JEE-Mathematics 8 . e tan1 x (1 x x2 ). d(cot1 x ) is equal to - (A) – e tan1 x + c (B) e tan1 x + c (C) –x. e tan1 x + c (D) x. e tan1 x + c 9 . e x (1 n.x n 1 x2n ) dx is equal to - (1 xn ) 1 x2n (A) ex 1 xn +c (B) ex 1 xn +c (C) –ex 1 xn +c (D) –ex 1 xn +c 1 xn 1 xn 1 xn 1 xn 1 0 . ex4 (x x3 2x5 ) e x2 dx is equal to - (A) 1 xex2 . ex4 c (B) 1 x2 ex4 c (C) 1 ex2 . ex4 c (D) 1 x2ex2. ex4 c 2 2 2 2 1 1 . Primitive of 3x4 1 w.r.t. x is - x4 x 1 2 x x x 1 c x 1 c (A) c (B) c (C) x4 x 1 (D) x4 x 1 x4 x 1 x4 x 1 dx 1 2 . x4 [x (x5 1)]1 / 3 equals - 3 x5 2/3 c 3 x5 2/3 c 3 x5 2/3 c 3 x5 1 2 / 3 c 2 10 4 5 1 1 1 (A) (B) (C) (D) x5 x5 x5 x5 1 3 . sin x dx is equal to - sin 4x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 (A) 1 n 1 2 sin x + 1 n 1 sin x + c (B) 1 n 1 2 sin x – 1 n 1 sin x + c 2 2 1 2 sin x 8 1 sin x 2 2 1 2 sin x 8 1 sin x (C) 1 n 1 2 sin x + 1 n 1 sin x + c (D) 1 n 1 2 sin x – 1 n 1 sin x + c 4 2 1 2 sin x 8 1 sin x 4 2 1 2 sin x 8 1 sin x 1 4 . The value of integral d can be expressed as irrational function of tan as - cos3 sin 2 (A) 2 tan2 5 tan c (B) 2 tan2 5 tan c 5 5 (C) 2 tan2 5 tan c (D) 2 tan2 5 tan c 5 5 15. If 3 sin x 2 cos x dx = ax + bn[2sinx + 3cosx| + c, then - 3 cos x 2 sin x 12 15 17 6 12 15 17 1 (A) a = – 13 , b = 39 (B) a = 13 , b = 13 (C) a = 13 , b = – 39 (D) a = – 13 , b = – 192 E 21
JEE-Mathematics 16. x 1 dx is equal to - x x 1 (A) n x x2 1 – tan–1x + c (B) n x x2 1 – tan–1x + c (D) n x x2 1 – sec–1x + c (C) n x x2 1 – sec–1x + c 17. dx is equal to - (1 x ) x x2 2( x 1) 2(1 x ) 2( x 1) 2(1 x ) (A) 1 x + c (B) 1 x + c (C) + c (D) x 1 + c x 1 18. Let f'(x) = 3x2.sin 1 1 1 = 0, then which of the following is/are not correct. x – xcos x , x 0, f(0) = 0, f (A) f(x) is continuous at x = 0 (B) f(x) is non-differentiable at x = 0 (C) f'(x) is discontinuous at x = 0 (D) f'(x) is differentiable at x = 0 1 9 . 1 n x 1 dx equals - x2 1 x 1 (A) 1 n2 x 1 c (B) 1 n2 x 1 c (C) 1 n2 x 1 c (D) 1 n2 x 1 c 2 x 1 4 x 1 2 x 1 4 x 1 2 0 . dx equals, where x 1 ,1 - x x2 2 (A) 2 sin 1 x c (B) sin 1 (2 x 1) c (D) cos1 2 x x2 c (C) c cos1 (2 x 1) (C) tan1 tan2 x c sin 2x NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 2 1 . dx is equal to - sin4 x cos4 x (A) cot1 cot2 x c (B) cot1 tan2 x c (D) tan1 cos 2 x c BRAIN TEASERS AANNSSWWEERR KKEEYY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. A B C D C C C C B C Que. 11 12 13 14 15 16 17 18 19 20 Ans. B B D C C D A B,C,D B,D A,B,C,D Que. 21 Ans. A,B,C,D 22 E
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS FILL IN THE BLANKS 1 .If4ex 6ex dx = Ax + B log(9e2x – 4) + C, then A = ......., B = ....... and C = ...... 9ex 4ex 2 . If the graph of the antiderivative F(x) of f(x) = log(logx) + (logx)–2 passes through (e, 1998 – e) then the term independent of x in F(x) is ....... 3 . Let F(x) be the antiderivative of f(x) = 3cosx – 2sinx whose graph passes through the point (/2, 1). Then F(/2) = ....... 4 . Let f be a function satisfying f\"(x) = x–3/2, f'(4) = 2 and f(0) = 0. Then f(784) is equal to ........ MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ONE statement in Column-II. 1 . The antiderivative of Column-I Column-II (A) f(x) = 1 is (p) 1 ta n –1 a tan x +c ab b 2 (a2 b2 ) (a2 b2 ) cos x 1 (q) a2 1 tan–1 tan x + c, = cos–1 b (B) f(x) = a2 sin2 x b2 cos2 x is sin sin a (C) f(x) = 1 is (r) 1 ta n –1 a tan x + c ab b a cos x b sin x (D) f(x) = 1 is ; (a2 > b2) (s) 1 log tan 1 x tan 1 a +c a2 b2 cos2 x a2 b2 2 b 2 . f(x) dx when Column-I Column-II NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 1 (p) c– 1 sin–1 a (A) f(x) = (a2 x2 )3 /2 a | x| x2 (q) a2 sin–1 x x a2 x2 + c (B) f(x) = 2 a – a2 x2 2 1 x (C) f(x) = (r) c – (x2 a2 )3 /2 1 a2 x2 a2 (D) f(x) = x x x2 a2 (s) + c a2 x2 a2 ASSERTION & REASON In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it . Of the statements mark the correct answer as (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explantion for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explantion for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. E 23
JEE-Mathematics f1 (x) f2 (x ) f3 (x ) 1 . If D(x) = a2 b2 c2 , where f1, f2, f3 are differentiable function and a2, b2, c2, a3, b3, c3 are constants. a3 b3 c3 f1 (x )dx f2 (x)dx f3 (x)dx Statement - I : D(x) dx = a2 b2 c2 + c a3 b3 c3 Because Statement - II : Integration of sum of several function is equal to sum of integration of individual functions. (A) A (B) B (C) C (D) D dx 2 . Statement - I : If a > 0 and b2 – 4ac < 0, then the values of integral ax2 bx c will be of the type µ tan–1 x A + c. where A, B, C, µ are constants. B Because Statement - II : If a > 0, b2 – 4ac < 0, then ax2 + bx + c can be written as sum of two squares. (A) A (B) B (C) C (D) D 3 . If y is a function of x such that y(x – y)2 = x. Statement - I : dx = 1 log[(x – y)2 – 1] 2 x3y Because Statement - II : dx = log(x – 3y) + c. x3y (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts u(x) v(x)dx = u(x) v1(x) – u'(x)v2(x) + u\"(x) v3(x) – ..... + (–1)n–1 un–1(x) vn(x) – (–1)n–1 un(x) vn(x) dx where v1(x) = v(x)dx, v2(x) = v1(x) dx ...., vn(x) = vn–1(x) dx NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when calculating Pn(x) Q(x) dx, where Pn(x), is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times. If1 . (x3 – 2x2 + 3x – 1)cos2x dx = sin 2x u(x) + cos 2 x v(x) + c, then - 48 (A) u(x) = x3 – 4x2 + 3x (B) u(x) = 2x3 – 4x2 + 3x (C) v(x) = 3x2 – 4x + 3 (D) v(x) = 6x2 – 8x 2 . If e2x .x4 dx e2x f(x ) C then f(x) is equal to - 2 (A) x 4 2x3 3x2 3x 3 1 (B) x4 – x3 + 2x2 – 3x + 2 2 2 (C) x4 – 2x3 + 3x2 – 3x + 3 (D) x4 – 2x3 + 2x2 – 3x 3 2 + 2 E 24
JEE-Mathematics Comprehension # 2 Integrals of class of functions following a definite pattern can be found by the method of reduction and recursion. Reduction formulas make it possible to reduce an integral dependent on the index n > 0, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas. (Add a constant in each question) 1 . dx then 12n . 1 If In = (x2 a2 )n In+1 + 2n a2 In is equal to - x 11 1x 11 (A) (x2 a2 )n (B) 2na2 (x2 a2 )n 1 (C) 2na2 . (x2 a2 )n (D) 2na2 . (x2 a2 ) sinn x n 1 2. If I = dx then I+ I is equal to- n, –m cosm x m 1 n–2, 2–m n, –m sinn1 x 1 sinn1 x 1 sinn1 x n 1 sinn1 x (A) cosm 1 x (B) (m 1) cosm1 x (C) (n 1) cosm1 x (D) m 1 cosm1 x 3 . If un = xn dx , then (n + 1)au + (2n + 1)bu + ncu is equal to - ax2 2bx c n+1 n n–1 (A) xn–1 ax2 2bx c x n 2 xn (D) xn ax2 2bx c (B) (C) ax2 2bx c ax2 2bx c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Fill in the Blanks 3 35 2. 1998 3. 1 4. 2240 1 . 2 , 36 , any real value Match the Column 1 . (A) p; (B) r; (C) s; (D) q 2 . (A) s; (B) q,; (C) r; (D) p Assertion & Reason 1 . (A) 2 . (A) 3 . (C) Comprehension Based Questions Comprehension # 1 : 1 . (B) 2 . (C) Comprehension # 2 : 1. (C) 2 . (B) 3 . (D) E 25
JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE Evaluate the following Indefinite integrals : dx 5x4 4x5 2. dx 3 . tan x. tan 2x. tan 3x dx 1 . sin(x a) sin(x b) (x5 x 1)2 4 . x2 1 n(x2 1) 2 n x x4 dx 5 . Int eg r at e 1 f '(x) w.r.t. x4 , wh er e f( x ) = tan 1 x n 1 x n 1 x 2 6 . cos ecx cot x . sec x dx (ax2 b) dx x2 cos ecx cot x 1 2 sec x 8. dx 7. x c2 x2 (ax2 b)2 (x sin x cos x)2 9. cos 2. n cos sin d 1 0 . xx ex n x dx xn x cos sin e x 1 1 . (x2 1)3 /2 dx x3 3x 2 13. 3 x2 1 dx dx 1 2 . dx [JEE 99] (x2 1)3 [JEE 84] (x2 1)2 (x 1) 1 4 . x2 (x4 1)3 / 4 dx 1 6 . (sin x)11 / 3 (cos x)1 / 3 dx 1 7. cos x sin x dx 7 9 sin 2x 1 5 . sin2 x sin 2x cos2 x 1 9 . tan x co t x dx 18. dx [JEE 89] 1 tan x (cos 2x)1 / 2 1 [JEE 87] 21. dx [JEE 92] 2 0 . dx 3 x4 x sin x CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . cos ec(b a). n sin(x b) c 2. x 1 c sin (x a) x5 x 1 3. – n(sec x) 1 n(sec 2 x ) 1 n(sec 3 x ) + c 4. (x2 1) x2 1 3 n 1 1 2 3 9x3 . 2 x2 5 . n 1 x4 c 6 . sin 1 1 sec2 x c sin 1 ax 2 b k sin x x cos x c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 2 2 cx x sin x cos x 7. 8. 9. 1 (sin2)n cos sin – 1 n(sec2) + c 10. x x e x c 11. arc secx – nx + c 2 cos sin e x 2 x2 1 1 2 . 3 tan 1 x 1 n(1 x ) 1 n(1 x2 ) x c x 14. – 1 1 1/4 c x4 22 4 1 x2 1 3 . C – (x2 1)2 1 n tan x c 3(1 4 tan2 x) c 1 n (4 3 sin x 3 cos x) c 15. 2 tan x 2 16. 17. 24 (4 3 sin x 3 cos x) 8 (tan x)8 / 3 1 8 . 1 n(cos + sinx) + x + 1 19. 2 tan–1 tan x cot x +c 2 4 8 (sin2x + cos2x) 2 1 2 1 tan2 x 20. log 2 – log(cotx + cot2 x 1 ) + c 1 tan2 x 2 2 1 . 3 x2/3 – 12 x7/12 + 2x1/2 – 12 x5/12 + 3x1/3 + 6x1/6 – 12x1/12 + 12log|x1/12 + 1| – 4x1/4 + c 27 5 26 E
EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1 . cos 8x cos7x dx 2. x x2 2 dx 3. sin(x a) dx cot x dx 1 2 cos 5x sin(x a) 4 . (1 sin x)(sec x 1) 5 . 1 x dx dx dx dx 1 x sec x cos ecx sin x sin(2x ) 6. 7. n cos x cos 2x 9 . ex 2 x2 8. dx dx sin2 x (1 x ) 1 x2 1 0 0 x x 1 0 . Let 6 2 4 0 x 2 = 2x x2 x R and f(x) is a differentiable function satisfying, 5 , 5 x x2 3 3 1 f(xy) = f(x) + x2 (y2 – 1) + x (y – 1) ; x, y R and f(1) = 3 . Evaluate x2 x dx f(x) 1 1 . cot x tan x dx 12. sin 1 x 1 3 sin 2x dx ax f(x)dx 1 3 . Let f(x) is a quadratic function such that f(0) = 1 and x2 (x 1)3 is a rational function, find the value of f'(0) 14. ecosx (x sin3 x cos x) dx x sin2 x 15. (7x 10 x2 )3 /2 dx BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 (2 sin 3x 3 sin 2x) c 1 x 3/2 2 6 x2 2 1. 2. 3 c 1/2 x x2 2 NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 3 . cos x a. 1 n tan x 1 sec2 x tan x c cos a.arc cos cos a sin n sin x sin2 x sin2 a c 4 . 2 24 2 2 5 . x 1 x 2 1 x arc cos x c 6. 1 sin x cos x 1 n tan x c 2 2 2 8 7. 1 n cot x cot cot2 x 2 cot cot x 1 c sin 8 . cos 2x x cot x. n e cos x cos 2x c 9. ex 1 x c sin x 1x HF KI FG JI1 0 . 3x – n x2 x 1 3 tan 1 2x 1 c 11. tan 1 2 sin 2x x c H K3 sin x cos x ax + c 13. 3 1 5 . 2(7x 20) + c 1 2 . (a + x) arc tan – a 9 7x 10 x2 1 4 . C – ecosx(x + cosecx) E 27
JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1. cos 2 x 1 dx = [AIEEE-2002] cos 2 x 1 (4) –x – cotx + C (1) tanx – x + C (2) x + tanx + C (3) x – tanx + C (log x) [AIEEE-2002] 2 . x2 dx 1 1 1 (4) log(x + 1) + C (1) (logx + 1) + C (2) – (logx + 1) + C (3) (logx – 1) + C 2 x x 3 . If sin x dx =Ax + B logsin(x –) + C then values of (A, B) is - [AIEEE-2004] sin(x ) (1) (sin, cos) (2) (cos, sin) (3) (–sin, cos) (4) (–cos, sin) 4. dx is equal to- [AIEEE-2004] cos x sin x (1) 1 log tan x +C (2) 1 log cot x +C 2 2 8 2 2 (3) 1 log tan x 3 +C (4) 1 log tan x 3 +C 2 2 8 2 2 8 5 . (log x 1) 2 dx is equals to - [AIEEE-2005] 1 (log x )2 log x x xex x (1) (log x)2 1 + C (2) x2 1 + C (3) 1 x2 + C (4) (log x)2 1 + C 6. dx equals- [AIEEE-2007] NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 cos x 3 sin x (1) 1 logtan x + C (2) 1 logtan x + C 2 2 12 2 2 12 (3) logtan x + C (4) logtan x +C 2 12 2 12 7. The value of sin x dx is - 2 [AIEEE-2008] 4 sin x (1) x + log cos x +c (2) x – log sin x +c (3) x + log sin x +c (4) x – log cos x +c 4 4 4 4 8 . If the integral 5 tan x dx = x + a ln|sin x – 2 cos x| + k then a is equal to : [AIEEE-2012] tan x 2 E (1) 2 (2) –1 (3) –2 (4) 1 28
9 . If ƒ(x)dx (x) , then x5 ƒ(x3 )dx is equal to : JEE-Mathematics [JEE (Main)-2013] 1 3 3 2 3 1 x3(x3 ) 3 x3(x3 )dx C 3 3 (1) x ( x ) x ( x ) dx C (2) 1 x3(x3 ) x2(x3 )dx C 1 x 3 (x3 ) x3 (x3 )dx C 3 3(4) (3) NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Qu e. 1 2 3 4 5 6 7 8 9 Ans 3 2 2 4 4 1 3 1 3 E 29
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] 1 . sin 1 2x 2 Evaluate : 4x2 8x 13 dx . [JEE 2001 (Mains) 5M out of 100] 2 . (x3m x2m xm )(2x2m 3xm 1 where 0 For any natural number m, evaluate + + + + 6) m dx x > z3 . x2 1 dx is equal to - [JEE 2002 5M out of 60] [JEE 2006, (3M, –1M) out of 184] x3 2x4 2x2 1 (A) 2x4 2x2 1 c (B) 2x4 2x2 1 c (C) 2x4 2x2 1 c (D) 2x4 2x2 1 c x2 x3 x 2x2 x (f f...f) (x). Then xn2 g(x)dx equals. [JEE 2007, 3M] Let f(x) = (1 xn )1 / n f occurs n times 4 . for n 2 and g(x) = (A) 1 (1 n x n 1 1 K (B) 1 (1 nx n 1 1 K (C) 1 (1 n x n 1 1 K (D) 1 (1 nx n 1 1 K n n n n ) ) ) ) n(n 1) n 1 n(n 1) n 1 5 . Let F(x) be an indefinite integral of sin2x. [JEE 2007, 3M] Statement-1 : The function F(x) satisfies F(x + ) = F(x) for all real x. because Statement-2 : sin2(x + ) = sin2x for all real x. (A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. ex ex 6 . Let I= e4x e2x dx , J = e 4 x e 2 x 1 dx . [JEE 2008, 3M, –1M] 1 Then, for an arbitrary constant c, the value of J – I equals (A) 1 e4x e2x 1 c (B) 1 e4x e2x 1 c 2 log e4x e2x 1 2 log e2x e2x 1 (C) 1 e2x ex 1 c (D) 1 e4x e2x 1 c NODE6\\ E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#06\\Eng\\01-INDEFINITE INTEGRATION.p65 2 log e2x ex 1 2 log e4x e2x 1 7. The integral sec2 x equals (for some arbitrary constant K) [JEE 2012, 3M, –1M] dx (sec x tan x)9 /2 (A) 1 x 11 / 2 1 1 sec x tan x 2 K (B) sec 1 x 11 / 2 1 1 sec x tan x 2 K tan 11 7 tan 11 7 sec x x (C) 1 x 11 / 2 1 1 sec x tan x 2 K (D) sec 1 x 11 / 2 1 1 sec x tan x 2 K tan 11 7 tan 1 1 7 sec x x PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1 . (x 1) tan1 2(x 1) 3 n(4x2 8x 13) c 2. m 1 34 6. 2 x3m 3 x2m 6 x m m 3. D 4. A 5. D C 6 (m 1) C 7. C 30 E
JEE-Mathematics INVERSE TRIGONOMETRIC FUNCTION 1. INTRODUCTION : The inverse trigonometric functions, denoted by sin–1x or (arc sinx), cos–1x etc., denote the angles whose sine, cosine etc, is equal to x. The angles are usually the numerically smallest angles, except in the case of cot–1x, and if positive & negative angles have same numerical value, the positive angle has been chosen. It is worthwhile noting that the functions sinx, cosx etc are in general not invertible. Their inverse is defined by choosing an appropriate domain & co-domain so that they become invertible. For this reason the chosen value is usually the simplest and easy to remember. 2 . DOMAIN & RANGE OF INVERSE TRIGONOMETRIC FUNCTIONS : S.No ƒ (x) Domain Range (1) sin–1x |x| 1 (2) cos–1x |x| 1 , (3) tan–1x x R 2 2 (4) sec–1x |x| 1 0, (5) cosec–1x |x| 1 , (6) cot–1x x R 2 2 0, or 0, , 2 2 2 , 0 2 2 (0, ) 3 . GRAPH OF INVERSE TRIGONOMETRIC FUNCTIONS : (a) f : , [1, 1] f 1 : [1, 1] [–/2, /2] 2 2 f–1(x) = sin–1(x) f(x) = sin x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 y y=x y y=arcsinx y=arc sinx /2 /2 1 y=sinx –1 01 –/2 –1 x x y=arc sinx /2 0 1 /2 y=sinx –1 y=x y=arcsinx /2 (Taking image of sin x about y = x to get sin–1x) (y = sin–1x) 46 E
JEE-Mathematics ( b ) f : [0, ] [–1, 1] f 1 : [1, 1] [0, ] f 1 (x) cos1 x f(x) = cos x y=x y y y=arc cosx /2 1 /2 /2 –1 0 x –1 y=cosx –1 O 1 x y=x (Taking image of cos x about y = x) (y = cos–1x) ( c ) f : (–/2, /2) R f1 : R (–/2, /2) f(x) = tan x f 1 (x) tan1 x y y y=tanx y=x /2 /2 y=arc tanx y=arc tanx /2 0 /2 x y=arc tanx 0 x y=arc tanx /2 –/2 y=arc cotx y=x x y=tanx (Taking image of tan x about y = x) (y = tan–1x) f 1 : R (0, ) ( d ) f : (0, ) R f 1 (x) cot1 x f(x) = cot x y y y=x /2 y=arc cotx y=arc cotx Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 /2 y=arc cotx 0 0 x –/2 y=x (y = cot–1x) y=cotx y (Taking image of cot x about y = x) ( e ) f : [0, /2) (/2, ] (, –1] [1, ) /2 f(x) = sec x f 1 : (, –1] [1, ) [0, /2) (/2,] f 1 (x) sec1 x –1 0 x E 47
JEE-Mathematics ( f ) f : [–/2, 0) (0, /2] (, –1] [1, ) y /2 f(x) = cosec x –1 0 /2 x f 1 : (, –1] [1, ) [–/2, 0) (0, /2] f 1 (x) cosec1 x From the above discussions following IMPORTANT points can be concluded. (i) All the inverse trigonometric functions represent an angle. (ii) If x 0, then all six inverse trigonometric functions viz sin–1 x, cos–1 x, tan–1 x, sec–1x, cosec–1x, cot–1x represent an acute angle. (iii) If x < 0, then sin–1x, tan–1x & cosec–1x represent an angle from /2 to 0 (IVth quadrant) (iv) If x < 0, then cos–1 x, cot–1x & sec–1x represent an obtuse angle. (IInd quadrant) (v) IIIrd quadrant is never used in inverse trigonometric function. Illustration 1 : The value of tan–1(1) + cos–1 1 + sin–1 1 is equal to 2 2 5 3 13 (A) 4 (B) 12 (C) 4 (D) 12 Solution : tan–1 ( 1) + cos–1 1 + sin–1 1 2 3 Ans.(C) 2 2 4 36 4 2 4 2n 2n Illustration 2 : If cos1 xi 0 then find the value of xi i 1 i 1 Solution : We know, 0 cos–1 x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Hence, each value cos–1x1, cos–1x2,cos–1x3,......,cos–1x2n are non-negative their sum is zero only when each value is zero. i.e., cos–1xi = 0 for all i xi = 1 for all i 2n xi x1 x2 x3 ...... x2n = {111......... 1} 2n {using (i)} i 1 2 n times 2n Ans. xi 2n i 1 Do yourself - 1 : ( i ) If are roots of the equation 6x2 + 11x + 3 = 0, then (A) both cos–1 and cos–1 are real (B) both cosec–1 and cosec–1 are real (C) both cot–1 and cot–1 are real (D) none of these ( i i ) If sin–1x + sin–1y = and x = ky, then find the value of 392k + 5k. 48 E
