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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
`int _(3)^(8) (2-3x)/(xsqrt(1+x))dx` is equal toA. `2log((3)/(2e^(3)))`B. `log((3)/(e^(3)))`C. `4log((3)/(e^(3)))`D. None of these |
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Answer» Correct Answer - A Let ` l = int _(3)^(a) (2-3x)/(xsqrt(1+x))dx ` Put ` 1+ x = t^(2) rArr dx = 2 t dt ` When x = 3, t= 2 and x = 8, t=3 ` :. l = 2 int _(2)^(3) (5- 3t^(2))/(t^(2)-1) dt = 2int_(2)^(3) (2/(t^(2)-1)-3) dt` ` = 2 [ 2/(2.1) log ((t-1)/(t+1)) - 3t ]_(2)^(3) = 2 log (3/(2e^(3)))` |
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| 152. |
`int _(0)^(a) (x^(4)dx)/((a^(2)+x^(2))^(4))` is equal toA. `(1)/(16a^(3))((pi)/(4)-(1)/(3))`B. `(1)/(16a^(3))((pi)/(4)+(1)/(3))`C. `(1)/(16)a^(3)((pi)/(4)-(1)/(3))`D. `(1)/(16)a^(3)((pi)/(4)+(1)/(3))` |
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Answer» Correct Answer - A Put ` x= a tan theta rArr dx = a sec^(2) theta d theta ` ` :. int _(0)^(a) (x^(4) dx)/((a^(2)+x^(2))) = int _(0)^(pi//4) (a^(4) tan^(4) theta . a sec^(2) theta d theta )/(a^(8) sec^(8) theta )` ` = 1/(a^(3)) int _(0)^(pi//4) sin^(4) cos^(2) theta d theta ` ` = 1/(a^(3)) = int _(0)^(pi//4) [ ((1-cos2theta)^(2))/4 - ((1-cos 2 theta )^(3))/8 ] d theta ` ` = 1/(8a^(3)) int _(0)^(pi//4) (1+ cos 2theta ) (1+ cos^(2) 2theta - 2 cos 2 theta ) d theta ` ` = 1/(8a^(3)) int _(0)^(pi//4) (1- cos 2 theta - cos^(2) 2theta + cos^(3) 2 theta ) d theta ` ` = 1/(32a^(3)) int _(0)^(pi//4) (2 - cos 2theta - cos 4 theta + cos 6theta ) d theta` ` = 1/ (32a^(3)) [ 2theta - (sin2theta )/2 - (sin 4theta )/2 + (sin 6 theta )/6]_(0)^(pi//4)` ` =1/(16a^(3)) (pi/4 - 1/3)` |
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| 153. |
The correct evalution of `int_(0)^(pi//2) sin x sin 2x dx ` isA. `(4)/(3)`B. `(1)/(3)`C. `(3)/(4)`D. `(2)/(3)` |
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Answer» Correct Answer - D Let ` l = int _(0)^(pi//2) sin x sin 2x dx = 2 int _(0)^(pi//2) sin^(2) x cos x dx ` Put ` t = sin x rArr dt = cos x dx ` ` :. L = 2 int _(0)^(1) t^(2) dt = 2/3 [ t^(3) ] _(0)^(1) = 2/3 ` |
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| 154. |
Evaluate:`int_1^2 1/((x+1)(x+2))dx`(ii) `int_1^2 1/(x(1+x^2))dx` |
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Answer» Let `(1)/(x(1+x^(2)))=(A)/(x)+(Bx+C)/((1+x^(2)))`. Then, `1-=A(1+x^(2))+(Bx+C)x`. Putting `x=0`, we get `A=1`. Comparing the coefficients of `x^(2)`, we get `A+B=0` or `B=-1`. Comparing coefficients f `x`, we get `C=0`. `:.(1)/(x(1+x^(2)))=[(1)/(x)-(x)/(1+x^(2))]` So, `int_(1)^(2)(dx)/(x(1+x^(2)))=int_(1)^(2)(dx)/(x)-(1)/(2)int_(1)^(2)(2x)/(1+x^(2))dx` `=[logx]_(1)^(2)-(1)/(2)[log(1+x^(2))]_(1)^(2)` `=[(3)/(2)(log2)-(1)/(2)(log5)]`. |
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| 155. |
`int_(a//4)^(3a//4)sqrt(x)/(sqrt(a-x)+sqrt(x)) dx=a/4` |
| Answer» Apply, `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`. | |
| 156. |
