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### Definite & Indefinite Integration Q.B..pdf

1. QUESTION BANK ON DEFINITE & INDEFINITE INTEGRATION XII (ALL) M A T H E M A T I C S Time Limit : 5 Sitting Each of 100 Minutes duration approx.
2. Question bank on Definite & Indefinite Integration There are 168 questions in this question bank. Select the correct alternative : (Only one is correct) Q.11/def The value ofthe definiteintegral,       1 1 x 3 1 x dx ) e e ( is (A*) 2 e 4  (B) e 4  (C)          e 1 tan 2 e 1 1 2 (D) 2 e 2  [Hint: I =     1 x 3 x ) e · e e · e ( dx =    1 2 x 2 x ) e e ( e dx e (multiplyNr and Dr byex) put ex = t  ex dx = dt I =    e 2 2 e t dt e 1 =   e 1 2 e t tan e 1 =          4 2 e 1 2 = 2 e 4  Ans. ] Q.22/def The value ofthedefinite integral, dx e x 2 · e cos 2 2 x 2 n 0 x               l is (A) 1 (B) 1 + (sin 1) (C*) 1 – (sin 1) (D) (sin 1) – 1 [Hint: put 2 x e = t; 2 x e · 2x dx = dt ; dt t cos 2 1   =   2 1 t sin  = 1 – (sin 1) ] Q.35/def Value ofthe definite integral        2 1 2 1 3 1 3 1 dx ) ) x 3 x 4 ( cos ) x 4 x 3 ( sin ( (A) 0 (B*) 2   (C) 2 7 (D) 2  [Hint: Note that in        2 1 , 2 1 , sin–1(3x – 4x3) = 3 sin-1x and cos–1(4x3 – 3x) = 2 – 3 cos–1x hence f (x) = 3 sin–1x – 2 + 3 cos–1x = 2    I = 2 dx 2 2 / 1 2 / 1        ] [Alternate: f (x) = sin–1 (3x – 4x3) – [ – cos–1(3x – 4x3 ) ] = – + (sin–1 (3x – 4x3) + cos–1 (3x – 4x3)) = 2   ] Q.46/def Let f (x) =   x 2 4 t 1 dt and g be the inverse off. Then the value ofg'(0) is (A) 1 (B) 17 (C*) 17 (D) noneofthese
3. [Sol. f ' (x) = 4 x 1 1  = dx dy now g ' (x) = dy dx = 4 x 1 when y = 0 i.e.   x 2 4 t 1 dt = 0 then x = 2 (think !) hence g '(0) = 16 1 = 17 ] Q.56/inde   x x 1 e ) e ( cot dx is equal to : (A) 2 1 ln (e2x + 1)  x x 1 e ) e ( cot + x + c (B) 2 1 ln (e2x + 1) + x x 1 e ) e ( cot + x + c (C*) 2 1 ln (e2x + 1)  x x 1 e ) e ( cot  x + c (D) 2 1 ln (e2x + 1) + x x 1 e ) e ( cot  x + c Q.69/def    k 0 x 1 0 k dx ) x 2 sin 1 ( k 1 Lim (A) 2 (B) 1 (C*) e2 (D)nonexistent [Hint: l = k dx ) x 2 sin 1 ( Lim k 0 x 1 0 k    differentiating Using L'opital rule l = k 1 0 k ) k 2 sin 1 ( Lim   = ) k 2 (sin k 1 Lim 0 k e  = e2 ] Q.711/def    5 n 0 x x x 3 e 1 e e l dx = (A*) 4   (B) 6   (C) 5   (D) None [Hint: put ex  1 = t2 ] Q.816/def If x satisfies the equation 2 1 0 2 x 1 cos t 2 t dt             – x dt 1 t t 2 sin t 3 3 2 2            – 2 = 0 (0 <  < ), then the value xis (A) ±   sin 2 (B) ±   sin 2 (C) ±   sin (D*) ±   sin 2
4. [Sol.    3 3 2 2 dt 1 t t 2 sin t = 0 as the integrand is an odd function. also     1 0 2 1 cos t 2 t dt = 1 0 1 sin cos t tan sin 1      =   sin 2 Thus the givenequationreduces to x2   sin 2 – 2 = 0  x = ±   sin 2 Ans.] Q.917/def If f (x) = eg(x) and g(x) = 2 x  t dt t 1 4  then f (2) has the value equalto : (A*) 2/17 (B) 0 (C) 1 (D) cannot be determined [Hint: f (x) = eg (x). g (x) ; g (x) = x x 1 4  hence f (2) = eg (2). g  (2) = e0. 2 17 = 2 17 ] Q.107/inde  etan  (sec – sin)d equals : (A)  etan  sin  + c (B) etan  sin  + c (C) etan  sec  + c (D*) etan  cos  + c [Hint: integrate  etan  . sin  d byparts. One integralcancels. Q.1118/def   0 (x · sin2x · cosx) dx = (A) 0 (B) 2/9 (C)  2/9 (D*) 4/9 [Hint:   0 2 dx ) x cos x (sin · x =     0 3 0 3 dx x sin 3 1 3 x sin · x ] Q.1222/def The value of         n 4 r 1 r 2 n n 4 r 3 r n Lim is equalto (A) 35 1 (B) 14 1 (C*) 10 1 (D) 5 1 [Sol. Tr = 2 4 n r 3 n · n r 1          S =           n 4 1 2 n r · 4 n r 3 1 n 1 =   4 0 2 ) 4 x 3 ( x dx
5. put 4 x 3  = t  dt dx x 1 2 3  =  10 4 2 t dt 3 2 =               10 1 4 1 3 2 t 1 3 2 4 10 = 10 1 40 6 · 3 2  ] Q.1323/def     c b c a dx ) c x ( f = (A*)  b a dx ) x ( f (B)   b a dx ) c x ( f (C)    c 2 b c 2 a dx ) x ( f (D)   b a dx ) c 2 x ( f [Hint: Put x + c = t ] Q.1425/def Let I1 =     2 / 0 dx x cos . x sin 1 x cos x sin ; I2 =   2 0 6 dx ) x cos ( ; I3 =     2 / 2 / 3 dx ) x (sin & I4 =         1 0 dx 1 x 1 n l then (A) I1 = I2 = I3 = I4 = 0 (B) I1 = I2 = I3 = 0 but I4  0 (C*) I1 = I3 = I4 = 0 but I2  0 (D) I1 = I2 = I4 = 0 but I3  0 Q.158/inde    ) x 1 ( x x 1 7 7 dx equals : (A) lnx + 2 7 ln (1 + x7) + c (B) lnx  2 7 ln (1  x7) + c (C*) lnx  2 7 ln (1 + x7) + c (D) lnx + 2 7 ln (1  x7) + c [Hint: I =   dx x x 1 7    x x 6 7 1   dx ] Q.1626/def    n 2 / 0 n nx tan 1 dx = (A) 0 (B*) n 4  (C) 4 n (D) n 2  [Hint: nx = t; I =    2 0 n ) t (tan 1 dt n 1 =    2 0 n n n dt ) t (cos ) t (sin ) t (cos n 1 ] Q.1728/def f(x) =    x 0 dt ) 2 t ( ) 1 t ( t takes onits minimumvalue when: (A) x= 0 , 1 (B) x = 1 , 2 (C*) x = 0 , 2 (D) x = 3 3 3 
6. [Hint: f ' (x) = x (x – 1) (x – 2) = 0 x = 0, 1, or 2 at x = 0 & 2, f ' (x) changes sign from – ve to + ve  minimum ] Q.1829/def   a a dx ) x ( f = (A*)      a 0 dx ) x ( f ) x ( f (B)      a 0 dx ) x ( f ) x ( f (C) 2  a 0 dx ) x ( f (D) Zero [Hint: I =   a a dx ) x ( f =    a a dx ) x ( f (using K)  2I =       a a dx ) x ( f ) x ( f =      a 0 dx ) x ( f ) x ( f 2 ( as integralis even)  result ] Q.1930/def Let f (x) be a function satisfying f ' (x) = f (x) with f (0) = 1 and g be the function satisfying f (x) + g (x) = x2. The value ofthe integral  1 0 dx ) x ( g ) x ( f is (A) e – 2 1 e2 – 2 5 (B) e – e2 – 3 (C) 2 1 (e – 3) (D*) e – 2 1 e2 – 2 3 [Sol. f ' (x) = f (x)  f (x) = C ex and since f (0) = 1  1 = f (0) = C  f (x) = ex and hence g (x) = x2 – ex Thus  1 0 dx ) x ( g ) x ( f =   1 0 x 2 x 2 dx ) e e x ( = 1 0 x 2 1 0 x 1 0 x 2 2 e dx xe 2 e x    = (e – 0) – 2 [ 1 0 x xe – 1 0 x e ] – 2 1 (e2 – 1) = (e – 0) – 2 [(e – 0) – (e – 1)] – 2 1 (e2 – 1) = e – 2 1 e2 – 2 3 ] Q.2011/inde   | x | n 1 x | x | n l l dx equals : (A*) | x | n 1 3 2 l  (lnx 2) + c (B) | x | n 1 3 2 l  (lnx+ 2) + c (C) | x | n 1 3 1 l  (lnx 2) + c (D) | x | n 1 2 l  (3 lnx 2) + c [Hint: Start : ln | x | = t  dx x 1 = dt    t 1 dt t ]