3. PROPERTIES OF INVERSE CIRCULAR FUNCTIONS : JEE-Mathematics P - 1 (i) y = sin (sin–1x) = x y x [–1,1], y [–1,1] 1 y = x (ii) y = cos (cos–1 x) = x 45° x [–1,1], y [–1,1] –1 O + 1 x –1 y 1 y = x 45° –1 O + 1 x –1 (iii) y = tan(tan–1 x) = x y x x R, y R y=x 45° O (iv) y = cot(cot–1 x) = x, y y=x x O x R; y R y = x y y = x y = x1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 (v) y = cosec (cosec–1 x) = x, –1 O x |x| > 1, |y| > 1 1 –1 (vi) y = sec(sec–1 x) = x y |x| > 1 ; |y| > 1 1 –1 O 1 x –1 y = x Note : All the above functions are aperiodic. E 49
JEE-Mathematics (ii) cos(tan–1 3/4) (iii) sin sin 1 1 2 2 Illustration 3 : Evaluate the following : (i) sin(cos–13/5) Solution : (i) Let cos–1 3/5 = . Then, cos = 3/5 sin = 4/5 sin(cos–1 3/5) = sin = 4/5 (ii) Let tan–1 3/4 = . Then, tan = 3/4 4 as cos2 1 1 cos = 5 tan2 cos(tan–1 3/4) = cos = 4/5 (iii) sin sin 1 1 sin = sin 2 3 Ans. 2 2 2 6 3 2 Do yourself - 2 : Evaluate the following : (i) ta n co s 1 8 (ii) sin 1 co s 1 4 (iii) cos sin 1 3 17 2 5 5 sin 1 1 cos sin 1 1 sin cos 1 3 (iv) sin 3 2 (v) 2 (vi) 5 P - 2 (i) y = sin–1 (sin x), x R, y , periodic with period 2 and it is an odd function. (ii) 2 2 y x , x y = –x 2 y = –(+ x) 2 y = 2 + x – y = x–2 y=x2 45° 3 sin 1 (sin x) x , x 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 2 2 –2 O – 3 – 2 2 x 2 – x x , x 2 2 2 E y = cos–1 (cos x), x R, y [0,], periodic with period 2 and it is an even function. y c o s 1 (cos x) x , x 0 y = x + 2 y = 2–x–y = x x , 0x y = –x 2 – 2 – – O 2 2 (iii) y = tan–1 (tan x) x R – (2n 1) ,n ; y , , periodic with period and it is an odd function. 2 2 2 50
JEE-Mathematics y x , 3 x y = x + , 22 y= x tan 1 (tan x ) x , x y = x– 2 x 22 x –2 – – O x 3 22 22 – 2 (iv) y = cot–1(cot x), x R – {n , n }, y (0, ), periodic with period and neither even nor odd function. y x , x 0 y = x + 2 cot1 (cot x) x y = x + , 0x y = x y = x – x , x 2 –2 – O 2 (v) y = cosec–1 (cosec x), x R – {n , n } y , 0 0, , is periodic with period 2 and it is an odd 2 2 function. y y = –x y = –(+ x) 2 y=x y = x–2 – 2 45° 3 2 – 3 – O 2 2 2 x – 2 (vi) y = sec–1 (sec x), y is periodic with period 2 y and it is an even function. y = x – Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 y = x + 2 y = 2 – x2 (2n , 0, y = –x x R – 1) 2 n y 2 2 , –2 – 3 – – O 3 2 x 22 22 Illustr ation 4 : The value of sin–1 (– 3 /2) + cos–1 (cos (7 / 6) is - (A) 5 / 6 (B) / 2 (C) 3 / 2 (D) none of these Solution : sin–1 3 / 2 = – sin–1 3 / 2 = – / 3 and cos–1 (cos (7 / 6) = cos–1 cos (2 – 5 / 6) = cos–1 cos ( 5 / 6) = 5/6 Hence sin–1 (– 3 /2 ) + cos–1 (cos 7 / 6) = 5 Ans.(B) 3 6 2 E 51
JEE-Mathematics (ii) co s 1 cos 7 6 Illustration 5 : Evaluate the following : (i) sin–1(sin/4) Solution : (i) sin–1(sin/4) = 4 (ii) cos–1 co s 7 7 , because 7 does not lie between 0 and . 6 6 6 Now, cos–1 cos 7 cos–1 co s 2 5 7 2 5 6 = 6 6 6 = co s 1 cos 5 = 5 Ans. 6 6 Illustration 6 : Evaluate the following : (i) sin–1(sin10) (ii) tan–1(tan (– 6)) (iii) cot–1(cot 4) (i) We know that sin–1(sin) = , if –/2 /2 Solution : Here, = 10 radians which does not lie between –/2 and /2 But, 3 – i.e., 3 – 10 lie between – 2 and 2 Also, sin(3 – 10) = sin 10 sin–1(sin 10) = sin–1 (sin (3 – 10)) = (3 – 10) (ii) We know that, tan–1(tan) , if –/2 < < /2. Here, = –6, radians which does not lie between –/2 and /2. We find that 2 – 6 lies between –/2 and /2 such that; tan (2 – 6) = –tan 6 = tan(–6) tan–1(tan(–6)) = tan–1 (tan(2 – 6)) = (2 – 6) (iii) cot–1(cot4) = cot–1(cot( + (4 – ))) = cot–1(cot(4 – )) = (4 – ) Ans. Illustration 7 : Prove that sec2(tan–12) + cosec2(cot–1 3) = 15 Solution : We have, Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 sec2 (tan–12) + cosec2 (cot–13) 22 2 3 2 2 sec 1 cos 1 cos ec cot1 3 1 co t 1 sec tan1 2 tan ec 22 2 2 sec sec1 5 cosec cosec –1 10 5 10 15 Illustration 8 : Find the number of solutions of (x, y) which satisfy |y| = cos x and y = sin–1(sin x), where |x| 3. Solution : Graphs of y = sin–1(sinx) and |y| = cosx meet exactly six times in [–3, 3]. y 5 3 –3 2 2O 2 3 x 2 5 –2 3 2 22 52 E
JEE-Mathematics Do yourself - 3 : Evaluate the following : co s 1 cos 1 3 ta n 1 ta n 7 ( i i i ) sin–1(sin2) sin 1 sin 5 6 6 6 (i) (ii) (iv) (v) tan 1 ta n 2 (vi) tan 1 tan 3 (vii) cos 1 cos 4 3 4 3 P - 3 (i) –1 < x < 1 sin–1 x + cos–1 x = x R (ii) |x| > 1 2 (iii) P - 4 (i) tan–1 x + cot–1 x = 2 (ii) (iii) (iv) cosec–1 x + sec–1 x = (v) (vi) 2 P - 5 (i) sin–1 (–x) = – sin–1 x , –1 < x < 1 (ii) cosec–1(–x) = – cosec–1 x, x < –1 or x > 1 (iii) tan–1 (–x) = – tan–1 x , x R cot–1 (–x) = – cot–1 x , x R cos–1 (–x) = – cos–1 x , –1 < x < 1 sec–1 (–x) = – sec–1 x , x <–1 or x > 1 1 x <–1, x >1 x <–1, x >1 cosec–1 x = sin–1 ; x 1 sec–1 x = cos–1 ; x tan 1 1 ; x 0 x ; x 0 cot 1 x 1 tan 1 x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Illustration 9 : Prove that tan 1 x tan 1 1 /2 , if x 0 x / 2 , if x 0 tan 1 1 cot1 x , for x 0 x , for x 0 Solution : We have , co t 1 x tan 1 x tan 1 1 tan 1 x cot 1 x /2 , if x 0 x cot1 x /2 , if x 0 tan 1 x / 2 Do yourself - 4 : (i) Prove the following : (a) c o s 1 5 tan 1 12 (b) sin 1 4 tan 1 4 c o s 1 3 13 5 5 3 5 1 ( i i ) Find the value of sin(tan–1a + tan–1 a ); a 0 E 53
JEE-Mathematics tan–1 xy where x > 0, y > 0 & xy < 1 1 xy x y P - 6 (i) (a) tan–1 x + tan–1 y = +tan–1 1 xy where x > 0, y > 0 & xy > 1 2 , where x > 0, y > 0 & xy = 1 xy (b) tan–1 x – tan–1 y = tan–1 1 xy where x > 0, y > 0 x y z xyz (c) tan–1x + tan–1y + tan–1 z = tan–1 1 xy yz zx if x > 0, y > 0, z > 0 & xy + yz + zx < 1 (ii) (a) sin –1 x + sin –1 y = sin 1 [x 1 y2 y 1 x2 ] where x > 0, y > 0 & (x2 + y2 ) 1 sin 1 [x 1 y2 y 1 x2 ] where x > 0, y > 0 & x2 y2 > 1 (b) sin–1 x – sin–1 y = sin–1 [x 1 y2 y 1 x2 ] where x > 0, y > 0 (iii) (a) cos–1 x + cos–1 y = cos–1 [xy 1 x2 1 y2 ] where x > 0, y > 0 cos1 xy 1 x2 1 y2 ; x y, x, y 0 (b) cos–1 x – cos–1 y = cos1 xy 1 x2 1 y2 ; x y, x, y 0 Note : In the above results x & y are taken positive. In case if these are given as negative, we first apply P-4 and then use above results. Illustration 10 : Prove that (i) 1 +tan–1 1 = tan–1 2 (ii) tan 1 1 tan 1 1 tan 1 1 tan 1 1 tan–1 7 13 9 5 7 3 84 Solution : (i) L.H.S. = tan–1 1 +tan–1 1 7 13 = tan 1 1 1 tan 1 x tan 1 y tan 1 xy ; if xy Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 7 1 1 xy 1 13 1 1 7 13 = tan 1 20 tan 1 2 R.H.S. 90 9 (ii) tan 1 1 tan 1 1 ta n 1 1 tan 1 1 5 7 3 8 tan 1 1 1 ta n 1 1 1 ta n 1 6 ta n 1 1 1 5 7 3 8 17 2 3 1 1 1 1 1 1 5 7 3 8 tan 1 6 11 tan 1 325 tan 1 (1) Ans. 17 325 4 6 23 E 11 17 1 23 54
JEE-Mathematics Illu stration 11 : Prove that sin 1 12 cot1 4 tan 1 63 13 3 16 Solution : We have, sin 1 12 cos 1 4 tan 1 63 13 5 16 tan 1 12 tan 1 3 tan 1 63 sin 1 12 tan 1 12 and co s 1 4 tan 1 3 5 4 16 13 5 5 4 tan 1 12 3 tan 1 63 tan 1 x tan 1 y tan 1 xy , if 5 4 16 xy 1 12 3 1 xy 1 5 4 tan 1 63 tan 1 63 16 16 = tan 1 63 tan 1 63 tan 1 (x ) tan 1 x 16 16 = Illustration 12 : Prove that : cos 1 12 sin 1 3 sin 1 56 13 5 65 Solution : We have, cos1 12 sin 1 3 sin 1 5 sin 1 3 cos1 12 sin 1 5 13 5 13 5 13 13 sin 1 5 1 3 2 3 1 5 2 sin 1 5 4 3 1 2 sin 1 56 13 5 5 13 13 5 5 1 3 65 Illustration 13 : If x = cosec(tan–1(cos(cot–1(sec(sin–1a))))) and y = sec(cot–1(sin(tan–1(cosec(cos–1a))))), where a [0, 1]. Find the relationship between x and y in terms of 'a' Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Solution : Here, x = cosec(tan–1(cos(cot–1(sec(sin–1a))))) Let sin = a sec = 1 = cosec(tan–1(cos(cot–1(sec)))) 1 a2 x = cosec ta n 1 co s c o t 1 1 Let co t 1 1 cot 1 1 a2 1 a2 1 a2 1 2 a2 = cosec(tan–1(cos)) therefore cos 1 1 1 2 a2 2 a2 2 a2 x = cosec ta n 1 Let, tan 1 tan = cosec therefore cosec 3 a2 x = 3 a2 ...... (i) E 55