The value of `int_(-3)^(3)(ax^()+bx^(3)+cx+k)dx`, where a,b,c,k are constants, depends only on. . . .A. a and kB. a and bC. a, b and cD. k |
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Answer» Correct Answer - D Integral depends upon because `ax^(5) + bx^(3) +cx ` is odd function . |
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| 157. |
`3a int_(0)^(1) ((ax-1)/(a-1))^(2) dx` is equal toA. `a-1+(a-1)^(-2)`B. `a+a^(-2)`C. `a-a^(2)`D. `a^(2)+(1)/(a^(2))` |
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Answer» Correct Answer - A ` 3a int _(0)^(1) ((ax-1)/(a-1))^(2) dx = (3a)/((a-1)^(2)) [ ((ax-1)^(3))/3xx1/a]_(0)^(1)` ` = 1/((a-1)^(2)) [ (a-1)^(3) +1]` ` = (a-1) +(a-1)^(-2)` |
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| 158. |
` int _(0)^(pi) cos^(3) x dx` is equal to |
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Answer» Correct Answer - A Let ` l = int _(0)^(pi) cos^(3) x dx " " ` …(i) ` rArr l = int _(0)^(pi) cos^(3) (pi-x) dx = - int _(0)^(pi) cos^(3) x dx " "` …(ii) On adding Eqs. (i) and (ii) , we get ` 2l = 0 rArr l=0` |
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| 159. |
` int _(0)^(pi) | cos x| dx ` is equal toA. `(1)/(2)`B. `-2`C. 1D. 2 |
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Answer» Correct Answer - D ` int _(0)^(pi) | cos x| dx = int _(0)^(pi//2) cos x dx - int _(pi//2)^(pi) cos x dx ` ` = [ sin x ] _(0)^(pi//2) - [ sin x] _(pi//2)^(pi)` ` = sin pi/2 - sin 0 - sin pi + sin. pi/2 ` ` = 2 sin. pi/2 = 2 ` |
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| 160. |
The value of ` int _(1)^(e^(2)) (dx)/(x(1+ log x)^(2) ) ` isA. `2/3`B. `1/3`C. `3/2`D. log 2 |
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Answer» Correct Answer - A Let `l = int_(1)^(e^(2))(dx)/(x(1+log x)^(2))` Put (1 + log x) `= t rArr dt = (1)/(x) dx` When x = 1, t = 1 When `x=e^(2), t= 3` `therefore " " l = int_(1)^(3)(dt)/(t^(2))=[(-1)/(t)]_(1)^(3)=-[(1)/(3)-1]=(2)/(3)` |
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| 161. |
` int _(-10)^(10) log ((a+x)/(a-x)) dx` is equal to |
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Answer» Correct Answer - A Let `f(x) = log ((a+x)/(a-x))` ` rArr f(-x) = log ((a-x)/(a+x))` ` = - log ((a+x)/(a-x)) = - f(x)` ` rArr f(x) ` is an odd function , ` :. int _(-10)^(10) f(x) dx = 0` |
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| 162. |
` int _(-pi)^(pi) (sin^(4)x)/(sin^(4) x + cos^(4) x )dx ` is equal toA. `(pi)/(4)`B. `(pi)/(2)`C. `(3pi)/(2)`D. `pi` |
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Answer» Correct Answer - D Let ` l = 2 int _(0)^(pi) (sin^(4)x)/(sin^(4) x + cos^(4)x) dx` ` = 4 int _(0)^(pi//2) (sin^(4)x)/(sin^(4)x + cos^(4) x ) dx" "` …(i) , ` = 4 int _(0)^(pi//2) (sin^(2) (pi/2-x)_(dx))/(sin^(4)(pi/2-x) + cos ^(4) (pi/2 -x))dx ` `l = 4 int _(0)^(pi//2) (cos^(4) x)/(sin^(4) x+ cos ^(4) x) dx " "` ...(ii) On adding Eqs. (i) and (ii) , we get ` 2l = 4 int _(0)^(pi//2) 1.dx = 2pi rArr l = x` |