7. Q.2131/def              2 1 3 2 1 dx 4 | x 1 | | 3 x | 2 1 equals: (A) 2 3  (B) 8 9 (C*) 4 1 (D) 2 3 Where {*} denotesthe fractionalpart function. [Hint:              2 1 3 2 1 dx 4 | x 1 | | 3 x | 2 1                 1 2 1 dx 4 x 1 x 3 2 1 +                3 1 dx 4 1 x x 3 2 1 +              2 7 3 dx 4 1 x 3 x 2 1 =      2 7 3 1 2 1 dx } 4 x { dx } x { =     2 7 3 1 2 1 dx } x { dx ) x 1 ( =      2 7 3 1 2 1 dx ) 3 x ( dx ) x 1 ( ] Q.2232/def   / 4 0        x 1 cos . x x 1 sin . x 3 2 dx has the value : (A) 8 2 3  (B) 24 2 3  (C*) 32 2 3  (D) None [Hint: Consider   4 0 2 dx x 1 sin x 3 and IBP taking x 1 sin as the 1st and 3x2 as the 2nd function. Two integrals cancel] [Sol.              dx x 1 cos x x 1 sin x 3 I 2 = 3 x · x 1 sin –         dx x x 1 x 1 cos 3 2 –  dx x 1 cos x 2 =  4 0 3 x 1 sin · x ] Q.2338/def                              3 4 n 6 ) 1 n ( sec ..... n 6 · 2 sec n 6 sec n 6 Lim 2 2 2 n has the value equalto (A*) 3 3 (B) 3 (C) 2 (D) 3 2 [Sol. Tr = n 6 r sec n 6 2   S =       n 1 r 2 r n 6 r sec n 6 T =    1 0 2 dx 6 x sec 6 = 1 0 6 x tan  = 3 1 = 3 3 ]
8. Q.2439/def Suppose that F(x) is an antiderivative of f(x) = x x sin , x > 0 then  3 1 dx x x 2 sin canbe expressed as (A*) F (6) – F (2) (B) 2 1 ( F (6) – F (2) ) (C) 2 1 ( F (3) – F (1) ) (D) 2( F (6) – F (2) ) [Hint: Given  dx x x sin Let 2x = t, dx = 2 dt  I =    6 2 dt t t sin 2 2 =  6 2 dt t t sin But  dx x x sin = F (x)   6 2 dt t t sin = F (6) – F (2) ] Q.2513/inde Primitive of 2 4 4 ) 1 x x ( 1 x 3    w.r.t. x is : (A) 1 x x x 4   + c (B*)  1 x x x 4   + c (C) 1 x x 1 x 4    + c (D)  1 x x 1 x 4    + c [Hint:   3 1 1 4 2 3 1 2 x x x x     =   3 1 2 2 3 1 2 x x x x      ] Q.2641/def   n Lim     2 1 2 2 2 1 2 n n n n n            cos cos ..... cos ( ) equal to (A*) 1 (B) 1 2 (C) 2 (D) none [Hint: Tr = n 2 r cos n 2   S =      1 n 0 r n r · 2 cos n 1 · 2 =    1 0 dx 2 x cos 2 = 1 ] Q.2742/def   log log x x n 2 2 2 2 2 4            dx = (A*) 0 (B) 1 (C) 2 (D) 4 [Hint: dx x n 2 n nx 2 n 4 2 2         l l l l if f(x) = x n 1 l x n 1 ) x ( ' f x 2 l     I = ln2 4 2 nx x       l =        2 n 2 4 n 4 2 n l l l = 0 ]
9. Q.2844/def If m & n are integers such that (m  n) is an odd integer then the value of the definite integral   0 dx nx ·sin mx cos = (A) 0 (B*) 2 2 m n n 2  (C) 2 2 m n m 2  (D) none [Hint: Note that ifm–nis odd m+nisalso odd.Split the integrandusing 2 sinA cosB as sumor difference.] Q.2945/def Let y={x}[x] where {x}denotes the fractional part of x & [x] denotes greatest integer  x, then  3 0 dx y = (A) 5/6 (B) 2/3 (C) 1 (D*) 11/6 Q.3015/inde If      2 2 4 1 x x 1 x dx =Aln x+ 2 x 1 B  + c , where c is the constant of integration then : (A) A = 1 ; B =  1 (B) A =  1 ; B = 1 (C*) A = 1 ; B = 1 (D) A =  1 ; B =  1 [Hint: add and subtract x2 inthe numerator] Q.3148/def      2 / x cos 1 x sin 1 dx = (A*) 1  ln2 (B) ln2 (C) 1 + ln2 (D) none [Hint:    dx 2 x sin 2 x sin 1 I 2 =         dx 2 x cot 2 x ec cos 2 1 2 = –x cot   2 2 x Q.3249/def Let f : R  R be a differentiable function & f(1) = 4, then the value of; 1 x Lim    ) x ( f 4 1 x dt t 2 is : (A) f(1) (B) 4f(1) (C) 2f (1) (D*) 8f(1) Q.3350/def If  ) x ( f 0 2 dt t = x cos x , then f ' (9) (A*) is equalto – 9 1 (B) is equalto – 3 1 (C) is equal to 3 1 (D) isnonexistent [Sol. ) x ( f 0 3 3 t = x cos x  [f (x)]3 = 3x cos x ....(1) [f (9)]3 = – 27  f (9) = – 3 also differentiating  ) x ( f 0 2 dt t = x cos x
10. [f (x)]2 · f ' (x) = cos x – x  sin  x  [f (9)]2 · f ' (9) = – 1  f ' (9) = –  2 ) 9 ( f 1 = – 9 1 f ' (9) = – 9 1  (A) ] Q.3451/def   3 / 1 ) 2 / ( 0 3 5 dx x ·sin x = (A) 1 (B) 1/2 (C) 2 (D*) 1/3 Q.3516/inde Integral of ) x ec cos x (cot x cot 2 1   w.r.t. x is : (A) 2 ln cos 2 x + c (B*) 2 ln sin 2 x + c (C) 2 1 ln cos 2 x + c (D) ln sin x  ln(cosec x  cot x) + c [Hint: I =    x cot 2 x cot x ec cos 2 1 2 =    dx x cot x cot ecx cos 2 x ec cos 2 2 =   dx ) x cot ecx (cos ] Q.3652/def If f (x) = x+ x1 + x 2, x  R then  3 0 dx ) x ( f = (A) 9/2 (B) 15/2 (C*) 19/2 (D) none Q.3756/def Number ofvalues ofx satisfying the equation           x 1 2 dt 4 t 3 28 t 8 =   1 x log 1 x ) 1 x ( 2 3    , is (A) 0 (B*) 1 (C) 2 (D) 3 [Hint: Integrating LHS a cubic in x ofthe formx(ax2 + bx + c) = 0 giving x=0, –3/2,–1/4. Onlyx = –1/4 satisfies] Q.3858/def   1 0 1 dx x x tan = (A)   4 / 0 dx x x sin (B)   2 / 0 dx x sin x (C*)   2 / 0 dx x sin x 2 1 (D)   4 / 0 dx x sin x 2 1 [Sol. I =   1 0 1 dx x x tan x = tan ; dx = sec2 d
11. I =       4 / 0 2 d tan sec . =       4 / 0 d cos sin =      4 / 0 d 2 sin 2 2 = y =   2 / 0 dy y sin y 2 1 =   2 / 0 dx x sin x 2 1  (C) ] Q.3961/def Domain ofdefinition ofthe function f(x) =   x 0 2 2 t x dt is (A) R (B) R+ (C) R+  {0} (D*) R – {0} Q.4017/inde If  e3x cos 4x dx = e3x (A sin4x + B cos 4x) + c then : (A) 4A = 3B (B) 2A = 3B (C*) 3A = 4B (D) 4B + 3A = 1 Q.4163/def If f (a + b x) = f(x) , then    b a dx ) x b a ( f . x = (A) 0 (B) 2 1 (C*)   b a dx ) x ( 2 b a f (D)   b a dx ) x ( 2 b a f Q.4265/def The set ofvalues of'a' which satisfythe equation   2 0 2 dt ) a log t ( = log2       2 a 4 is (A) a  R (B*) a  R+ (C) a < 2 (D) a > 2 [Hint: 2 0 2 2 t · a log 2 t  = 2 – log2(a2) (2 – 2 log2a) = 2 – 2 log2a 2 log2a = 2 log2a  a  R+ ] Q.4366/def The value ofthe definite integral dx ) 5 x 4 ( 5 x 2 ) 5 x 4 ( 5 x 2 3 2             = (A) 7 3 3 5 3 2  (B) 4 2 (C) 4 3 + 4 3 (D*) 7 7 2 5 3 2  [Hint: Put 4x  5 = 5t2  4dx = 10t dt or better will be 5(4x – 5) = t2 ] I = dt t t 5 ) t 1 ( 2 5 t 5 ) t 1 ( 2 5 2 5 5 7 5 3 2 2       =   dt t | ) 1 t ( | | 1 t | 2 5 5 7 5 3 2 / 3           =                                dt t ) 1 t ( ) 1 t ( dt t | ) t 1 ( | ) t 1 ( 2 5 5 7 1 1 5 3 2 / 3