JEE-Mathematics and y = sec(cot–1(sin(tan–1(cosec(cos–1 a))))) Let cos–1a = cos = a cosec = 1 1 a2 = sec(cot–1(sin(tan–1(cosec)))) y = sec co t 1 sin tan 1 1 Let, tan 1 1 tan 1 a2 1 a2 1 a2 1 1 = sec(cot–1(sin())) sin 2 a2 y = sec co t 1 1 Let cot1 1 cot 1 sec 3 a2 2 a2 2 a2 2 a2 = sec y = 3 a2 ...... (ii) from (i) and (ii), x = y = 3 a2 . Ans. Do yourself - 5 : Prove the following : (i) sin 1 3 sin 1 8 c o s 1 36 (ii) tan 1 3 tan 1 3 tan 1 8 5 17 85 4 5 19 4 (iii) tan 1 2 tan 1 7 tan 1 1 11 24 2 4 . SIMPLIFIED INVERSE TRIGONOMETRIC FUNCTIONS : y /2 2 tan 1 x if | x| 1 D I (a) y f(x) sin 1 2x if x 1 x 1 x2 2 tan 1 x if x 1 2 ta n 1 x) –1 0 1 ( D I Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 –/2 y y f(x) co s 1 1 x2 2 tan 1 x if x 0 D 2 I 1 x2 if (b) 2 tan 1 x x0 –1 0 1 X y /2 2 tan 1 x if | x| 1 I I (c) y f(x) tan 1 2x 2 tan 1 x if x 1 –1 1 x x2 if x 1 I I 1 2 tan 1 x) E ( –/2 56
( d ) y = f(x) = sin–1 (3x – 4x3) JEE-Mathematics y /2 ( 3 sin 1 x ) if 1 D ID 1 x – 3/2 –1/2 +1/2 x 2 –1 3/2 1 3 sin 1 x if 1 x 1 I 22 1 x 1 2 3 sin 1 x if –/2 ( e ) y f(x) cos1 (4 x3 3x) y 3 cos1 x 2 if 1 D ID 1 x –1 – 3/2 2 2 3 cos1 x if 1 x 1 /2 if 2 2 x 1 x 1 2 3 cos1 x I –1/2 +1/2 3/2 1 2 sin –1 x 1 y 1 x /2 2 1 – ( f ) sin1 2x 1 x2 2 sin1 x 1 x 1 x 22 2 1 2 2 sin 1 x 1 x 1 /2 2 y –1x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 2 cos1 x 0 x 1 2cos 2cos–1 x 1 x 0 (g) c o s 1 2 x2 1 – 2 2 2 cos 1 x –1 0 1x tan 2 tan 1 1 ta n 1 cos 1 5 Illustration 14 : Evalulate : (i) 5 (ii) 2 4 3 1 21 tan 2 tan 1 5 ta 5 Solution : (i) tan n 1 ta n 1 1 2 tan 1 x tan 1 2x , if | x| 1 1 1 1 x2 E 4 25 5 1 tan ta n 1 5 tan 1 1 tan tan 1 12 5 tan tan 1 7 7 12 12 17 17 1 57
JEE-Mathematics (ii) Let cos-1 5 5 = . Then, cos = , 0 < < /2 33 Now, ta n 1 c o s 1 5 2 3 tan 1 cos 1 5 3 5 (3 5 )2 (3 5 )2 3 5 2 1 cos 3 3 5 (3 5 )(3 5 ) 9 5 2 1 5 3 Illu stration 15 : Prove that : 2 tan 1 1 tan 1 1 tan 1 31 2 7 17 Solution : We have, 2 tan 1 1 tan 1 1 27 tan 1 2 1 tan 1 1 2 tan 1 2x , if 2 7 1 x2 1 tan 1 1 2 x 1 x 1 2 tan 1 4 tan 1 1 tan 1 4 1 tan 1 31 3 7 3 4 17 7 1 1 3 7 Illustration 16 : Prove that tan 1 x 1 c o s 1 1 x , x 0,1 2 1 x We have, 1 1 x 1 1 x 2 1 2 1 x 2 x 2 Solution : co s 1 co s 1 1 2 2 tan 1 x tan 1 x. Alter : Putting x = tan , we have 0, Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 4 RHS = 1 c o s 1 1 x 1 c o s 1 1 tan2 1 cos 1 (cos 2) 2 0, 2 1 x 2 tan2 2 2 1 tan 1 x LHS Illustration 17 : Prove that : (i) 4 tan 1 1 tan 1 1 tan 1 1 5 70 99 4 (ii) 2 tan 1 1 sec 1 5 2 2 tan 1 1 57 84 Solution : (i) 4 tan 1 1 tan 1 1 tan 1 1 2 2 tan 1 1 tan 1 1 tan 1 1 5 70 99 5 70 99 58 E
JEE-Mathematics 2 tan 1 x 2 t a n 1 2 1/ 5 tan 1 1 tan 1 1 2x 1 (1 / 5 )2 70 99 tan 1 ,if| x| 1 1 x2 1 1 70 2 tan 1 5 tan 1 1 tan 1 1 tan 1 2 5 /12 tan 1 . 1 99 12 70 99 1 (5 / 12)2 1 1 70 99 120 1 119 tan 1 120 tan 1 29 tan 1 120 tan 1 1 tan 1 120 239 tan 1 1 119 6931 119 239 1 1 4 119 239 (ii) 2 tan 1 1 sec1 5 2 2 tan 1 1 2 tan 1 1 tan 1 1 sec 1 52 5 7 8 5 8 7 2 tan 1 1 1 tan 1 2 sec 1 x tan 1 x2 5 1 8 1 5 2 1 1 7 1 5 8 2 tan 1 13 tan 1 1 2 tan 1 1 tan 1 1 39 7 37 tan 1 2 1/ 3 tan 1 1 2 tan 1 x tan 1 1 2x , if| x| 1 1 (1 / 3)2 7 x2 tan 1 3 tan 1 1 tan 1 3 1 tan 1 1 4 7 47 4 3 1 1 4 7 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Do yourself - 6 : Prove the following results : (i) 2 tan 1 1 tan 1 1 tan 1 4 (ii) 2 sin 1 3 tan 1 17 5 8 7 5 31 4 6 . EQUATIONS IN VOLVING IN VERSE TRIGONOMETRIC FUNCTIONS : Illustration 18 : The equation 2cos–1x + sin–1x = 11 has 6 (A) no solution (B) only one solution (C) two solutions (D) three solutions Solution : Given equation is 2 cos–1 x + sin–1x = 11 6 E cos–1x + ( cos–1 x + sin–1x ) = 11 cos–1 x + 11 cos–1 x = 4 / 3 6 2 6 which is not possible as cos–1 x [ 0 , ] Ans.(A) 59
JEE-Mathematics Illu stration 19 : If (tan–1 x)2 + (cot–1 x )2 = 52 / 8 , then x is equal to- (A) –1 (B) 0 (C) 1 (D) none of these Solution : The given equation can be written as (tan–1 x + cot–1 x )2 – 2 tan–1 x cot–1 x = 52 / 8 Since tan–1 x + cot–1 x = /2 we have (/2)2 – 2tan–1 x (/2 – tan–1 x ) = 52 / 8 2(tan–1 x)2 – 2 (/2) tan–1 x – 32 / 8 = 0 tan–1 x = – / 4 x = –1 Ans. (A) Illustration 20 : Solve the equation : tan1 x 1 tan1 x 1 x 2 x2 4 Solu ti on : tan 1 x 1 tan 1 x 1 x 2 x2 4 taking tangent on both sides tan tan 1 x 1 tan tan 1 x 1 x 2 x 2 x 1 x 1 1 ta n ta n 1 x 2 tan 1 x 2 1 1 tan tan 1 x 1 tan tan 1 x 1 x 2 x 2 x 1 x 1 x 2 x2 (x 1)(x 2) (x 2)(x 1) 1 1 x 1. x 1 1 x2 4 (x2 1) 1 2x2 – 4 = – 3 x = ± 2 x 2 x 2 1 Now verify x = 2 1 1 1 1 2 1 2 1 2 2 1 2 2 1 2 tan 1 2 tan 1 tan 1 = tan 1 1 1 = 2 2 2 2 = tan 1 2 2 1 2 1 2 2 1 2 1 = tan 1 6 tan 1 (1) 2 2 12 2 1 2 1 2 6 4 1 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 x = – 2 1 1 1 1 2 1 2 1 tan 1 2 2 1 2 1 tan 1 2 tan 1 2 tan 1 = 1 1 = {same as above} 2 2 2 2 2 = tan1 (1) Ans. 4 1 x = ± are solutions 2 Illustration 21 : Solve the equation : 2 tan–1(2x + 1) = cos–1x. Solution : Here, 2 tan–1(2x + 1) = cos–1x 1 tan2 or cos(2tan–1(2x +1)) = x W e kn ow cos 2 1 tan2 60 E
JEE-Mathematics 1 (2x 1)2 1 (2x 1)2 x (1 – 2x – 1)(1 + 2x + 1) = x(4x2 + 4x + 2) – 2x . 2(x + 1) = 2x(2x2 + 2x + 1) 2x(2x2 + 2x + 1 + 2x + 2) = 0 2x(2x2 + 4x + 3) = 0 x= 0 or 2x2 + 4x + 3 = 0 {No solution} Verify x = 0 2tan–1(1) = cos–1(1) 22 x = 0 is only the solution Ans. Do yourself - 7 : Solve the following equation for x : (i) sin 1 1 c o s 1 1 (ii) cos1 x sin 1 x sin 5 x 26 ( i i i ) cot1 x cot1 (x 2) , w h e r e x > 0 . 12 7 . INEQUATIONS IN VOLVING IN VERSE TRIGONOMETRIC FUNCTION : Illustration 22 : Find the complete solution set of sin–1(sin5) > x2 – 4x. Solution : sin–1(sin5) > x2 – 4x sin–1[sin(5 – 2)] > x2 – 4x x2 – 4x + (2 – 5) < 0 x2 – 4x < 5 – 2 2 9 2 x 2 9 2 x (2 9 2, 2 9 2) Ans. Illustration 23 : Find the complete solution set of [cot–1x]2 – 6[cot–1x] + 9 0, where [.] denotes the greatest integer function. Solution : [cot–1x]2 – 6[cot–1x] + 9 0 ([cot–1x] – 3)2 0 [cot–1x] = 3 3 cot–1x < 4 x (–, cot3] Illustration 24 : If cot–1 n , n N , then the maximum value of n is - 6 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 (A) 1 (B) 5 (C) 9 (D) none of these Solution : cot–1 n 6 cot co t 1 n cot n 3 6 n 3 n < 5.5 (approx) Ans. (B) n = 5 (n N) Do yourself - 8 : (i ) Solve the inequality tan–1x > cot–1x. ( i i ) Complete solution set of inequation (cos–1x)2 – (sin–1x)2 > 0, is (A) 0, 1 (B) 1, 1 (C) (1, 2 ) (D) none of these 2 2 E 61
JEE-Mathematics 8 . SUMMATION OF SERIES : Illustration 25 : Prove that : tan 1 c1 x y tan 1 c 2 c1 tan 1 c 3 c2 ... tan 1 c n cn1 tan 1 1 tan 1 x c1 y x 1 c2 c1 1 c3c2 1 c n c n1 cn y Solution : L.H.S. tan1 c1 x y tan1 c2 c1 tan1 c3 c 2 ... tan1 cn cn1 tan1 1 c1 y x 1 c2c1 1 c3c2 1 c n c n 1 cn x1 1 1 1 1 1 1 tan1 tan1 c1 1 c2 tan1 c2 1 c3 ... tan1 cn tan1 1 y c1 c1 1 c2 1 cn1 1 1 x . 1 1 . 1 . 1 1 . cn y c1 c2 c3 cn1 cn tan 1 x tan 1 1 ta n 1 1 tan 1 1 ta n 1 1 tan 1 1 y c1 c1 c2 c2 c3 +...... ta n 1 1 tan 1 1 tan 1 1 c n 1 cn c n tan 1 x R.H.S. y Do yourself - 9 : ( i ) 2 Evaluate : 1)(2r tan 1 1 (2r 1) r 1 Miscellaneous Illustrations : Illustration 26 : If tan–1 y = 4 tan–1 x, | x| tan , find y as an algebraic function of x and hence prove that 8 Solution : tan is a root of the equation x4 – 6x2 + 1 = 0. 8 (as |x| < 1) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 We have tan–1 y = 4 tan–1 x 2x tan–1 y = 2 tan–1 1 x2 4x = tan 1 1 x2 = tan 1 4 x(1 x2 ) 2x 4x2 x4 6x2 1 as 1 x2 1 1 (1 x2 )2 y 4 x(1 x2 ) x4 6x2 1 If x = tan 8 tan 1 y 4 tan 1 x y is not defined x4 – 6x2 + 1 = 0 Ans. 2 E Illustration 27 : If A = 2 tan–1 2 2 1 and B = 3 sin–1(1/3) + sin–1(3/5), then show A > B. Solution : We have, A = 2tan–1 2 2 1 = 2tan–1(1.828) A > 2tan–1 3 A 2 ....... (i) 3 62
