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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.
| 401. |
`int(dx)/(sqrt((1-x)(x-2)))` is equal toA. `sin^(-1)(2x-3)+C`B. `sin^(-1)(2x+5)+C`C. `SIN^(-1)(3-2X)+C`D. `sin^(-1)(5-2x)+C` |
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Answer» Correct Answer - A Let `I=int(dx)/(sqrt((1-x)(x-2)))` `=int(dx)/(sqrt(-x^(2)+3x-2))int(dx)/(sqrt((1)/(4)-(x-(3)/(2))^(2)))` `=sin^(-1)(((x-(3)/(2)))/((1)/(2)))+C=sin^(-1)(2x-3)+C` |
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| 402. |
`int(x^2+1)/(x^4+x^2+1)`dx |
| Answer» Correct Answer - `tan^(-1) (x-(1)/(x))+c` | |
| 403. |
`int(x^(2))/((xsin x+cosx)^(2))dx` is equal toA. `(sinx+cosx)/(x sin x+cosx)+C`B. `(x sin x-cosx)/(x sin x+cosx)+C`C. `(sinx-x cos x)/(x sin x+cosx)+C`D. None of these |
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Answer» Correct Answer - C `"Since, "(d)/(dx)(x sinx +cosx)=x cos x` `therefore" "l=int(x^(2)dx)/((x sin x+cosx)^(2))` `=int(x)/(cosx).(x cosx)/((x sinx+cosx)^(2))dx` `=(x)/(cosx).((-1)/(x sinx+cosx))-int(cosx-x(-sinx))/(cos^(2)x).(-1)/((x sin x+cosx))dx` `(-x)/(cosx(x sin x+cosx))+int sec^(2)dx` `=(-x)/(cosx(x sinx +cosx))+tanx +C` `=(-x +sinx(x sinx+cosx))/(cosx(x sin x+cosx))+C` `=((sinx-x cosx)/(cosx+x sinx))+C` |
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| 404. |
`d/(dx)[atan^(-1)x+blog((x-1)/(x+1))]=1/(x^4-1)=>a-2b=`A. 1B. `-1`C. 0D. 2 |
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Answer» Correct Answer - B Given `(d)/(dx)[ a tan^(-1)x+b log((x-1)/(x+1))]=(1)/(x^(4)+1)` On intergrating both sides, we get `a tan^(-1)x+b log((x-1)/(x+1))=(1)/(2)int[(1)/(x^(2)-1)-(1)/(x^(2)+1)]dx` `rArr" " a tan^(-1)x+b log((x-1)/(x+1))=(1)/(4)log((x-1)/(x+1))-(1)/(2)tan^(-1)x` `rArr" "A=-(1)/(2), b=(1)/(4)` `therefore" "a-2b=-(1)/(2)-2((1)/(4))=-1` |
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| 405. |
`int(1)/(sqrt(8+2x-x^(2)))dx` is equal toA. `(1)/(3)sin^(-1)((x-1)/(3))+c`B. `sin^(-1)((x+1)/(3))+c`C. `(1)/(3)sin^(-1)((x+1)/(3))+c`D. `sin^(-1)((x-1)/(3))+c` |
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Answer» Correct Answer - D `int(dx)/(sqrt(8+2x-x^(2)))=int(dx)/(sqrt(8-(x^(2)-2x+1)+1))` `=int(dx)/(sqrt(3^(2)-(x-1)^(2)))=sin^(-1)((x-1)/(3))+c` |
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| 406. |
`int(dx)/(x^2+2x+2)`equals(A) `xtan^(-1)(x+1)+C` (B) `tan^(-1)(x+1)+C`(C) `(x+1)tan^(-1)x+C` (D) `tan^(-1)x+C`A. `tan^(-1) (x+1) +C`B. `(x+1) tan^(-1) x+C`C. `tan^(-1) x+C`D. |
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Answer» Correct Answer - b `int (1)/(x^(2) +2x+2)dx = int (1)/((x^(2)+2x+1)+1)dx` `= int (1)/((x+1)^(2)+1^(2))dx` `=tan^(-1) (x+1)+c` |
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| 407. |
Evaluate:`int(sqrt(tanx)+sqrt(cotx))dx` |
| Answer» Correct Answer - `sqrt(2) sin ^(-1) (sin x +cos x ) +c` | |
| 408. |
`int(sqrt(tanx)+sqrt(cotx))dx` is equal toA. `sqrt2 tan^(-1)((tanx)/(sqrt(2tanx)))+C`B. `sqrt2 tan^(-1)((tanx-1)/(sqrt(2tanx)))+C`C. `(tanx)/(sqrt2).tan^(-1)((cotx+1)/(sqrt(2tanx)))+C`D. `(tanx)/(sqrt2).tan^(-1)((cotx+1)/(sqrt(2tanx)))+C` |