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| 163. |
` int _(0)^(3) (3x+1)/(x^(2)+9) dx = `A. `log(2sqrt(2))+pi/12`B. `log(2sqrt(2))+pi/2`C. ` log (2sqrt(2)) +pi/6`D. ` log (2sqrt(2))+pi/3` |
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Answer» Correct Answer - A `int_(0)^(3)(3x+1)/(x^(2)+9)dx=(3)/(2)int_(0)^(3)(2x)/(x^(2)+9)dx+int_(0)^(3)(dx)/(x^(2)+9)` `=[(3)/(2)log(x^(2)+9)+(1)/(3)tan^(-1)((x)/(3))]_(0)^(3)` `=(3)/(2)(log18-log9)+(1)/(3)((pi)/(4))` `=(3)/(2)log2+(pi)/(12)=log(2sqrt2)+(pi)/(12)` |
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| 164. |
Value of the integral `int _(-pi//2)^(pi//2) cos x dx ` isA. 4B. 2C. 0D. 1 |
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Answer» Correct Answer - B Let `l = int _(-pi//2)^(pi//2) cos x dx ` Since, `cos (-x) = cos x` ` :. L = 2 int _(0)^(pi//2) cos x dx = 2 [ sin x ] _(0)^(pi//2) =2` |
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| 165. |
` int _(-1)^(2) |x|^(3) dx` is equal toA. `5/4`B. `17/4`C. `15/4`D. `4/5` |
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Answer» Correct Answer - B `int_(-1)^(2)|x|^(3)dx=int_(-1)^(0)-x^(3)dx+int_(0)^(2)x^(3)dx=-[(x^(4))/(4)]_(-1)^(0)+[(x^(4))/(4)]_(0)^(2)` `=-(1)/(4)[0-1]+(1)/(4)[2^(4)]=(1)/(4)+4=(17)/(4)` |
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| 166. |
The value of `l = int _(-pi//2)^(pi//2) | sin x | dx ` is |
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Answer» Correct Answer - B ` int _(-pi/2)^(pi/2) |sin x| dx =n 2x = 2 int _(0)^(pi//2) sin x dx = -2 [ cos x] _(0)^(pi/2)` ` = - 2(cos.pi/2 - cos 0) = 2 ` ` :. int _(0)^(2a) f(x) dx = int _(0)^(a) [ f(2a - x) + f(x)] dx` |
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| 167. |
Let a, b and c be non - zero real numbers such that ` int _(0)^(3) (3ax ^(2) + 2bx +c) dx = int _(1)^(3) (3ax^(2) + 2bx +c) dx` , thenA. `a+b+c= 3`B. `a+b+c=1`C. `a+b+c=0`D. `a+b+c=2` |
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Answer» Correct Answer - C `int_(0)^(3)(3ax^(2)+2bx+c)dx=int_(1)^(3)(3ax^(2)+2bx+c)dx` `rArr int_(0)^(1)(3ax^(2)+2bx+c)dx+int_(1)^(3)(3ax^(2)+2bx+c)dx` `=int_(1)^(3)(3ax^(2)+2bx+c)dx` `rArr" "int_(0)^(1)(3ax^(2)+2bx+c)dx=0` `rArr" "[(3ax^(3))/(3)+(2bx^(2))/(2)+cx]_(0)^(1)=0` `rArr" "a+b+c=0` |
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| 168. |
The value of `int _(2)^(4) { |x-2|+|x-3|} dx` isA. 1B. 2C. 3D. 5 |
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Answer» Correct Answer - C Let `l=int_(2)^(4){|x-2|+|x-3|}dx` `=int_(2)^(3){(x-2)+(3-x)}dx+int_(3)^(4){(x-2)+(x-3)}dx` `=int_(2)^(3)dx+int_(3)^(4)(2x-5)dx=[x]_(2)^(3)+[x^(2)-5x]_(3)^(4)` `=3-2+[16-20-(9-15)]=1+2=3` |
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| 169. |
If `int_0^af(2a-x)dx=m and int_0^af(x)dx=n,` then `int_0^(2a) f(x) dx` is equal toA. `2lambda-mu`B. `lambda+mu`C. `mu-lambda`D. `lambda-2mu` |
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Answer» Correct Answer - B Using properties of definite integral ` int _(0)^(2a) f(x) dx = int _(0)^(a) f(x) dx + int _(0)^(a) f(2a - x) dx ` ` rArr int _(0)^(2a) f(x) dx = lambda +mu` |