JEE-Mathematics also we have, sin 1 1 sin 1 1 sin 1 1 3 2 3 6 3 sin 1 1 3 2 1 1 1 3 23 3 3. 3 3 27 also, 3 sin 1 sin 1 4 = sin 1 = sin–1(0.852) 3sin–1(1/3) < sin–1 3 / 2 3sin–1(1/3) < /3 also, sin–1(3/5) = sin–1 (0.6) < sin–1 3 / 2 sin–1(3/5) < /3 Hence, B = 3sin–1 (1/3) + sin–1 (3/5) < 2 ........ (ii) 3 From (i) and (ii), we have A > B. Illu stration 28 : Solve for x : If [sin–1cos–1sin–1tan–1x] = 1, where [.] denotes the greatest integer function. Solution : We have, [sin–1cos–1sin–1tan–1x] = 1 1 sin–1 . cos–1 . sin–1 . tan–1x sin1 cos–1 . sin–1 . tan–1x 1 2 sin cos sin1 tan–1x sin cos1 cos sin1 sin–1 . tan–1x cos1 tan sin cos sin1 x tan sin cos1 Hence, x [tan sin cos 1, tan sin cos sin1] Ans. Illustration 29 : If = tan–1(2 tan2) 1 sin 1 3 sin 2 then find the sum of all possible values of tan. 2 4 cos2 5 Solution : = tan–1(2 tan2) 1 sin 1 3 sin 2 = tan–1(2 tan2) 1 sin 1 6 tan 2 4 cos 2 2 9 tan2 5 1 2 1 tan 2 1 2 3 2 3 = tan–1(2 tan2) sin 1 = tan–1(2 tan2) tan 1 2 tan 1 1 3 tan = tan–1(2 tan2) tan 1 1 tan ........ (i) 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 taking tangent on both sides 6 tan2 tan 2tan4 – 6tan2 + 4tan = 0 tan 3 2 tan3 2tan(tan3 –3 tan + 2) = 0 2tan(tan –1)2 (tan + 2) = 0 tan = 0, 1, – 2 which satisfy equation (i) sum = 0 + 1 – 2 = –1 Ans. Illustration 30 : Transform sin–1x in other inverse trigonometric functions, where x (–1, 1) – {0} Solution : Case-I : 0 x 1 Let sin–1x = 0, 2 Now, cos 1 sin2 cos1 1 x2 1 x sin 1 x cos1 1 x2 se c 1 1 1 x2 1– x2 E 63
JEE-Mathematics x tan 1 x2 tan 1 x sin 1 x tan 1 x 1 x2 1 x2 sin 1 x tan 1 x 1 x2 cot1 x 1 x2 Hence, sin 1 x cos1 1 x2 sec1 1 tan 1 x co t 1 1 x2 cosec 1 1 , 0 x 1 1 x2 x2 x x 1 Case-II : 1 x 0 , 0 Let sin 1 x 2 Then x = sin cos 1 x2 cos 1 x2 cos1 1 x2 sin1 x cos1 1 x2 se c 1 1 Again, tan x 1 x2 1 x2 tan 1 x sin 1 x tan 1 x 1 x2 1 x2 sin 1 x tan 1 x 1 x2 tan 1 x c o t 1 1 , x cot1 x 0 1 x2 x Hence, sin 1 x cos1 1 x2 sec1 1 x cot1 1 x2 cosec 1 1 , 1 x 0 1 x2 tan 1 1 x2 x x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 ANSWERS FOR DO YOURSELF 1 : ( i ) C (ii) 1526 2 : (i) 15 1 4 (iv) 1 (v) 34 8 ( i i ) 10 (iii) 5 (vi) 25 3 : (i) (iii) – 2 6 (ii) 6 (iv) 6 (v) 2 (vi) (vii) 3 4 3 1, if a 0 4 : ( i i ) 1, if a 0 7 : (i) 1 (ii) 1 (iii) 3 (ii) B 8 : (i ) 5 9 : (i ) (1, ) /4 64 E
EXERCISE - 01 JEE-Mathematics CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The value of sin–1 ( 3 / 2) is - (A) –/3 (B) –2/3 (C) 4/3 (D) 5/3 (D) cos(4 cot1 4) 2. cos 2 tan 1 1 equals - (D) –1 7 (A) sin(4 cot1 3) (B) sin(3 cot1 4) (C) cos(3 cot1 4) 3. The value of sec 1 sin 50 co s 1 cos 31 is equal to - sin 9 9 (A) sec 10 (B) sec (C) 1 9 9 4. cos c o s 1 cos 8 tan 1 tan 8 has the value equal to - 7 7 (A) 1 (B) –1 (D) 0 (C) cos 7 5 . (sin 1 x )2 (sin 1 y )2 2(sin 1 x)(sin 1 y ) 2 , then x2+y2 is equal to - (A) 1 (B) 3/2 (C) 2 (D) 1/2 6 . cot–1 [ (cos )1/2 ] – tan–1 [ (cos)1/2 ] = x , then sin x = (A) tan 2 (B) cot 2 (C) tan (D) cot 2 2 2 7 . tan(cos–1 x) is equal to x 1 x2 1 x2 (D) 1 2x (A) 1 x2 (B) (C) x x 1 1 1 x If x = 2cos–1 2 + sin–1 2 + tan–1 2 2 8 . 3 and y = cos sin 1 sin then which of the following state- ments holds good ? Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 3 5 (C) x 4 cos1 y (D) none of these (A) y cos (B) y cos 16 16 9. If x = tan–1 1–cos–1 1 sin 1 1 ; y cos 1 cos 1 1 then - 2 2 2 8 (A) x = y (B) y = x (C) tan x = –(4/3)y (D) tan x = (4/3)y 1 0 . tan–12+ tan–13 = cosec–1x, then x is equal to - (A) 4 (B) 2 (C) 2 (D) none of these 1 1 . The number k is such that tan {arc tan(2) + arc tan(20k)}=k. The sum of all possible values of k is - 19 21 (C) 0 1 (A) (B) (D) 40 40 5 12. If sin–1 x + cot–1 1 = , then x is - 2 2 (A) 0 1 2 3 (B) 5 (C) 5 (D) E 65 2
JEE-Mathematics 1 3 . If tan(cos–1x) = sin (cot–1 1/2) then x is equal to - (A) 1/ 5 (B) 2 / 5 (C) 3 / 5 (D) 5 / 3 1 4 . sin 1 (2 x 1 x2 ) 2 sin 1 x is true if - (B) 1, 1 1 1 3, 3 2 2 2 2 (A) x [0,1] 2 (C) , (D) 2 1 5 . Domain of the explicit form of the function y represented implicitly by the equation (1+x) cosy – x2 = 0 is - 1, 1 5 1 5 ,1 5 1 5 2 2 0, 2 (A) (–1,1] (B) (C) 2 (D) 1 6 . If cos1 x cos1 y , then 4x2 – 4xy cos + y2 is equal to - 2 (A) –4sin2 (B) 4sin2 (C) 4 (D) 2 sin 2 (D) x2+ y2+ z2+ 2xyz = 1 1 7 . If cos–1 x + cos–1 y + cos–1 z = , then - (A) x2+ y2+ z2+ xyz = 0 (B) x2+ y2+ z2+ xyz = 1 (C) x2+ y2+ z2+ 2xyz = 0 1 8 . If tan 1 x , x N , then the maximum value of x is - (D) none of these 3 (A) 2 (B) 5 (C) 7 1 9 . The solution of the inequality (tan 1 x)2 3 tan 1 x 2 0 is - (A) , tan1 tan 2, (B) , tan1 (C) , tan1 tan 2, (D) tan 2, 2 0 . The set of values of x, satisfying the equation tan2(sin–1x) > 1 is - 2, 2 (1,1) 2, 2 2, 2 2 2 (D) [–1,1] 2 2 (A) [–1,1] (B) 2 (C) 2 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 21. If numerical value of tan c o s 1 4 tan 1 2 is a , then - 5 3 b (A) a + b = 23 (B) a – b = 11 (C) 3b = a + 1 (D) 2a = 3b 1 cos 1 co s 1 4 2 5 22. The value of cos is/are - (A) cos 7 (B) sin (C) cos 2 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 5 10 5 (D) –cos 5 23. tan 1 1 tan 1 2 equals to 4 9 (A) 1 c o s 1 3 (B) 1 sin 1 3 (C) 1 tan 1 3 (D) tan 1 1 2 5 2 5 2 5 2 2 4 . sin 1 3x sin 1 4 x sin 1 x , then roots of the equation are - 55 (A) 0 (B) 1 (C) –1 (D) –2 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 23 4 5 6 7 8 9 10 11 12 13 14 15 Ans. A AD BCACACDABDBC Que. 16 17 18 19 20 21 22 23 24 Ans. B DB B C A,B,C B,C,D A, D A,B,C 66 E
EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . cos–1x = tan–1x then - 5 1 5 1 (A) x2 2 (B) x2 2 (C) s in (c o s 1 x) 5 1 (D) ta n (co s 1 x) 5 1 2 2 2. The value of sin 1 c o t 1 3 cos 1 c o t 1 3 is/are equal to - 2 4 2 4 3. 4. (A) 1 32 5. (B) 6. 10 7. 8. (C) 2 sin 1 c o t 1 3 c o t 1 (1) (D) 2 sin tan 1 (1) 1 tan 1 4 9. 2 4 2 3 E The value of tan 1 1 tan 2 A tan 1 (cot A) tan 1 (cot3 A) for 0 < A < (/4) is - 2 (A) 4 tan–1(1) (B) 2 tan–1(2) (C) 0 (D) none For the equation 2x = tan(2tan–1a) + 2tan(tan–1a+tan–1a3), which of the following is/are invalid ? (A) a2x + 2a = x (B) a2 + 2ax +1= 0 (C) a 0 (D) a 1, 1 1 sin 1 a tan 1 sin 1 a 1 The value of tan 4 2 b 2 b , where ( 0 < a < b), is - 4 b a (C) b2 a2 (D) b2 a2 (A) (B) 2b 2a 2a 2b Identify the pair(s) of functions which are identical - (A) y = tan (cos–1x) ; y = 1 x2 1 x (B) y = tan (cot–1x) ; y = x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 x (D) y = cos (arc tan x) ; y = sin (arc cot x) (C) y = sin (arc tan x) ; y = 1 x2 Which of the following, satisfy the equation 2 cos1 x co t 1 2x2 1 4x2 4x4 (A) (–1, 0) (B) (0, 1) (C) – 1 , 1 (D) [–1, 1] 2 2 The solution set of the equation sin 1 1 x2 is - 1 x2 cos1 x cot1 x – sin–1x (A) [–1, 1]–{0} (B) (0,1] U {–1} (C) [–1,0) U {1} (D) [–1,1] If 0 < x < 1, then tan–1 1 x2 is equal to - 1x (A) 1 cos1 x (B) cos1 1 x (C) sin 1 1 x (D) 1 tan 1 1 x 2 2 2 2 1x 67
JEE-Mathematics 1 0 . The number of real solutions of tan–1 x(x 1) sin 1 x2 x 1 is - [JEE 99] 2 (A) zero (B) one (C) two (D) infinite 1 1 . If [sin–1x] + [cos–1x] = 0, where ‘x’ is a non negative real number and [.] denotes the greatest integer function, then complete set of values of x is - (A) (cos1, 1) (B) (–1, cos1) (C) (sin1, 1) (D) (cos1, sin1) 1 2 . Value of k for which the point (, sin–1)( > 0) lies inside the triangle formed by x + y = k with co-ordinate axes is - (A) 1 , (B) 1 , 1 (C) , 1 (D) (–1–sin1, 1+sin1) 2 2 2 2 13. Solution set of the inequality sin 1 2x2 3 5 is - sin x2 1 2 (A) ( , 1) (1 , ) (B) [–1, 1] (C) (–1, 1) (D) ( , 1] [1 , ) 14. Consider two geometric progressions a ,a ,a .......a & b , b , b ,.....b with a = 1 2r1 and another se- 123 n 123 n r br n quence t ,t ,t .......t such that t = tr 123 n r cot–1 (2a + b ) then lim is - r r n r 1 (A) 0 (B) / 4 (C) tan–12 (D) / 2 1 5 . The sum of the infinite terms of the series - co t 1 12 3 co t 1 2 2 3 co t 1 32 3 ........... is equal to - 4 4 4 (A) tan–1(1) (B) tan–1(2) (C) tan–1(3) (D) 3 tan 1 3 4 BRAIN TEASERS ANSWER KEY EXERCISE-2Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Que. 1 2 3 456 7 8 9 10 Ans. A,C B,C,D A B,C C A,B,C,D B C A,B,C C Que. 11 13 14 15 Ans. D 12 B E A B B,D 68
