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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.
| 251. |
`int_(pi//6)^(pi//3)(1)/((1+sqrt(tanx)))dx=(pi)/(12)` |
| Answer» Apply, `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`. | |
| 252. |
` int _(1)^(2)e^(x) (1/x - 1/(x^(2)))dx ` is qual toA. ` e-(e^(2))/2`B. ` (e^(2))/2 - e`C. ` (e^(2))/2 + e`D. ` (e^(2))/2 - 2 ` |
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Answer» Correct Answer - B `int_(1)^(2)e^(x)((1)/(x)-(1)/(x^(2)))dx=[(1)/(x)e^(x)]_(1)^(2)+int_(1)^(2)(e^(x))/(x^(2))dx-int_(1)^(2)(e^(x))/(x^(2))dx` `= (e^(2))/(2)-e` |
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| 253. |
`int_(0)^(pi//2)(1)/((1+tanx))dx=?`A. `pi`B. `pi//2`C. `pi//3`D. `pi//4` |
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Answer» Correct Answer - D Given, `l = int_(0)^(pi//2)(dx)/(1+tan x)` `l = int_(0)^(pi//2) (cos x)/(sin x + cos x)` ….(i) `l = int_(0)^(pi//2)(cos((pi)/(2)-x))/(sin((pi)/(2)-x)+cos((pi)/(2)-x))dx` `= int_(0)^(pi//2)(sin x)/(cos x+ sin x)dx` ….(ii) On adding Eqs. (i) and (ii), we get `2 l = int_(0)^(pi//2)((sin x + cos x)/(sin x + cos x))dx` `= int_(0)^(pi//2)dx = (pi)/(2)rarr l = (pi)/(4)` |
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| 254. |
The value of the integral ` l = int_(0)^(1) x(1-x)^(n) dx` isA. `(1)/(n+1)`B. `(1)/(n+2)`C. `(1)/(n+1)-(1)/(n+2)`D. `(1)/(n+1)+(1)/(n+2)` |
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Answer» Correct Answer - C ` l = int _(0)^(1) x (1-x)^(n) dx` , Put ` 1- x = z rArr -dx = dz` ` :. L = int_(1)^(0) (1-z) z^(n) (-dz) = int_(0)^(1) (z^(n) - z^(n+1)) dz` , ` = [ (z^(n+1))/(n+1) - (z^(n+2))/(n+2)]_(0)^(1) = 1/(n+1) - 1/(n+2)` |
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| 255. |
The value of `int_(0)^(a) sqrt((a-x)/x )` dx isA. `(a)/(2)`B. `(a)/(4)`C. `(pia)/(2)`D. `(pia)/(4)` |
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Answer» Correct Answer - C Let ` l = int_(0)^(a) sqrt(a-x)/xdx` Put x = a `sin^(2) theta rArr dx = 2a sin theta cos theta d theta ` ` :. i = int _(0)^(pi//2) sqrt((cos^(2)theta)/(sin^(2)theta)). 2a sin theta cos theta d theta ` ` = 2a int_(0)^(pi//2) cos^(2) theta d theta ` ` = 2a int _(0)^(pi//2) [ (1+cos 2theta)/2] d theta = 2a [ theta/2 + (sin 2theta)/4]_(0)^(pi//2)` ` = 2a xx1/2 xx pi/2 = (pia)/2` |
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| 256. |
The value of `int_(2)^(3)(x+1)/(x^(2)(x-1))dx` isA. `log.(16)/(9)+(1)/(6)`B. `log.(16)/(9)-(1)/(6)`C. `2log2-(1)/(6)`D. `log.(4)/(3)-(1)/(6)` |
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Answer» Correct Answer - B Let ` l = int _(2)^(3) (x+1)/(x^(2) (x-1)) dx = int _(2)^(3) ((-2)/x - 1/(x^(2)) + 2/(x-1))dx` , ` = [ -2 log x +1/x +2 log (x-1) ]_(2)^(3)` , ` = 2 log. 4/3 - 1/6 = log. 16/9 -1/6` |
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| 257. |
The value of the integral ` int_(-a)^(a) (xe^(x^(2)))/(1+x^(2))dx` isA. `e^(a^(2))`B. 0C. `e^(-a^(2))`D. a |
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Answer» Correct Answer - B Since, ` (xe^(x^(2)))/(1+x^(2)) ` is an odd function ` int _(-a)^(a) (xe^(x^(2)))/(1+x^(2)) dx = 0 ` |