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| 170. |
If `int_0^af(2a-x)dx=m and int_0^af(x)dx=n,` then `int_0^(2a) f(x) dx` is equal toA. `2m+n`B. `m+2n`C. `m-n`D. `m+n` |
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Answer» Correct Answer - D ` :. int _(0)^(2a) f(x) dx = int _(0)^(2a) { f ( 2a-x) + f(x)}dx` ` = int _(0)^(2a) f(2a - x) dx + int _(0)^(a) f(x) dx = m+n` |
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| 171. |
The value of integral ` sum _(k=1)^(n) int _(0)^(1) f(k - 1+x) dx ` isA. ` int _(0)^(1) (x) dx`B. `int _(0)^(2) f(x) dx`C. ` int _(0)^(n) f(x) dx`D. ` n int _(0)^(1) f(x) dx` |
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Answer» Correct Answer - C Let `l=int_(0)^(1)f(k-1+x)dx` `rArr" "l=int_(k-1)^(k)f(t)dt," where "t=k-1+x` `rArr" "l=int_(k-1)^(k)f(x)dx` `therefore sum_(k=1)^(n)int_(k-1)^(k)f(x)dx=int_(0)^(1)f(x)dx+int_(1)^(2)f(x)dx+…+int_(n-1)^(n) f(x)dx=int_(0)^(n)f(x)dx` |
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| 172. |
The value of ` int _(-2)^(4) | x+1| dx` is equal toA. 12B. 14C. 13D. 16 |
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Answer» Correct Answer - C Let ` l = int _(-2)^(4) | x+1| dx = int _(-2)^(-1) - (x+1) dx + int _(-1)^(4) (1+x) dx ` ` = - [ (x^(2))/2 + x]_(-2)^(-1) + [ x + (x^(2))/2]_(-1)^(4)` ` = - [ 1/2 -1-(2-2)]+[4+8 -(-1+1/2)]` ` = 1/2 + (25/2) = 13` |
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| 173. |
`int_(0)^(pi//2) sec x log (sec x+tan x) dx` is equal toA. `(1)/(2)[log(1+sqrt2)]^(2)`B. `[log(1+sqrt2)]^(2)`C. `(1)/(2)[log(sqrt2-1)]^(2)`D. `[log(sqrt2-1)]^(2)` |
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Answer» Correct Answer - A Let ` int _(0)^(pi//4) sec x log (sec x + tan x) dx ` , Put log `(sec x + tan x) = t rArr sec x dx = dt` ` :. l = int _(0)^(log(sqrt(2)+1))t dt = [ (t^(2))/2]^(log(sqrt(2)+1))= ([log (sqrt(2)+1)]^(2))/2 ` |
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| 174. |
If `int_(log" "2)^(x)(du)/((e^(u)-1)^(1//2))=(pi)/(6)`, then `e^(x)` is equal toA. 1B. 2C. 4D. `-1` |
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Answer» Correct Answer - C Let ` l = int _(log2)^(x) e^(u)/(e^(u)(e^(u)-1)^(1//2) )du ` Put `e^(u) - 1 = t^(2) rArr e^(u) du = 2t dt` ` :. l = int _(1)^(sqrt(e^(x)-1))dt/((1+t^(2))) = 2 [ tan^(-1) t]_(1)^(sqrt(e^(x)-1))` ` = - 2 [ tan^(-1) sqrt(e^(x)-1) -pi/4 ] = pi/6` [ Given ] ` rArr tan^(-1) sqrt(e^(x)-1) = pi/12 + pi/4 = pi/3 ` ` rArr sqrt(e^(x)-1) = tan (pi/3) = sqrt(3)` ` rArr e^(x) = 3+1 = 4` |
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| 175. |
The value of the integral ` int _(0)^(2) | x^(2)-1| dx` is |
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Answer» Correct Answer - B ` int _(0)^(2) | x^(2) -1| dx = int _(0)^(1) - (x^(2) -1) dx + int _(1)^(2) (x^(2) -1)dx ` ` = - int _(0)^(1) (x^(2) -1) dx + int _(1)^(2) (x^(2) -1) dx = ((x^(3))/3-x)_(0)^(1) + ((x^(3))/3-x)^(2)` , ` = ((x^(3))/3 - x)_(0)^(1) + ((x^(3))/3 - x)_(1)^(2) = 2 - 1/3 + 1 + 8/3 -2 -1/3 + 1 = 2` |