EXERCISE - 03 JEE-Mathematics MISCELLANEOUS TYPE QUESTIONS FILL IN THE BLANKS 1. tan 1 1 tan 1 1 = .................. 2. cos (tan 1 2) = ................... cos 2 3 3. tan sin 1 3 co t 1 3 = ................... 4. cos s 1 3 = ................... 5 2 co 2 6 5. sin 1 3 cos1 11 cot1 3 = .............. 73 146 6. tan 1 1 sin 1 1 c o s 1 1 cot 1 1 2 2 5 10 1 2 = ................. 7. sin sin 1 3 = ................. 2 2 8. co s 1 1 cos1 1 co s 1 10 1 4 cot 1 1 = ................. 3 6 3 2 9. tan 1 3 sin 2 tan 1 tan , where = ................... 5 3 cos2 4 2 2 1 0 . The number of roots of the equation sin x cos1 (cos x) is ................... MATCH THE COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E s t a t e m e n t i n C o l u m n - I I . 1 . Column-I Column-II (p) –2/7 (A) sin–1 sin 33 (q) 2/7 7 (r) 3/7 (s) 4/7 (B) cos –1 co s 4 6 7 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 (C) t a n – 1 ta n 33 7 (D) cot–1 co t 46 7 2 . Column-I Column-II (A) sin(tan–1x) (B) cos(tan–1x) (p) x x (C) sin(cot–1(tan(cos–1x ))). x (0,1] (D) sin(cosec–1(cot(tan–1x))) ; x (0,1] (q) x2 1 E 1 (r) x2 1 (s) 1 x2 69
JEE-Mathematics Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E O R M O R E s t a t e m e n t ( s ) i n C o l u m n - I I . 3 . x > 0, y > 0, z > 0 and tan–1x + tan–1y + tan –1z = k, the possible value(s) of k, if Column-I Column-II (A) xy + yz + zx = 1, then (B) x + y + z = xyz, then (p) k = (C) x2 + y2 + z 2 = 1 and x + y + z = 3 , then 2 (D) x = y = z and xyz 3 3 , then (q) k = (r) k = 0 7 (s) k = 6 ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1. Statement-I : Range of cos sec 1 1 cos ec 1 1 tan 1 x is 1, 1 x x 2 2 Because Statement-II : Range of sin–1 x + tan–1 x + cos–1 x is , 3 . 4 4 (A) A (B) B (C) C (D) D 2 . Statement-I : If r, s & t be the roots of the equation : x(x – 2)(3x – 7) = 2, then tan–1r + tan–1s + tan–1t = 3/4. Because Statement-II : The roots of the equation x(x – 2)(3x – 7) = 2 are real & negative. (A) A (B) B (C) C (D) D 2n nn n sin1 xi n, n N . Then 2 3 3 . x i x i Statement-I : If xi i 1 i 1 i 1 i 1 Because Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 Statement-II : sin 1 x , x [1, 1] . 22 (A) A (B) B (C) C (D) D 4 . Let : R [0, /2) defined by (x) = tan–1(x2 + x + a), then Statement-I : The set of values of a for which (x) is onto is 1 , . 4 Because 1 Statement-II : Minimum value of x2 + x + a is a – . 4 (A) A (B) B (C) C (D) D 5. Statement-I : cosec–1 cose c 9 9 . 5 5 Because Statement-II : cosec–1(cosecx) = – x ; x , 3 {} 2 2 (A) A (B) B (C) C (D) D 70 E
JEE-Mathematics COMPREHENSION BASED QUESTIONS Comprehension # 1 cos1 x (ii) sin 1 x 0 Consider the two equations in x ; (i) sin y 1 cos y The sets X1, X2 [1, 1] ; Y1, Y2 I {0} are such that X : the solution set of equation (i) 1 X : the solution set of equation (ii) 2 Y : the set of all integral values of y for which equation (i) possess a solution 1 Y : the set of all intergral values of y for which equation (ii) possess a solution 2 Let : C be the correspondence : X Y such that x C y for x X , y Y & (x, y) satisfy (i). 1 1 1 1 1 1 C be the correspondence : X Y such that x C y for x X , y Y & (x, y) satisfy (ii). 2 2 2 2 2 2 On the basis of above information, answer the following questions : 1 . The number of ordered pair (x, y) satisfying correspondence C is 1 (A) 1 (B) 2 (C) 3 (D) 4 2 . The number of ordered pair (x, y) satisfying correspondence C is 2 (A) 1 (B) 2 (C) 3 (D) 4 3. C : X Y is a function which is - 1 1 1 (A) one-one (B) many-one (C) onto (D) into Comprehension # 2 Let h (x) = sin–1(3x – 4x3) ; h (x) = cos–1(4x3 – 3x) & f(x) = h (x) + h (x) 12 12 1 f(x) = a cos–1x + b ; a, b Q when x [–1, ] ; let 2 h1(x) = p sin–1x + q ; p, q Q h (x) = r cos–1x + s ; r, s Q 2 Let C be the circle with centre (p, q) & radius 1 & C be the circle with centre (r, s) & radius 1. 1 2 On the basis of above information, answer the following questions : 1 . p + r + 2q – s = (A) 0 (B) 1 (C) 2 (D) 4 2 . If b.log |p + q|= k.a, then value of k is - |s| 9 (B) 6 3 (D) none of these (A) 2 (C) 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 3 . Radical axis of circle C & C is - 12 (A) 12x – 2y – 3 = 0 (B) 12x + 2y – 3 = 0 (C) –12x + 2y – 3 = 0 (D) none of these MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 Fill in the Blanks 1 1 17 4. –1 5 6. – 1 8. 0 1. 2. 3. 5. 7. 3 5 6 12 2 9. 1 0 . infinite many solutions Match the Column 1. (A) (q), (B) (s), (C) (q), (D) (r) 2. (A) (q), (B) (r), (C) (p), (D) (p) 3. (A) (p), (B) (q,r), (C) (p), (D) (q,s) Assertion & Reason 1. D 2. C 3. A 4. D 5. A Comprehension Based Questions Comprehension # 1 : 1. B 2. D 3. A,C Comprehension # 2 : 1. A 2. C 3. A E 71
JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the domain of definition the following functions. (Read the symbols [ * ] and { * } as greatest integers and fractional part functions respectively) (a) f(x) cos1 2 (b) f(x) 1 2arc sin x 1 2 sin x x x2 (c) e cos1 x co t 1 x 1 1 ln{x} (d) f(x) sin 1 x 3 log10 4 x 2 2 2 (e) f(x) 1 sin x c o s 1 1 {x} log5 (1 4x2 ) (f) f(x) 3 x co s 1 3 2x log6 2| x| 3 sin 1 log2 x 5 2 . Find the domain and range of the following functions. (Read the symbols [ * ] and { * } as greatest integers and fractional part function respectively) (a) y cot1 (2x x2 ) (b) f(x) = sec1 (log3 tan x log tan x 3) co s 1 sin x 1 2 cos x 3 (d) f(x) = tan 1 log 4 (5 x2 8 x 4) (c) f(x) = 2 2 5 3 . Draw the graph of the following functions : (a) f(x) = sin–1(x + 2) (b) g(x) = [cos–1x], where [ ] denotes greatest integer function. (c) h(x) = –|tan–1(3x)| 4. Express f(x) = arc cos x + arc cos x 1 3 3 x2 in simplest form and hence find the values of 2 2 (a) f 2 (b) f 1 3 3 5. If cos1 x cos1 y then prove that x2 2xy y2 sin2 . a b a2 cos b2 ab 6 . Prove that : sin 1 3 sin 1 8 sin 1 77 5 17 85 7 . Prove that : cos1 x 2 sin 1 1 x 2 cos1 1 x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 22 8 . Prove that : tan 1 2 1 tan 1 12 32 5 9 . Prove that : 3 tan 1 1 tan 1 1 tan 1 1 4 20 4 1985 10. If sin2x + sin2y < 1 for all x, y R then prove that sin–1 (tanx . tany) , 2 2 11. Prove that : co t 1 1 ab co t 1 1 bc co t 1 1 ca , (a > b > c > 0) a b b c c a 1 2 . Let cos–1x + cos–1(2x) + cos–1(3x) = . If x satisfies the cubic ax3 + bx2 + cx – 1 = 0, then find the value of a + b + c. 13. If 2 tan 1 1 x & sin 1 1 x2 for 0 < x < 1 then prove that . What is the value of 1 x 1 x2 will be if x > 1 ? 72 E
JEE-Mathematics 1 4 . Solve the following equations : (a) sin 1 x sin 1 2x 3 (b) tan 1 1 tan 1 1 1 tan 1 2 1 2x 4x x2 (c) tan1 (x 1) tan1 (x) tan1 (x 1) tan1 (3x) (d) sin 1 x cos1 x sin 1 3x 2 (e) sin 1 x sin 1 1 x cos1 x (f) 2 tan 1 x cos1 1 a2 cos1 1 b2 a > 0, b > 0 1 a2 1 b2 (g) c o s 1 x2 1 tan 1 2x 2 x2 1 x2 1 3 1 5 . Find the sum of the series : (a) tan 1 1 tan 1 2 ........ tan 1 2n 1 .......... 3 9 22n 1 1 (b) cot1 7 cot1 13 cot1 21 cot1 31 ....... to n terms. (c) tan 1 x2 1 tan 1 x2 1 tan 1 x2 1 tan 1 x2 1 +...... to n terms. x 3x 5x 7x 1 3 7 13 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . (a) 2n , (2n 1) ; n I (b) (not defined for any real x) (c) ( –1, 1) –{0} (d) 1 x 4 (e) x (–1/2, 1/2), x 0 (f) 3 , 2 2 2 . (a ) D : x R R : [ / 4, ) (b) D : x 2 n, 2n (2n 1) , 2n 3 x| x 2n or 2n 5 n I ; R : , 2 2 2 4 4 3 3 2 or n < x < /2 + n x /4 + n Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 ( c ) 2 / 6 , 2 (d) D : x R ; R : , 2 4 y y /2 3 2 0x 3 . (a) –3 –2 –1 0 x (b) /2 (c) 1 /2 –1 cos3 cos2 0 cos1 1 x /2 1 12. 26 13. – 4 . (a) (b) 2cos–1 3 – 3 3 13 11 11 ab 14. (a) x = (b) x = 3(c) x = 0, , (d) x = 1, (e) x = 0, (f) x = 27 22 22 1 ab (g) x = 2 – 3 3 2n 5 arc 15. (a) 4 (b) arc cot (c) arc tan(x + n) – tanx n E 73
