.b) aˆ} rr r uur = 3[aˆ b b] - 3(aˆ.b)[aˆ b aˆ ] = 0 p Þ ÐR = . 2 ur r r Also | p |= 3(aˆ ´ b) = 3 | aˆ | | b | sin q r = 3 | b | sin q , r q is the angle between aÌ‚ and b and | aˆ |= 1 r r r r r and | q |= {b - ( aˆ.b)aˆ}.{b - ((aˆ.b ) aˆ} r r r = | b |2 -2(aˆ.b)2 + (aˆ.b)2 (aˆ )2 r r r = | b |2 + | b |2 cos 2 q. - 2 | b |2 cos 2 q r = | b | sin q ® c q A 2 uur uur Þ b.c = 8- l - l . 2 4 Furthermore, uur uur uur uur uur uur uur uur uur b - 2c = l a Þ b.c - 2 c.c = l a.c Þ AD = when a , b and c are non-collinear coplanar vectors. uuur uuur uuur r r DB = AB - AC = a - b . 2 a 2 - ab cos q = a - uur uur 1ö æ Þ 16 - 2 b . c = l ç l + ÷ è 2ø [from (1)] p–q ® a B uur uur uur (a) Given | a | = 1, | c | = 1 and | b | = 4 uur uur 1 1 Þ a . c = 1.1. = . 4 4 uur uur uur uur uur uur uur Again, b - 2 c = l a Þ a . b - 2 a . c = l a 2 R ® q ® p Q ® r P MATHEMATICS 190 \ ur ur | p| | p| r = tan P Þ P = p . But r = 3 |q| 3 |q| pù é êëQ ÐR = 2 úû 10. (c) Vol. of paralleopiped formed by r r µ = ai$+ $ u = $i + a$j + $ k , v = $j + ak$, w k is 1 a 1 r r ur V = [u v w] = 0 1 a p æ 1 ö p = tan -1( 3) , Q = = tan -1 ç ÷ è 3ø 6 3 p and R = = cot -1 (0) . 2 p [Note that as soon as we get R = , only one \ option has the value of an angle 8. 2 p = cot -1 (0) , 2 hence (b) is the correct answer]. r r (c) For the vector a and b to be inclined at an rr obtuse angle, we must have a. b < 0 for all x Î (0, ¥) Þ c (log 2 x)2 - 12 + 6c log 2 x < 0 for all x Î (0, ¥) Þ cy 2 + 6cy - 12 < 0 for all y Î R , where y = log 2 x Þ c < 0 and 36c 2 + 48c < 0 , (using ax 2 + bx + c < 0, "x Î R if a < 0 and D < 0 ) æ 4 ö Þ c < 0 and c (3c + 4) < 0 Þ c Î ç - , 0÷ . è 3 ø ur r 9. (b) p . q = ab + bc + ca = a 2 + b 2 + c 2 b 2 + c 2 + a 2 cos q Þ cos q = ab + bc + ca . (a 2 + b2 + c 2 ) Þ Now (a - b)2 + (b - c )2 + (c - a) 2 ³ 0 Þ Also a 2 + b 2 + c 2 ³ ab + bc + ca ab + bc + ca £ 1. a 2 + b2 + c 2 (a + b + c )2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) ³ 0 Þ Þ a 0 1 P= ab + bc + ca ³ -1/ 2 a 2 + b2 + c 2 1 - £ cos q £ 1 Þ q Î[0, 2p / 3]. 2 = 1(1 - 0) - a(0 - a 2 ) + 1(0 - a) = 1 + a 3 - a 11. For V to be min dV =0 da Þ 3a 2 - 1 = 0 Þ a=± 1 3 . (b) We observe that rr rr r r æ b.a ö r r r r r r a.b1 = a.b - ç r 2 ÷ a.a = a.b - a.b = 0 è|a| ø r ur rr rr r æ r c.a r c.b r ö a.c 2 = a. ç c - r a - uur 1 b1 ÷ çè | a |2 | b1 |2 ÷ø r ur rr r r r a.c r 2 c.b r r = a.c - c. r | a | - uur 1 (a.b1 ) | a |2 | b1 |2 rr rr rr = a.c - a.c - 0 = 0 [ Q a.b1 = 0 ] r ur rr r r r æ r c.a r c.b r ö and b1 .c 2 = b1. ç c - r a - uur 1 b1 ÷ çè | a |2 | b1 |2 ÷ø r r ur r r ur r r (c.a)(b .a ) c.b r r 1 = b1.c - uur 1 b1.b1 r | a |2 | b1 |2 r r r r r = b1.c - 0 - b1.c = 0 (Using b1.a = 0 ) rr r uur r r Hence a.b1 = a.c2 = b1.c 2 = 0 . r uur r Þ (a, b1, c 2 ) is a set of orthogonal vectors. 12. (d) Let xi$ + y $j + zk$ be the required unit vector. Since a$ is perpendicular to (2$i - $j + 2k$ ) . \ 2x – y + 2z = 0 ......... (i) $ $ $ Since vector xi + y j + zk is coplanar with the vector $i + $j - k$ and 2$i + 2 $j - k$ . \ xi$ + y $j + zk$ = p ( $i + $j - k$ ) + q ( 2$i + 2 $j - k$ ), Solutions 191 where p and q are some scalars. Þ xi$ + y $j + zk$ = ( p + 2q)i$ + ( p + 2q ) $j - ( p + q )k$ Þ x = p + 2q, y = p + 2q, z = – p – q Now from equation (i), 2p + 4q – p – 2q – 2p – 2q = 0 Þ –p=0Þp=0 \ x = 2q, y = 2q, z = –q. Since vector xi$ + y $j + zk$ is a unit vector,, therefore 2 2 2 | xi$ + y $j + zk$ | = 1 Þ x + y + z = 1 Þ x2 + y2 + z2 = 1 Þ 4q2 + 4q2 + q2 = 1 Þ 9q2 = 1 Þ q = ± 1 3 2 1 2 1 , then x = , y = , z = – . 3 3 3 3 2 1 2 1 When q = – , then x = – , y = – , z = . 3 3 3 3 When q = Here required unit vector is 2$ 2 $ 1 $ i+ j- k 3 3 3 2 2 1 or - $i - $j + k$ . 3 3 3 13. (b) Let A be the first term and D be the common difference of the corresponding AP. Then, 1 1 = A + ( p - 1) D, = A + ( q - 1) D , a b 1 = A + (r - 1) D c Þ a -1 (q - r ) + b-1 (r - p) + c -1 ( p - q) = 0 r r r r Þ v × u = 0 Þ u ^ v. r r Hence u and v are orthogonal vectors. r r 14. (b) We have p × q = 0 r r r r Þ (5a - 3b ) × (-a - 2b ) = 0 r r r r ...(1) Þ 6 | b |2 -5 | a |2 -7a × b = 0 r r Also r × s = 0 r r r r Þ (- 4a - b ) (-a + b ) = 0 r r r r ...(2) Þ 4 | a |2 - | b |2 -3a × b = 0 Now xr = 1 ( pr + rr + sr ) 3 r r 1 r r r r r = (5a - 3b - 4a - b - a + b ) = -b 3 r 1 r r 1 r r and y = ( r + s ) = ( -5a ) = - a 5 5 r Angle between xr and y , i.e., r r r r x×y a ×b cos q = r r = r r ...(3) | x || y | | a || b | From (1) and (2), r r 25 r r 43 r r | a |= a × b and | b |= a ×b 19 19 r r 25 ´ 43 r r \ | a || b |= × a ×b 19 æ 19 ö q = cos -1 ç ÷ è 5 43 ø r r r r r rrr 15. (d) Let d × a = (cos y )[a b c ] = -d × (b + c ) r r r r [as d × (a + b + c ) = 0] r r r d × (b + c ) . Þ cos y = - r r r ...(1) [a b c ] r r r d × (a + b ) Similarly, sin x = - r r r ...(2) [ a b c ] r r r d × (a + c ) 2=- rrr . ...(3) [a b c ] Adding above equations, we get sin x + cos y + 2 = 0 Þ sin x + cos y = -2 Þ sin x = -1, cos y = -1 Þ x = (4n - 1)p / 2, y = (2n + 1); n Î Z x = (4n - 1)p / 2, y = (2n + 1) p . Since we want minimum value of x2 + y2, 2 p so x = - , y = ±p Þ x 2 + y 2 = 5p . 2 4 r r 16. (d) As angle between a and b is obtuse, r r a ×b < 0 Þ (2l2iˆ + 4lˆj + kˆ) × (7iˆ - 2 ˆj + lkˆ) < 0 Þ 14l 2 - 8l + l < 0 Þ l(2l - 1) < 0 Þ0<l< 1 2 ...