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M1-Allens Made Maths Theory + Exercise [II]

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JEE-Mathematics 8 .  e tan1 x (1  x  x2 ). d(cot1 x ) is equal to - (A) – e tan1 x + c (B) e tan1 x + c (C) –x. e tan1 x + c (D) x. e tan1 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 + bn[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) cos1 2 x  x2  c (C) c  cos1 (2 x  1)  (C) tan1 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) cot1 cot2 x  c  (B)  cot1 tan2 x  c (D)  tan1 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  6ex dx = Ax + B log(9e2x – 4) + C, then A = ......., B = ....... and C = ...... 9ex  4ex 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  x3y Because Statement - II : dx = log(x – 3y) + c.  x3y (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. If1 . (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 12n . 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 sinn1 x 1 sinn1 x 1 sinn1 x n 1 sinn1 x (A) cosm 1 x (B) (m 1) cosm1 x (C) (n 1) cosm1 x (D) m 1 cosm1 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 .  xx   ex  n x dx  xn 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 x4 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  = 2x  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 ax 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 1x 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 xn2 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 ex  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) tan1 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)  cos1 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 f1 : R  (–/2, /2) f(x)  =  tan  x f 1 (x)  tan1 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)  cot1 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)  sec1 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)  cosec1 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  cos1 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   =  {111......... 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 , 0x   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 cot1 (cot x)  x y =  x +  , 0x 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 cot1 3       1 co t 1 sec tan1 2  tan  ec 22          2 2  sec sec1 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    cot1 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    cot1 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 xy   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 xy (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  cos1 xy  1  x2 1  y2 ; x  y, x, y  0  (b) cos–1  x  –  cos–1  y  =    cos1 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  xy  ; 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  cot1 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  xy , 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,  cos1 12  sin 1 3  sin 1 5  sin 1 3  cos1 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 cot1 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 x0 –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)  cos1 (4 x3  3x)        y  3 cos1 x  2 if 1 D ID  1  x      –1 – 3/2 2    2  3 cos1 x if 1  x  1       /2   if 2 2      x  1 x 1 2  3 cos1 x I  –1/2 +1/2 3/2 1     2 sin –1 x 1 y  1  x   /2  2 1 –  ( f ) sin1 2x 1  x2  2 sin1 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 cos1 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    21  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  sec1 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  =  52  /  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  =  52  /  8 Since        tan–1  x  +  cot–1  x  =  /2    we  have (/2)2  –  2tan–1  x  (/2  –  tan–1  x  )  =  52  /  8              2(tan–1  x)2  –  2  (/2)  tan–1  x –  32 /  8  = 0          tan–1  x  =  –    /  4     x  = –1      Ans.  (A) Illustration  20  : Solve  the  equation  :  tan1 x  1  tan1 x  1   x 2 x2 4 Solu ti on  : tan 1 x  1  tan 1 x  1   x 2 x2 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 x2 (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 12 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  =  tan1 (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) cos1 x  sin 1 x   sin  5  x 26   ( i i i ) cot1 x  cot1 (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  cn1   tan 1  1   tan 1  x   c1 y  x   1 c2 c1   1 c3c2   1 c n c n1   cn   y                 Solution  : L.H.S.  tan1  c1 x  y   tan1  c2  c1   tan1  c3  c 2   ...  tan1  cn  cn1   tan1  1   c1 y  x   1 c2c1   1  c3c2   1 c n c n 1   cn             x1  1  1  1  1   1 1             tan1   tan1  c1 1 c2   tan1  c2 1 c3  ...  tan1  cn   tan1 1  y c1  c1 1  c2 1  cn1 1           1  x . 1 1  . 1  .  1  1 . cn  y c1 c2 c3  cn1 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     cos1 1  x2 1 x  sin 1 x  cos1 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   cot1    x  1  x2 Hence,  sin 1 x  cos1 1  x2  sec1 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     cos1 1  x2  sin1 x   cos1 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      cot1           x 0 1  x2  x    Hence,  sin 1 x   cos1 1  x2   sec1 1  x     cot1  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 cot1 4) 2. cos  2 tan 1  1   equals  - (D)  –1   7   (A)   sin(4 cot1 3) (B)   sin(3 cot1 4) (C)   cos(3 cot1 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 cos1 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  cos1 x  cos1 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 cos1 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  cos1 x  cot1  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  - 1x (A)  1 cos1 x (B)  cos1 1  x (C)  sin 1 1  x (D)   1 tan 1 1  x 2 2 2 2 1x 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  2r1   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  cos1 11  cot1 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  cos1 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  cos1 (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 sin1 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  cos1 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)  cos1 2 (b) f(x)  1  2arc sin x  1 2  sin x x x2 (c) e cos1 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  cot1 (2x  x2 ) (b) f(x)  =  sec1 (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  cos1 x  cos1 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  :  cos1 x  2 sin 1 1  x  2 cos1 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) tan1 (x  1)  tan1 (x)  tan1 (x  1)  tan1 (3x) (d) sin 1 x  cos1 x  sin 1 3x  2 (e) sin 1 x  sin 1 1  x   cos1 x (f) 2 tan 1 x  cos1 1  a2  cos1 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)  cot1 7  cot1 13  cot1 21  cot1 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 ab 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) sin1 cos(sin1 x)  cos1 sin(cos1 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   +  ......+  sin1 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  K2 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  (cos1 x )3 . 1 2 . Find  the  number  of  roots  of  the  following  equations  :  (b)  1  x 1  (a) 1  cos 2x  2 sin1 sin x  sin sin 1 (log 1 x )  2 cos  sin  2   0 2 (c) | y| cos x and y  cot1 (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 73  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 i1 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 sin1 1  x2  sin tan1 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”. xa 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). xa 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). xa h 0 Limit of a function f(x) is said to exist as, x  a when Lim f(x)  Lim f(x) = Finite quantity. xa  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  h0 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 h0 –1 Lim f(x)  Lim f(1  h)  f(1 )  0 x 1  h0 Lim f(x)  Lim f(2  h)  f(2_ )  1 x 2  h0 Fig. 1 Lim f(x)  0 and Lim f(x) does not exist. x0 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 cos1 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  x6  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. x6  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 x2 y= (x) (c) lim f (x )  2 (d) lim f (x)  2 2 x2 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) xc (h) lim f (x ) exists at every c  (1, 3) xc (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 : xa xa ( 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 xa b gxa g x m 2E