JEE-Mathematics EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Find the domain of definition the following functions. (a) f(x) log10 (1 log7 (x2 5x 13)) c o s 1 3 sin 2 9 x 2 cos 1 2 sin x 1 2 2 sin x (b) f(x) sin(cos x) n(2 cos2 x 3 cos x 1) e 2 . Prove that : m 1 (a) sin1 cos(sin1 x) cos1 sin(cos1 x) , x 1 (b) tan 1 m tan 1 n m 4 2 n n m 3 n m 1 4 n 3. Prove that : sin 1 1 sin 1 2 1 + ......+ sin1 n n 1 ......... = 2 6 n(n 1) 2 4 . If arc sin x + arc siny + arc sinz = then prove that : (x , y , z > 0 ) (a) x 1 x2 y 1 y2 z 1 z2 2xyz (b) x4 y 4 z4 4 x2 y2 z2 2 x2 y2 y2z2 z2 x2 5 . Find the integral values of K for which the system of equations ; cos x (arc sin y )2 K2 arc 4 possesses solutions & find those solutions. sin y )2 (arc cos x) 4 (arc 16 6. Express 3 cosec2 1 tan 1 3 sec2 1 tan 1 as an integral polynomial in . 2 2 2 2 7 . Solve the following inequalities : (a) arc cot2x – 5 arc cot x + 6 > 0 (b) arc sin x > arc cos x (c) 4 arc tan2x – 8 arc tan x + 3 < 0 & 4 arc cot x – arc cot2x – 3 0 8 . Find all the positive integral solutions of, tan–1x + cos–1 y3 = sin–1 . 1 y2 10 9. Let f(x) = cot–1(x2 + 4x + 2 –) be a function defined R 0, then find the complete set of real values of 2 for which f(x) is onto. 10. Find all values of k for which there is a triangle whose angles have measure tan 1 1 , ta n 1 1 k and 2 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 tan 1 1 2k . 2 1 1 . Find the range of the function f(x) = (sin 1 x )3 (cos1 x )3 . 1 2 . Find the number of roots of the following equations : (b) 1 x 1 (a) 1 cos 2x 2 sin1 sin x sin sin 1 (log 1 x ) 2 cos sin 2 0 2 (c) | y| cos x and y cot1 (cot x ) in 3 , 5 2 2 BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 21 25 2 2 1 . (a) , (b) 2n ; n 5 . K = 2 ; cos , 1 & cos , –1 6. (2 + 2)( + ) 99 6 44 7. 2 ; (c) tan 1 , cot 1 8. x = 1; y = 2 & x = 2; y = 7 9. 1 17 (a) (cot2, )(–, cot3) ; (b) 2 ,1 2 2 11 3 73 12. (a) Infinite ; (b) zero ; (c) 2 10.k 1 1 . 32 , 8 4 74 E
EXERCISE - 05 [A] JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . The value of cos–1(–1) – sin–1(1) is- [AIEEE-2002] (1) 3 (4) 3 (2) (3) 2 2 2 2 . The trigonometric equation sin–1 x = 2 sin–1a, has a solution for- [AIEEE-2003] (4) |a| < 1 1 11 (3) all real values of a (1) |a| (2) < |a| < 2 2 2 2 [AIEEE-2005] 3 . If cos–1x – cos–1 y = , then 4x2 – 4xy cos + y2 is equal to - (4) –4 sin2 2 (1) 2 sin 2 (2) 4 (3) 4 sin2 4. If sin–1 x + cosec–1 5 = then a value of x is- [AIEEE-2007] 5 4 2 (1) 1 (2) 3 (3) 4 (4) 5 5. The value of co t co s e c 1 5 tan 1 2 is equal to- [AIEEE-2008] 3 3 5 6 3 4 (4) 17 (1) 17 (2) 17 (3) 17 [JEE (Main)-2013] 6 . If x, y, z are in A.P. and tan–1x, tan–1y and tan–1z are also in A.P., then (4) 6x = 4y = 3z (1) x = y = z (2) 2x = 3y = 6z (3) 6x = 3y = 2z PREVIOUS YEARS QUESTIONSNode-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 ANSWER KEY EXERCISE-5 [A] Qu e. 1 2 34 5 6 Ans 2 132 1 1 E 75
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] x2 1 [JEE 2002 (Mains), 5] 1 . Prove that cos tan–1sin cot –1x = x2 2 2 . Domain of f (x) = sin 1 (2x) is - 6 (A) 1 , 1 (B) 1 , 3 (C) 1 , 1 (D) 1 , 1 2 2 4 4 4 4 4 2 3 . If sin (cot–1 (x + 1)) = cos (tan–1x), then x = [JEE 2003 (screening), 3] [JEE 2004 (screening)] 1 (B) 0 1 (D) 1 (A) (C) 2 2 FGH IKJ4. 1 t , then find the value of tan(t). [JEE 2006, 1½ ] tan 1 i1 2i2 5 . Let (x, y) be such that sin–1 (ax) + cos–1 (y) + cos–1 (bxy) = [JEE 2007, 6] 2 Match the statements in column-I with statements in column-II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrixgiven in the ORS. Column-I Column-II Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 (A) If a = 1 and b = 0, then (x, y) (p) lies on the circle x2 + y2 = 1 (B) If a = 1 and b = 1, they (x, y) (q) lies on (x2 – 1) (y2 – 1) = 0 (C) If a = 1 and b = 2, then (x, y) (r) lies on y = x (D) If a = 2 and b = 2, then (x, y) (s) lies on (4x2 – 1) (y2 – 1) = 0 6 . If 0 < x< 1, then 1 x2 [{xcos(cot–1x) + sin(cot–1x)}2 – 1]1/2 = [JEE 2008, 3] x (B) x (C) x 1 x2 (D) 1 x2 (A) 1 x2 23 1 n cot cot 1 7 . 2k is [JEE (Advanced) 2013, 2] The value of n 1 k 1 23 25 23 24 (A) 25 (B) 23 (C) (D) 23 24 E 76
JEE-Mathematics 8 . Match List-I with List-II and select the correct answer using the code given below the lists. List-I List-II P. 2 1/2 1 cos tan 1 y y sin tan 1 y y4 15 tan sin 1 y 1. y2 cot sin 1 y takes value 23 Q. If cosx + cosy + cosz = 0 = sinx + siny + sinz then 2. 2 1 possible value of cos x y is 2 3. 2 R. If cos x cos2x + sinx sin2x secx = cosx sin2x secx+ 4 cos x cos2x then possible value of secx is 4 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#03\\ENG\\02.INVERSE\\INVERSE.p65 S. If cot sin1 1 x2 sin tan1 x 6 , x 0 , 4. 1 [JEE-Advanced 2013, 3, (–1)] then possible value of x is Codes : P QR S (A) 4 3 1 2 (B) 4 3 2 1 (C) 3 4 2 1 (D) 3 4 1 2 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 2 . D 3 . A 4 . 1 5 . (A) - p, (B) - q, (C) - p, (D) - s 6 . C 7. B 8. B E 77
JEE-Mathematics LIMIT 1. INTRODUCTION : The concept of limit of a function is one of the fundamental ideas that distinguishes calculus from algebra and trigonometry. We use limits to describe the way a function f varies. Some functions vary continuously; small changes in x produce only small changes in f(x). Other functions can have values that jump or vary erratically. We also use limits to define tangent lines to graphs of functions. This geometric application leads at once to the important concept of derivative of a function. 2 . DEFINITION : Let f(x) be defined on an open interval about ‘a’ except possibly at ‘a’ itself. If f(x) gets arbitrarily close to L (a finite number) for all x sufficiently close to ‘a’ we say that f(x) approaches the limit L as x approaches ‘a’ and we write Lim f(x) L and say “the limit of f(x), as x approaches a, equals L”. xa This implies if we can make the value of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. 3. LEFT HAND LIMIT AND RIGHT HAND LIMIT OF A FUNCTION : E The value to which f(x) approaches, as x tends to ‘a’ from the left hand side (x a–) is called left hand limit of f(x) at x = a. Symbolically, LHL = Lim f(x)= Lim f(a –h). xa h 0 The value to which f(x) approaches, as x tends to ‘a’ from the right hand side (x a+) is called right hand limit of f(x) at x = a. Symbolically, RHL = Lim f(x)= Lim f(a + h). xa h 0 Limit of a function f(x) is said to exist as, x a when Lim f(x) Lim f(x) = Finite quantity. xa x a Example : Graph of y = f(x) Lim f(x) Lim f(1 h) f(1 ) 1 x 1 h 0 y Lim f(x) Lim f(0 h) f(0 ) 0 x 0 h 0 1 Lim f(x) Lim f(0 h) f(0 ) 0 x 0 h0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 x Lim f(x) Lim f(1 h) f(1 ) 1 –1 0 1 2 x 1 h0 –1 Lim f(x) Lim f(1 h) f(1 ) 0 x 1 h0 Lim f(x) Lim f(2 h) f(2_ ) 1 x 2 h0 Fig. 1 Lim f(x) 0 and Lim f(x) does not exist. x0 x 1 Important note : In Lim f ( x ) , x a necessarily implies x a . That is while evaluating limit at x = a, we are not concerned x a with the value of the function at x = a. In fact the function may or may not be defined at x = a. Also it is necessary to note that if f(x) is defined only on one side of ‘x = a’, one sided limits are good enough to establish the existence of limits, & if f(x) is defined on either side of ‘a’ both sided limits are to be considered. As in Lim cos1 x 0 , though f(x) is not defined for x >1, even in it’s immediate vicinity. x 1 1
JEE-Mathematics y 4 Illustration 1 : Consider the adjacent graph of y = ƒ (x) 3 Find the following : 2 1 (a) lim ƒ(x) (b) lim ƒ(x) (c) lim ƒ(x) 0 1 2345 6 x 0 x 0 x 1 –1 (d) lim ƒ(x) (e) lim ƒ(x) (f) lim ƒ(x) x x 1 x 2 x 2 (g) lim ƒ(x) (h) lim ƒ(x) (i) lim ƒ(x) x 3 x 3 x 4 (j) lim ƒ(x) (k) lim ƒ(x) 2 (l) lim ƒ(x) x 4 x x6 Solution : (a) As x 0– : limit does not exist (the function is not defined to the left of x = 0) (b) As x 0+ : ƒ (x) –1 lim ƒ(x) = –1. (c) As x 1– : ƒ (x) 1 lim ƒ(x) = 1. x 0 x 1 (d) As x 1+ : ƒ (x) 2 lim ƒ(x) = 2. (e) As x 2– : ƒ (x) 3 lim ƒ(x) = 3. x 1 x 2 (f) As x 2+ : ƒ (x) 3 lim ƒ(x) = 3. (g) As x 3– : ƒ (x) 2 lim ƒ(x) = 2. x 2 x 3 (h) As x 3+ : ƒ (x) 3 lim ƒ(x) = 3. (i) As x 4– : ƒ (x) 4 lim ƒ(x) = 4. x 3 x 4 (j) As x 4+ : ƒ (x) 4 lim ƒ(x) = 4. (k) As x : ƒ (x) 2 lim ƒ(x) = 2. x 4 x (l) As x 6– ,ƒ (x) lim ƒ(x) limit does not exist because it is not finite. x6 Do yourself - 1 : ( i ) Which of the following statements about the function y = (x) graphed here are true, and which are false ? (a) lim f (x) 1 (b) lim f (x) does not exist y x 1 x2 y= (x) (c) lim f (x ) 2 (d) lim f (x) 2 2 x2 x 1 1 (e) lim f (x ) does not exist (f) lim f (x) lim f(x) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 x 1 x 0 x 0 –1 0 123 x (g) lim f (x ) exists at every c (–1, 1) xc (h) lim f (x ) exists at every c (1, 3) xc (i) lim f (x) = 0 (j) lim f (x) does not exist. x 1 x 3 4 . FUNDA MENTAL THEOREMS ON LIMITS : b g b gLet Lim f x l & Lim g x m. If l & m exists finitely then : xa xa ( a ) Sum rule : Lim f x g x l m ( b ) Difference rule : Lim f x g x l m x a x a b g b g( c ) Product rule : Lim f x .g x l.m b g( d ) Quotient rule : Lim f x l , provided m 0 xa b gxa g x m 2E
