Explore topic-wise InterviewSolutions in .

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.

201.

Resolve `(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6)` into partial fractions:

Answer» Here the given function is an improper rational function. On dividing we get
`(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6) = x-1+(-x+4)/(x^(2)-5x+6)`…………..(i)
We have, `(-x+4)/(x^(2)-5x+6)= (-x+4)/((x-2)(x-3))`
So, let `(-x+4)/((x-2)(x-3)) = A/(x-2)=B/(x-3)`, then
`-x+4=A(x-3)+B(x-2)`...........(ii)
Putting `x-3=0` or `x=3` in (ii), we get `(-x+4)/((x-2)(x-3))=A/(x-2)+B/(x-3)`, then
`-x+4=A(x-3)+B(x-2)`...............(2)
puttting x-2=0 or x=2 in eq. (ii), we get
`2=A(2-3)rArr A=-2`
`therefore (-x+4)/((x-2)(x-3))=-2/(x-2)+1/(x-3)`
Hence `(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6)=x-1-2/(x-2)+1/(x-3)`
Discussion
202.

Evaluate: `intxsinx dx`

Answer» Correct Answer - `-xcosx+sinx+C`
Discussion
203.

Evaluate: `int(sec^(2)x)/(1+tanx)dx`

Answer» Correct Answer - `"ln"|1+tanx|+C`
Discussion
204.

Evaluate: `int(sin("ln"x))/(x)`dx

Answer» Correct Answer - `cos("ln"x)+C`
Discussion
205.

What is `int (dx)/(sqrt(4 + x^(2)))` equal to ?A. `l n |sqrt(4 + x^(2)) + x| + c`B. `l n |sqrt(4 + x^(2)) -x| + c`C. `sin^(-1) ((x)/(2)) + c`D. None of these

Answer» Correct Answer - A
`int (dx)/(sqrt(4 + x^(2))) = int (dx)/(sqrt(2^(2) + x^(2))) = ln |sqrt(4 + x^(2)) + x| + C`
Discussion
206.

Evaluate : `intsin4x.e^(tan^(2)x)dx`

Answer» Correct Answer - `-2e^(tan^(2)x)cos^(4)x+C`
`intsin4x.e^(tan^(2)x)dx=4intsinx cosx 2x e^(tan^(2)x)dx`
`=4inttanx.sec^(2)x.cos^(4)x.cos2xe^(tan^(2)x)dx`
`2int(1)/((1+t)^(2))(1-t)/(1+t)e^(t)dt`
(Putting, `tan^(2)x=t rArr 2 tan x. sec^(2)xdx=dt`)
`=-2int((t+1)-2)/((1+t)^(3))e^(t)dt`
`=-2inte^(t)((1)/((1+t)^(2))+(-2)/((1+t)^(3)))dt`
`=(-2e^(t))/((1+t)^(2))+C`
`=-2e^(tan^(2)x)cos^(4)+C`
Discussion
207.

`int(dx)/(sqrt(2e^(x)-1))=`A. `2sec^(-1)sqrt(2e^(x))+c`B. `-2"tan"^(-1)(1)/(sqrt(2e^(x)-1))+c`C. `2sec^(-1)(sqrt(2)e^(x))+c`D. `2tan^(-1)sqrt(2e^(x)-1)+c`

Answer» Correct Answer - A::B::D
`I=int(dx)/(sqrt(2e^(x)-1))`
`"Let " 2e^(x)-1=t^(2)`
` :. 2e^(x)dx=2t dt`
` :. I=int(t dt)/((t^(2)+1)/(2)*t)`
`=2tan^(-1)t+C`
`=2tan^(-1)sqrt(2e^(x)-1)+C`
`=-2"tan"^(-1)(1)/(sqrt(2e^(x)-1))+C`
`"Also, " tan^(-1)sqrt(2e^(x)-1)=sec^(-1)sqrt(2e^(x))`
Discussion
208.

Evaluate: `intxsqrt(1+x-x^(2))`dx

Answer» Let `x=lambda. d/(dx)(1+x-x^(2))+mu`
`rArr x=lambda(1-2x)+mu`
Comparing the coefficients of like powers of x, we get
`1=-2lambda` and `lambda+mu=0 rArr lambda=-1/2` and `mu=1/2 therefore x=-1/2(1-2x)+1/2`
So, `intxsqrt(1+x-x^(2))dx`
`=int{-1/2(1-2x)+1/2}sqrt(1+x-x^(2)dx) = -1/2int(1-2x)sqrt(1+x-x^(2)dx+1/2intsqrt(1+x-x^(2))dx`
`=-1/2intsqrt(1+x-x^(2))d(1+x-x^(2))+1/2intsqrt((sqrt(5)/2)^(2)-(x-1/2)^(2))dx`
`=-1/3(1+x-x^(2))^(3//2)+1/2[1/2(x-1/2)sqrt((sqrt(5)/(2))^(2)-(x-1/2)^(2))+1/2(sqrt(5)/(2))^(2)sin^(-1)(x-1//2)/(sqrt(5)//2)]+C`
`=-1/3(1+x-x^(2))^(3//2)+1/2[(x-1/2)sqrt(1+x-x^(2))+5/8sin^(-1)(2x-1)/(sqrt(5))]+C`
Discussion
209.