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Answer» Correct Answer - B Let `l=int((sin x+cosx))/(sqrt(sinx.cosx))dx` `=int(sqrt2(sinx+cosx))/(sqrt(2sinx.cosx))dx=sqrt2 int(sinx+cosx)/(sqrt(sin2x))dx` `"Put "sinx-cosx=t` `rArr" "(cosx+sinx)dx=dt` `"Also, "sin 2x=(1-t^(2))` `therefore" "l=sqrt2 int(dt)/(sqrt(1-t^(2)))=sqrt2 sin^(-1)t+C` `=sqrt2 sin^(-1)(sinx - cosx)+C` `=sqrt2 sin^(-1)(sinx-cosx)+C` `=sqrt2 tan^(-1)((tanx-1)/(sqrt(2tanx)))+C` |
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| 409. |
`inte^(tanx)(sec^(2)x+sec^(3)x sinx)dx` is equal toA. `sec x e^(tanx)+C`B. `tanx e^(tanx)+C`C. `e^(tanx)+tanx+C`D. `(1+tanx)e^(tanx)+C` |
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Answer» Correct Answer - B Let `l=int e^(tanx)(sec^(2)x+sec^(3)x sinx)dx` `=inte^(tanx)(1+tanx)sec^(2)xdx` Put `tan x= t rArr sec^(2)x dx=dt` `l=int e^(t)(1+t)dt=e^(t)+intte^(t)dt` `=e^(t)+te^(t)-e^(t)+C=te^(t)+C= tanx e^(tanx)+C` |
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| 410. |
Evaluate: `inte^xsecx (1+tanx) dx`A. `e^(x) sec x+C`B. `e^(x) sin x+C`C. `e^(x) tan x+C`D. |
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Answer» Correct Answer - b `" Let " I= int e^(x) sec x(1+tanx) dx` `rArr I= int e^(x) sec x dx + int e^(x) sec x tan x dx " " .....(1)` `" Now " int e^(x) sec x dx` ` = sec x int e^(x) dx- int ((d)/(dx) sec x int e^(x) dx)dx` ` =e^(x) sec x- int sec x tan x e^(x) dx" " .....(2)` Put the value from equation (2) to equation (1) ` I=e^(x) sec x- int e^(x) sec x tan xdx` `+ int sec x tan xe^(x) dx+C` `rArr I=e^(x) sec x+C` |
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| 411. |
`int secx. log (sec x+ tan x ) dx ` |
| Answer» Correct Answer - `(1)/(2) [log (sec x+tan x)]^(2)+c` | |
| 412. |
`int(secx- tan x)/(sec x+ tan x) dx` |
| Answer» Correct Answer - `2tan x-2 sec x-x+c` | |
| 413. |
`int x^2(ax + b)^-2 dx` is equal toA. `(2)/(a^(2))(x-(b)/(a)log(ax+b))+C`B. `(2)/(a^(2))(x-(b)/(a)log(ax+b))-(x^(2))/(a(ax+b))+C`C. `(2)/(a^(2))(x+(b)/(a)log(ax+b))+(x^(2))/(a(ax+b))+C`D. `(2)/(a^(2))(x+(b)/(a)log(ax+b))-(x^(2))/(a(ax+b))+C` |
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Answer» Correct Answer - B Let `l=int(x^(2))/((ax+b)^(2))dx` Put `ax+b=t rArr dx=(1)/(a)dt` and `" "x=((t-b)/(a))` `therefore" "l=(1)/(a^(3))int((t-b)^(2))/(t^(2))dt=(1)/(a^(3))int(1+(b^(2))/(t^(2))-(2b)/(t))dt` `=(1)/(a^(3))(t-(b^(2))/(t)-2b log t)+C` `=(1)/(a^(3))[ax+b-(b^(2))/(ax+b)-2blog(ax+b)]+C` `=(1)/(a^(3))[(a^(2)x^(2)+b^(2)+2axb-b^(2))/((ax+b))-(a^(2)x^(2))/(ax+b)-2b log(ax+b)]+C` `=(1)/(a^(3))[2ax-(a^(2)x^(2))/(ax+b)-2blog(ax+b)]+C` `=(2)/(a^(2))[x-(b)/(a)log(ax+b)]-(x^(2))/(a(ax+b))+C` |
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| 414. |
`int(x^(4)+x+1)/(x^(2)-x+1)dx` is equal toA. `(x^(3))/(3)-(x^(2))/(2)+x+C`B. `(x^(3))/(3)+(x^(2))/(2)+x+C`C. `(x^(3))/(3)-(x^(2))/(2)-x+C`D. `(x^(3))/(3)+(x^(2))/(2)-x+C` |
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Answer» Correct Answer - B `int(x^(4)+x^(2)+1)/(x^(2)-x+1)dx=int(x^(2)+x+1)dx` `=(x^(3))/(3)+(x^(2))/(2)+x+C` |
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| 415. |
`int"cosec"^(4)x dx ` is equal toA. `cotx+(cot^(3))/(3)+C`B. `tanx+(tan^(3)x)/(3)+C`C. `-cotx-(cot^(3)x)/(3)+C`D. `-tanx-(tan^(3)x)/(3)+C` |