(1) MATHEMATICS 192 r Angle between b and kÌ‚ is acute and less than p . 6 r r b × kˆ = | b | × | kˆ | cos q Þ cos q > 53 + l 2 B bˆ 60° A a uuur uuur From the figure, OA = aˆ, OC = bˆ uuur uuur uuur OA + OC = OB = aˆ + bˆ . O Þ l = 53 + l 1× cos q l aˆ C bÌ‚ 2 Þ cos q = 18. (3) \ q< p p Þ cos q > cos 6 6 3 3 l Þ > 2 2 2 53 + l 2 Þ 4l 2 - 3(53 + l 2 ) > 0 Þ l > 159 ...(2) Þ l < - 159 From Eqs. (1) and (2), l = f \ Domain of l is null set. 17. (c) Given, r r r r cos q = (a ´ iˆ) × (b ´ iˆ) + (a ´ ˆj ) × (b ´ ˆj ) r r ...(1) +(a ´ kˆ) × (b ´ kˆ) Consider, r r r r r r (a ´ iˆ) × (b ´ iˆ) = [(a ´ iˆ)b iˆ] = ((a ´ iˆ) ´ b ) × iˆ r r r r ((a × b )iˆ) - ((iˆ × b )a )iˆ = (a × b)(iˆ × iˆ) - (iˆ × bˆ)(a × iˆ) r r rr = a × b - a1b1 . r r r r r r Similarly, (a ´ ˆj ) × (b ´ ˆj ) = a × b - a2b2 r r r r r r and (a ´ kˆ)(b ´ kˆ) = a × b - a3b3 \ From Eq. (i), we get r r cos q = 3a × b - (a1b1 + a2b2 + a3b3 ) r r r r = 3a × b - a × b r r a ×b r r r r 1 Þ r r = 2 a × b Þ | a || b |= . | a || b | 2 r r Now, use AM ³ GM on | a |, | b | r 1 r r |a| +|b | 2 r r r \ ³ (| a | × | b |) 2 Þ| a | + | b | ³ 2 2 r r Þ| a| +|b |³ 2 . Now | aˆ + bˆ |= 1 \ DOAB is equilateral. \ In DAOC, cos120° = OA2 + OC 2 - AC 2 2OA × OC r r 1 1 + 1 - AC 2 \ AC = | a - b |= 3 = k . = 2 2 Þ k = 3. \- 19. (3) ( xˆ + yˆ + zˆ)2 ³ 0 Þ 3 + 2 S xˆ. yˆ ³ 0 Þ 2 2 2 S xˆ. yˆ ³ -3 2 Now, xˆ + yˆ + yˆ + zˆ + zˆ + xˆ = 6 + 2 S xˆ. yˆ ³ 6 + ( -3) Þ 20. (3) 2 2 2 xˆ + yˆ + yˆ + zˆ + zˆ + xˆ ³ 3 . r F = aaˆ + bbˆ + gcˆ r Fˆ × (bˆ ´ cˆ) \ F × (bˆ ´ cˆ) = a[aˆ bˆ cˆ] \ a = [aˆ bˆ cˆ] Now, bˆ ´ cˆ = bˆ ´ (aˆ ´ bˆ) = (bˆ × bˆ)aˆ - (bˆ × aˆ )bˆ 1 = aˆ - bˆ 3 ˆ ˆ] = (aˆ ´ bˆ)2 = | aˆ |2 | bˆ |2 - (aˆ × bˆ)2 [aˆ bc = 1- 1 8 = 9 9 9 3 9ì 1 ü \a = í Fˆ × aˆ - Fˆ × bˆ ý \ k1 = , k2 = 8 8 8î 3 þ Þ k1 + k2 = 12 3 = \ 2(k1 + k 2 ) = 3. 8 2 Solutions 193 r r 21. (5) a . b = 0 Þ x1 + x2 + x3 = 0 We have to obtain the number of integral solution of this equation Þ Coefficient of x 0 in ( x -3 + x -2 + x -1 + x 0 + x + x 2 )3 æ 1 + x + x 2 + x3 + x 4 + x5 ö = Coeff. of x in ç ÷ x3 è ø 3 0 = Coeff. of x 9 in (1 - x 6 )3 (1 - x ) -3 = 11C9 - 3. 5C3 = 25 . uuur uuur 22. (2) AB = 2iˆ + ˆj + kˆ, AC = (t + 1)iˆ + 0 ˆj - kˆ iˆ uuur uuur AB ´ AC = 2 ˆj kˆ 1 1 t + 1 0 -1 = -iˆ + (t + 3) ˆj - (t + 1)kˆ = 1 + (t + 3)2 + (t + 1)2 = 2t 2 + 8t + 11 . 1 uuur uuur Area of DABC = | AB ´ AC | 2 1 = 2t 2 + 8t + 1 2 1 2 (2t + 8t + 1) 4 f '(t ) = 0 Þ t = -2 At t = –2, f '' (t) > 0. So D is minimum at t = –2. r r r r r r r r 23. (0 ) Let V1 = (a ´ b ) ´ (c ´ d ) = (a ´ b ) ´ p 2 Let f (t ) = D = (say) r r r r r r r r r r r r r r = (a × p)b - (b × p)a = a × (c ´ d )b - b × (c ´ d )a r r r r r r r r ...(1) = [a c d ]b - [b c d ]a r r r r r r r r Similarly V2 = (a ´ c ) ´ (d ´ b ) = q ´ (d ´ b ) r r r r r r r r ...(2) = [a c b ]d - [a c d ]b r r r r r r r r And V3 = (a ´ d ) ´ (b ´ c ) = (a ´ d ) ´ r r r r r rrr r ...(3) = [a b c ]d - [d b c ]a r r r r rr r r r r r r \ A = V1 + V2 + V3 = -2[b c d ]a ¹ 0 as b , c , d are non-coplanar and which is a vector parallel to ar r r r r r r r r Þ A ´ a = -2[b c d ](a ´ a ) = 0 r r Þ| A ´ a |= 0 . r r æ é 0 -1/ 2ù ö 24. (2) vn +1 - vn = ç ê ÷ 0 úû ø è ë1/ 2 r r é 0 -1/ 2 ù r v2 - v1 = ê v0 0 úû ë1/ 2 n +1 r v0 3 r r é 0 -1/ 2 ù r v3 - v2 = ê v0 0 úû ë1/ 2 n é 0 -1/ 2 ù r r r vn - vn -1 = ê v0 0 úû ë1/ 2 Adding all the equations, r r r vn - v0 = ( A + A2 + A3 + .....An )v0 where é 0 -1/ 2 ù r r A=ê Þ vn = ( I + A + A2 + ....)v0 . ú 0 û ë1/ 2 25. (7) Each side subtends an angle of 2p/8 = p/4 at the centre of the octagon. Let 'O' be the centre of the octagon and r the radius of the circumcircle of the octagon. Therefore uuur uuur pör æ OA1 ´ OA2 = ç r 2 sin ÷ n 4ø è r where n is the vector perpendicular to the plane of the polygon such that from the side of nr , the points A1, A2, A3, ...., An are in counterclock sense. Hence uuur uuur æ pö r OA2 ´ OA3 = ç r 2 sin ÷ n 4ø è uuur uuur pör æ OA3 ´ OA4 = ç r 2 sin ÷ n , ..... etc. 4ø è Therefore 7 uuur uuur æ pör å (OA j ´ OA j +1 ) = 7 çè r 2 sin 4 ÷ø n j =1 uuur uuur = 7(OA1 ´ OA2 ) . MATHEMATICS 194 CHAPTER Three Dimensional Geometry 27 1. (b) The given equations are 3l + m + 5n = 0...(i) Þ cos 2q = 0 or cos2q = – 1 and 6mn - 2nl + 5lm = 0 From (i), we have m = –3l – 5n. Putting m = –3l – 5n in (ii), we get Þ 2q = p/2 or 2q = p ...(ii) Þ q = p/4 or q = 6(-3l - 5n )n - 2nl + 5l (-3l - 5n) = 0 3. Þ (n + l )(2n + l ) = 0 Þ either l = –n or l = –2n. If l = –n, then putting l = –n in (i), we obtain m = –2n. If l = –2n, then putting l = –2n in (i), we obtain m = n. Thus, the direction ratios of two lines are –n, –2n, n and –2n, n, n i.e., 1, 2, –1 and –2, 1, 1. Hence, the direction cosines are 1 6 , 2 6 , -1 6 or -2 6 , 1 6 , 1 6 ép pù p Þq= ê , ú ë 4 2û 2 (b) Let OA and OB be two lines with DC's l1, m1, n1 and l2, m2, n2. Let OA = OB = 1. Then co-ordinates of A and B are (l1, m1, n1) and (l2, m2, n2) respectively. Let OC Z be the bisector of ÐAOB such that C is the midpoint of AB and so its co-ordinates are æ l1 + l2 m1 + m2 n1 + n2 ö , ç 2 , 2 2 ÷ø è l + l m + m2 n1 + n2 , \ DR's of OC are 1 2 , 1 2 2 2 . The angle q \ We have between the lines is given by cos q = -2 2 1 -1 1 æ -1 ö Þ q = cos -1 ç ÷ . è 6 ø 2. 