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| 176. |
` int _(-pi//2)^(pi//2) sin | x | dx ` is equal to |
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Answer» Correct Answer - C Let `l = int _(-pi//2)^(pi//2)sin |x| dx` , ` = 2 [ -cos x ] _(0)^(pi//2) = 2 ` |
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| 177. |
The value of `int_(-pi//2)^(pi//2) log""((2-sintheta)/(2+sintheta))d theta` is |
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Answer» Correct Answer - A Let `f(theta) log ((2-sin theta)/(2+sin theta))` , Now, ` f(-theta) = log ((2+sin theta)/(2-sin theta) ) = - log ((2-sin theta)/(2+sin theta)) = - f(theta)` ` = [ -(x^(3))/3 + x]_(0)^(1) + [ (x^(3))/3-x]_(1)^(2) ` ` = - 1/3 + 1 + 8/3 - 2 - 1/3 + 1 = 2 ` ` :. f(theta) ` is an odd function . ` int _(-pi//2)^(pi//2) log ((2-sin theta )/(2+sin theta)) d theta = 0` |
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| 178. |
If `int_(0)^(pi) x f (sin x) dx = A int _(0)^(pi//2) f(sin x) dx ,` then A is equal to |
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Answer» Correct Answer - B Let ` l = int _(0)^(pi) x f ( sin x) dx " "` …(i) ` = int _(0)^(pi) (pi-x) f[ sin (pi-x) ] dx ` ` rArr l = int _(0)^(pi) (pi-x) f(sin x) dx " "` ...(ii) ` 2l = int _(0)^(pi) pi f (sin x) dx = 2pi int _(0)^(pi//2) f(sin x) dx ` ` rArr l = pi int _(0)^(pi//2) pif(sin x) dx = 2pi int_(0)^(pi//2) f(sin x) dx ` ` rArr l = pi int_(0)^(pi//2) f(sin x) dx = A int_(0) ^(pi//2) f(sin x) dx ` ` :. A = pi` |
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| 179. |
If f(x) = `f(a+x) and int_(0)^(a) f(x) dx = k, "then" int _(0)^(na) f(x)` dx is equal toA. nkB. `(n-1)k`C. `(n+1)k`D. 0 |
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Answer» Correct Answer - A Given, `f(x)=f(a+x) and int_(0)^(a)f(x)dx=k` `int_(a)^(na)f(x)dx=nint_(0)^(a)f(9x)=nk" "[because "f(x) is periodic function"]` |
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| 180. |
The value of the integral ` int _(0)^(pi//2) log | tan x| dx `isA. ` pi log 2`B. 0C. `-pi log 2`D. None of these |
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Answer» Correct Answer - B Let `l=int_(0)^(pi//2)log|tanx|dx" …(i)"` `rArr" "l=int_(0)^(pi//2)log|tan((pi)/(2)-x)|" …(ii)"` On adding Eqs. (i) and (ii), we get `2l=int_(0)^(pi//2)log|tanxcotx|dx` `=int_(0)^(pi//2)log1dx=0` `rArr" "l=0` |
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| 181. |
The value of `I=overset(0)underset(-2)int{x^(3)+3x^(2)+3x+3+(x+1)cos(x+1)cos(x+1)}dx`, is |
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Answer» Correct Answer - C `l =int_(-2)^(0)[x^(3)+3x^(2)+3x+3+(x+1)cos(x+1)]dx` `=int_(-2)^(0)[(x+1)^(3)+2+(x+1)cos(x+1)]dx` Put `x+1=trArrdx=dt` `therefore l =int_(-1)^(1)t^(3)dt+2int_(-1)^(1)dt+int_(-1)^(1)tcos t dt` `=0+2[1-(-1)]+0` `rArr l =4` `[{:(becauset^(3) "and t cos t are odd functions".,),(therefore int_(-1)^(1)t^(3)dt=int_(1)^(1)t cos t dt =0,):}]` |