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Answer» Correct Answer - C `int"cosec"^(4)xdx=int"cosec"^(2)x."cosec"^(2)xdx` `=int"cosec^(2)x(1+cot^(2)x)dx` `=int"cosec"^(2)xdx+intcot^(2)x."cosec"^(2)xdx` `=-cotx-(cot^(3)x)/(3)+C` |
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| 416. |
`int(xtan^-1x)/(1+x^2)^(3/2)dx`A. `(x-tan^(-1)x)/(1-x^(2))+C`B. `(x+tan^(-1)x)/(sqrt(1-x^(2)))+C`C. `(x-tan^(-1)x)/(sqrt(1+x^(2)))+C`D. `(x+sqrt(1-x^(2)))/(sqrt(1+x^(2)))+C` |
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Answer» Correct Answer - C Let `l=int(x tan^(-1)x)/((1+x^(2))^(3//2))dx` `"Put "x= tan theta` `rArr" "dx=sec^(2) theta d theta` `therefore" "l=int(tan theta tan^(-1)(tan theta))/((1+tan^(2) theta)^(3//2))sec^(2) theta d theta` `rArr" "l=int(theta tan theta sec^(2) theta)/(sec^(3) theta)d theta= int theta sin theta d theta` By using integration by parts, we get `rArr" "l=theta(-cos theta)+int cos theta d theta` `= - theta cos theta + sin theta+C` `rArr" "l=-tan^(-1)x.(1)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+C` `rArr" "l=((x-tan^(-1)x))/(sqrt(1+x^(2)))+C` |
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| 417. |
Evaluate `int(secx)/((secx+tanx)) dx`.A. `tan x - sec x +C`B. `log(1+sec x)+C`C. `sec x+tan x+C`D. `log sin x+log cos x+C` |
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Answer» Correct Answer - A Let `I=int(secx)/(secx+tanx)dx` `=int(secx(secx-tanx))/(sec^(2)x-tan^(2)x)dx` `=int(sec^(2)x-secxtanx)dx` `=tanx-secx+C` |
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| 418. |
`int{1+2tanx(tanx+secx)}^(1//2)dx` is equal toA. `log(secx+tanx)+C`B. `log(secx+tanx)^(1//2)+C`C. `log secx(secx+tanx)+C`D. None of the above |
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Answer» Correct Answer - C Let `l=int{1+2 tan x(tanx+secx)}^(1//2)dx` `=int{1+2 tan^(2)x+2 tanx secx}^(1//2)dx` `=int{sec^(2)x+tan^(2)x+2tanx secx}^(1//2)dx` `=int(secx+tanx)dx` `=log(secx+tanx)+log secx+C` `=log secx(secx+tanx)+C` |
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| 419. |
`inte^(-x)(1-tanx)secx dx` is equal toA. `e^(x)cos x+C`B. `e^(x)sec x+C`C. `e^(x)sinx+C`D. `e^(x)tanx+C` |
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Answer» Correct Answer - B Let`" "l=int e^(x) sec x (1+tanx)dx` `rArr" "l=int e^(x)sec x dx+int e^(x)sec x tanx dx" …(i)"` Now, `int e^(x)sec x dx` On applying integration by parts, we get `=secx inte^(x) dx-int[(d)/(dx)secx int e^(x)dx]dx` `=e^(x)sec x-int sec x tanx e^(x)dx" ...(ii)"` `l=e^(x) sec x - int e^(x)secx tanx dx+ int sec x tan x e^(x)dx+C` `rArr" "l=e^(x)sec x +C` |
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| 420. |
Evaluate `int_0^(pi/4)(sqrt(tanx)+sqrt(cotx))dx`A. `-(pi)/(sqrt(2))`B. `(pi)/(2)`C. `-(pi)/(2)`D. |
| Answer» Correct Answer - A | |
| 421. |
`int " x tan"^(-1) " x dx "` |
| Answer» Correct Answer - `(1)/(2) (x^(2) +1) tan^(-1) x-(x)/(2)+c` | |
| 422. |
`intsinsqrt(x) dx` |
| Answer» Correct Answer - `2[sin sqrt(x) -sqrt(x) cos sqrt(x)]+c` | |
| 423. |
Evaluate: (i) `intsecxlog(secx+tanx) dx`(ii) `intcos e c xlog(cos e c x-cotx) dx` |
| Answer» Correct Answer - `(1)/(2) [log (sec x+ tan x) ]^(2)+c` | |