12. =                      dt t dt t 2 2 5 2 5 7 1 1 5 3 2 / 3 ] Q.4467/def Number ofordered pair(s) of(a, b) satisfying simultaneouslythesystemofequation 0 dx x b a 3   and 3 2 dx x b a 2   is (A) 0 (B*) 1 (C) 2 (D) 4 [Hint: a = – 1 and b = 1 ] Q.4518/inde        x cot x tan x cot x tan 1 1 1 1 dx is equal to : (A)  4 x tan1 x +  2 ln (1 + x2)  x + c (B)  4 x tan1 x   2 ln (1 + x2) + x + c (C)  4 x tan1 x +  2 ln (1 + x2) + x + c (D*)  4 x tan1 x   2 ln (1 + x2)  x + c Q.4668/def Variable x and y are related by equation x=   y 0 2 t 1 dt . The value of 2 2 dx y d is equalto (A) 2 y 1 y  (B*) y (C) 2 y 1 y 2  (D) 4y [Hint: x = ln         2 y 1 y ] Q.4769/def Let f (x) =      h x x 2 0 h t 1 t dt h 1 Lim , then ) x ( · x Lim x f    is (A) equalto 0 (B) equalto 2 1 (C) equalto 1 (D*) nonexistent [Hint: f (x) = x x 1 2   ;            x x 1 x Lim 2 x  –   DNE ] Q.4873/def If the primitive of f (x) =  sin x + 2x 4, has the value 3 for x = 1, thenthe set of x for which the primitive off(x) vanishesis : (A) {1, 2, 3} (B) (2, 3) (C*) {2} (D) {1, 2, 3, 4}
13. Q.4974/def If f & g are continuous functions in [0, a] satisfying f (x) = f(a  x) & g (x) + g (a  x) = 4 then  a 0 dx ) x ( g . ) x ( f = (A)  a 0 2 1 f(x)dx (B*)  a 0 2 f(x)dx (C)  a 0 f(x)dx (D)  a 0 4 f(x)dx [Sol. I = dx ) x ( g . ) x ( f a 0  ....(1) I = dx ) x a ( g . ) x a ( f a 0    I = dx ) ) x ( g 4 ( . ) x ( f a 0   =    a 0 a 0 ) x ( g ) x ( f ) x ( f 4 ....(2) (1) + (2) 2I =  a 0 ) x ( f 4  I = dx ) x ( f 2 a 0   (B) ] Q.5020/inde  x . 2 2 x 1 x 1 x n          l dx equals : (A*) 2 x 1 ln         2 x 1 x  x + c (B) 2 x . ln2         2 x 1 x  2 x 1 x  + c (C) 2 x . ln2         2 x 1 x + 2 x 1 x  + c (D) 2 x 1 ln         2 x 1 x + x + c [Hint: use I.B.P. taking ln         2 x 1 x as the first functionand x x 1 2  as the second function ] Q.5175/def If f (x) =      2 x 1 ) 6 x 7 ( 1 x 0 x 1 3 1        , then  2 0 dx ) x ( f is equal to (A) 6 31 (B) 21 32 (C) 42 1 (D*) 42 55 Q.5277/def The value ofthe definite integral   1 0 x e dx ) e · x 1 ( e x is equalto (A*) ee (B) ee – e (C) ee – 1 (D) e [Hint: start with ex = t and use     dx ) x ( ' f ) x ( f ex = ex f(x) ]
14. Q.5379/def         2 2 / 1 dx x 1 x sin x 1 has the value equalto (A*) 0 (B) 4 3 (C) 4 5 (D) 2 [Sol. x = t 1  dx = 2 t 1 dt I =                2 / 1 2 2 dt t 1 t t 1 sin t =         2 / 1 2 dt t 1 t sin t 1 = –         2 2 / 1 dt t 1 t sin t 1 = – I  2I = 0  I = 0 Alternatively : put x = et  I =     2 n 2 n t t dt ) e e sin( l l = 0 (odd function) ] Q.5481/def The value of the integral   0 e2x (sin 2x + cos2x) dx = (A) 1 (B) 2 (C*) 1/2 (D)zero [Hint: Put  2x = t and use  ex ( f(x) + f (x) ) d x = ex f(x) + c ] Q.5582/def The value ofdefinite integral      0 z 2 z dz e 1 e z . (A*) – 2 n 2 l  (B) 2 n 2 l  (C) –  ln 2 (D)  ln 2 [Hint: Put e–z = sin  ] Q.5622/indeAdifferentiable function satisfies 3f 2(x) f '(x) = 2x. Given f (2) = 1 then the value of f (3) is (A) 3 24 (B*) 3 6 (C) 6 (D) 2 [Hint: Integrating both sides we gets f3(x) = x2 + C ; f (2) = 1  C = – 3 f 3(x) = x2 – 3  f 3(3) = 6  f (3) = 3 6 ] Q.5786/def For In =  e 1 (lnx)ndx, n  N;which of the following holds good? (A) In + (n + 1) In + 1 = e (B) In + 1 + n In = e (C*) In + 1 + (n +1) In = e (D) In + 1 + (n – 1) In = e [Hint: In = e – nIn – 1 ]
15. Q.5889/def Let fbe acontinuous functions satisfying f ' (lnx) =   1 x x for 1 x 0 for 1    and f(0) = 0 thenf (x) can be defined as (A) f(x) =    0 if x e 1 0 if x 1 x    (B) f(x) =    0 if x 1 e 0 if x 1 x    (C) f (x) =    0 if x e 0 if x x x   (D*) f(x) =    0 if x 1 e 0 if x x x    [Sol. f ' (ln x) =    1 x x for 1 x 0 for 1    put ln x = t  x = et for x > 1 ; f ' (t) = et for t > 0 integrating f (t) = et + C ; f (0) = e0 + C  C = – 1  f (t) = et – 1 for t > 0 (corresponding to x > 1)  f (x) = ex – 1 for x > 0 ....(1) again for 0 < x  1 f ' (ln x) = 1 f ' (t) = 1 for t  0 f (t) = t + C f (0) = 0 + C  C = 0  f (t) = t for t  0  f (x) = x for x  0] Q.5991/def Let f: R  R be a differentiable functionsuch that f(2) = 2. Then the value ofLimit x  2 4 2 3 2 t x f x   ( ) dt is (A) 6 f (2) (B) 12 f (2) (C*) 32 f (2) (D) none Q.6093/def    2 / 0 2 2 x sin a 1 dx has the value : (A*)  2 1 2 a (B)  1 2 a (C) 2 1 2  a (D) none [Hint: I =     2 / 0 2 2 2 2 x tan a x tan 1 dx x sec =     2 / 0 2 2 2 1 x tan ) a 1 ( dx x sec ] Q.6124/inde Let f(x) =       x e x n x 1 l then its primitive w.r.t. x is (A) 2 1 ex – ln x + C (B) 2 1 ln x – ex + C (C*) 2 1 ln2x – x + C (D) x 2 ex + C [Sol.  dx e x n x 1 x l =   dx ) ne nx ( x 1 x l l
16. =   dx x x nx l =          dx x x 1 dx x n x 1 l (put ln x = u ; du dx x 1  ) =  dx u –  dx 1 = 2 1 ln2x – x + C ] Q.6294/def      n 1 k 2 2 2 n x k n n Lim , x > 0 is equal to (A) x tan–1(x) (B) tan–1(x) (C*) x ) x ( tan 1  (D) 2 1 x ) x ( tan Q.6397/def Let f(x) = 2 2 2 2 0 2 2 cos sin( ) sin sin sin cos sin cos x x x x x x x x   then 0 2 /  [f(x) + f (x)] dx = (A*)  (B) /2 (C) 2  (D)zero [Hint: Use c2  c2 + 2 cos x c3  f(x) = 2  f (x) = 0 ] Q.64100/def The absolute value of sinx x 1 8 10 19   is less than : (A) 10 10 (B) 1011 (C*) 10 7 (D) 109 [Hint: sinx x 1 8 10 19    sinx x 1 8 10 19   dx  dx x 1 8 10 19   < dx x8 10 19  = x        7 10 19 7 =  1 7 [19 7  10 7] = 1 7 [10 7  19 7] < 10 7 ] Q.65101/def The value ofthe integral     (cos px  sin qx)2 dx where p, q are integers, is equalto : (A)  (B) 0 (C)  (D*) 2 [Hint : I =      dx ) qx sin px (cos 2 I =      dx ) qx sin px (cos 2 (UsingKing) 2I = 2      dx ) qx sin px (cos 2 2 I =    0 2 2 dx ) qx sin 2 px cos 2 ( =      0 dx ) qx 2 cos 1 ( ) px 2 cos 1 ( = 2 (Boththe integrals vanish)