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| 258. |
The value of integral `int_(-1)^(1) (|x+2|)/(x+2)dx` isA. 1B. 2C. 0D. `-1` |
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Answer» Correct Answer - B Let ` l = int _(-1)^(1) (|x+2|)/(x+2) dx ` For `-1 le xle 1, |x+2| = 2 +x` ` :. l = int _(-1)^(1)(x+2)/(x+2) dx = int _(-1)^(1) 1 dx ` ` [ x] _(-1)^(1)= 1 - (-1)=2` |
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| 259. |
The value of the integral `int_(0)^(1)(x^(3))/(1+x^(8))dx` isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(8)`D. `(pi)/(16)` |
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Answer» Put `x^(4)=t` and `4x^(3)dx=dt`. `[x=0impliest=0]` and `[x=1impliest=1]` `:. I=(1)/(4)int_(0)^(1)(dt)/((1+t^(2)))=(1)/(4)[tan^(-1)t]_(0)^(1)=(1)/(4)[tan^(-1)-tan^(-1)0]=((1)/(4)xx(pi)/(4))=(pi)/(16)` |
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| 260. |
The value of `int_(-pi)^(pi) sin^(3) x cos^(2) ` x dx is equal toA. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - D Let `f(x)=sin^(3)x cos^(2)x` `f(-x)=-sin^(3)xcos^(2)x=-f(x)` `rArr (f(x))` is an odd function `therefore " " int_(-pi)^(pi)sin^(3)x cos^(2)x dx = 0` |
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| 261. |
The value of `int_(0)^(pi) x sin^(3) x dx ` isA. `(4pi)/3`B. `(2pi)/3`C. 0D. None of these |
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Answer» Correct Answer - B Let `l = int_(0)^(pi)x sin^(3)x dx` ….(i) Also, `l = int_(0)^(pi)(pi-x)sin^(3) x dx` …(ii) On adding Eqs. (i) and (ii), we get `2l=pi int_(0)^(pi)sin^(3)x dx` `= (pi)/(4)int_(0)^(pi)(3 sin x - sin 3x)dx` `= (pi)/(4)[-3 cos x + (cos 3x)/(3)]_(0)^(pi)=(4pi)/(3)` Hence, `l = (2pi)/(3)` |
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| 262. |
`int_(pi//6)^(pi//3)(1)/((1+sqrt(tanx)))dx=(pi)/(12)`A. `pi/12`B. `pi/2`C. `pi/6`D. `pi/4` |
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Answer» Correct Answer - A Let `l=int_(pi//6)^(pi//3)(dx)/(1+sqrt(tan x))= int_(pi//6)^(pi//3)((sqrt(cos x))/(sqrt(sin x)+sqrt(cos x)))dx` `rArr l = int_(pi//6)^(pi//3)(sqrt(sin x))/(sqrt(cos x)+sqrt(sin x))dx` ….(ii) `[because int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx]` On adding Eqs. (i) and (ii), we get `2l=int_(pi//6)^(pi//3)1dx=[x]_(pi//6)^(pi//3)=(pi)/(6)` `rArr " " l = (pi)/(12)` |
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| 263. |
The value of integral `int_(0)^(1)sqrt((1-x)/(1+x))`dx isA. `(pi)/(2)+1`B. `(pi)/(2)-1`C. `-1`D. 1 |
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Answer» Correct Answer - B Let ` l = int_(0)^(1) sqrt((1-x)/(1+x))` Put `x = cos 2 theta rArr dx = -2 sin 2 theta d theta` ` :. l = int_(pi//4)^(0) sqrt((1- cos 2 theta)/(1+cos 2theta)) . 2 sin 2 theta d theta ` ` = 2 int _(0)^(pi//4) (sin theta)/(cos theta) . 2 sin theta cos theta d theta` ` = 2 int_(0)^(pi//4) (1+ cos 2 theta) d theta = 2 [ theta - (sin 2 theta)/2 ] _(0)^(pi//4) = pi/2 -1 ` |
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| 264. |
Evaluate the following integral: `int_0^(pi//4)(sqrt(t a n x)+sqrt(cotx))dx` |