JEE-Mathematics ( e ) Constant multiple rule : Lim kf x k Lim f x ; where k is constant. xa x a ( f ) Power rule : If m and n are integers then Lim f(x )m / n lm / n provided lm / n is a real number. xa b g FH b gKI b g( g ) Lim f g x f Lim g x f m ; provided f(x) is continuous at x = m. xa xa For example : Lim n(g(x)) n[Lim g(x)] x a x a = n (m); provided nx is continuous at x = m, m = lim g(x) . x a 5 . INDETERMINATE FORMS : 0 , 0 , 1 , 00, 0 . 0, , Initially we will deal with first five forms only and the other two forms will come up after we have gone through differentiation. Note : (i) Here 0,1 are not exact, infact both are aproaching to their corresponding values. (ii) We cannot plot on the paper. Infinity ( ) is a symbol & not a number It does not obey the laws of elementary algebra, (a) (b) (c) (d) 0 0 6 . GENER AL METHODS TO BE USED TO EVALUATE LIMITS : (a) Factorization : Important factors : (i) xn – an = (x – a)(xn–1 + axn–2 + ........... + an–1), n N (ii) xn + an = (x + a)(xn–1 – axn–2 + ........... + an–1), n is an odd natural number. Note : Lim xn an na n1 xa x a Illustration 2 : Evaluate : lim 1 2(2x 3) x 2 x3 3x2 2x x 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 Solution : We have lim 1 2(2x 3) = lim 1 2(2x 3) = lim x(x 1) 2(2x 3) x 2 x3 3x2 2x x 2 x(x 1) (x 2) x 2 x 2 x 2 x(x 1) (x 2) x2 5x 6 (x 2)(x 3) x3 1 x(x 1) =– 2 = lim x(x 1) (x 2) = lim x(x 1) (x 2) = lim Ans. x 2 x 2 x 2 Illustration 3 : lim 2x 23x 6 is equal to (C) 2 (D) none of these Solution : x2 2 x / 2 21x (B) 4 (A) 8 22x 23 6.2x 2x 42x 2 2x 23x 6 2x 22x 6.2x 8 lim lim 12 lim lim x2 2x / 2 21x x 2 2x/2 2x x 2 2x/2 2 x 2 2x / 2 2 x 2 2.4 2 8 lim 2 2 2 2x 2 Ans. (A) x 2 E3
JEE-Mathematics Illustration 4 : Evaluate : lim xP1 (P 1)x P x 1 (x 1)2 Solution : lim xP1 (P 1)x P 0 form x1 (x 1)2 0 = lim xP1 Px x P = lim x(xP 1) P(x 1) x 1 x 1 (x 1)2 (x 1)2 Dividing numerator and denominator by (x –1), we get x(xP 1) P = lim (x x2 x3 .... xP ) P lim x 1 x 1 (x 1) x1 (x 1) (x x2 x3 .... xP ) (1 1 1 .........upto P times) = lim x1 (x 1) (x 1) (x2 1) (x3 1) ...... (xP 1) = lim x 1 ( x 1) (x 1) (x 1) (x 1) = 1 + 2(1)2–1 + 3(1)3–1 +.......+ P(1)P–1 = 1 + 2 + 3 + .......+ P = P(P 1) Ans. 2 Do yourself - 2 : (i) Evaluate : lim x 1 x1 2 x2 7x 5 (b) Rationalization or double rationalization : 4 15x 1 Illustration 5 : Evaluate : lim x1 2 3 x 1 Solution : lim 4 15x 1 = lim (4 15x 1 )(2 3x 1 )(4 15x 1) 2 3x 1 x 1 (2 3x 1 )(4 15x 1 )(2 3x 1) x 1 = lim (15 15x) 2 3x 1 5 Ans. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 (3 3x) × 4 = x 1 15x 1 2 Illustration 6 : x2 8 10 x2 Evaluate : lim x1 x2 3 5 x2 Solution : This is of the form 33 0 = if we put x = 1 22 0 0 To eliminate the factor, multiply by the conjugate of numerator and the conjugate of the denominator 0 Limit = lim ( x2 8 – ( x2 8 10 x2 ) ( x2 3 5 x2 ) 5 x2 ) x 1 10 x2 ) x2 8 10 x2 ) × ( x2 3 5 x2 )( x2 3 ( = lim x2 3 5 x2 (x2 8) (10 x2 ) x2 3 5 x2 22 2 Ans. x 1 × = lim × 1 = = x2 8 10 x2 (x2 3) (5 x2 ) x 1 x2 8 10 x2 33 3 4E
JEE-Mathematics Do yourself - 3 : (i) Evaluate : lim px px (ii) Evaluate : lim a 2x 3x , a 0 qx qx xa 3a x 2 x x 0 (iii) If G(x) = – 25 x2 , then find the lim G ( x) G (1) x 1 x 1 ( c ) Limit when x : (i) Divide by greatest power of x in numerator and denominator. (ii) Put x = 1/y and apply y 0 Illustration 7 : Evaluate : Lim x2 x 1 Solution : x 3x2 2x 5 Illustration 8 : Lim x2 x 1 , form x 3x2 2x 5 Put x = 1 y 1 y y2 1 Limit = Lim Ans. y0 3 2y 5y2 3 HGF IKJIflim x3 1 (ax b) 2 , then x2 1 x (A) a = 1, b = 1 (B) a = 1, b = 2 (C) a = 1, b = –2 (D) none of these F Ilim HG JKx Solution : x3 1 (ax b) 2 lim x 3 (1 a ) bx 2 ax (1 b) 2 x2 1 x2 1 x lim x (1 a) b a (1 b) 2 1 – a = 0, – b = 2a = 1, b = – 2 Ans. (C) x x2 x 1 1 x2 Do yourself - 4 : Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 (i) Evaluate : lim n 2 n 1 (ii) Evaluate : lim (n n2 n ) n n 2 n 1 n (d) Squeeze play theorem (Sandwich theorem) : b g b g b gStatement : If f x g x h x ; x in the neighbourhood at x = a and b g b g b gLim f x Lim h x then Lim g x , y xa xa xa y=x2 E x . 1 Lim x2 sin 1 0 , y=x2sin 1 x0 x x sin 1 lies between –1 & 1 0x x x2 x2 sin 1 x2 x Lim x2 sin 1 0 as Lim(x2 ) Lim x2 0 y=–x2 xx 0 x 0 x 0 E5
JEE-Mathematics y 1 Ex .2 lim x sin 0 x0 x sin 1 lies between –1 & 1 –21 1 y=xsin 1 x 2 x x x sin 1 x – 1 x x 1 Lim x sin 1 0 as Lim(x) Lim x 0 x0 x x 0 x 0 Illustration 9 : Evaluate : lim [x] [2x] [3x] .....[nx] Where [.] denotes the greatest integer function. n n2 Solution : We know that x – 1 < [x] x n x + 2x + .....nx – n < [rx] x 2x + ........+ nx r 1 n [rx] x.n(n 1) x 1 1 1 1 n x xn 2 n – < n2 2 1 1 [rx] n (n + 1) – n < n 2 r 1 2 r 1 Now, lim x 1 1 x and lim x 1 1 – 1x n 2 n = n 2 n = 2 n2 Thus, lim [x] [2x] ...... [nx] x Ans. n2 = n 2 7 . LIMIT OF TRIGONOMETRIC FUNCTIONS : sin x tan x tan 1 x sin 1 x Lim 1 Lim Lim Lim [where x is measured in radians] x0 x x0 x x0 x x0 x (a) If Lim f(x) 0 , then Lim sin f(x) =1, e.g. Lim sin ( n x ) 1 xa f(x) xa x 1 ( n x ) x3 cot x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 Illustration 10 : Evaluate : lim x0 1 cos x Solution : lim x3 cos x lim x3 cos x 1 cos x x3 .cosx(1 + cos x) = 2 Ans. = lim x 0 s in x 1 cos x x 0 sin x.sin2 x x Ans. x0 sin3 E (2 x) sin(2 x) 2 sin 2 Illustration 11 : Evaluate : lim x0 x 2(sin(2 x) sin 2) x sin(2 x) 2 .2 . cos 2 x sin x lim = 2 2 Solution : x0 x lim sin(2 x ) x x 0 2 cos 2 x sin x 2 2 = lim lim sin(2 x) = 2cos2 + sin 2 x0 x x 0 2 6
JEE-Mathematics sin a Illustration 12 : Evaluate : lim n n tan b n 1 Solution : 1a As n , n 0 and n also tends to zero sin a an sin should be written as a so that it looks like lim sin n 0 n sin a b n The given limit = lim a n 1 . a(n 1) n tan n b n.b n 1 sin a b a a n ×1= b = lim n 1 . a 1 1 = 1 × 1 × Ans. a tan n b b n b n n GF JI FG JIIllustration 13 : lim x cos sin is equal to - 1 H K H Kx 4x 4x (A) (B) (C) 1 (D) none of these Ans. (B) F I F Ix G J G Jlim 2 sin cos lim sin x Solution : H K H Kx 2 4x 4x x 2 2x FG IJsin H K lim lim sin y , where y 2x 2x x 44 y0 y 4 2x Do yourself - 5 : (i) Evaluate : (a) lim sin x (b) lim sin2 x sin2 y ( c ) lim (a h)2 sin(a h) a2 sin a x0 tan x xy x2 y2 h0 h Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 ( b ) Using substitution Lim f(x ) Lim f(a h ) or Lim f(a h ) i.e. by substituting x by a – h or a + h xa h 0 h0 Illustration 14 : Evaluate : Lim(sec x tan x) x 2 Solution : Lim(sec x tan x);( form) x 2 Lim 1 sin x ; n o w in 0 form cos x 0 x 2 Put x = h 2 1 sin h 1 cosh 2 sinh Limit = Lim = Lim h0 2 h0 cos h E7
JEE-Mathematics = Lim 2 sin2 h h = Lim sin h 0 Ans. 2 2 2 h0 2 h0 sin h cos cos h 2 2 8 . LIMIT USING SERIES EXPANSION : Expansion of function like binomial expansion, exponential & logarithmic expansion, expansion of sinx, cosx, tanx should be remembered by heart which are given below : ( a ) a x 1 x na x2n2a x3n3a ..........a 0 ( b ) e x 1 x x2 x3 .......... 1! 2! 3! 1! 2! 3! x3 x5 x7 b g(c) n x2 x3 x4 1 x x .......for 1 x 1 ( d ) sin x x .......... 234 3! 5! 7! ( e ) cos x 1 x2 x4 x6 .......... ( f ) tan x x x3 2x5 ....... 2! 4! 6! 3 15 ( g ) tan1 x x x3 x5 x7 ....... 357 ( h ) sin1 x x 12 x3 12.32 x5 12.32.52 x7 ....... 3! 5! 7! ( i ) sec1 x 1 x2 5x4 61x6 ....... 2! 4! 6! ( j ) (1 +x)n = 1 + nx + n(n 1) x2 + ........... n Q 2! ex ex 2x Illustration 15 : lim x0 x sin x x2 x3 x2 x3 lim 1 x 2 ! 3 ! ...... 1 x 2 ! 3 ! ..... 2x Solution : ex ex 2x x0 x3 x5 lim x0 x sin x x x 3 ! 5 ! ..... x3 x5 x3 1 1 x2 ..... 2. 2. ...... 3 60 6 5! 1/3 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#04\\Eng\\01.LIMIT\\01.THEORY.p65 lim x3 x5 lim 1 1 ..... 1/6 Ans. x 0 x0 6 120 x3 x2 ..... 6 5! Do yourself - 6 : (i) Evaluate : Lim x sin x (ii) Evaluate : Lim x tan 1 x x0 sin(x3 ) xx 0 3 9 . LIMIT OF EXPONENTIAL FUNCTIONS : ( a ) Lim a x 1 n a (a 0) In particular Lim ex 1 1 . x0 x x0 x In general if Lim f(x) 0 ,then af(x) 1 na, a 0 Lim xa xa f(x) 8 E
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