Evaluate:`int("log"(tanx/2)/(sinx)dx`

Answer» `(d)/(dx)[log("tan"(x)/(2))]=((1)/(2)"sec"^(2)(x)/(2))/("tan"(x)/(2))=(1)/(sinx)`
Now `int(log("tan"(x)/(2)))/(sinx)dx=intlog("tan"(x)/(2))(d)/(dx)[log("tan"(x)/(2))]dx`
`=([log("tan"(x)/(2))]^(2))/(2) +C`
Discussion
210.

Evaluate:`inttan^4x dx`

Answer» `int tan^(4)xdx=int tan^(2)x tan^(2)xdx=int tan^(2)x(sec^(2)x-1)dx`
`=int tan^(2)x sec^(2)xdx-int tan^(2)xdx=(sec^(3)x)/(3)-int(sec^(2)x-1)dx`
`=(sec^(3)x)/(3)-tanx+x+C`
Discussion
211.

Evaluate: `int((x-1)^2)/(x^4+x^2+1) dx`

Answer» Correct Answer - `(1)/(sqrt3)tan^(-1)(x-1//x)/(sqrt3)-(2)/(sqrt3)tan^(-1)((2x^(2)+1)/(sqrt3))+C`
`int((x-1)^(2))/(x^(4)+x^(2)+1)dx`
`=int(x^(2)+1)/(x^(4)+x^(2)+1)dx-int(2x)/(x^(4)+x^(2)+1)dx`
`=int(1+(1)/(x^(2)))/((x-(1)/(x))^(2)+3)dx-int(2x)/((x^(2)+(1)/(2))^(2)+(3)/(4))dx`
`=(1)/(sqrt3)tan^(-1).(x-1//x)/(sqrt3)-(2)/(sqrt3)tan^(-1)((2x^(2)+1)/(sqrt3))+C`
Discussion
212.

Evaluate : `int(x^(2)-4)/(x^(4)+9x^(2)+16)dx`

Answer» Correct Answer - 5
`I=int(x^(2)-4)/(x^(4)+9x^(2)+16)dx`
Divide both numberator and denominator by `x^(2)`
`I=int((1-(4)/(x^(2))))/(x^(2)+9+(16)/(x^(2)))dx=int((1-(4)/(x^(2))))/((x+(4)/(x))^(2)+1)dx`
`"Put "u=x+(4)/(x),du=(1-(4)/(x^(2)))dx`
`I=int(du)/(u^(2)+1)=tan^(-1)(u)+B`
`=tan^(-1)(x+(4)/(x))+C`
Discussion
213.

Evaluate:`int(2x+3)/(sqrt(x^2+4x+1))dx`

Answer» `int(2x+3)/sqrt(x^(2)+4x+1)dxint((2x+4)-1)/sqrt(x^(2)+4x+1)dx=int(2x+4)/sqrt(x^(2)+4x+1)x-int1/sqrt(x^(2)+4x+1)`dx
`=int(dt)/sqrt(t) -int1/((x+2)^(2)-(sqrt(3))^(2))dx`, where `t=(x^(2)+4x+1)` for Ist integral
`=2sqrt(t)-ln|(x-2)+sqrt(x^(2)+4x+1)|+C=2sqrt(x^(2)+4x+1)-ln|x+2+sqrt(x^(2)+4x+1)|+C`
Discussion
214.

Evaluate :`int(2x-3)/(x^2+ 3x-18) dx`

Answer» Let `2x-3=lambdad/(dx)(x^(2)+3x-18)+mu`
Then `2x-3=lambdad/(dx)(x^(2)+3x-18)+mu`
Comparing the coefficients of like power of x, we get.
`2lambda=2`, and `3lambda+mu=-3 rArr lambda=1` and `mu=-6`
So, `int(2x-3)/(x^(2)+3x-18)dx = int(2x+3-6)/(x^(2)+3x+9/4-9/4-18dx)= int(2x+3)/(x^(2)+3x-18)dx - 6int1/(x^(2)+3x-18)dx`
`=ln|x^(2)+3x-18|-6int1/(x^(2)+3x+9/4-9/4-18)dx = ln|x^(2)+3x-18|-6int1/((x+3/2)^(2)-(9/2)^(2))dx`
`ln|x^(2)+3x-18|-6.1/(2(9/2)ln|(x+3/2-9/2)/(x+3/2+9/2)|+C) = ln|x^(2)+3x-18|-2/3ln|(x-3)/(x+6)|+C`
Discussion
215.