17. Q.6625/inde Primitive of f (x) = ) 1 x ( n 2 2 · x  l w.r.t. x is (A) ) 1 x ( 2 2 2 ) 1 x ( n 2   l + C (B) 1 2 n 2 ) 1 x ( ) 1 x ( n 2 2    l l + C (C*) ) 1 2 n ( 2 ) 1 x ( 1 2 n 2    l l + C (D) ) 1 2 n ( 2 ) 1 x ( 2 n 2   l l + C [Sol. I =   dx 2 x ) 1 x ( n 2 l let x2 + 1 = t ; x dx = 2 dt Hence I =  dt 2 2 1 t n l =  dt t 2 1 2 n l = 1 2 n t · 2 1 1 2 n   l l + C = 1 2 n ) 1 x ( · 2 1 1 2 n 2    l l + C  (C) ] Q.67103/def            2 0 n n dt 1 n t 1 Lim is equal to (A) 0 (B) e2 (C*) e2 – 1 (D) does not exist [Sol.            2 0 n n dt 1 n t 1 Lim = 2 0 1 n n 1 n t 1 Lim                    = 1 1 n 2 1 Lim 1 n n             = e2 – 1               function linear a is 1 n t 1 ] Q.68106/def Limit h  0   n t dt n t dt h a x h a x 2 2     = (A) 0 (B*) ln2 x (C) 2nx x (D) does not exist Q.69108/def Let a, b, c be nonzero real numbers such that ; 0 1  (1 + cos8x) (ax2 + bx + c) dx = 0 2  (1 + cos8x) (ax2 + bx + c) dx , then the quadratic equation ax2 + bx+ c = 0 has : (A) no root in (0, 2) (B*) atleast one root in (0, 2) (C) a double root in (0, 2) (D) none [Hint:   1 0 8 dx ) x ( f ) x cos 1 ( =   2 0 8 dx ) x ( f ) x cos 1 ( =      2 1 8 1 0 8 dx ) x ( f ) x cos 1 ( dx ) x ( f ) x cos 1 ( Hence   2 1 8 dx ) x ( f ) x cos 1 ( = 0  (1+cos8x) f(x) = 0 at least once in (1,2) but 1 + cos8x 0  f(x) = ax2 + bx + c vanishes at least once in (1,2) ]
18. Q.70110/def Let In = 0 4 /  tann x dx, then 1 1 1 2 4 3 5 4 6 I I I I I I    , , ,.... are in : (A*)A.P. (B) G.P. (C) H.P. (D) none [Hint: In = 0 4 /  tann  2 x (sec2 x  1) dx = 1 1 n   In  2  In + In  2 = 1 1 n   I2 + I4 = 5 1 ; I5 + I3 = 6 1 and I6 + I4 = 7 1  result ] Q.7127/inde Let g (x) be an antiderivative for f (x). Then ln     2 ) x ( g 1 is an antiderivative for (A)  2 ) x ( 1 ) x ( ) x ( 2 f g f  (B*)  2 ) x ( 1 ) x ( ) x ( 2 g g f  (C)  2 ) x ( 1 ) x ( 2 f f  (D) none [Sol. Given  dx ) x ( f = g (x)  g ' (x) = f (x) now   ) x ( g 1 ( n dx d 2  l = ) x ( g 1 ) x ( ' g ) x ( g 2 2  = ) x ( g 1 ) x ( ) x ( 2 2  g f  (B)] Q.72114/def 0 4 /  (cos 2x)3/2. cos x dx = (A) 3 16  (B) 3 32  (C*) 3 16 2  (D) 3 2 16  [Hint: I = 0 4 /  (1  2 sin2 x)3/2 cos x dx. Put 2 sin x = sin   I = 1 2 0 2 /  cos4  d = 3 16 2  ] Q.73116/def The value ofthe definite integral     2 1 0 2 2 2 ) x 1 1 ( x 1 dx x is (A) 4  (B) 2 1 4   (C*) 2 1 4   (D) none [Sol. I =     2 1 0 2 2 2 ) x 1 1 ( x 1 dx x put x = sin ;dx = cos d =         4 0 2 ) cos 1 ( cos d cos sin =      4 0 d ) cos 1 ( = =   4 0 sin     =         2 1 4 Ans. ]
19. Q.74117/def The value ofthe definite integral      37 19 2 dx ) x 2 (sin 3 } x { where { x} denotes the fractionalpart function. (A) 0 (B*) 6 (C) 9 (D) cannot be determined Q.75122/def The value ofthe definite integral   2 0 dx x tan , is (A)  2 (B*) 2  (C)  2 2 (D) 2 2  [Sol. I =   2 0 dx x tan ....(1) ; I =   2 0 dx x cot ....(2) adding (1) and (2), we get 2I =      2 0 dx x cot x tan =    2 0 dx x 2 sin x cos x sin 2 =      2 0 2 dx ) x cos x (sin 1 x cos x sin 2 =    1 1 2 t 1 dt 2 =   1 0 2 t 1 dt 2 2 =  2 (where sin x – cos x = t)  I = 2  Ans. ] Q.7629/inde Evaluate the integral:  dx x ) x 6 ( n 2 l (A) 3 2 )] x 6 ( n [ 8 1 l + C (B*) )] x 6 ( n [ 4 1 2 2 l + C (C) )] x 6 ( n [ 2 1 2 l + C (D) 4 2 )] x 6 ( n [ 16 1 l + C [Hint: ln(6x2) = t ] Q.77123/def               6 5 6 2 2 d ) sin 1 ( 2 1 ) sin 3 ( 2 1 (A)  – 3 (B*)  (C)  – 3 2 (D)  + 3 Q.78125/def Let l =    x 2 x x t dt Lim and m =   x Lim  x 1 dt t n x n x 1 l l then the correct statement is (A*) l m = l (B) l m= m (C) l = m (D) l > m
20. [Hint: l =   x Lim ln 2x – ln x = ln 2 ; m = x n x t n x 1 l l  =   x Lim x n x 1 · x x n l l  =   x Lim x n 1 x n l l  = 1 Hence l × m = ln 2 · 1 = ln 2 = l ] Q.79126/def If f (x) = e–x + 2 e–2x + 3 e– 3x +...... +  , then  3 n 2 n dx ) x ( f l l = (A) 1 (B*) 2 1 (C) 3 1 (D) ln 2 [Sol.  3 n 2 n l l (e–x + 2 e–2x + 3 e– 3x +...... +  ) dx = –   3 n 2 n x 4 x 3 x 2 x ...... e e e e l l         =                          ....... 2 1 2 1 2 1 ....... 3 1 3 1 3 1 3 2 3 2 = 2 1 2 1 1 3 / 1 1 3 / 1 2 / 1 1 2 / 1       Ans: ] Q.80127/def If I = n x (sin ) / 0 2   dx then n x x (sin cos ) / /      4 4 dx = (A*) I 2 (B) I 4 (C) I 2 (D) I [Hint: I1 =     / / 4 4 ln (sin x + cos x) dx =     / / 4 4 ln (cos x  sin x) dx (usingking)  2 I1 =     / / 4 4 ln cos 2x dx = 2 0 4 /  ln (cos 2x) = 0 2 /  ln (cos t) dt where 2x = t 0 2 /  ln (sin t) dt = I  I1 = I/2 ] Q.81129/def The value of                        1 0 n 1 k n 1 r dx k x 1 ) r x ( equals (A) n (B) n ! (C) (n + 1) ! (D*) n · n ! [Hint: The givenintegrand is perfect differentialcoeff. of    n 1 r ) r x (  I = 1 0 n 1 r ) r x (          = (n + 1)! – n! = n · n! ]
21. Q.8233/inde    x sin x sin x cos x cos 4 2 5 3 dx (A) sinx  6 tan1 (sinx) + c (B) sin x  2 sin1 x + c (C*) sinx  2 (sinx)1  6 tan1 (sin x) + c (D) sinx  2 (sinx)1 + 5 tan1 (sinx) + c [Hint: sin x = t ; I =          2 2 2 2 t 1 t t 2 t 1 dt =     ) y 1 ( y ) 2 y ( ) 1 y ( dy = 1 + ) 1 y ( y ) y 2 1 ( 2   ; y = t2 = 1 + 6         1 y 1 y 3 1 =            2 2 t 1 6 t 2 1 dt ] Q.83130/def 0 3  1 4 4 4 4 2 2 x x x x              dx = (A) ln 5 2 3 2  (B)ln 5 2 3 2  (C*) ln 5 2 5 2  (D) none [Hint: I =     3 0 dx ) | 2 x | | 2 x | 1 ( =   3 0 dx 2 x dx +   3 0 dx | 2 x | ] Q.84132/def The value ofthe function f(x) = 1+ x+ 1 x  (ln2t + 2lnt) dt where f (x) vanishes is : (A) e1 (B) 0 (C) 2 e1 (D*) 1 + 2 e1 [Hint: f (x) = 1 + ln2 x + 2 ln x = 0  (1 + ln x)2 = 0  x = 1 e Hence f       e 1 = 1 + e 1 +   e 1 1 2 dt ) nt 2 t n ( l l = e 1 1 2 t ln t e 1 1      = e 1 e 1 1   = 1+ 2e–1  [D] Q.85134/def Limit n   1 1 1 2 3 3 1 n n n n n n n n n n                   ....... ( ) has the value equalto (A) 2 2 (B) 2 2  1 (C*) 2 (D) 4 [Hint : Tr = 1 n n n r   S = 1 n     3 n 3 0 r r n n = 1 1 0 3   x dx ] Q.86136/def Let a function h(x) be defined as h(x) = 0, for all x  0. Also     dx ) x ( · ) x ( f h = f (0), for every function f (x). Thenthe valueofthe definite integral     dx x sin · ) x ( ' h , is (A) equalto zero (B) equalto 1 (C*) equalto – 1 (D)nonexistent [Sol. I =       dx x sin · ) x ( ' I II h =    ) x ( h · x sin –     dx ) x ( h · x cos = 0 – cos 0 = – 1  (A) note that here cos x = f (x) ]