JEE-Mathematics ( e ) Constant multiple rule : Lim kf x  k Lim f x  ; where k is constant. xa 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. xa 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. xa xa 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 n1 xa 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)   x3  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  23x  6 is equal to (C) 2 (D) none of these Solution : x2 2 x / 2  21x (B) 4 (A) 8  22x  23  6.2x 2x  42x  2 2x  23x  6 2x 22x  6.2x  8 lim  lim 12  lim  lim x2 2x / 2  21x 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 xP1  (P  1)x  P x 1 (x  1)2 Solution : lim xP1  (P  1)x  P  0 form  x1 (x  1)2  0  = lim xP1  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) x1 (x  1) (x  x2  x3  ....  xP )  (1  1  1  .........upto P times) = lim x1 (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 x1 2 x2  7x  5 (b) Rationalization or double rationalization : 4  15x  1 Illustration 5 : Evaluate : lim x1 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   x1  x2  3  5  x2  Solution : This is of the form 33 0 = if we put x = 1 22 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  22 2 Ans. x 1 × = lim  × 1 = = x2  8  10  x2 (x2  3)  (5  x2 ) x 1  x2  8  10  x2  33 3 4E

JEE-Mathematics Do yourself - 3 : (i) Evaluate : lim px  px (ii) Evaluate : lim a  2x  3x , a  0 qx  qx xa 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. y0 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 = 2a = 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 xa xa xa y=x2 E x . 1 Lim x2 sin 1  0 , y=x2sin 1 x0 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 x0 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 x0 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] x0 x x0 x x0 x x0 x (a) If Lim f(x)  0 , then Lim sin f(x) =1, e.g. Lim sin ( n x ) 1 xa f(x) xa 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 x0 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. x0 sin3 E (2  x) sin(2  x)  2 sin 2 Illustration 11 : Evaluate : lim x0 x 2(sin(2  x)  sin 2)  x sin(2  x)  2 .2 . cos  2  x sin x  lim =   2  2  Solution : x0 x lim  sin(2  x )  x x 0 2 cos  2  x  sin x  2  2 = lim  lim sin(2  x) = 2cos2 + sin 2 x0 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 y0 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 x0 tan x xy x2  y2 h0 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 xa h 0 h0 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    h0   2   h0   cos  h E7

JEE-Mathematics = Lim  2 sin2 h h  = Lim  sin h  0 Ans. 2 2 2 h0  2  h0    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  x2n2a  x3n3a ..........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 ) tan1 x  x  x3  x5  x7 ....... 357 ( h ) sin1 x  x  12 x3  12.32 x5  12.32.52 x7 ....... 3! 5! 7! ( i ) sec1 x  1  x2  5x4  61x6 ....... 2! 4! 6! ( j ) (1 +x)n = 1 + nx + n(n 1) x2 + ........... n  Q 2! ex  ex  2x Illustration 15 : lim x0 x  sin x x2 x3  x2 x3   lim 1  x  2 !  3 !  ......  1  x  2 !  3 !  .....  2x Solution : ex  ex  2x x0  x3 x5  lim x0 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 x0  6 120 x3  x2  ..... 6 5! Do yourself - 6 : (i) Evaluate : Lim x  sin x (ii) Evaluate : Lim x  tan 1 x x0 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 . x0 x x0 x In general if Lim f(x)  0 ,then af(x) 1  na, a  0 Lim xa xa f(x) 8 E

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