Evaluate:`int1/(x^4+1)dx`

Answer» Correct Answer - `(1)/(2sqrt(2))"tan"^(-1)((x^(2)-1)/(sqrt(2)x))-(1)/(4sqrt(2))"log"|(x^(2)-sqrt(2)x+1)/(x^(2)+xsqrt(2)+1)|+C`
`I=int (1)/(x^(4)+1)dx=int((1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int((2)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int((1+(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))-(1-(1)/(x^(2)))/(x^(2)+(1)/(x^(2))))dx`
`=(1)/(2)int(1+(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx-(1)/(2)int(1-(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int(1+(1)/(x^(2)))/((x-(1)/(x))^(2)+2)dx-(1)/(2)int(1-(1)/(x^(2)))/((x+(1)/(x))^(2)-2)dx`
` "Putting "x-(1)/(x)=u" in first integral and "x+(1)/(x)=v " in second integral." `
we get
`I=(1)/(2)int(du)/(u^(2)+(sqrt(2))^(2))-(1)/(2)int(dv)/(v^(2)-(sqrt(2))^(2))`
`=(1)/(2sqrt(2))"tan"^(-1)((u)/(sqrt(2)))-(1)/(2)xx(1)/(2sqrt(2))"log"|(v-sqrt(2))/(v+sqrt(2))|+C`
`=(1)/(2sqrt(2))"tan"^(-1)((x-1//x)/(sqrt(2)))-(1)/(4sqrt(2))"log"|(x+1//x-sqrt(2))/(x+1//x+sqrt(2))|+C`
`=(1)/(2sqrt(2))"tan"^(-1)((x^(2)-1)/(sqrt(2)x))-(1)/(4sqrt(2))"log"|(x^(2)-sqrt(2)x+1)/(x^(2)+xsqrt(2)+1)|+C`
Discussion
216.

Evaluate:`int(dx)/(sqrt(x)+sqrt(x-2))`

Answer» Correct Answer - `(1)/(3){x^(3//2)-(x-2)^(3//2)}+C `
`int(dx)/(sqrt(x)+sqrt(x-2))=int(sqrt(x)-sqrt(x-2)dx)/(x-(x-2)) " (rationalizing) " `
`=(1)/(2)int(sqrt(x)-sqrt(x-2))dx `
`=(1)/(3){x^(3//2)-(x-2)^(3//2)}+C `
Discussion
217.

Evaluate:`int(x^3)/(x+1)dx`

Answer» Correct Answer - `(x^(3))/(3)+x-(x^(2))/(2)-" In "|x+1|+C`
`x^(3)/(x+1)=(x^(3)+1-1)/(x+1)=(-1)/((x+1))+(x^(2)+1-x)`
Thus, the given integral is
`int(x^(2)+1-x-(1)/(1+x))dx=(x^(3))/(3)+x-(x^(2))/(2)-" In "|x+1|+C`
Discussion
218.

Evaluate:`int(1+2x+3x^2+4x^3+)dx ,(

Answer» Correct Answer - `(1-x)^(-1)+C`
`int(1+2x+3x^(2)+4x^(3)+...)dx=int(1-x)^(-2)dx `
`=(1-x)^(-1)+C`
Discussion
219.

Evaluate `int(dx)/(1-x-x^(2))`

Answer» Correct Answer - ` -(1)/(sqrt(5))log_(e)|(2x+1-sqrt(5))/(2x+1+sqrt(5))|+C`
`I= -int(dx)/(x^(2)+x-1)= -int(dx)/((x+(1)/(2))^(2)-(5)/(4))`
`=-(1)/(2*(sqrt(5))/(2))"log"_(e)|((x+(1)/(2))-(sqrt(5))/(2))/((x+(1)/(2))+(sqrt(5))/(2))|+C`
`= -(1)/(sqrt(5))log_(e)|(2x+1-sqrt(5))/(2x+1+sqrt(5))|+C`
Discussion
220.

`int(a+bsinx)/((b+asin x)^(2))dx`

Answer» `-(cosx)/(b+asinx)+C`
Discussion
221.

Evaluate `int(x^(2)+x+1)/(x^(2)-1)dx`

Answer» Correct Answer - `(x^(2))/(2)+log|x^(2)-1|+(1)/(2)"log"|(x-1)/(x+1)|+C`
`int(x^(2)+x+1)/(x^(2)-1)dx=int(x+(2x+1)/(x^(2)-1))dx`
`=int x dx+int(2x)/(x^(2)-1)dx+int(1)/(x^(2)-1)dx`
`=(x^(2))/(2)+log|x^(2)-1|+(1)/(2)"log"|(x-1)/(x+1)|+C`
Discussion
222.

Evaluate `int (x^(2)-sqrt(3x)+1)/(x^(4)-x^(2)+1)dx`

Answer» Correct Answer - `2 tan^(-1)(2x+sqrt(3))+C`
`int (x^(2)-sqrt(3x)+1)/(x^(4)-x^(2)+1)dx=int (x^(2)-sqrt(3x)+1)/((x^(2)-1)^(2)-3x^(2))dx`
`=int (x^(2)-sqrt(3x)+1)/((x^(2)-sqrt(3)x+1)(x^(2)+sqrt(3)x+1))dx`
`=int(1)/(x^(2)+sqrt(3)x+1)dx`
`=int (1)/((x+(sqrt(3))/(2))^(2)+(1)/(4))dx`
`=(1)/(1//2)"tan"^(-1)(x+(sqrt(3))/(2))/((1)/(2))+C`
`=2 tan^(-1)(2x+sqrt(3))+C`
Discussion
223.

Evaluate `int(dx)/((x-a)sqrt((x-a)(x-b))`

Answer» `-2/(alpha-beta)sqrt(alpha-beta)/(x-alpha)+C`
Discussion
224.

Evaluate : `int((cos2x)^(1//2))/(sinx)`dx

Answer» `sqrt(log)[(sqrt(cot^(2)x-1)+sqrt(2cot^(2)x))/sqrt(cot^(2)x+1)]-log[(cotx+sqrt(cot^(2)x-1))]+c`
Discussion
225.