2 (c) It makes q with x and y-axes. l = cosq, m = cosq, n = cos (p – 2q) we have l2 + m2 + n2 = 1 Þ cos2q + cos2q + cos2 (p – 2q) = 1 Þ 2 cos2q + (–cos2q)2 = 1 Þ 2 cos2q – 1 + cos22q = 0 Þ cos2q – [1 + cos2q] = 0 2 æ l + l ö æ m + m2 ö æ n1 + n2 ö OC = ç 1 2 ÷ + ç 1 ÷ +ç ÷ 2 è 2 ø è ø è 2 ø -1 ´ + ´ + ´ = 6 6 6 6 6 6 6 1 = 2 1 (l12 + m12 + n12 ) + (l22 + m22 + n22 ) + 2(l1l2 + m1m2 + n1n2 ) 2 = 1 2 + 2 cos q [Q cos q = l1l2 + m1m2 + n1n2 ] 2 = 1 æqö 2(1 + cos q) = cos ç ÷ . 2 è2ø l1 + l2 m1 + m2 n1 + n2 \ DCs of OC are 2(OC ) , 2(OC ) , 2(OC ) i.e., l1 + l2 m + m2 n +n , 1 , 1 2 2 cos q / 2 2 cos q / 2 2 cos q / 2 Solutions 195 Y B æ l1 + l2 m1 + m2 n1 + n2 ö , ç 2 , 2 2 ÷ø è C A A(l , m , n ) 1 X' O 1 = 1 a 2 + b 2 + c2 X E æ l1 - l2 m1 - m2 n1 - n2 ö , ç 2 , 2 2 ÷ø è D (- l2 , - m2 , - n2 ) 4. a(cos q + 3) + (b 2)sin q + c(cos q - 3) = a 2 + b 2 + c2 2 + 6 (d) The given lines are y + 3 z -1 = =s -l l and 2 x = y - 1 = In order that cos a in independent of q ..........(i) z-2 =t -1 a + c = 0 and b = 0 .........(ii) \ cos a = The lines are coplanar, if 6. 0 - ( -1) -1 - 3 -2 - ( -1) 1 -l l =0 1 1 -1 2 +(cos q - 3)2 (a + c) cos q + b 2 sin q + (a - c) 3 Y' x -1 = (cos q + 3) 2 + 2sin 2 q 2a 3 a 2 ×2 2 = 3 p Þa= 2 6 (c) Given one vertex A(7, 2, 4) and line x + 6 y + 10 z + 14 = = 5 3 8 General point on above line, B º (5l - 6, 3l - 10, 8l - 14) Direction ratios of line AB are 1 -5 -1 C2 ® C2 + C3 ; 1 0 l =0 1 2 0 -1 l Þ 5( -1 - ) = 0 Þ l = -2 2 5. (a) Both the lines pass through origin. Line L1 is parallel to the vector r V1 = (cos q + 3)ˆi + ( 2 sin q)ˆj + (cos q - 3)kˆ and L2 is parallel to the vector r V2 = aˆi + bˆj + ckˆ r r V1 × V2 r r \ cos a = | V1 || V2 | < 5l - 13, 3l - 12, 8l - 18 > Direction ratios of line BC are <5, 3, 8> Since, angle between AB and BC is cos p = 4 p . 4 (5l - 3)5 + 3(3l - 12) + 8(8l - 18) 52 + 32 + 82 × (5l - 13) 2 + (3l - 12) 2 +(8l - 18)2 Squaring and solving, we have l = 3, 2. Hence, equation of lines are x-7 y-2 z-4 = = and -3 2 6 x-7 y-2 z-4 = = . 3 6 2 MATHEMATICS 196 7. D= (d) 1 ˆ ( j + l kˆ) ´ (iˆ + lkˆ) 2 z C 1 1 = - kˆ + l iˆ + lˆj = 2l 2 + 1 2 2 P 9 1 33 Þ £ (2l 2 + 1) £ 4 4 4 O Þ 4 £ l 2 £ 16 Þ 2 £| l |£ 4. 8. x y B A (c) The line has +ve and equal direction 1 cosines, these are 3 1 , 3 1 , 3 or direction ratios are 1, 1, 1. Also the lines passes through P (2, – 1, 2). \ Equation of line is x - 2 y +1 z - 2 = = = l (say) 1 1 1 h k l x + y + z = p or hx + ky + lz = p 2 p p p æ ö æ æ p2 ö p2 p2 ö B 0, , 0 C 0, 0, ç ÷ ç ÷ , , A , 0, 0 ÷ \ çç ç ÷ ç ÷ k l ÷ø è ø è è h ø Now, Area of DABC , D = Let Q (l + 2, l - 1, l + 2) be a point on this line where it meets the plane 2 where, Axy is area of projection of DABC on xy plane = area of DAOB 2x + y + z = 9 Then Q must satisfy the eqn of plane 2(l + 2) + l - 1 + l + 2 = 9 Þ l = 1 i.e. 2 2 2 Axy + Ayz + Azx \ Q has coordinates (3, 0, 3) Hence the length of line segment p2 / h 0 1 Now, Axy = 2 0 2 Similarly, Ayz = p4 p4 and Azx = 2 | kl | 2 | lh | 0 1 p /k 1 = 0 1 p4 2 | hk | = (2 - 3)2 + (-1 - 0)2 + (2 - 3) 2 = 3 9. 2 2 2 (d) Here OP = h + k + l = p \ DRs of OP are : h h2 + k 2 + l 2 or , k h2 + k 2 + l2 , l h2 + k 2 + l2 h k l , , p p p Since OP is normal to the plane, therefore, equation of plane is 2 2 , + A2yz + Azx \ D 2 = Axy D= p5 2hkl 10. (b) A plane containing the given lines is 2 x + 3 y + 4 z - 6 + l ( x + y + z - 3) = 0 … (i) This plane is perpendicular to plane z = 0 if 4 + l = 0 Þ l = –4 So, the equation (i) becomes –2 x - y + 6 = 0 Þ 2 x + y - 6 = 0 … (ii) Solutions 197 Equation of the projection will be the line of intersection of plane (2) and the plane z = 0. If the line has d.c. proportional to l, m, n then 2l + m = 0 and n = 0 Þ l : m : n = 1 : –2 : 0. Obviously (0, 6, 0) is a point on both the planes, hence lies on the line as well. x y -6 z = \ Equation of the line is = 1 -2 0 x y z + + = 1 which meets the axes at a b c A (a, 0, 0), B (0, b, 0) and C (0. 0, c). æ a b cö Centroid of D ABC is ç , , ÷ è 3 3 3ø and it satisfies the relation 1 1 1 9 9 9 + 2 + 2 =k Þ + 2 + 2 =k 2 x y z a b c 2 Þ 1 a 2 + 1 b + 2 1 z 2 = k 9 ...(i) Also given that the distance of plane Þ 1 1 1 Þ a2 + 1 b2 + 1 c2 \ a = 5k, b = –k, c = –k On putting the value of a, b and c in equation (i), 5(x – 1) – (y – 2) – (z – 3) = 0 Þ 5x – y – z = 0 ....... (iv) when x = 1, y = 0 and z = 5; then L.H.S. of equation (iv) = 5x – y – 2 =5×1–0–5 =0 = R.H.S. of equation (iv) Hence coordinates of the point (1, 0, 5) satisfy the equation plane represented by equations (iv), Therefore the plane passes through the point (1,0,5). 13. (c) The position vectors of two given points r = (5iˆ + 2 ˆj - 7kˆ) + 9 = 0 or r × n + d = 0 =1 We have a × n + d = (iˆ - ˆj + 3kˆ) × (5iˆ + 2 ˆj - 7kˆ) + 9 =1 From (i) and (ii), we get ....... (iii) the equation of the given plane is + + a2 b2 c 2 1 \ a.1 + b.1 + c.4 = 0 i.e., a + b + 4c = 0 From (ii) and (iii), are a = iˆ - ˆj + 3kˆ and b = 3iˆ + 3 ˆj + 3kˆ and x y z + + = 1 from (0, 0, 0) is 1 unit. a b c 1 x y z = = 1 1 4 a b c = = = k (let) 8 - 3 3 - 4 1- 2 11. (d) Let the eqn of variable plane be \ where a.1 + b.2 + c.3 = 0 i.e., a + 2b + 3c = 0 ....... (ii) Since the plane (i) parallel to the line ...