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Answer» We have `I=int_(0)^(pi//2){(sqrt(sinx))/(sqrt(cosx))+(sqrt(cosx))/(sqrt(sinx))}dx=int_(0)^(pi//2)((sinx+cosx))/(sqrt(sinx+cosx))dx` `=sqrt(2)*int_(0)^(pi//2)((sinx+cosx))/sqrt(2sinxcosx)dx=sqrt(2)*int_(0)^(pi//2)((sinx+cosx))/(sqrt(1-(sinx-cosx)^(2)))dx` Put `(sinx-cosx)=t` and `(cosx+sinx)dx=dt`. Also, `[x=0impliest=-1]` and `[x=(pi)/(2)impliest=1]`. `:.I=sqrt(2)*int_(-1)^(1)(dt)/(sqrt(1-t^(2)))=sqrt(2)[sin^(-1)t]_(-1)^(1)` `=sqrt(2){sin^(-1)(1)-sin^(-1)(-1)}=sqrt(2){2sin^(-1)(1)}` `=(sqrt(2)xx2xx(pi)/(2))=sqrt(2)pi`. |
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| 265. |
`int_-1^1 (17x^5-x^4+29x^3-31x+1)/(x^2+1)dx` is equal to (A) `4/5` (B) `5/4` (C) `4/3` (D) `3/4`A. `4/5`B. `5/4`C. `4/3`D. `3/4` |
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Answer» Correct Answer - C Let `l=int_(-1)^(1)(17x^(5)+29x^(3)-31x)/(x^(2)+1)dx-int_(-1)^(1)(x^(4)-1)/(x^(2)+1)dx` `=0-2int_(0)^(1)((x^(2)-1)(x^(2)+1))/((x^(2)+1))dx` `=-2[((x^(3))/(3)-x)]_(0)^(1)=(4)/(3)` |
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| 266. |
`Ifint_0^1cot^(-1)(1-x+x^2)dx=lambdaint_0^1tan^(-1)x dx ,t h e nlambdai se q u a lto`1 (b)2 (c) 3(d) 4A. `pi- log2`B. `pi+log2`C. `pi/2 +log2`D. `pi/2 - log2` |
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Answer» Correct Answer - D Let `l=int_(0)^(1)cot^(-1)(1+x+x^(2))dx` `=int_(0)^(1)tan^(-1)((1)/(x^(2)-x+1))dx` `=int_(0)^(1)tan^(-1)((x-(x-1))/(1+x(x-1)))dx` `=int_(0)^(1)tan^(-1)xdx-int_(0)^(1)tan^(-1)(x-1)dx` `=int_(0)^(1)tan^(-1)xdx-int_(0)^(1)tan^(-1)(1-x-1)dx` `=2int_(0)^(1)tan^(-1)xdx` `=2[x tan^(-1)x-int(x)/(1+x^(2))dx]_(0)^(1)` `=2[x tan^(-1)x-(1)/(2)log(1+x^(2))]_(0)^(1)` `=2[(1tan^(-1)1-(1)/(2)log2)-(0-(1)/(2)log1)]` `=(pi)/(2)-log2` |
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| 267. |
`int_(-pi)^(pi)tanxdx=?`A. `2`B. `(1)/(2)`C. `-2`D. `0` |
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Answer» `f(x)=tanximpliesf(-x)=tan(-x)=-tanx=-f(x)`. `:.f(x)` is odd and hence `I=0`. |
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| 268. |
`int_(0)^(pi//3)tanxdx` |
| Answer» `int tan x dx=log sec x`. | |
| 269. |
The value of the integral `int_3^6 sqrtx/(sqrt(9-x)+sqrtx)dx` isA. `3/2`B. 2C. 1D. `1/2` |
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Answer» Correct Answer - A Let `l=int_(3)^(6)(sqrtx)/(sqrt(9-x)+sqrtx)dx" …(i)"` `=int_(3)^(6)(sqrt(9-x))/(sqrt(9-9+x)+sqrt(9-x))dx` `rArr" "l=int_(3)^(6)(sqrt(9-x))/(sqrtx+sqrt(9-x))dx" …(ii)"` On adding Eqs. (i) and (ii), we get `2l=int_(3)^(6)1dx=[x]_(3)^(6)=6-3` `rArr" "l=(3)/(2)` |
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| 270. |
if `int_0^k (dx)/(2+8x^2)=pi/16` then find the value of `k`A. 1B. `(1)/(2)`C. `(1)/(4)`D. None of these |
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Answer» Correct Answer - B ` int _(0)^(k) 1/(2+8x^(2)) dx = 1/2 int_(0)^(k) (dx)/(1+(2x)^(2))` `1/4 int _(0)^(2k) (dt)/(1+t^(2)) = 1/4 [ tan^(-1) t ] _(0)^(2k) = 1/4 tan^(-1) 2k ` Comparing it with the given value , we get ` tan ^(-1) 2k = pi/4 rArr 2k = 1 rArr k = 1/2 ` |
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| 271. |
`int_0^picos^3x dx=` |
| Answer» `cos3x=4cos^(3)x-3cosximpliescos^(3)x=(1)/(4)(cos3x+3cosx)`. | |