22. Q.87137/def 0 4 /  (tann x + tann  2 x)d(x  [x]) is : ( [. ] denotes greatest integer function) (A*) 1 1 n  (B) 1 2 n  (C) 2 1 n  (D) noneofthese Q.88139/def               1 1 0 0 dx ) x 1 ( Lim is equal to (A) 2 ln 2 (B*) e 4 (C) ln e 4 (D) 4 [Sol.               1 1 0 0 dx ) x 1 ( Lim =                   1 1 0 1 0 1 ) x 1 ( Lim =                 1 1 0 1 1 2 Lim (1 form) =                    1 1 1 2 1 Lim 1 0 e =                   ) 1 ( 2 2 Lim 1 0 e =               1 ) 1 2 ( 2 Lim 0 e = e2 ln 2 – 1 =       e 4 n e l = e 4 ] Q.8934/inde Which one ofthe following is TRUE. (A) C | x | n x x dx . x    l (B*) Cx | x | n x x dx . x    l (C) C x tan dx x cos . x cos 1    (D) C x dx x cos . x cos 1    [Hint: ) C | x | n ( x x dx . x    l = x ln | x | + Cx ] Q.90140/def 0   x2n + 1·e x  2 dx is equal to (n  N). (A) n ! (B) 2 (n !) (C*) n ! 2 (D) 2 )! 1 n (  [Sol. I =    0 x n 2 dx e x . ) x ( 2 put x2 = t  x dx = – dt/2 =    0 t n dt e t 2 1 =                 0 t 1 n 0 t n dt e t n e t 2 1 =              0 t 1 n dt e t n 0 2 1 Hence I = 2 ! n ]
23. Q.91141/def The true set of values of'a' for which the inequality  0 a (32x  2. 3x) dx  0 is true is: (A) [0, 1] (B) ( , 1] (C) [0, ) (D*) ( ,  1]  [0, ) [Sol.      0 a x x 0 dx ) 2 3 ( 3 put 3–x = t 3–x ln3 dx = –dt     a 3 1 0 dt ) 2 t ( 3 ln  0 t 2 2 t a 3 1 2       0 2 2 1 3 . 2 2 3 a a 2                     3–2a – 4×3–a + 3 > 0 (3–a – 3) (3–a – 1) > 0 3–a > 31  a < 1 or 3–a < 30  a > 0 Hence ) , 0 [ ) 1 , ( a      ] Q.92142/def If (2, 3) then number ofsolution of the equation   0 cos(x+ 2) dx = sin is : (A) 1 (B*) 2 (C) 3 (D) 4. [Sol.    0 2 ) x sin( = sin sin(2 + ) – sin2 = sin 2 cos(2 +/2) sin/2 = sin now proceed and get 2 ,    1 1 8 2   2 solutions ] Q.93143/def If x · sin x =  2 x 0 dt ) t ( f where f is continuous functions then the value of f(4) is (A*) 2  (B) 1 (C) 2 1 (D) cannot be determined [Hint: differentiate w.r.t. x, 2x f (x2) = sin x + x cos x put x = 2, 4· f (4) = 0 + 2  f (4) = 2  ; when x = – 2, (–4) · f (4) = – 2, f (4) = 2  ] Q.9436/inde     dx ) 1 x 4 x ( ) 1 x 2 ( 2 / 3 2 (A) C ) 1 x 4 x ( x 2 / 1 2 3    (B*) C ) 1 x 4 x ( x 2 / 1 2    (C) C ) 1 x 4 x ( x 2 / 1 2 2    (D) C ) 1 x 4 x ( 1 2 / 1 2   
24. [Hint:     dx ) 1 x 4 x ( 1 x 2 2 / 3 2 =           dx x 1 x 4 1 x 1 x 2 2 / 3 2 3 =             dx x 1 x 4 1 x x 2 2 / 3 2 3 2 now put 2 2 t 1 x 4 x 1    ] Q.95144/def Ifthe value ofthe integral ex2 1 2  dxis , thenthe value of nx e e4  dxis : (A) e4  e  (B*) 2 e4  e  (C) 2 (e4  e)  (D) 2 e4 – 1 –  [Hint : Put ln x = t2  x =et2  dx = 2t et2 dt. Hence I = 1 2  t. 2 t et2 dt. Now I.B.P. takingt as the first functionand 2 tet2 as the second function ] Q.96145/def          2 x 1 x 2 1 tan dx d 2 1 3 0 equals (A*) 3  (B) 6   (C) 2  (D) 4  [Hint: tan–1        2 x 1 x 2 =    1 x if x tan 2 1 x 0 if x tan 2 1 1        ] Q.97146/def Let A= 0 1  e d t t t 1  then      a 1 a t 1 a t dt e has the value (A)Aea (B*) Aea (C)  aea (D)Aea [Hint: I =      a 1 a t dt 1 a t e put t = a–1+ y(so that lower limit becomes zero) I =     1 0 y a 1 dy 2 y e (nowusingking) I =       1 0 y 1 a 1 dy 2 y 1 e =     1 0 y a dy y 1 e e = – e–a A (B) ] Q.98148/def sin / 2 0 2    sin d is equal to : (A) 0 (B*) /4 (C) /2 (D) 
25. [Hint: I = sin / 2 0 2    . cos  d 2 I = sin / 2 0 2    (cos  + sin ) d =   1 2 0 2    sin cos /    . (cos  + sin ) d Put sin  cos  = t ] Q.9937/inde dx 4 x 2 x 4 2    is equal to (A) C x 2 2 x tan 2 1 2 1    (B) C ) 2 x ( tan 2 1 2 1    (C) C 2 x x 2 tan 2 1 2 1    (D*) C x 2 2 x tan 2 1 2 1    Q.100150/def If  + 2 x e x 2 0 1 2   dx = e x   2 0 1 dx then the value of  is (A*) e1 (B) e (C) 1/2e (D) cannot be determined [Hint:  + 0 1  x. 2 xe x  2 d x = 0 1  e x  2 d x  +     xe x2 0 1  0 1  e x  2 d x = 0 1  e x  2 d x  = 1 e ] Q.101151/defAquadratic polynomialP(x) satisfies the conditions, P(0) = P(1) = 0 & 0 1  P(x) dx= 1.The leading coefficient ofthequadratic polynomialis : (A) 6 (B*) 6 (C) 2 (D) 3 [Hint : Take P(x) =Ax (x  1) and now compute Ausing , A 0 1  x (x  1) dx = 1 ] Q.102152/def Whichone ofthe following functions is not continuous on (0,)? (A) f(x)= cotx (B) g(x) =  x 0 dt t 1 sin t (C) h (x) =            x 4 3 x 9 2 sin 2 4 3 x 0 1 (D*) l (x) =               x 2 , ) x sin( 2 2 x 0 , x sin x [Sol. g (x) = dt t 1 sin t x 0  g (x) = x sin(1/x) which is diff  cont. in(0, ) l (x) =             x 2 / x sin 2 / 2 / x 0 x sin x obvious discontinuityat x = /2  (D) ]
26. Q.103153/def If f (x) =    0 2 2 t sin x tan 1 dt t sin t for 0 < x < 2  (A) f (0+) = –  (B) 8 4 f 2          (C*) fis continuous and differentiablein        2 , 0 (D) fis continuous but not differentiable in        2 , 0 [Sol. f (x) = dt t sin x tan 1 t sin t 0 2 2    Using king and add. f (x) = dt t sin x tan 1 t sin 2 0 2 2     = dt ) t cos 1 ( x tan 1 t sin 2 / 0 2 2      = dt t cos x tan x sec t sin 2 / 0 2 2 2     =    1 0 2 2 2 y . x tan x sec dy =    1 0 2 2 y x ec cos dy x tan = x tan x ) x (sin sin x tan ecx cos y sin x tan 1 1 0 1              ] Q.10441/inde Consider f(x) = x x 2 3 1 ; g(t) = f t dt ( )  . Ifg(1) = 0 theng(x) equals (A) 1 3 1 3 n x ( )  (B*) 1 3 1 2 3 n x        (C) 1 2 1 3 3 n x        (D) 1 3 1 3 3 n x        Q.105158/def The value ofthe definite integral,  100 0 x dx e x 2 is equalto (A) 2 1 (1 – e–10) (B) 2(1 – e–10) (C) 2 1 (e–10 – 1) (D*) 2 1         4 10 e 1 [Hint: put x2 = t; x dx = 2 dt I = 2 1  4 10 0 t e dt = – 2 1  4 10 0 t e = 2 1         4 10 e 1 Ans. ] Ans. ]
27. Q.106159/def 0   [2 ex] dx where [x] denotes the greatest integer function is (A) 0 (B*) ln 2 (C) e2 (D) 2/e [Hint: for 0 < x < ln 2, [2ex] = 1, otherwise zero  I = 0 2 n  dx + n2   0 dx = ln 2 alternatively, put ex = t ; – x = ln t ; dx = dt t 1  1 0 dt t 1 ] t 2 [ –  1 0 t dt ] t 2 [ ;  2 1 0 dt 0 +  2 1 2 1 t dt = 1 2 1 t n l =  – ln 2 1 = ln 2 Ans.] Ans.] Q.107160/def The value of   1 1 | x | dx is (A) 2 1 (B) 2 (C*) 4 (D) undefined [Sol.  1 0 x dx 2 = 1 0 1 2 1 1 2 1 x               = 4  1 0 x = 4  (C) ] Q.108161/def x n x dx l 1 2 0 1         = (A*) 3 4 1 2 3 2        ln (B) 3 2 7 2 3 2  ln (C) 3 4 1 2 1 54  ln (D) 1 2 27 2 3 4 ln  [Sol. I =         1 0 dx 2 2 x n x l =    1 0 dx ) 2 n ) 2 x ( n ( x l l =     1 0 1 0 dx x 2 n dx ) 2 x ( n x l l          1 0 2 1 0 2 2 2 n dx 2 x x 2 x . ) 2 x ( n l l =       1 0 2 2 2 n dx 2 x 4 4 x 3 n 2 1 l l             1 0 dx 2 x 4 ) 2 x ( 2 3 n 2 1 l now proceed] Q.10942/inde The evaluation of p x q x x x dx p q q p q p q         z 2 1 1 2 2 2 1 is (A) – x x C p p q    1 (B) x x C q p q    1 (C*)     x x C q p q 1 (D) x x C p p q    1