Evaluate:`1/(sqrt(1-e^(2x)))dx`

Answer» Correct Answer - `-log|e^(-x)+sqrt(e^(-2x)-1)|+C`
`I=int(1)/(sqrt(1-e^(2x)))dx=int(1)/(sqrt(1-(1)/(e^(-2x))))dx=int(e^(-x))/(sqrt(e^(-2x)-1))dx`
`=int(e^(-x))/(sqrt((e^(-x))^(2)-1^(2)))dx`
`"Let " e^(-x)=t " or " -e^(-x)dx=dt`
` :. I= -int(dt)/(sqrt(t^(2)-1^(2)))= -log|t+sqrt(t^(2)-1)|+C`
`= -log|e^(-x)+sqrt(e^(-2x)-1)|+C`
Discussion
226.

Evaluate:`int((cos2x)^(1/2))/(sinx)dx`

Answer» Correct Answer - `-"log"|"cot x" + sqrt("cot"^(2)x-1|)+(1)/(sqrt(2))"log"|(sqrt(2)+sqrt(1- "tan"^(2)x))/(sqrt(2)-sqrt(1- "tan"^(2)x))|+c`
Let `I=int(sqrt(cos2x))/(sinx)dx=int sqrt((cos^(2)x-sin^(2)x)/(sin^(2)x))dx`
`=intsqrt(cot^(2)x-1)dx`
Put `cot x = sec theta rArr - "cosec"^(2)x dx = sec theta tan theta d theta`
`therefore I=int sqrt(sec^(2)theta-1)*(sec theta*tan theta)/(-(1+sec^(2)theta))d theta`
`=-int(sec theta*tan^(2)theta)/(1+ sec^(2)theta)d theta`
`=-int (sin^(2)theta)/(cos theta + cos^(3)theta)d theta`
`=-int(1-cos^(2)theta)/(cos theta(1+cos^(2)theta))d theta`
`=-int((1+cos^(2)theta)-2cos^(2)theta)/(cos theta(1+cos^(2)theta))d theta`
`=-intsec theta d theta +2 int(cos theta)/(1+cos^(2)theta)d theta`
`=-log|sec theta+tan theta|+2 int(cos theta)/(2-sin^(2)theta)d theta`
`=-log|sec theta+tan theta|+ int(dt)/(2-t^(2)), "where sin" theta=t`
`=-log|sec theta + tan theta|+2*(1)/(2 sqrt(2))log|(sqrt(2)+sin theta)/(sqrt(2)-sin theta)|+c`
`=-log|cot x +sqrt(cot^(2)x-1)|+(1)/(sqrt(2))log|(sqrt(2)+sqrt(1-tan^(2)x))/(sqrt(2)-sqrt(1-tan^(2)x))|+c`
Discussion
227.

Evaluate:`int(x^2)/(x^6-a^6)dx`

Answer» Correct Answer - `(1)/(3)(1)/(2a^(3))"log"_(e)|(x^(3)-a^(3))/(x^(3)+a^(3))|+C`
`I=int(x^(2))/(x^(6)-a^(6))dx`
` "let "x^(3)=t`
` :. 3x^(2)dx=dt`
` :. I=(1)/(3)int(dt)/(t^(2)-a^(6))=(1)/(3)(1)/(2a^(3))"log"_(e)|(t-a^(3))/(t+a^(3))|+C`
`=(1)/(3)(1)/(2a^(3))"log"_(e)|(x^(3)-a^(3))/(x^(3)+a^(3))|+C`
Discussion
228.

Evaluate: `int(sinx)/(cos2x) dx`

Answer» Correct Answer - ` -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C`
`I=int(sinx)/(cos2x)dx=int(sinx)/(2cos^(2)x-1)dx`
` "Let " cosx=t`
` :. -sinx dx=dt`
` :. I=int(-dt)/(2t^(2)-1)= -(1)/(2)int(dt)/(t^(2)-(1)/(2))`
`=-(1)/(2)*(1)/(2(1)/(sqrt(2)))"log"_(e)|(t-(1)/(sqrt(2)))/(t+(1)/(sqrt(2)))|+C`
`= -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C`
Discussion
229.

Evaluate:`int(e^x)/(e^(2x)+6e^x+5)dx`

Answer» Correct Answer - `(1)/(4)"log"|(e^(x)+1)/(e^(x)+5)|+C`
`I=int(e^(x))/(e^(2x)+6e^(x)+5)dx=int(e^(x))/((e^(x))^(2)+6e^(x)+5)dx`
` "Let "e^(x)=t " or " e^(x)dx=dt`
` :. I=int(dt)/(t^(2)+6t+5)`
`=int(1)/((t+3)^(2)-2^(2))dt`
`=(1)/(2xx2)"log"|(t+3-2)/(t+3+2)|+C=(1)/(4)"log"|(e^(x)+1)/(e^(x)+5)|+C`
Discussion
230.