(ii) = 5 – 2 – 21 + 9 < 0 and b × n + d = (3iˆ + 3 ˆj + 3kˆ) × (5iˆ + 2 ˆj - 7kˆ) + 9 k = 1 i.e. k = 9 9 12. (b) Equation of the plane containing the line x -1 y - 2 z - 3 is = = 1 2 3 a (x – 1) + b (y – 2) + c (z – 3) = 0 ....... (i) = 15 + 6 – 21 + 9 > 0 So, the points a and b are on the opposite sides of the plane. 14. (c) 2 x 2 - 2 y 2 + 4 z 2 + 6 xz + 2 yz + 3xy = 0 or 2 x 2 + x(6 z + 3 y ) - 2 y 2 + 4 z 2 + 2 yz = 0 MATHEMATICS 198 \x = -(6 z + 3 y) ± 36 z 2 + 9 y 2 + 36 yz - 8( -2 y 2 + 4 z 2 + 2 yz ) 4 -(6 z + 3 y) ± (2 x + 5 y)2 \x = 4 \x = -(6 z + 3 y ) ± (2 z + 5 y ) 4 or 2 x - y + 2 z = 0, x + 2 y + 2 z = 0 \ Angle between planes q = cos -1 (2)(1) + ( -1)(2) + (2)(2) 2 (2) + (-1) 2 + (2) 2 (1) 2 + (2) 2 + (2) 2 æ4ö = cos -1 ç ÷ è9ø 15. (b) PR : PQ = 1: 3 Þ 3PR = PQ 1 P R (–3, 1, 1) 2 Q (3, 4, 2) s2 : ax + by + cz = 0 s3 : a2x + b2y + c2z = 0 D= 1 a 1 b 1 c a2 b2 c2 So, for unique solution, D ¹ 0 Þ D = (a – b) (b – c) (c – a) ¹ 0 Þ a ¹ b, b ¹ c, c ¹ a rr 17. (a) The equation of plane is r. a = 5 r r r r Q r -b + r-c =4 r Þ sum of distances of a point (r ) from two r r fixed points with position vector b and c is constant. Þ such points lies on ellipsoid. r r Now points with position vector b and c r r satisfies the equation of plane r . a = 5, then r r r r b . a = 5 and c . a = 5 Þ 3PR = PR + RQ Þ 2 PR = RQ Therefore, PR : RQ = 1 : 2. Hence B(b) C(c) Plane r × a = 5 4ö æ -6 + 3 2 + 4 2 + 2 ö æ R =ç , , ÷ = ç -1, 2, ÷ 3 3 ø è 3ø è 1+ 2 The normal to the required plane is PQ = (6, 3, 1). Hence, the equation of the Area in the plane constitutes an ellipse r r Distance between b and c required plane is = 2 × (semi major axis) × e = 14 4ö æ 6( x + 1) + 3( y - 2) + 1ç z - ÷ = 0 3ø è 2ae = 14 Þ 18 x + 9 y + 3z - 4 = 0 16. (c) s1 is perpendicular to (iˆ + ˆj + kˆ) s2 is perpendicular to (aiˆ + bjˆ + ckˆ) and s3 is perpendicular to (a 2 iˆ + b2 ˆj + c 2 kˆ) Then, the planes are s1 : x + y + z = 0 ...(i) Sum of distance = constant = major axis = 4 2a = 4 ...(ii) From eqn (i) and (ii) e= 1 14 Þ b = a (1 - e2 ) = (semi 2 4 minor axis) Solutions 199 Area of ellipse = p.a.b. = p.2. 18. (3) L2 : L1 : 20. (7) 21. (7) A plane containing line of intersection of the given planes is 1 = 2 p » 4.443 2 x - y - z - 4 + l ( x + y + 2 z - 4) = 0 x - 0 y +1 z - 0 = = =l 1 1 1 i.e., (l + 1) x + (l - 1) y + (2l - 1) z - 4(l + 1) = 0 vector normal to it V = (l + 1)iˆ + (l - 1) ˆj + (2l - 1)kˆ x +1 y - 0 z - 0 = = =m 2 1 1 Hence any point on L1 and L2 can be Now the vector along the line of intersection of the planes (l, l - 1, l) and (2m - 1, m, m) , respectively.. 2x + 3 y + z - 1 = 0 According to the question, and x + 3 y + 2 z - 2 = 0 is given by 2m - 1 - l m - l + 1 m - l = = 2 1 2 iˆ On solving, we get m = 1 and l = 3 \ A = (3, 2, 3) B = (1, 1, 1) \ AB = 3. 19. (2) The equation of any plane passing through given line is : ( x + y + 2z - 3) + l(2x + 3y + 4z - 4) = Þ (1 + 2l)x + (1 + 3l) y + (2 + 4l)z - (3 + 4l) = 0 ....(i) If this plane is parallel to z-axis whose direction cosines are 0, 0, 1; then the normal to the plane will be perpendicular to z-axis \ (1 + 2l)(0) + (1 + 3l)(0) + (2 + 4l)(1) = 0 Þl=- 1 2 n = 2 3 1 = 3(iˆ - ˆj + kˆ) 1 3 2 As n is parallel to the plane (i), therefore, n ×V = 0 (l + 1) - (l - 1) + (2l - 1) = 0 Þ 2 + 2l - 1 = 0 Þ l = -1 2 Hence, the required plane is x 3y - 2z - 2 = 0 2 2 Þ x - 3 y - 4z - 4 = 0 Hence, | A + B + C - 4 |= 7 Put in eq (i), the required plane is 1 (x + y + 2z - 3) - (2x + 3y + 4z - 4) = 0 2 ...(ii) Þ y+2 =0 \ S.D. = distance of any point say (0, 0, 0) on z-axis from plane (ii) = ˆj kˆ 2 (1) 2 =2 22. (7) Under the given conditions the possible situation is f(–2) = 2; f(0) = 3; f(1) = 1. {where f(–2) = 1 is false, f (0) ¹ 2 is true and f (1) ¹ 1 is false}. The triangle formed is with vertices A(–2, 1, 0), B(2, 1, 3) and O(0, 0, 0) iˆ ˆj kˆ 1 Area of DAOB = 2 1 3 2 -2 1 0 MATHEMATICS 200 1 1 | -3iˆ - 6 ˆj + 4 kˆ |= 61 square units 2 2 k = ; so k = 61. 2 23. (6) The required plane is of the form = ( x + 2 y + 3z - 4) + l(2 x + y - z + 5) = 0 whose normal is (1 + 2l, 2 + l, 3 - l). This plane is perpendicular to the plane 5 x + 3 y + 6 z + 8 = 0. So we have 5(1 + 2l) + 3(2 + l) + 6(3 - l) = 0 Þ [bc + bl, (c - 2at ) - cb - cbl ]2 = k a (-bl )(ba + bl)(a - at 2 ) - (ab - abl ) Þ [bc + blc - 2ablt - cb - cbl )2 = -kbal(ba + abl - ablt 2 - ab - abl) Þ 4a 2 b2 l 2 t 2 = -kbal (-ablt 2 ) Þ 4a 2 b 2 l 2 t 2 = k 2b 2 a 2 l 2 t 2 Þ k = 4. 25. (4) | x |£ 8 Þ x Î [-8, 8] similarly for y and z. This represent a cube of side 16 units with centre at origin. -29 Þ 7 l = -29 Þ l = 7 Therefore, the required plane is Now, -8 £ x + y + z £ 8 gives space between two panes namely Z 29 ( x + 2 y + 3 z - 4) - (2 x + y - z + 5) = 0 7 (0,0,8) Þ 51x + 15 y - 50 z + 173 = 0 Comparing this with ax + by + cz + 173 = 0 we get a = 51, b = 15, c = –50 Y' (0,–8,0) 24. (4) Any point on parabola z2 = 4ax, y = 0 is given by Q(at2, 0, 2at) Now equation of line joining P(a, b, c) and Q (at2, 0, 2at) is given by = 2 y -b z-c = = l (say) b - 0 c - 2at a - at Þ x = a + l(a - at 2 ); y = b + bl; z = c + l; z = c + l(c - at ) by given condition point Q lies on (bz - cy )2 = k a (b - y )(bx - ay ) (at ,0,2at) Q 0) , (a,0 A (8,0,0) X Z' x + y + z + 8 = 0 and x + y + z – 8 = 0 subject to the limits of x, y, z Î [-8, 8] æ1 1ö This will take out ç + ÷ volume of the cube è4 4ø 1 3 Þ The required volume = ´ (16) = 2018. 