28. [Sol. p x qx x p q q p q       z 2 1 1 2 1 ( ) dx = p x qx x x dx p q p q       z 1 1 2 ( ) taking xq as x2q common fromDenominator and take it in Nr ] Q.110162/def x x x x 3 2 1 1 1 2 1       | | | | dx = a ln 2 + b then : (A) a = 2 ; b = 1 (B*) a = 2 ; b = 0 (C) a = 3 ; b =  2 (D) a = 4 ; b =  1 [Hint: I =     1 1 2 3 1 x 2 x x dx +   x x     1 1 2 1 1 dx  2 dx x 1 0 1   = 2 ln 2 ] odd  vanishes even ] Q.111163/def a b  [x] dx + a b  [x] dxwhere [. ] denotes greatest integer function is equalto : (A) a + b (B) b  a (C*) a  b (D) a b  2 [Hint: [x] + [ x] = 0 1 if x I if x I        I = a b   dx = a  b ] Q.112166/def If 0 2  375 x5 (1 + x2) 4 dx = 2n then the value of n is : (A) 4 (B*) 5 (C) 6 (D) 7 [Sol. n 2 0 4 2 5 2 dx ) x 1 ( x 375    ; n 2 0 4 2 8 5 2 dx ) x 1 ( x x 375     ; n 2 0 4 2 3 2 dx ) x 1 ( x 375      simplifying n 2 0 3 2 2 2 x 1 x 6 375               n 2 5 . 5 . 5 4 . 4 . 4 6 375   2n = 32  n = 5 ] Q.113167/def     2 / 1 0 2 x 1 x 1 n x 1 1  dx is equal to : (A*) 3 1 n 4 1 2  (B) 2 1 ln2 3 (C)  4 1 ln2 3 (D) cannot be evaluated. [Hint: Put ln (1 + x)  ln (1  x) = t  dx x 1 2  = 1 2 dt I = 1 2 0 3 n  t dt = 1 4 ln2 3 = 1 4 ln2 1 3 ]
29. Q.11443/inde If    dx e ) 5 x 2 x ( x 3 2 3 = e3x (Ax3 + Bx2 + Cx+ D) then the statement which is incorrect is (A) C + 3D = 5 (B) A + B + 2/3 = 0 (C*) C + 2B = 0 (D) A+ B + C = 0 [Hint: A = 3 1 , B = – 1, C = 3 2 ; D = 9 13 ] Q.115169/def Given     2 / 0 x cos x sin 1 dx = ln 2, then the value of the def. integral.     2 / 0 x cos x sin 1 x sin dx is equalto (A) 2 1 ln2 (B) 2   ln 2 (C*) 4  – 2 1 ln 2 (D) 2  + ln 2 [Hint: I = cos sin cos / x x x 1 0 2      2 I = sin cos sin cos / x x x x       1 1 1 0 2  dx  2 I =  2  ln 2  I =  4 1 2  ln 2 ] Q.116171/def Afunction f satisfying f (sinx) = cos2 x for allx and f(1) = 1 is : (A) f(x) = x + 3 1 3 x3  (B) f(x) = 3 2 3 x3  (C*) f(x) = x  3 1 3 x3  (D) f(x) = x  3 1 3 x3  [Hint: Put sin x = t  f (t) = 1  t2. Now integrate ] Q.117172/def For 0 < x <  2 , 1 2 3 2 / /  ln (ecos x). d (sinx) is equal to : (A*) 12  (B) 6  (C)       1 sin 3 sin 1 3 4 1    (D)       1 sin 3 sin 1 3 4 1    [Hint : I =   / / 6 3  cos x cos x dx =   / / 6 3  cos2 x dx =   / / 6 3  sin2 x dx  2 I =   / / 6 3  dx =  6  I =  12 ] Q.118173/def      0 2 x sin 1 x cos x dx is equal to : (A) 2 (B)  (2 + ) (C) zero (D*) 2  [Hint: Take x as the first and   cos sin x x 1 2  as the second function. I B P]
30. Q.11944/inde   dx x x x e x  (A*) 2   1 x x e x   + C (B)   1 x 2 x e x   (C)   C x x e x   (D)   C 1 x x e x    [Sol.   dx x x x e x  ; put x = t2 ; dx = 2t dt =   dt ) t t ( e 2 t = et (At2 + Bt + C) (Let) Diffrentiate boththe sides et (t2 + t) = et (2At + B) + (At2 + Bt + C) et Oncomparingcoefficient we get A = 1 ; B = – 1 ; C = 1 ] Q.120174/def dx x x cos sin / 6 6 0 2    is equal to : (A) zero (B*)  (C) /2 (D) 2  [Hint: I = dx x x 1 3 2 2 0 2   sin cos /  = dx x 1 2 3 4 2 0 2   sin /  = 2 dt t 4 3 2 0   sin  where 2x = t = 4 dt t 4 3 2 0 2   sin /  etc. ] Q.121178/def The true solution set ofthe inequality,      x 0 2 2 x 6 x 5 dz > x 0   sin2 x dxis : (A) R (B) (1, 6) (C) ( 6, 1) (D*) (2, 3) [Hint: 5 6 2 x x   + 2 x  > x  2  5x  6  x2 > 0 ] Q.122179/def If   1 0 2 x 1 x n  dx = k 0   ln (1 + cosx) dx then the value of k is : (A) 2 (B*) 1/2 (C)  2 (D) 1/2 [Hint: Put x = sin   L H S = 0 2 /  ln (sin ) d = k 0   ln (1  cos x) = 2 k 0 2 /  ln (sin x) dx  k = 1/2 ] Q.123182/def Let a, b and c be positive constants. The value of 'a' in terms of 'c' if the value of integral     1 0 5 b 3 3 1 b dx ) bx a acx ( is independent of b equals (A*) 2 c 3 (B) 3 c 2 (C) 3 c (D) c 2 3
31. [Sol. I =     1 0 5 b 3 3 1 b dx ) bx a acx ( ; ac· 1 0 6 b 3 3 2 b 6 b 3 bx a 2 b x         = ) 2 b ( 3 b a 2 b ac 3    I(b) = ) 2 b ( 3 1  [3ac + a3b]  ) 2 b ( 3 a c 3 b a 2 3         Ifthis is independent of b, then 2 a c 3 2  Alternatively: Ifthe integralis independent ofb, then I'(b) = 0  I'(b) = 2 ) 2 b ( ac   +                  2 b 2 2 b · 3 a D 3 = 2 ) 2 b ( ac   +          2 3 ) 2 b ( 2 3 a = 2 ) 2 b ( 1           ac 3 a 2 3 = 3 a 2 3 = ac  a = 2 c 3 Ans.] Ans.] Q.12451/inde       d ) tan (sec sec 2 2 (A) C )] tan (sec tan 2 [ 2 ) tan (sec          (B) C )] tan (sec tan 4 2 [ 3 ) tan (sec          (C*) C )] tan (sec tan 2 [ 3 ) tan (sec          (D) C )] tan (sec tan 2 [ 2 ) tan (sec 3          [Hint: put sec  + tan  = t] Q.125184/def    2 1 4 2 1 x 1 x dx is equal to: (A) 1 2 tan1 2 (B*) 1 2 cot1 2 (C) 1 2 tan1 1 2 (D) 1 2 tan1 2 Q.126186/def 1 x x Limit  1 x x x   x x1 f(t) dt is equal to : (A)   f x x 1 1 (B*) x1 f(x1) (C) f(x1) (D) does not exist [Sol.          x x x dt ) t ( f Limit 1 x x x x 1 1 = 1 2 x x x x . ) x ( f Limit 1  (using Lopital's rule) = x1 f(x1)  (B) ]