Evaluate `int xsin^(2)x dx`

Answer» Correct Answer - `(1)/(4)x^(2)-(x)/(4)sin2x-(1)/(8)cos 2x+C`
`int xsin^(2)x dx`
`=int x{(1-cos2x)/(2)}dx=(1)/(2)intxdx-(1)/(2)int underset(I)(x)underset(II)(cos)2x dx`
`=(1)/(2)((x^(2))/(2))-(1)/(2)[x{intcos 2x dx}-int{(d)/(dx)(x)int cos2x}dx]`
`=(1)/(4)x^(2)-(1)/(2){(x)/(2)sin2x-int1xx(sin2x)/(2)dx}`
`=(x^(2))/(4)-(1)/(2){(x)/(2)sin 2x-(1)/(2)int sin2xdx}`
`=(x^(2))/(4)-(1)/(2){(x)/(2)sin 2x-(1)/(2)(-(1)/(2)cos2x)}+C`
`=(1)/(4)x^(2)-(x)/(4)sin2x-(1)/(8)cos 2x+C`
Discussion
231.

Evaluate: `intcosx cos2x`dx

Answer» `intcosxcos2x dx= 1/2int2cosx cos2x dx = 1/2int(cos3x+cosx)dx=1/2(sin(3x)/(3)+sinx)+C`
Discussion
232.

`int sqrt(x)(1+x^(1//3))^(4)dx " is equal to "`A. `2{x^(2//3)+(4)/(11)x^(11//6)+(6)/(13)x^(13//6)+(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`B. `6{x^(2//3)-(4)/(11)x^(11//6)+(6)/(13)x^(13//6)-(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`C. `6{(1)/(9)x^(3//2)+(4)/(11)x^(11//6)+(6)/(13)x^(13//6)+(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`D. non of these

Answer» Correct Answer - C
`"Let " x=t^(6) " or " dx=6t^(5)dt`
`:. I=int t^(3)(1+t^(2))^(4)6t^(5)dt `
`=6int t^(8)(1+4t^(2)+6t^(4)+4t^(6)+t^(8))dt`
`=6int(t^(8)+4t^(10)+6t^(12)+4t^(14)+t^(16))dt`
`=6{t^(9)/(9)+(4t^(11))/(11)+(6t^(13))/(13)+(4t^(15))/(15)+t^(17)/(17)}+C`
`=6{(1)/(9)x^(3//2)+(4)/(11)x^(11//6)+(6)/(13)x^(13//6)+(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+C`
Discussion
233.

Consider two differentiable functions `f(x),g(x)` satisfying `6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c and 2 int(g(x)dx)/(f(x))=x^(2)+c, " where " f(x) gt 0 AA x in R.` `int (g(x)-f(x))dx` is equal toA. `(x^(4))/(4)-(x^(2))/(2)+x+c`B. `(x^(4))/(4)+(x^(2))/(2)-(x^(3))/(3)+x+c`C. `(x^(4))/(4)-(x^(3))/(3)+(x^(2))/(2)-x+c`D. `(x^(4))/(4)+(x^(3))/(3)+c`

Answer» Correct Answer - B
We have, `6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c`
Differentiating both sides w.r.t.x, we get
`6f(x)g(x)=6x^(5)+12x^(3)+6x`
` :. f(x)g(x)=x^(5)+2x^(3)+x " ...(1)" `
`2 int (g(x)dx)/(f(x))=x^(2)+c`
Differentiating both sides w.r.t.x, we get
`(g(x))/(f(x))=x " ...(2)" `
Multiplying (1) and (2), we get
`(g(x))^(2)=x^(6)+2x^(4)+x^(2)=(x^(3)+x)^(2)`
` :. g(x)=x^(3)+x`
` :. f(x)=x^(2)+1 " " ("as"f(x) gt 0).`
Now ` int (g(x)-f(x))dx=int (x^(3)+x-x^(2)-1)dx`
`=(x^(4))/(4)+(x^(2))/(2)-(x^(3))/(3)-x+c`
`underset (x to 0)(lim)(log(f(x)))/(g(x))=underset(x to 0)(lim)(log(1+x^(2)))/(x+x^(3))`
`=underset(x to 0)(lim)(log(1+x^(2)))/(x^(2))*(x^(2))/(x+x^(3))`
`=underset(x to 0)(lim)(x)/(1+x^(2))=0`
Discussion
234.

Consider two differentiable functions `f(x),g(x)` satisfying `6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c and 2 int(g(x)dx)/(f(x))=x^(2)+c, " where " f(x) gt 0 AA x in R.` `lim_(x to 0) (log(f(x)))/(g(x))=`A. eB. 2C. 1D. 0

Answer» Correct Answer - D
We have, `6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c`
Differentiating both sides w.r.t.x, we get
`6f(x)g(x)=6x^(5)+12x^(3)+6x`
` :. f(x)g(x)=x^(5)+2x^(3)+x " ...(1)" `
`2 int (g(x)dx)/(f(x))=x^(2)+c`
Differentiating both sides w.r.t.x, we get
`(g(x))/(f(x))=x " ...(2)" `
Multiplying (1) and (2), we get
`(g(x))^(2)=x^(6)+2x^(4)+x^(2)=(x^(3)+x)^(2)`
` :. g(x)=x^(3)+x`
` :. f(x)=x^(2)+1 " " ("as"f(x) gt 0).`
Now ` int (g(x)-f(x))dx=int (x^(3)+x-x^(2)-1)dx`
`=(x^(4))/(4)+(x^(2))/(2)-(x^(3))/(3)-x+c`
`underset (x to 0)(lim)(log(f(x)))/(g(x))=underset(x to 0)(lim)(log(1+x^(2)))/(x+x^(3))`
`=underset(x to 0)(lim)(log(1+x^(2)))/(x^(2))*(x^(2))/(x+x^(3))`
`=underset(x to 0)(lim)(x)/(1+x^(2))=0`
Discussion
235.