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| 182. |
The value of `underset(0)overset(1)int tan^(-1) ((2x-1)/(1+x-x^(2)))dx` isA. 1B. 0C. `-1`D. None of these |
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Answer» Correct Answer - B Let `= int_(0)^(1)tan^(-1)((2x-1)/(1+x-x^(2)))dx` `=int_(0)^(1)tan^(-1)x dx+int_(0)^(1)tan^(-1)(x-1)dx` `= int_(0)^(1)tan^(-1)x dx+int_(0)^(1)tan^(-1)(1-x-1)dx` `= int_(0)^(1)tan^(-1)x dx - int_(0)^(1)tan^(-1)x dx =0` |
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| 183. |
The value of the integral ` int _(0)^(pi//2) (sin ^(100)x - cos^(100)x) dx ` isA. `(1)/(100)`B. `(100!)/((100)^(100))`C. `(pi)/(100)`D. 0 |
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Answer» Correct Answer - D ` l = int _(0)^(pi//2) ( sin ^(100) x - cos^(100)x) dx ` ` rArr l = uint _(0)^(pi//2) [ sin^(100) (pi/2 - x ) - cos ^(100) (pi/2 - x) dx ] ` , ` rArr l = int _(0)^(pi//2) ( cos ^(100) x - sin ^(100) x) dx ` ` rArr 2l = 0` ` rArr l = 0 ` |
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| 184. |
`overset(3)underset(0)int |x^(3)+x^(2)+3x|dx` is equal toA. `(171)/(2)`B. `(171)/(4)`C. `(170)/(4)`D. `(170)/(3)` |
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Answer» Correct Answer - B ` int _(0)^(3) | x^(3) +x^(2) + 3x | dx = int _(0)^(3) (x^(3) +x^(2) + 3x ) dx ` ` = [ (x^(4))/4 + (x^(3))/3 + (3x^(2))/2 ] _(0)^(3)` ` = 81/4 + 27/3 + 27/2 = 171/4` |
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| 185. |
`int_(0)^(2)((x^(4)+1))/((x^(2)+1))dx` |
| Answer» `((x^(4)+1))/((x^(2)+1))={x^(2)-1+(2)/((x^(2)+1))}`. | |
| 186. |
`int_(0)^(pi//2)sqrt(1+sinx)dx` |
| Answer» `sqrt(1+sinx)={cos^(2)((x)/(2))+sin^(2)((x)/(2))+2sin((x)/(2))cos((x)/(2))}^(1//2)=(cos.(x)/(2)+sin.(x)/(2))`. | |
| 187. |
`int_(-pi//4)^(pi//4)(dx)/((1+sinx))` |
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Answer» `I=int(1)/((1+sinx))xx((1-sinx))/((1-sinx))dx=int((1-sinx))/(cos^(2)x)dx=int((1)/(cos^(2)x)-(sinx)/(cos^(2)x))dx` `=int(sec^(2)x-secxtanx)dx`. |
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| 188. |
`int_0^(pi/2) sin^3x dx` |
| Answer» `sin3x=3sinx-4sin^(3)ximpliessin^(3)x=(1)/(4)(3sinx-sin3sx)`. | |
| 189. |
`int_(pi//4)^(pi//2)((1-3cosx))/(sin^(2)x)dx` |
| Answer» `I=int((1)/(sin^(2)x)-(3cosx)/(sin^(2)x))dx=int(cosec^(2)x-3cosecxcotx)dx`. | |
| 190. |
`int_(0)^(pi//8) (sec^(2) 2x)/2 dx` is equal toA. `(1)/(4)`B. `(1)/(3)`C. `(1)/(2)`D. None of these |
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Answer» Correct Answer - A ` 1/2 int_(0)^(logx)sec^(2) 2x dx = 1/4 [ tan 2x ] _(0)^(pi/8) = 1/4 (1) = 1/4` |
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| 191. |
`int_0^1(dx)/(sqrt(x+1)-sqrt(x))` 1 dxA. `(2sqrt2)/(3)`B. `(4sqrt2)/(3)`C. `(8sqrt2)/(3)`D. None of these |
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Answer» Correct Answer - B Let ` l = int _(0)^(1) (dx)/(sqrt(1+x)-sqrt(x))=int_(0)^(1) ((sqrt(1+x)+sqrt(x))dx)/((1+sqrt(1+x)-sqrt(x))(sqrt(1+x)+sqrt(x)))` ` l = int _(0)^(1) ((sqrt(1+x)+sqrt(x)))/(1+x -x) dx = int _(0)^(1) sqrt(1+x)dx + int _(0)^(1) sqrt(x)dx = (4sqrt(2))/3` |