2 Þ t = 2048 Þ B(0,b,0) O L(a,b,0) Y Y (0,0,–8) Z 2 (0,8,0) O so that, b - 9(a + c) = 15 - 9 = 6. x-a X' t =4 512 Solutions 201 CHAPTER Probability-2 28 1. (c) Total number of functions from A to B is (20)10. The number of non-decreasing function is the number of non-negative integral solutions of the equation x1 + x2 +¼+ x20 = 10 = 29 C10 1 = p n - 2 + p n -3 2 29 1 = p n -3 + p n -4 2 ................................. ................................. C19 (20)10 (c) When two dice is thrown then sample space has 6 × 6 = 36 elements so n(S) = 36. Now consider the event of getting 9 is (3, 6), (4, 5), (5, 4) and (6, 3). So probability of getting 9 when two dice is thrown is 4/36 = 1/9. If Sanchita starts the game then the probability that she wins is æ 1 ö æ 8 öæ 8 öæ 1 ö ç 9 ÷ + ç 9 ÷ç 9 ÷ç 9 ÷ +..... ¥ = è ø è øè øè ø 3. 1 1 p n + p n -1 = pn -1 + pn - 2 2 2 (Here xi is the pre-image of number i, i = 1, 2, 3, ....,20) Þ the probability = 2. 1 ( pn - 2 + pn -1 ) 2 Now = 1 9 = 1 ´ 81 = 9 64 9 17 17 181 And if Raj starts the game then probability that Sanchita wins the game is 1– 9/17 = 8/17 (d) As usual H denotes head and T denotes tail so that 1 = P (T) 2 Let En denote the event that the score is n. One can easily see that En = (En–2 Ç H) È (En–1 Ç T) Therefore Pn = P(En) = P(En–2 Ç H) + P(En–1 Ç T) = P(En–2) P(H) + P(En–1)P(T) P (H) = 1 = p 2 + p1 2 Since p1 = P ( T ) = 1 2 and p 2 = P ( ( T Ç T ) È H ) = P ( T ) P (T ) + P ( H) 1 1 1 3 ´ + = 2 2 2 4 We have = 1 1 3 1 pn + pn -1 = p2 + p1 = + = 1 2 2 4 4 Therefore pn - 2 1 1 = - p n -1 3 3 2 2 1æ 2ö 1 2 = - ç p n -1 - ÷ = æç - ö÷ æç p n - 2 - ö÷ 2è 3ø è 2ø è 3ø 3 2ö æ 1ö æ = ç - ÷ ç pn -3 - ÷ ..... 3ø è 2ø è æ -1 ö =ç ÷ è 2 ø n -1 2ö æ ç p1 - 3 ÷ è ø MATHEMATICS 202 æ 1ö = ç- ÷ è 2ø Hence pn = 2 = 4. n -1 2 æ 1ö +ç- ÷ 3 è 2ø n +1 5 4 4 2 4 3 5 3 3 5 4 3 = ´ ´ + ´ ´ + ´ ´ + ´ + 7 7 7 7 7 7 7 7 7 7 7 7 æ1 2ö ç - ÷ è2 3ø + ( -1) n -1 æ 1 ö 2 1æ 1 ö ç- ÷ = + ç- ÷ è 6ø 3 3è 2ø 6. 80 + 24 + 45 + 60 209 = 7´ 7´ 7 343 (d) 7. (b) The graph of y = 16 x 2 + 8(a + 5) = n n x - 7a - 5 is strictly above the x-axis Þ y>0 " x Î R 3.2n (c) The number of determinants formed = 16. Observe that the determinant is non-zero when exactly once (–1) appears as shown 1 1 Þ 16 x 2 + 8(a + 5) x - 7a - 5 > 0 " x Î R The above inequality holds if discriminant < 0 [Q coefficient of x 2 > 0 ] = 2 Þ 4 ways -1 1 Similarly the determinant is non-zero when (–1) is used exactly three times as shown Þ 64(a + 5)2 -4.16( -7 a - 5) < 0 -1 -1 Þ ( a + 2 ) (a + 15) < 0 -1 1 Þ a 2 + 17 a + 30 < 0 = -2 Þ 4 ways. Þ - 15 < a < - 2 So non-zero determinant can be obtained in 8 ways. Similarly determinant will be zero in 8 determinants. 1 2 (b) Let E1, E2 and E3 be the events of the critics giving favourable remarks. Then Þ 5. S E = ( E1 Ç E 2 Ç E3 ) È ( E1 Ç E 2 Ç E3 ) È ( E1 Ç E 2 Ç E 3 ) È ( E1 Ç E 2 Ç E3 ) Hence P(E) = P ( E1 ) P ( E 2 ) P ( E 3 ) + P ( E1 ) P ( E 2 ) P ( E 3 ) + P ( E1 ) P ( E 2 ) P ( E 3 ) + P ( E1 ) P ( E 2 ) P ( E 3 ) 5 4 æ 3ö æ 5ö 4 3 5 = ´ ´ ç1 - ÷ + ç1 - ÷ ´ ´ + 7 7 è 7ø è 7ø 7 7 7 æ 4ö 3 5 4 3 ´ ç1 - ÷ ´ + ´ ´ è 7ø 7 7 7 7 –15 – + –2 Given -20 £ a £ 0 and favourable cases -15 < a < -2 \ Required probability P(E) = 5 4 3 P ( E1 ) = , P ( E 2 ) = and P ( E 3 ) = 7 7 7 Let E be the event that majority reviewed favourably. Therefore + = 8. length of interval (-15, - 2) - 2 - (-15) 13 = = length of interval (-20, 0) 0 - (-20) 20 (a) Let G, C, K, and A are events where G = Guess answer, C = Copy answer, K = Know the answer, and A = Answer correctly P(G) = Probability that the Candidate Guess the answer = 1/3 (given) P(C) = Probability that the Candidate Copy the answer = 1/6 (given) P(K) = Probability that the Candidate Know the answer = ? Now, G,C and K are mutually exclusive and exhaustive events Therefore, P(G) + P(C) + P(K) = 1 P(K) = 1 – 1/3 – 1/6 = 1/2. Say G(Candidate guesses) has occurred. As, there are four choices out of which only one is correct, then the probability that the candidate made a guess is 1/4. Solutions 9. 203 P(A/G) = 1/4 P( A/C) = 1/8 ( given) then P( A/K) = Answer correctly that the candidate know = 1 Applying Baye’s Theorem we have, P(K/A) = [ P(K).P(A/K)]/[P(G) × P(A/G) + P(C) × P(A/C) + P(K) × P(A/K) = 24/29. (b) Let the sides be x, y, l – (x + y) Since the triangle will be formed when sum of two sides is larger than the third. Y 1 4 and P ( E 2 ) = 5 5 Let E be the event of the boy falls asleep. Again by hypothesis P ( E1 ) = P ( E / E1 ) = 3 1 and P ( E / E 2 ) = 4 4 Now, B E = E Ç ( E1 È E2 ) = ( E1 Ç E ) È ( E 2 Ç E ) so that, y l /2 11. (c) Let E1 and E2 be the events of the boy watching DOORDARSHAN and TEN SPORTS, respectively. It is given that Q P ( E ) = P ( E1 ) P ( E / E1 ) + P ( E 2 ) P ( E / E 2 ) By Bayes' theorem P l A X O l /2 l i.e. l – y > y, l – x > x and x + y > l – (x + y) P ( E1 / E ) = R Þ 0< y< l l l , 0 < x < and < (x + y) < l 2 2 2 DPQR 1 = DOAB 4 (as OQ = 1/2 OB) 10. (b) Let x and y be the two quantities. When the sum of two non-negative quantities is fixed, the product will be maximum when they are equal. So, the greatest product = xy = 10000 where x = y = 100 Hence required probability = 3 Now, xy ³ ´10000 4 Þ xy ³ 7500 Þ x ( 200 - x ) ³ 7500 Þ x 2 - 200x + 7500 £ 0 P ( E1 ) P ( E / E1 ) P ( E1 ) P ( E / E1 ) + P ( E 2 ) P ( E / E 2 ) = (1/ 5 ) ´ ( 3/ 4 ) (1/ 5) ´ ( 3 / 4 ) + ( 4 / 5 ) ´ (1/ 4 ) = 3 7 12. (c) Here, Pn = apn, n > 1 and P0 = 1 – ap (1 + p + p2 + ...) Consider the following events : Ej = There are j children in the family; j = 0, 1, 2,....., n A = There are exactly k boys in the family We have, P(Ej) = pj = apj; j = 0, 1, 2, ...., n Þ æ A ö jC Pç ÷ = k , j ³ k ç Ej ÷ 2j è ø ¥ ( Now, A = U A Ç E j j= k Þ ) æ ¥ ö P (A) = P ç U A Ç E j ÷ ç j= k ÷ è ø ( ) Þ ( x - 50 )( x - 150 ) £ 0 Þ 50 £ x £ 150 So, favourable number of ways = 101 Total number of ways = 201 ¥ ¥ æAö \ P (A) = å P A Ç E j = å P E j P ç ÷ ç Ej ÷ j= k j= k è ø 101 201 Hence, (b) is the correct answer. j ¥ ¥ æ jC ö æpö = å ap j ç k ÷ = a å ç ÷ . jC k ç 2j ÷ j= k j= kè 2 ø è ø So, required probability = ( ) ( ) MATHEMATICS 204 ¥ =aå k +r j= k æpö Cr ç ÷ è2ø æpö P (A) = a ç ÷ è2ø published) = P(X = 1) + P(X = 2) k+r k ¥ å 2 k +r r=0 æpö Cr ç ÷ è2ø r = 2´ We have, k+rCr = (–1)r–(k+1)Cr k \ ¥ æpö æ pö - k +1 P ( A ) = a ç ÷ å ( ) Cr ç - ÷ è 2 ø r=0 è 2ø k æpö æ pö = a ç ÷ ç1 - ÷ è2ø è 2ø -( k +1) k k +1 2 æpö = aç ÷ . è 2 ø ( 2 - p )k +1 3 1 = 6 2 \ Required probability = 2n 3 æ1ö æ1ö + 2n +1C3. ç ÷ . ç ÷ è2ø è2ø 2n - 2 ( ) æ 1ö = 2n +1 C1 + 2n +1C3 + ..... + 2n +1C2n +1 ç ÷ è 2ø 1 1 = 2 2 8 æ1 1ö \ The binomial distribution is ç + ÷ è2 2ø Also, |x – 4| < 2 Þ –2 < x – 4 < 2 Þ 2 < x < 6 \ p(|x – 4| < 2) = p(x = 2) + p(x = 3) + p(x = 4) + p(x = 5) + p(x = 6) 6 3 2n +1 E = ( G È G ') Ç E = ( G Ç E ) È ( G 'Ç E ) 1 2 ; P ( E / G ') = 3 4 1 P ( G ) = = P ( G ') 2 1 2 1 1 11 ´ + ´ = 2 3 2 4 24 Further, X denotes the number of books published. Then P (at least one book will be 4 3 6 æ1ö æ1ö æ1ö æ1ö + 8C5 ç ÷ ç ÷ + 8C6 ç ÷ ç ÷ è2ø è2ø è2ø è 2ø 2n +1 2n +1 5 æ1ö æ1ö æ1ö æ1ö æ1ö æ1ö = 8C2 ç ÷ ç ÷ + 8C3 ç ÷ ç ÷ + 8C4 ç ÷ ç ÷ è2ø è2ø è 2ø è 2ø è 2ø è 2ø 5 1 æ1ö = 22n. ç ÷ = 2 è2ø 14. (a) Let G = Event of good book G' = Event of not a good book E = Event of publication Then Therefore, P ( E ) = 1 ,n=8 2 q = 1 – p = 1- \ 2 æ1ö +.... + 2n +1C2n +1 ç ÷ è2ø Now, P ( E / G ) = 2 407 11 13 æ 11 ö ´ +ç ÷ = 576 24 24 è 24 ø 15. (b) Here, p = r 13. (d) The probability of showing an even number in a throw 1 æ1ö = 2n +1C1. . ç ÷ 2 è2ø 0 æ 11 öæ 13 ö æ 11 ö æ 13 ö = 2C1 ç ÷ç ÷ + 2C2 ç ÷ ç ÷ è 24 øè 24 ø è 24 ø è 24 ø = 8 2 C2 + 8C3 + 8C4 + 8C5 + 8C6 28 238 119 = 256 128 16. (a) If both have all 4 cards of the same color, then there are two possibilities at the end. Posibility 1: Ravi holds 4 red cards and Rashmi 4 black cards. Probability of this possibility = P(Ravi picks red in 1st pick 'AND' Rashmi picks black in 1st pick 'AND' Ravi picks red in 2nd pick' AND 'Rashmi pick black in 2nd pick') All 4 picks are independent = 2 2 1 1 1 ´ ´ ´ = 4 5 4 5 100 Posibility 2: Ravi holds 4 black cards and Rashmi 4 red cards similarly probability 1 = 100 = 4 Solutions 205 [A wins his serve then B wins his serve or A loses his serve then B also loses his serve] So, probability that ‘A’ wins the game after n deuces Hence, required probability P= 1 1 2 + = 100 100 100 1 = 0.02 = 2% 50 17. (a) We have 30 males and 20 females. n = P(males) = 20 2 = 50 5 ¥ Disease 2 1 . = 5 9 2 = 0.50 19 19. (0.25) Let l be the length of the chord AB of the given circle of radius a and r be the distance of the mid point D of the chord from the centre C, then r = a cos q and l = 2a sin q . According to given condition : 3/5 2/5 1/2 4/5 4/5 (1) No disease Disease No disease not +ve +ve not +ve +ve not +ve 1/5 1/5 4/5 4/5 1/5 1/5 4/5 (3) (4) (5) 1 Female 1/5 not +ve +ve (2) n æ 5ö 2 1 = å çè 9 ÷ø . 3 . 3 = n= 0 Male +ve wins then B serves and loses] \ Required probability of ‘A’ Winning the game 30 3 = 50 5 P(females) = 1/2 æ 5ö 2 1 = ç ÷ . . [After nth deuce, A serves and è 9ø 3 3 (6) (7) C a q (8) 15 50 = 75 = 15 16 107 + 50 125 18. (0.50) Let us assume that A wins after n deuces, n = 0, 1, 2, 3 ........... 2 2 1 1 5 Probability of a deuce = . + . = 3 3 3 3 9 B A (1) + ( 3) Required probability = (1) + ( 3) + ( 5) + ( 7 ) 3 1 4 3 1 1 ´ ´ + ´ ´ 5 2 5 5 2 5 = 3 1 4 3 1 1 2 1 4 2 4 1 ´ ´ + ´ ´ + ´ ´ + ´ ´ 5 2 5 5 2 5 5 5 5 5 5 5 r D 2 5 (2a) < 2a sin q < (2a ) 6 3 Þ 2 5 5 11 < sin q < < cos q < 3 6 Þ 6 3 11 5 a <r < a 6 3 \ The given condition is satisfied if the mid point of the chord lies within the region Þ between the concentric circles of radii and 5 a. 3 11 a 6 MATHEMATICS 206 Hence, the required probability 2 = æ 5 ö æ 11 ö pç a÷ - p ç a÷ è 3 ø è 6 ø For maximum probability; 2 2 = 1 = 0.25 4 Þ- pa 20. (0.60) Let A denotes the event that the runner succeeds exactly 3 times out of five and B denotes the event that the runner succeeds on the first trial. æ B ö P ( B Ç A) Pç ÷ = P (A) èAø 2 2 3 æ B ö 6p (1 - p ) = = 0.60 Thus, P ç ÷ = è A ø 10p3 (1 - p )2 5 21. (6) Let x shell are fixed at point I. Define the following events E1 : The target is at point I Þ P(E1 ) = 8 9 E 2 : The target is at point II Þ P ( E2 ) = 1 9 A : The target is hit The target will be hit if at least one shell hits the target. P ( A / E1 ) = 1 – None of the shells hit when the x æ 1ö target is at point I = 1 - ç ÷ and è 2ø æ 1ö P ( A / E2 ) = 1 - ç ÷ è 2ø \ P (A) = x -3 21- x ù 1 é æ 1ö æ 1ö = 1- ê ç ÷ +ç ÷ ú è 2ø 9 êë è 2ø úû æ 1ö ln 2 - ç ÷ è 2ø x-3 ù ln2 ú = 0 úû Þ x = 12 2 Also, d P ( A) dx 2 1 é æ 1ö ê 9 ê çè 2 ÷ø x -3 21- x ù (ln2)2 ú < 0 úû ë \ P(A) is maximum when x = 12 Þ k = 6 22. (2) Clearly all the solutions of f(x) = x are also solutions of f(f(x)) = x. First, we solve f(x) = x f(x) = x Þ x2 – 3x + 3 = x Þ x2 – 4x + 3 = 0 Þ (x – 1) (x – 3) = 0 Þ x = 1, 3 Therefore x = 1, 3 are also solutions of f(f(x)) = x. We want to seek if there are any more solutions of f(f(x)) = x other than 1 and 3 f(f(x)) = x Þ f(x2 – 3x + 3) = x Þ (x2 – 3x + 3)2 – 3(x2 – 3x + 3) + 3 = 0 Þ x4 – 6x3 + 12x2 – 9x + 3 = 0 Þ (x2 – 4x + 3)(x2 – 2x + 1) = 0 Þ (x – 1) (x – 3) (x – 1)2 = 0 Þ x = 1, 3 In this case we have no additional solutions. Therefore the probability that x satisfies equation æ 1ö (ln2)2 + ç ÷ è 2ø 23. 