32. Q.127187/def Which ofthe following statements could be true if, f(x) = x1/3. I II III IV f(x) = 28 9 x7/3 + 9 f (x) = 28 9 x7/3  2 f (x) = 4 3 x4/3 + 6 f(x) = 4 3 x4/3  4 (A) I only (B) III only (C) II & IV only (D*) I & III only [Sol. f   (x) = x1/3  f  (x) = 3/4 x4/3 + C1 .....(1) f (x) = 2 1 3 / 7 C x C x 3 7 . 4 3   = 2 1 3 / 7 C x C x 4 7    I if C1 = 0 ] Q.128191/def The value ofthe definite integral 0 2 /  sinx sin2x sin3x dx is equalto : (A) 1 3 (B)  2 3 (C)  1 3 (D*) 1 6 [Hint: Use 0 a  f(x) = 0 a  (a  x) dx and add two integrals ] Q.12952/inde dx x 1 x 1 cos x 1 sec ) x 1 ( e 2 2 1 2 2 1 2 x 1 tan                                (x > 0) (A) C x tan . e 1 x 1 tan    (B)   C 2 x tan . e 2 1 x 1 tan    (C*) C x 1 sec . e 2 2 1 x 1 tan                 (D) C x 1 ec cos . e 2 2 1 x 1 tan                 [Hint: note that sec–1 2 x 1 = tan–1x ; cos–1           2 2 x 1 x 1 = 2 tan–1x for x > 0 I =  dx x tan 2 ) x (tan x 1 e 1 2 1 2 x tan 1       put tan–1x = t = dt ) t 2 t ( e 2 t   = et . t2 =  2 1 x 1 tan x tan e   + C ] Q.130193/def Number of positive solution of the equation,     t t x   2 0 dt = 2 (x  1) where { } denotes the fractionalpart functionis : (A) one (B*) two (C) three (D) morethanthree [Hint: consider 0 < x < 1 ; 1  x < 2 ; 2  x < 3 etc. Ans. x = 1 and 5/2 ]
33. Q.131194/def Iff(x) = cos(tan–1x) then the value of the integral dx ) x ( ' ' f x 1 0  is (A) 2 2 3 (B) 2 2 3 (C) 1 (D*) 2 2 3 1 [Sol. f(x) = cos (tan–1x) f  (x) = 2 1 x 1 ) x sin(tan    I = dx ) x ( ' ' f x 1 0  =  dx ) x ( ' f ) x ( ' f x 1 0 1 0   = f  (1) – 1 0 ) x ( f = f  (1) – [ f (1) – f (0) ] = f  (1) – f (1) + f (0) f (0) = 1 ; f  (1) = – 2 2 1 ; f (1) = 2 1 ] Q.1322/inde If   2 x sin 1 dx =Asin         4 4 x then value ofA is: (A) 2 2 (B) 2 (C) 1 2 (D*) 4 2
34. Q.133201/def For Un = 0 1  xn (2  x)n dx; Vn = 0 1  xn (1  x)n dx n  N, which of the following statement(s) is/are ture? (A) Un = 2n Vn (B) Un = 2 n Vn (C*) Un = 22n Vn (D) Un = 2  2n Vn [Sol. Given Un =   1 0 n n dx ) x 2 ( . x ; Vn =   1 0 n n dx ) x 1 ( . x put x = 2t  dx = 2dt  Un =   2 / 1 0 n n n n dt ) t 1 ( 2 t . 2 2 ....(1) Now Vn =   2 / 1 0 n n dx ) x 1 ( x 2 (Using Queen) .....(2) From (1) and (2) Un = 22n. Vn  (C) ] Q.13454/inde               x 1 x tan ) 1 x 3 x ( dx ) 1 x ( 2 1 2 4 2 = ln | f (x) | + C then f (x) is (A) ln        x 1 x (B*) tan–1        x 1 x (C) cot–1        x 1 x (D) ln                 x 1 x tan 1 [Hint:  dx x 1 x tan 3 x 1 x x 1 1 1 2 2 2                  ;  dx x 1 x tan 1 x 1 x dx x 1 1 1 2 2                                ] Q.135202/def Let f(x) be integrable over (a, b) , b > a > 0. If I1 =   / / 6 3  f (tan  + cot ). sec2  d & I2 =   / / 6 3  f (tan  + cot ). cosec2  d , thenthe ratio I I 1 2 : (A*) is a positive integer (B) isa negative integer (C) is anirrationalnumber (D) cannot be determined. [Hint: Using king in I1 , I1 =   / / 6 3  f(tan  + cot ). cosec2  d = I2  I I 1 2 = 1 ]
35. Q.136203/def f(x) = cos sin x x  (1  t + 2 t3) d t has in [ 0, 2  ] (A) a maximumat  4 & aminimumat 3 4  (B*) amaximumat 3 4  & aminimumat 7 4  (C) a maximumat 5 4  &a minimumat 7 4  (D) neither amaxima nor minima [Hint : f (x) = (1. cosx  sinx cosx + 2 sin3 x. cos x)  (1. sinx + cosx sinx  2 cos3 x sin x) = cosx + sinx – 2 sinx · cosx + 2 sinx · cosx = cosx + sinx ] Q.137206/def Let S (x) = x x 2 3  ln t d t (x > 0) and H(x) =  S x x ( ) . Then H(x) is : (A) continuousbut not derivable inits domain (B*) derivable andcontinuous inits domain (C) neither derivable nor continuous inits domain (D) derivablebut not continuous inits domain. [Hint: S  (x) = ln x3. 3x2  ln x2. 2x = 9x2 ln x  4x ln x = x l n x (9 x  4). Hence  S x x ( ) = ln x (9 x  4). Now it is obvious that  S x x ( ) iscontinuous and derivable inits domain. ] Q.138211/def Number ofsolution ofthe equation d dx  x sin x cos dt t 1 2  = 2 2 in [0, ] is (A) 4 (B) 3 (C*) 2 (D) 0 [Hint: LHS = secx + cosecx =2 2  x =  4 and 11 12  ] Q.13955/inde Let f (x) = x cos 1 x sin 2 2  + x sin 1 ) 1 x sin 2 ( x cos   then     dx ) x ( ' f ) x ( f ex (where c is the constant ofintegeration) (A*) ex tanx + c (B) excotx + c (C) ex cosec2x + c (D) exsec2x + c [Hint: x sin 1 ) x sin 2 1 ( x cos   – x cos x sin x cos 2 2  = ) x sin 1 ( x cos ) x sin x )(cos x sin 1 ( ) x sin 2 1 ( x cos 2 2 2      = ) x sin 1 ( x cos x sin x cos x sin 3 2    = ) x sin 1 ( x cos ) x sin 1 ( x sin x cos x sin 2 2    = x cos x sin ) x sin 1 ( x sin 2   = tan x ]
36. Q.140212/def The value ofx that maximises the value ofthe integral t t dt x x ( ) 5 3    is (A) 2 (B) 0 (C*) 1 (D) none [Sol. F (x) = dx ) t 5 ( t 3 x x    F  (x) = (x + 3) (2 – x) – x (5 – x) = 6 – x – x2 – 5x + x2 = 6 – 6x = 0  x = 1 F  (x) = – 6  x =1 is maxima  (C) ] Q.141213/def For a sufficiently large value of n the sum of the square roots of the first n positive integers i.e. 1 2 3     ...................... n is approximatelyequalto (A) 1 3 3 2 n / (B*) 2 3 3 2 n / (C) 1 3 1 3 n / (D) 2 3 1 3 n / [Hint: Limit n n n n      1 2 3 .................... = x dx 0 1  = 2 3  Sn = 2 3 3 2 n / ] Q.142216/def The value of  2 0 2 ) x 1 ( dx is (A) –2 (B) 0 (C) 15 (D*) indeterminate [Sol.   2 0 2 ) x 1 ( dx =        2 1 2 1 0 2 ) x 1 ( dx ) x 1 ( dx = 2 1 1 0 x 1 1 x 1 1            = ( – 1) + (–1) – (– )  indeterminant Note that the shaded area is divergent ] Q.143217/def If          8 / 0 a 0 d 2 sin tan 2 x a x dx , then the value of 'a' is equal to (a > 0) (A) 4 3 (B) 4  (C) 4 3 (D*) 16 9 [Sol.      a 0 dx x a x a 1 =       d sec . tan 2 tan 2 2 8 / 0 (sin2 =    2 tan 1 tan 2 ) =   a 0 2 / 3 2 / 3 x ) a x ( 3 2 . a 1  = 8 / 0 tan   =     1 2 a a ) a 2 ( a 3 2 2 / 3 2 / 3 2 / 3     =   1 2 1 2 a 2 . a 3 2 2 / 3     1 a 3 4   16 9 a  ]
37. Q.14459/inde The value ofthe integral      dx 1 x ) x 2 2 ( n sin l is (A*) – cos ln (2x + 2) + C (B) ln       1 x 2 sin + C (C) cos       1 x 2 + C (D) sin       1 x 2 + C [Hint: ln 2(1 + x) = t ; dx ) x 1 ( 1  = dt] Q.145218/def If f(x) =Asin        2 x + B , f       2 1 = 2 and  1 0 f(x) dx =  A 2 , Then the constantsAand B are respectively. (A) 2 & 2   (B)   3 & 2 (C)   4 & 0 (D*) 0 & 4  Q.146219/def Let I1 =    2 0 x dx ) x sin( e 2 ; I2 =    2 0 x dx e 2 ; I3 =     2 0 x dx ) x 1 ( e 2 and considerthe statements I I1 < I2 II I2 < I3 III I1 = I3 Whichofthefollowing is(are) true? (A) I only (B) II only (C) Neither I nor II nor III (D*) BothI and II [Sol. since 0 < sin x < 1 and 1 + x > 1 in (0, /2) hence I3 > I2 > I1  A and B are correct  (D) ] Q.147220/def Let f (x) = x x sin , then           2 0 dx x 2 f ) x ( f = (A*)    0 dx ) x ( f 2 (B)   0 dx ) x ( f (C)    0 dx ) x ( f (D)    0 dx ) x ( f 1 [Hint: I =           2 0 dx x 2 x x cos x sin =     2 0 dx ) x 2 ( x x 2 sin ; put 2x = t I =     0 dt ) t ( t t sin =               0 dt ) t ( t sin t t sin 1 =    0 dt t t sin 2 ] Q.148222/def Let u =    1 0 2 dx 1 x ) 1 x ( n l and v =   2 0 dx ) x 2 (sin n l then (A) u = 4v (B*) 4u + v = 0 (C) u + 4v = 0 (D) 2u + v = 0