Match the following lists:

Answer» Correct Answer - `a to p,q; b to r,s; c to p; d to p,q`
a. `" Let " I=int(2^(x))/(sqrt(1-4^(x)))dx=(1)/(log2)int(1)/(sqrt(1^(2)-t^(2)))dt`
`"Putting "2^(x)=t, 2^(x)log2dx=dt, " we get "`
`I=(1)/(log2)sin^(-1)((t)/(1))+C=(1)/(log 2)sin^(-1)(2^(x))+C`
` :. k=(1)/(log2)`
b. ` int(dx)/((sqrt(x))^(2)+(sqrt(x))^(7))=int(dx)/((sqrt(x))^(7)(1+(1)/((sqrt(x))^(5))))`
`"Put " (1)/((sqrt(x))^(5))=y, (dy)/(dx)=-(5)/(2(sqrt(x))^(7))`
` :. I=int(-2dy)/(5(1+y))=-(2)/(5)In|1+y|+C=(2)/(5)In((1)/(1+(1)/((sqrt(x))^(5))))`
` :. a=(2)/(5), k=(5)/(2)`
c. Add and subtract ` 2x^(2)` in the numerator. Then `k=1` and ` m=1 `
d. `I=int(dx)/(5+4cosx)`
`=int(dx)/(5("sin"^(2)(x)/(2)+"cos"^(2)(x)/(2))+4("cos"^(2)(x)/(2)-"sin"^(2)(x)/(2)))`
`=int(dx)/(9"cos"^(2)(x)/(2)+"sin"^(2)(x)/(2))=int("sec"^(2)(x)/(2))/(9+"tan"^(2)(x)/(2))dx`
`"Let " t="tan"(x)/(2) " or " 2dt="sec"^(2)(x)/(2)dx`
` :. I=int(2dt)/(9+t^(2))=(2)/(3)"tan"^(-1)((t)/(3))+C`
`=(2)/(3)"tan"^(-1)(("tan"((x)/(2)))/(3))+C`
` :. k=(2)/(3), m=(1)/(3)`
Discussion
236.

`"Let " k(x)=int((x^(2)+1)dx)/(root(3)(x^(3)+3x+6)) " and " k(-1)=(1)/(root(3)(2)). " Then the value of " k(-2) " is "-.`

Answer» Correct Answer - 2
`k(x)=int((x^(2)+1)dx)/((x^(3)+3x+6)^(1//3))`
`"Put " x^(3)+3x+6=t^(3)`
`"or " 3(x^(2)+1)dx=3t^(2)dt`
`k(x)=int(t^(2)dt)/(t)=(t^(2))/(2)+C=(1)/(2)(x^(3)+3x+6)^(2//3)+C`
`k(-1)=(1)/(2)(2)^(2//3)+C " or " C=0`
` :. k(x)=(1)/(2)(x^(3)+3x+6)^(2//3),f(-2)=(1)/(2)(-8)^(2//3)`
`=(1)/(2)[(-2)^(3)]^(2//3)=2`
Discussion
237.

If `int((2x+3)dx)/(x(x+1)(x+2)(x+3)+1)=C-(1)/(f(x))` where f(x) is of the form of `ax^(2)+bx+c`, then the value of f(1) isA. 4B. 5C. 6D. none

Answer» Correct Answer - B
`I=int(2x+3)/((x^(2)+3x)(x^(2)+3x+2)+1)dx`
Put `x^(2)+3x=t" "rArr(2x+3)dx=dt`
`rArr`
`I=int(dt)/(t(t+2)+1),int (dt)/((t+1)^(2))=C-(1)/(t+1)=C-(1)/(x^(2)+3x+1)`
`rArr" "f(1)=5`
Discussion
238.

Match the following lists:

Answer» Correct Answer - ` a to p, q, r; b to p, q, r; c to p, q, r, s; d to p, q, r, s`
a. `int(x^(2)-x+1)/(x^(3)-4x^(2)+4x)dx=int[(A)/(x)+(B)/(x-2)+(C)/((x-2)^(2))]dx`
b. `int(x^(2)-1)/(x(x-2)^(3))dx=int[(A)/(x)+(B)/(x-2)+(C)/((x-2)^(2))+(D)/((x-2)^(3))]dx`
c. `int(x^(3)+1)/(x(x-2)^(2))dx=int[((x^(3)+1)/(x(x-2)^(2))-1)+1]dx`
`=int[((x^(3)+1-x(x-2)^(2))/(x(x-2)^(2)))+1]dx`
`=int[((A)/(x)+(B)/(x-2)+(C)/((x-2)^(2)))+1]dx`
d. `int(x^(5)+1)/(x(x-2)^(3))dx=int[x+k+(g(x))/(x(x-2)^(3))]dx, `
where k is constant ` ne 0 ` and `g(x)` is a polynomial of degree less than 4.
Discussion
239.