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| 192. |
`int_(pi//3)^(pi//2)cosecxdx=?`A. `(1)/(2)log2`B. `(1)/(2)log3`C. `-log2`D. none of these |
| Answer» `I=int_(pi//3)^(pi//2)cosecxdx=[logtan"(x)/(2)]_(pi//3)^(pi//2)=(log1-log.(1)/(sqrt(3)))=(1)/(2)log3`. | |
| 193. |
Evaluate the definiteintegrals`int0pi/2cos^2x""""dx`A. `(pi)/(2)`B. `pi`C. `(pi)/(4)`D. `1` |
| Answer» `I=(1)/(2)int_(0)^(pi//2)2cos^(2)xdx=(1)/(2)int_(0)^(pi//2)(1+cos2x)=(1)/(2)[x+(sin2x)/(2)]_(0)^(pi//2)=(pi)/(4)`. | |
| 194. |
Evaluate: `int_0^(pi//4)tan^2x dx`A. `(1-(pi)/(4))`B. `(1+(pi)/(4))`C. `(1-(pi)/(2))`D. `(1+(pi)/(2))` |
| Answer» `I=int_(0)^(pi//2)(sec^(2)x-1)dx=[tanx-x]_(0)^(pi//2)=(1-(pi)/(4))`. | |
| 195. |
`int_0^pixsin^3x dx` |
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Answer» Let `I=int_(0)^(pi) x sin^(3)xdx`…….`(i)` Then, `I=int_(0)^(pi)(pi-x)sin^(3)(pi-x)dx` or `I=int_(0)^(pi)(pi-x)sin^(3)xdx`……..`(ii)` Adding `(i)` and `(ii)` , we get `2I=int_(0)^(pi)pisin^(3)xdx=piint_(0)^(pi)sin^(2)x*sinxdx` `=pi int_(0)^(pi)(1-cos^(2)x)sinxdx` `=-pi int_(1)^(-1)(1-t^(2))dt`, where `cosx=t` [`x=0impliest=1` and `x=piimpliest=-1`] `=piint_(-1)^(1)(1-t^(2))dt=pi[t-(t^(3))/(3)]_(-1)^(1)=(4pi)/(3)`. Hence, `I=(2pi)/(3)impliesint_(0)^(pi)xsin^(3)xdx=(4pi)/(3)`. |
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| 196. |
`int_(0)^(2pi)|sinx|dx=?` |
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Answer» We know that `sinx` is positive, when `o le x le pi` and `sinx` is negative when `pi le x le 2pi`. `:.|sinx|={{:(sinx,"when" 0 le x le pi),(-sinx,"when" pi le x le 2pi):}` `:.int_(0)^(2pi)|sinx|dx=int_(0)^(pi)|sinx|dx+int_(pi)^(2pi)|sinx|dx` `=int_(0)^(pi)sinxdx+int_(pi)^(2pi)(-sinx)dx` `=[-cosx]_(0)^(pi)[cosx]_(pi)^(2pi)=(2+2)=4`. |
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| 197. |
`int_0^(pi/4)tan^2x dx` |
| Answer» `tan^(2)x=(sec^(2)x-1)`. | |
| 198. |
`int_(pi//4)^(pi//2)cotxdx=?`A. `log2`B. `2log2`C. `(1)/(2)log2`D. none of these |
| Answer» `I=int_(pi//4)^(pi//2)cotxdx=[logsinx]_(pi//4)^(pi//2)=[log1-log"(1)/(sqrt(2))]=(1)/(2)log2`. | |
| 199. |
`int_(-a)^asqrt((a-x)/(a+x))dx` |
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Answer» Put `x=acostheta ` so that `dx=-asin theta d theta`. Also, `(x=-aimpliestheta=pi)` and `(x=aimpliestheta=0)`. `:.int_(-a)^(a)sqrt((a-x)/(a+x))dx=int_(pi)^(0)sqrt((1-costheta)/(1+costheta))*(-asintheta)d theta` `=aint_(0)^(pi)sqrt((2sin^(2)(theta//2))/(2cos^(2)(theta//2)))*2sin(theta//2)cos(theta//2)d theta` `=a int_(0)^(pi)2sin^(2)(theta//2)d theta=a int_(0)^(pi)(1-costheta)d theta` `=a int_(0)^(pi)d theta-a int_(0)^(pi)cos theta d theta` `=a*[theta]_(0)^(pi)-a[sintheta]_(0)^(pi)=api`. |
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| 200. |
Evaluate :`int_(pi/4)^(pi//2)cotxdx` |
| Answer» `cot^(2)x=(cosec^(2)x-1)`. | |