38. [Sol. u =    1 0 2 dx 1 x ) 1 x ( n l put x = tan  =      4 0 d ) tan 1 ( n l =               4 0 d tan 1 tan 1 1 n l =      4 0 d tan 1 2 n l = 4  ln 2 – u  u = 8  ln 2  4u = 2  ln 2 ....(1) again v =   2 0 dx ) x 2 (sin n l (put 2x = t) ; v =   0 dt ) t (sin n 2 1 l =   2 0 dx ) t (sin n l v = – 2  ln 2 ....(2) (1) + (2)  4u + v = 0  (C) ] Q.149223/def If          d . cos 1 ·sin x sin x 2 16 / 2 x 2 f then the value of f '        2 , is (A*)  (B) –  (C) 2 (D) 0 [Sol.          d . cos 1 sin x sin x 2 16 / 2 x 2 f ; f ' (x) = x cos d cos 1 sin 0 x 2 · x cos 1 x sin x sin 2 2 x 16 2 2                          f '        2 = 1  Ans. ] Q.150225/def The value of the definite integral,   2 0 dx x sin x 5 sin is (A) 0 (B*) 2  (C)  (D) 2 [Sol. sin nx– sin(n – 2)x = 2 cos(n – 1)x sin x        dx x sin x ) 2 n sin( dx ) 1 n cos( 2 dx x sin nx sin          2 0 2 0 2 0 dx x sin x 3 sin dx x 4 cos 2 dx x sin x 5 sin =0 +   2 0 dx x sin x 3 sin =   2 0 dx = 2  Ans. ] Select the correct alternatives : (More than one are correct) Q.151502/def dx x sgn b a  = (where a, b R) (A*) | b | – | a | (B) (b–a) sgn (b–a) (C*) b sgnb – a sgna (D) | a | – | b |
39. Q.152503/inde   x cos 4 5 dx =  tan1       2 x tan m + C then : (A*)  = 2/3 (B*) m = 3 (C)  = 1/3 (D) m = 2/3 Q.153503/def Which ofthe following are true ? (A*) x f x a a . (sin )   dx =  2 . f x a a (sin )    dx (B*) f x a a ( )2   dx = 2. f x a ( )2 0  dx (C*)   f x n cos2 0   dx = n.   f x cos2 0   dx (D*) f x c b c ( )    0 dx = f x c b ( )  dx Q.154505/def The value of   2 3 3 1 2 2 2 2 0 1 x x x x x       ( ) dx is : (A*)  4 + 2 ln2 tan1 2 (B)  4 + 2 ln2tan1 1 3 (C*) 2 ln2 cot1 3 (D*)   4 + ln4+ cot1 2 [Hint : Numerator = 2 (x2 + 2 x + 2)  (x + 1) ] Q.155505/inde x x x 2 2 2 1    cos cosec2 x dx is equal to : (A) cot x  cot 1 x + c (B*) c  cot x + cot 1 x (C*)  tan1 x  cos sec ec x x + c (D*) e n x  tan1  cot x + c where 'c' is constant ofintegration . Q.156506/def Let f(x) = sint t x 0  dt (x > 0) then f(x) has : (A*) Maxima if x = n where n = 1, 3, 5,..... (B*) Minima if x = n where n = 2, 4, 6,...... (C) Maxima if x = n where n = 2, 4, 6,...... (D)Thefunctionismonotonic [Hint: f (x) = sinx x ; f (x) = x x x x cos sin  2 Now f (2 m) we have f (x) > 0 & f ((2 m  1) ) we have f (x) < 0 ] Q.157508/def IfIn =   dx x n 1 2 0 1   ;n  N, thenwhich of the following statements hold good ? (A*) 2n In + 1 = 2 n + (2n  1) In (B*) I2 =  8 1 4  (C) I2 =  8 1 4  (D) I3 =  16 5 48  [Hint: I.B.P. taking 1 as the 2nd and 1 1 2 ( ) x n as the 1st function ]
40. Q.158509/inde 1 1 1 1 2 x n x x dx    z  equals : (A) 1 2 ln2 x x   1 1 + c (B*) 1 4 ln2 x x   1 1 + c (C) 1 2 ln2 x x   1 1 + c (D*) 1 4 ln2 x x   1 1 + c [Hint: put ln (x  1)  ln (x + 1) = t ] Q.159510/def IfAn = 0 2 /  sin ( ) sin 2 1 n x x  d x ; Bn = 0 2 /  sin sin nx x       2 d x ; for n  N , then : (A*) An + 1 = An (B) Bn + 1 = Bn (C) An + 1  An = Bn + 1 (D*) Bn + 1  Bn = An + 1 [Hint: Consider An + 1  An = 0 2 /  sin ( ) sin ( ) sin 2 1 2 1 n x n x x   = 0 2 /  2 cos 2 n x d x = 0  An + 1 = An Again consider Bn + 1  Bn = 0 2 /  sin ( ) sin sin 2 1 2 2 n x x x   d x = 0 2 /  sin ( ) sin 2 1 n x x  dx = A An + 1 ] Q.160511/def 0   x x x ( ) ( ) 1 1 2   d x : (A*)  4 (B)  2 (C*) is same as 0   dx x x ( ) ( ) 1 1 2   (D) cannot be evaluated [Hint: Put x = 1/t and add the two integrals ] Q.161512/inde 1   cscx dx equals (A*) 2 sin1 sinx + c (B) 2 cos 1 cosx + c (C) c  2 sin1 (1  2 sin x) (D*) cos1 (1  2 sin x) + c [Hint: I =       1 1 1    sin sin sin sin x x x x =   cos sin sin x x x 1 =   cos sin x x 1 4 1 2 2   =      dt t 1 2 2 2 ] Q.162512/def If f(x) = 0 2 /  n x ( ) sin sin 1 2 2    d  , x  0 then : (A*) f (t) =    t   1 1 (B*) f (t) =  2 1 t  (C) f(x) cannot be determined (D) none ofthese. [Sol.         2 / 0 2 2 2 d ) sin x 1 ( sin sin dx dI =       2 / 0 2 sin x 1 d dx dI MultiplyNr. and Dr. bysec2 and proceed ]
41. Q.163514/def If a, b, c  R and satisfy 3 a + 5 b + 15 c = 0 , the equation ax4 + bx2 + c = 0 has : (A*) atleast one root in (1, 0) (B*) atleast one root in (0, 1) (C*) atleast two roots in (1, 1) (D) no root in (1, 1) [Hint: 0 1  f(x) d x = a b 5 3  + c = 1 15 (3 a + 5 b + 15 c) = 0  B Since f(x) is even  A  C ] Q.164515/def Let u =     0 2 4 1 x 7 x dx & v =     0 2 4 2 1 x 7 x dx x then : (A) v > u (B*) 6 v =  (C*) 3u + 2v = 5/6 (D*) u + v = /3 [Hint: put x = 1/t in u or v  u = v. Now consider u + v ] Q.165513/inde If  eu . sin2x dx can be found in terms ofknown functions of x then u can be : (A*) x (B*) sin x (C*) cos x (D*) cos 2x Q.166518/def Iff(x) = n t t x 1 1   dt where x > 0 thenthe value(s) ofx satisfying the equation, f(x) + f(1/x) = 2 is : (A) 2 (B) e (C*) e 2 (D*) e2 [Hint: f(x) = 2 x n2  = 2  C, D ] Q.167519/def Apolynomialfunction f(x) satisfying the conditions f(x) = [f(x)]2 & 0 1  f(x) dx = 12 19 canbe: (A) 4 9 x 2 3 4 x2   (B*) 4 9 x 2 3 4 x2   (C) 4 x2  x + 1 (D*) 4 x2 + x + 1 Q.168520/def Acontinuous anddifferentiable function 'f' satisfies the condition , 0 x  f(t) dt = f2 (x)  1 for all real 'x'. Then : (A*) 'f' is monotonic increasing  x R (B) 'f' is monotonic decreasing  x R (C) 'f' is non monotonic (D*) the graphofy= f(x) is a straight line. [Sol. Differentiating f(x) = 2 f(x). f  (x)  f (x) = 1 2 ( f(x))  0) Hence f(x) = x 2 + c. Put x = 0 ; f (0) = c ; but f2 (0) = 1  f (0) = ± 1 Hence f (x) = x 2 ± 1 ]