Let `g(x)=int(1+2cosx)/((cosx+2)^2)dxa n dg(0)=0.`then the value of `8g(pi/2)`is __________

Answer» Correct Answer - 0.5
`g(x)=int(cosx(cosx+2)+sin^(2)x)/((cosx+2)^(2))dx`
`=int underset(II)(underbrace(cosx))*(1)/(underset(I)(underbrace((cosx+2))))dx+int(sin^(2)x)/((cosx+2)^(2))dx`
`=(1)/(cosx+2)*sinx-int(sin^(2)x)/((cosx+2)^(2))dx+int(sin^(2))/((cosx+2)^(2))dx`
`:. g(x)=(sinx)/(cosx+2)+C`
`g(0)=0 " or " C=0`
`:. g(x)=(sinx)/(cosx+2) " or " g((pi)/(2))=(1)/(2)`
Discussion
240.

Consider `int(x^(3)+3x^(2)+2x+1)/(sqrt(x^(2)+x+1))dx` `=(ax^(2)+bx+c)sqrt(x^(2)+x+1)+lambda int(dx)/(sqrt(x^(2)+x+1))` Now, match the following lists and then choose the correct code. Codes: `{:(,a,b,c,d),((1),q,p,s,r),((2),s,p,q,r),((3),r,q,p,s),((4),q,s,p,r):}`

Answer» Correct Answer - 4
` int(x^(3)+3x^(2)+2x+1)/(sqrt(x^(2)+x+1))dx`
`=(ax^(2)+bx+c)sqrt(x^(2)+x+1)+lambda int(dx)/(sqrt(x^(2)+x+1))`
Differentiating both side, we get
`(x^(3)+3x^(2)+2x+1)/(sqrt(x^(2)+x+1))=((ax^(2)+bx+c)(2x+1))/(2sqrt(x^(2)+x+1))+(2ax+b)sqrt(x^(2)+x+1)+lambda(1)/(sqrt(x^(2)+x+1))`
`=(6ax^(3)+(5a+4b)x^(2)+(4a+3b+2c)x+(c+2b+2lambda))/(2sqrt(x^(2)+x+1))`
Now, compare coefficients on both sides and solve equations.
Discussion
241.

If A is a square matrix and `e^a` is defined as `e^A=1+A^2/(2!)+A^3/(3!)...........oo=1/2[f(x) ,g(x) and g(x) ,f(x)],` where `A=[(x,x),(x,x)].` and I being the identity matrix then `int (g(x))/(f(x))dx=`A. `(1)/(2sqrt(e^(x)-1))-"cosec"^(-1)(e^(x))+c`B. `(2)/(sqrt(e^(x)-e^(-x)))-"sec"^(-1)(e^(x))+c`C. `(1)/(2sqrt(e^(2x)-1))+"sec"^(-1)(e^(x))+c`D. none of these

Answer» Correct Answer - C
`A=[(x,x),(x,x)]`
`impliesA^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))],A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+ … +`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ... ,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ..., 1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...)]`
`=[((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)+(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)-(1)/(2)),((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)-(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)+(1)/(2))]`
`=(1)/(2)[(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
` :. f(x)=e^(2x)+1 and g(x)=e^(2x)-1`
`int(e^(2x)+1)/(sqrt(e^(2x)-1))dx=int(e^(2x))/(sqrt(e^(2x)-1))dx+(1)/(sqrt(e^(2x)-1))dx`
`=int(e^(2x))/(sqrt(e^(2x)-1))dx+int (e^(x))/(e^(x) sqrt(e^(2x)-1))dx`
`=(1)/(2sqrt(e^(2x)-1))+sec^(-1)(e^(x))+C`
Discussion
242.

If A is a square matrix and `e^a` is defined as `e^A=1+A^2/(2!)+A^3/(3!)...........oo=1/2[f(x) ,g(x) and g(x) ,f(x)],` where `A=[(x,x),(x,x)].` and I being the identity matrix then `int (g(x))/(f(x))dx=`A. `(e^(x))/(2)(sinx-cosx)`B. `(e^(2x))/(5)(2sinx-cosx)`C. `(e^(x))/(5)(sin2x-cos2x)`D. none of these

Answer» Correct Answer - B
`A=[(x,x),(x,x)]`
`impliesA^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))],A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+ … +`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ... ,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ..., 1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...)]`
`=[((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)+(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)-(1)/(2)),((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)-(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)+(1)/(2))]`
`=(1)/(2)[(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
` :. f(x)=e^(2x)+1 and g(x)=e^(2x)-1`
`int(g(x)+1)sinx dx=inte^(2x) sinx dx=(e^(2x))/(5)(2sinx-cosx)`
Discussion
243.

If `int(dx)/(x^(2)+ax+1)=f(g(x))+c, ` thenA. `f(x)` is inverse trigonometric function for `|a| lt 2`B. `f(x)` is logarithmic function for `|a| gt 2`C. `g(x)` is quadratic function for `|a| lt 2`D. `g(x)` is rational function for `|a| gt 2`

Answer» Correct Answer - A::B::D
`int(dx)/(x^(2)+ax+1)=int(dx)/((x+(a)/(2))^(2)+(1-(a^(2))/(4)))`
Discussion
244.