2 . Therefore m = 2. 9 (0) Let E1, E2, E3, E4, E5 and E6 be the events of occurrence of 1, 2, 3, 4, 5 and 6 on the dice respectively and let E be the event of getting a sum of numbers equal to 9. \ P ( E1 ) = 1- k 1 + 2k 1- k ; P (E2 ) = ; P ( E3 ) = 6 6 6 P ( E4 ) = 1+ k 1 - 2k 1+ k ; P ( E5 ) = ; P ( E6 ) = 6 6 6 f(f(x)) = x is 21- x 8 é æ 1 ö x ù 1 é æ 1 ö 21- x ù ê 1- ç ÷ ú + ê 1 ú 9 ëê è 2ø ûú 9 ëê çè 2 ÷ø ûú 21- x ë =- But P ( B Ç A ) = P (clearing succeeding in the first trial and exactly once in two other trials) P (4C2 p2 (1 – p2)) = 6p2 (1 – p)2 and P(A) = 5C3 p3 (1 – p)2 = 10p3 (1 – p)2 1 é æ 1ö ê 9 ê çè 2 ÷ø dP( A) =0 dx and 1 2 £ P (E) £ 9 9 Solutions 207 Then, E º {(3, 6), (6, 3), (4, 5), (5, 4)} Hence, P(E) = P(E3E6) + P(E6E3) + P(E4E5) + P(E5E4) = P(E3) P(E6) + P(E6) P(E3) + P(E4) P(E5) + P(E5) P(E4) 6 æ 1ö æ j = 2è 1 [2 - k - 3k 2 ] 18 Since, 1 2 £ P (E) £ 9 9 Þ 1 1 é 2 £ 2 - k - 3k 2 ù £ û 9 9 18 ë Þ 2 £ 2 - k - 3k 2 £ 4 Þ 2 2 £ 2 - k - 3k 2 and 2 - k - 3k £ 4 Þ 1ö æ 3k ç k + ÷ £ 0 and 3k2 + k + 2 > 0 3ø è Þ 1 - £ k £ 0 and k Î R 3 1 - £k £0 3 Hence, integral value of k is 0. 24. (5) The sportsman's chance of missing when r = ja is \ 1- a2 =1 - 1 ( j = 2,3, 4,5, 6 ) 2 j2 a 2 j The animal escapes when the sportsman misses in all the five shots. Therefore the probability of animal escaping to jungle is 1 öæ 1 öæ 1 ö éæ 1 ö æ 1 ö æ 1 ö æ 1 ö æ 1 ö ù = êç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ú ëè 2 ø è 3 ø è 4 ø è 5 ø è 6 ø û éæ 1 öæ 1 ö æ 1 ö æ 1 ö æ 1 ö ù ´ êç1 + ÷ç1 + ÷ ç1 + ÷ ç1 + ÷ ç1 + ÷ ú ëè 2 øè 3 ø è 4 ø è 5 ø è 6 ø û [since E1, E2, E3, E4, E5 and E6 are independent] = 1 öæ ø = 2P(E3) P(E6) + 2P(E4) P(E5) æ 1 - k öæ 1 + k ö æ 1 + k öæ 1 - 2k ö = 2ç ÷ç ÷ + 2ç ÷ç ÷ è 6 øè 6 ø è 6 øè 6 ø 1 öæ 1 - ÷ç1 - ÷ç1 - ÷ç1 - ÷ Õ çç1 - j2 ÷÷ = çè1 - 22 ÷ç øè 23 øè 2 4 øè 25 øè 26 ø æ 1 2 3 4 5 öæ 3 4 5 6 7 ö = ç ´ ´ ´ ´ ÷ç ´ ´ ´ ´ ÷ è 2 3 4 5 6 øè 2 3 4 5 6 ø 1 7 7 p = ´ = = 6 2 12 q Therefore, q – p = 12 – 7 = 5 25. (3) Let S be the sample space, then n(S) = Total number of determinants that can be made with 0 and 1 = 2 × 2 × 2 × 2 = 16 Q a b c d , each element can be replaced by two types i.e., 0 and 1 and let E be the event that the determinant made is non-negative. Also, E' be the event that the determinant is negative. \ \ ïì 1 1 0 1 0 1 ïü E' = í , , ý îï 1 0 1 1 1 0 ïþ P(E') = 3 n ( E ') 3 then P(E') = n S = 16 ( ) Hence, the required probability, P(E) = 1 – P(E') = 1 - 3 = 13 = m [given] 16 16 n Þ m = 13 and n = 16, then n – m = 3 MATHEMATICS 208 CHAPTER Properties of Triangle 29 1. (c) Sum of the roots of the equation is given by c(a + b) sin A + sin B = c2 = (c1 + c2 )2 - 2c1c2 (1 + cos 2 A) a+b = c = (2b cos A) 2 - 2(b2 - a 2 )(2cos 2 A) sin A + sin B sin C = 1 sin C Þ the triangle is right angled. (c) If D is the diameter of the circumscribed circle of D ABC, then a = D sin A, b = D sin B, c = D sin C = 2. \ a 3 + b3 + c 3 å sin3 A =7Þ ( D3 å sin 3 A å sin3 A 5. )=7 \ D=37. Since no side of triangle can exceed the diameter of the circle, the maximum possible value of a 3. Þ 6. b p C a = sin B sin C : sin C sin A : sin A sin B B 1 1 1 : : sin A sin B sin C \ sin A, sin B, sin C are in A.P. = 4. (d) cos A = b2 + c2 - a 2 2bc Þ c 2 - (2b cos A)c + b2 - a 2 = 0 . It is a quadratic in c, whose roots are c1 and c2, so c1 + c2 = 2b cos A and c1c2 = b2 - a 2 \ c12 + c22 - 2c1c2 cos 2 A (a + b + c )2 ³3Þ P³3 ab + bc + ca \ 3 £ P < 4 or P Î[3, 4) (a) We have, b+c ³ bc = l 2 Þ b + c ³ 2l 2 a b c b+c Now, = = = sin A sin B sin C sin B + sin C Þ \ p : q : r = b sin C : c sin A : a sin B c (a + b + c )2 <4ÞP<4 ab + bc + ca Again (a - b)2 + (b - c )2 + (c - a) 2 ³ 0 is 3 7 . (b) Let the altitudes from A, B, C be p, q, r respectively. Then, p = b sin C, q = c sin A, r = a sin B A = 4b 2 cos 2 A - 4b 2 cos 2 A + 4 a 2 cos 2 A = 4a2 cos2A. (b) a, b, c are sides of a triangle \ a + b > c, b + c > a, c + a > b \ a > | b – c |, b > | c – a |, c > |a – b | square and add a2 + b2 + c2 < 2 (ab + bc + ca) Þ a2 + b2 + c2 + 2 (ab + bc + ac) < 4 (ab + bc + ca) Þ a b+c = A A B+C B-C 2 sin cos 2 sin cos 2 2 2 2 A 2 Þa= æ B -Cö cos ç è 2 ÷ø (b + c )sin B-C £ 1 and 2 A b + c ³ 2l Þ a ³ 2l sin . 2 Q 0 £ cos Solutions 7. (c) L.H.S. = 209 Clearly a and b < 1 but c > 1 as sin a > 0 and cos a > 0 \ c is the greatest side and greatest angle is C 1 2 (a (b + c – a) + b2 (c + a – b) + c2 2 (a + b – c) 1 (a (b2 + c2 – a2) + b (c2 + a2 – b2) 2 + c (a2 + b2 – c2)) 1 = (2abc cos A + 2abc cos B + 2abc cos C) 2 A B Cö æ = abc ç1 + 4 sin sin sin ÷ 2 2 2ø è \ cos C = = A B Cö æ = 4RD ç1 + 4 sin sin sin ÷ . 