If A is a square matrix and `e^a` is defined as `e^A=1+A^2/(2!)+A^3/(3!)...........oo=1/2[f(x) ,g(x) and g(x) ,f(x)],` where `A=[(x,x),(x,x)].` and I being the identity matrix then `int (g(x))/(f(x))dx=`A. `log(e^(x)+e^(-x))+c`B. `log|e^(x)-e^(-x)|+c`C. `log|e^(2x)-1|+c`D. none of these

Answer» Correct Answer - A
`A=[(x,x),(x,x)]`
`impliesA^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))],A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+ … +`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ... ,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ..., 1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...)]`
`=[((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)+(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)-(1)/(2)),((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)-(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)+(1)/(2))]`
`=(1)/(2)[(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
` :. f(x)=e^(2x)+1 and g(x)=e^(2x)-1`
`int(e^(2x)-1)/(e^(2x)+1)dx=(e^(x)-e^(-x))/(e^(x)+e^(-x))dx=log|e^(x)-e^(-x)|+C`
Discussion
245.

If `int(1-x^(7))/(x(1+x^(7)))dx=alog_(e)|x|+blog_(e)|x^(7)+1|+c,` thenA. `a=1`B. `a= -1`C. `b=(2)/(7)`D. `b= -(2)/(7)`

Answer» Correct Answer - A::D
`I=int(1-x^(7))/(x(1+x^(7)))dx=a log_(e)|x|+b log_(e)|1+x^(7)|+c`
Differentiating both sides w.r.t. x, we get
`(1-x^(7))/(x(1+x^(7)))=(a)/(x)+b.(7x^(6))/(1+x^(7))`
`implies1-x^(7)=a(1+x^(7))+7bx^(7)`
`impliesa=1, a+7b= -1`
`implies b= -2//7`
Discussion
246.

If `int (3sin x+2 cosx)/(3cosx+2sinx)dx=ax+blog_(e)|2 sinx+3 cosx|+c` thenA. `a= -(12)/(13)`B. `b=(6)/(13)`C. `a=(12)/(13)`D. `b= -(15)/(39)`

Answer» Correct Answer - C::D
Differentiating both sides w.r.t. x, we get
`(3sin x+2 cosx)/(3cosx+2sinx)=a+(b(2cosx-3sinx))/((2sinx+3cosx))`
`=(sinx.(2a-3b)+cosx.(3a+2b))/((3cosx+2 sinx))`
Comparing like terms on both sides, we get
`3=2a-3b, 2=3a+2b`
`implies a=(12)/(13), b= -(15)/(39)`
Discussion
247.

Evaluate: (i) `int(e^x)/(sqrt(4-e^(2x))) dx`(ii) `int(x^2)/(sqrt(1-x^6)) dx`

Answer» `I=int(e^(x))/(sqrt(4-e^(2x)))dx=int(e^(x))/(sqrt(2^(2)-(e^(x))^(2)))dx`
Let `e^(x)=t " or " e^(x)dx=dt`
`:. I=int(dt)/(sqrt(4-t^(2)))=int(dt)/(sqrt(2^(2)-t^(2)))`
`=sin^(-1)((t)/(2))+C=sin^(-1)((e^(x))/(2))+C`
Discussion
248.

Evaluate `int(3x^(2)"tan"(1)/(x)-x"sec"^(2)(1)/(x))dx`.

Answer» `int(3x^(2)"tan"(1)/(x)-x"sec"^(2)(1)/(x))dx`
`=int3x^(2)"tan"(1)/(x)dx-int x"sec"^(2)(1)/(x)dx`
`="tan"(1)/(x)x^(3)-int("sec"^(2)(1)/(x))(-(1)/(x^(2)))x^(3)dx- int x "sec"^(2)(1)/(x)dx`
`=x^(3)"tan"(1)/(x)+c`
Discussion
249.

Evaluate:`int(sec^2x)/(sqrt(tan^2x+4))dx`

Answer» Since derivative of `tanx ` is `sec^(2)x.`
Let `tanx=t` or `sec^(2)x dx=dt`
`:. int(sec^(2)x)/(sqrt(tan^(2)x+4))dx=int(dt)/(sqrt(t^(2)+2^(2)))`
`=log|t+sqrt(t^(2)+4)|+C`
`=log|tanx+sqrt(tan^(2)x+4)|+C`
Discussion
250.

What `int sec x^(@) dx` is equal to ?A. `log (sec x^(@) + tan x^(@)) + c`B. `(pi log tan ((pi)/(4) + (pi)/(2)))/(180^(@)) + c`C. `(180^(@) log tan ((pi)/(4) + (x)/(2)))/(pi) + c`D. `(180^(@) log tan ((pi)/(4) + (x)/(360^(@))))/(pi) + c`

Answer» Correct Answer - A
`int sec x^(@).dx = int (sec x^(@).(sec x^(@) + tan x^(@)))/(sec x^(@) + tan x^(@)) .dx`
Let `u = sec x^(@) + tan x^(@)`
`rArr (du)/(dx) = sec x^(@) + tan x^(@) + sec^(2) x^(@)`
`rArr du = (sec x^(@) xx tan x^(@) + sec^(2) x^(@)) .dx`
`:. in (sec x^(@).(sec x^(@) + tan x^(@)))/(sec x^(@) + tan x^(@)) dx`
`= int (sec^(2)x^(@) + sec x^(@) tan x^(@))/(sec x^(@) + tan x^(@)).dx`
`= int (du)/(u) = log (u) + C = log (sec x^(@) + tan x^(@)) + C`
Discussion