2 2 2ø è 8. (b) a tan q + b sec q = c Þ b2 sec 2 q = (c - a tan q)2 Þ b 2 (1 + tan 2 q) = c 2 - 2ca tan q + a 2 tan 2 q Þ (a 2 - b2 ) tan 2 q - 2ca tan q + c 2 - b2 = 0 ...(i) Roots of equation (i) are tan a and tan b , where a and b are the two angles of the triangle. 2ca We have tan a + tan b = 2 and a - b2 tan a. tan b = c 2 - b2 a 2 - b2 2ca \ tan(a + b) = 9. a 2 - b 2 = 2ca c2 - b2 a 2 - c2 1- 2 a - b2 a2 + b 2 - c 2 2ab 1 sin 2 a + cos 2 a - 1 - sin a cos a =2 sin a cos a 2 \ C = 120° = 11. (d) sin2 A + sin2 C = 1001 sin2 B Þ a2 + c2 = 1001 b2 (using sine rule) Now, = 2 (tan A + tan C) × tan 2 B tan A + tan B + tan C 2 (tan A + tan C) × tan 2 B tan A × tan B × tan C æ cot A + cot C ö = 2ç ÷ø cot B è = 2 (cos A sin C + sin A cos C) sin B sin A × sin C × cos B = 2 sin (p - B) × sin B 2 sin 2 B = sin A sin C cos B sin A sin C cos B = 2 ´ 2b 2 2 ´ 2b 2 1 2 ´ 2b 2 = = = 2 2 2 2 250 2ac × cos B a + c - b 1000b 12. (b) sin A sin B sin C c b a + + = + + c sin B c b ab ac bc 3p 2ca \ tan æç p - ö÷ = Þ a 2 - c 2 = 2ca è 4 ø a 2 - c2 or a sin B sin C c b a + + = + + bc c b ab ac bc (b) We know that A + B + C = 180° A + C – B = 180° – 2B Þ or sin B sin C c b + = + c b ab ac é1 ù Now 2ac sin ê ( A - B + C ) ú ë2 û or b sin B + c sin C c 2 + b2 = bc abc = 2ac sin (90° - B ) = 2ac cos or a= 2ac (a 2 + c 2 - b2 ) = a 2 + c2 - b2 2ac 10. (c) Let a = sin a, b = cos a and B= c = 1 + sin a cos a = Þ b2 + c2 b sin B + c sin C b (2R sin B) + c (2R sin C) =2R b sin B + c sin C ÐA = p 2 MATHEMATICS 210 13. (a) A cos C = 4p D b p B C a 1 1 ab = p (4p) Þ ab = 4p2 2 2 Also, a2 + b2 = c2 = 16p2 \ (a – b)2 = a2 + b2 – 2ab = 8p2 Also, (a + b)2 = a2 + b2 + 2ab = 24p2 A-B a-b C 1 = cot = ´1 2 a+b 2 3 A-B or = 30° or A – B = 60° 2 BD AD 14. (a) In DBAD, = sin q sin 60° In DCAD, BD CD sin 45° sin 60° = sin q sin (75° - q) 2p öæ 4p öæ p ö æ sin 2 sin 2 sin 2 ÷ ç ÷ç ÷ç 7 17 17 ç1 ÷ç ÷ç ÷ 2 p ÷ç 2 2p ÷ç 2 4p ÷ ç sin sin sin 7 øè 7 øè 7 ø è = a2b2c2 17. (a) Since 4 sin A cos B = 1, so A and B can not be p 2 Þ sin q BD sin 60° 1 = = sin (75° - q) CD sin 45° 6 so, C = p p ÞB= -A 2 2 A æp ö Þ 4sin A cos ç - A ÷ = 1 è2 ø 75°–q Þ sin 2 A = 60°+ q 60° 1 D 3 45° p p 1 1 Þ sin A = Þ A = Þ B = 4 2 6 3 so angles are C 2æ A - Bö 15. æ b2 öæ c2 öæ a2 ö 1 - 2 ÷ç 1 - 2 ÷ = a2b2c2 çç 1 - 2 ÷ç a ÷ç b ÷ç c ÷ø è øè øè = a2b2c2 [As if B = Þ B p 2p 4p , ÐB = , ÐC = 7 7 7 p p , then cos B = 0 and if A = , then 2 2 tan A is not defined] CD AD = sin (75° - q) sin 45° q 16. (a) ÐA = (a2 – b2) (b2 – c2) (c2 – a2) D= tan a 2 + b 2 - c2 1 = Þ c= 6 2ab 8 1 - tan ç ÷ è 2 ø = 31 (c) cos (A – B) = æ A - B ö 32 1 + tan 2 ç ÷ è 2 ø 1 æA -Bö Þ tan ç = ÷ è 2 ø 3 7 C 1 æA -Bö a -b tan ç cot Þ cos C = ÷= 2 8 è 2 ø a+b 18. (b) Þ Þ p p p , , which are in A.P.. 6 3 2 2(b / a) b p+b = tan a = , tan 2a = 2 a a 1 - (b / a) 2ba 2 a -b 2 = p+b a 2ba 2 - a 2 b + b3 Þ p= a2 - b2 P p =p b( a 2 + b 2 ) (a 2 - b 2 ) O a a b a Solutions 211 19. (c) d = h cot 30° – h cot 60° and time = 3 min. 21. (4) A h(cot 30° - cot 60°) \ Speed = per minute 3 x y z b c h 60° 30° d B It will travel distance h cot 60° in h cot 60° ´ 3 = 1.5minute h(cot 30° - cot 60°) 20. (d) Let x and y be the heights of the flagstaffs at P and Q respectively Then, AP = x cot 60° = x 3 , AQ = y cot 30° = y 3 y BP = x cot 45° = x, BQ = y cot 60° = Þ AB = BP - AP = x - x 3 [Q AB = 30 m] 3 Þ 30 3 = ( 3 - 1) x Þ x = 15(3 + 3) S R x y 60° Q 45° B 30° 30 m 60° A 1 ö æ Similarly, 30 = y ç 3 ÷ Þ y = 15 3 è 3ø so that PQ = BP + BQ = x + y 3 = 15(3 + 3) + 15 = (60 + 15 3)m P a/3 D a/3 E a/3 a AD = 3sin x sin B 2a AE = 3sin(x + y) sin B Dividing Eq. (1) by Eq. (2), we get sin(x + y) AD = 2sin x AE 2a AD = 3sin (y + z) sin C C ...(1) ...(2) ...(3) ...(4) a AE = 3sin z sin C Dividing Eq. (5) by Eq. (4), we get ...(5) sin(y + z) AE = 2sin z AD Multiplying Eq. (3) and Eq. (6), we get ...(6) sin (x + y) sin(y + z) \ =4 sin x × sin z 22. (3) The given equation is 4 sin A sin B + 4 sin B sin C + 4 sin C sin A = 9 Þ 2 cos (A – B) – 2 cos (A + B) + 2 cos (B – C) –2 cos (B + C) + 2 cos (C – A) – 2 cos (C + A) = 9 or 2[cos(A – B) + cos(B – C) + cos(C – A)] 3 = 9 – 2 (cos A + cos B + cos C) ³ 9 – 2 × = 6 2 or cos (A – B) + cos (B – C) + cos (C – A) ³ 3 But cos (A – B) £ 1, cos (B – C) £ 1, cos (C – A) £ 1 or cos (A – B) = 1, cos (B – C) = 1, cos (C – A) = 1 Thus, A = B = C, i.e., triangle ABC is an equilateral triangle. Hence, D = 3. MATHEMATICS 212 23. (2) x B ID ID = = AD 2bc cos A b+c 2 P x 3 Q x c A 9 = c2 + x2 – 2cx cos B c But cos B = 3x c2 Þ 9= x + ...(1) 3 Similarly using cosine rule is DACQ, we get 2 16 = x2 + b2 3 ...(2) Adding (1) and (2), we get 25 = 2x2 + 2abc \ ID ID a = = AD AI + ID a + b + c \ ID IE IF a + b + c + + = =1 AD BE CF a + b + c \ a(b + c) sec \ k= 2 25. (7) A B ID + b(a + c) sec IE 2 2 C + c(a + b) sec IF = 2abc 2 B b 2 + c2 3 (3x)2 \ 25 = 2x2 + 3x2 3 \ x2 = 5 \ BC = 3x = 3 5 \ 25 = 2x2 + F B 100 m 30° A 24. (2) A I D A × ID 2 Now AI = AB = c = b + c ac ID BD a b+c 4 C b Let BP = PQ = QC = x In DABP, using cosine rule a(b + c) sec 60° d x D C 100 100 In DCBD, tan 60° = Þ x= x 3 100 In DACB, tan30° = x+d x + d = 100 3 . E C Þ d = 100 3 - 100 3 = 200 3 = 200 3 m 3