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101.

If `u=inte^(ax)sin " bx dx" and v=int^(e^(ax))cos " bx dx"`,then `tan^(-1)((u)/(v))+tan^(-1)((b)/(a))` equalsA. bxB. 2 bxC. `b^(2)x^(2)`D. `sqrt(bx)`

Answer» Correct Answer - a
We have, `u=(e^(ax))/(a^(2)+b^(2))( asin bx - b cos bx)`
and `v=(e^(ax))/(a^(2)+b^(2))(a cos bx+b sinbx)`
`:. (u)/(v)=(a sin bx-bcosbx)/(a cosbx+b sinx)`
`rArr (u)/(v)=(sin(bx-theta))/(cos(bx-theta))`, where `a=r costheta,b=r=sinthetaandtantheta=(b)/(a)`
`rArr (u)/(v)=tan(bx-theta)`
`rArr "tan"^(-1)(u)/(v)=bx-"tan"^(-1)(b)/(a)rArr"tan"^(-1)(u)/(v)+"tan"^(-1)(b)/(a)=bx`
Discussion
102.

Let the equation of a curve passing through the point (0,1) be given b `y=intx^2e^(x^3)dx`. If the equation of the curve is written in the form `x=f(y)`, then f(y) isA. `sqrt(log_(e)(3y-2))`B. `root(3)(log_(e)(3y-2))`C. `root(3)(log_(e)(2-3y))`D. none of these

Answer» Correct Answer - b
We have , `y=intx^(2)e^(x^(3))dx=(1)/(3)inte^(x^(3))d(x^(3))=(1)/(3)e^(x^(3))+C`.
It passes throught ( 0 , 1) Therefore ,
`1=(1)/(3)+CrArrC=(2)/(3)`
`thereforey=(1)/(3)e^(x^(3))+(2)/(3)`
`rArr3y=e^(x^(3))+2`
` rArr e^(x^(3))=3y=2rArrx^(3)-log_(e)(3y-2)rarrc=root(3)(log_(e)(3y-2))`
Discussion
103.

`inte^(tan^(-1)x)(1+(x)/(1+x^(2)))dx` is equal toA. `xe^(tan^(-1)x)+C`B. `x^(2)e^(tan^(-1)x)+C`C. `(1)/(x)e^(tan^(-1)x)+C`D. none of these

Answer» Correct Answer - a
Discussion
104.

If `I=int(e^x)/(e^(4x)+e^(2e)+1) dx. J=int(e^(-x))/(e^(-4x)+e^(-2x)+1) dx.` Then for an arbitrary constant c, the value of `J-I` equal toA. `(1)/(2)log((e^(4x)-e^(2x)+1)/(e^(4x)+e^(2x)+1))+C`B. `(1)/(2)log((e^(2x)+e^(x)+1)/(e^(2x)-e^(x)+1))+C`C. `(1)/(2)log((e^(2x)+e^(x)+1)/(e^(2x)+e^(x)+1))+C`D. `(1)/(2)log((e^(2x)+e^(2x)+1)/(e^(2x)+e^(2x)+1))+C`

Answer» Correct Answer - c
We have , `J-I=int(e^(3x))/(e^(4x)+e^(2x)+1)-(e^(x))/(e^(4)+e^(2x)+1)dx`
`rArrJ-I=int(e^(2x)-1)/(e^(4x)+e^(2x)+1)d(e^(x))`
`rArrJ-I=int(t^(2)-1)/(t^(4)+t^(2)+1)dt` where `t=e^(x)`
`rArrJ-I=int(1)/((t+(1)/(t))^(2)-1)d(t-(1)/(t))`
`rArrJ-I=(1)/(2)log((t+(1)/(t)-1)/(t+(1)/(t)+1))+C`
`rArrJ-I=(1)/(2)log((t^(2)-t+1)/(t^(2)+t+1))+C=(1)/(2)log((e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1))+C`
Discussion
105.

Evaluate `int(cosx-sinx)/(cosx+sinx)(2+2sin2x)dx`A. `sin2x+C`B. `cos2x+C`C. `tan2x+C`D. none of these

Answer» Correct Answer - a
Let `I=int(cosx-sinx)/(cosx+sinx)(2+2sin2x)dx`
`rArrI=2int(cosx-sinx)/(cosx+sinx)xx(cosx+sinx)^(2)dx`
`rArrI=int(cos^(2)x-sin^(2)x)dx=2intcos 2x dx=sin2x+C`
Discussion
106.

Evaluate:`int(x^(5/2))/(sqrt(1+x^7))dx`A. `(2)/(7)log|x^(7//2)+sqrt(1+x^(7))|+C`B. `(1)/(2)log|(x^(7)+1)/(x^(7)-1)|+C`C. `2sqrt(1+x^(7))+C`D. none of these

Answer» Correct Answer - a
Let `=int(x^(5//5))/(sqrt(1+x^(7)))dx`
`rArrI=(2)/(7)int(1)/(sqrt(1^(2)+(x^(7//2))^(2))^(7/(2)x(5)/(2)))dx`
`rArrI=(2)/(7)int(1)/(sqrt(1^(2)+(x^(7//2))^(2)))d(x^(7//2))=(2)/(7)log|x^(7//2)+sqrt(1+x^(7))|+C`
Discussion
107.

`int 5^(5^(5^x)) * 5^(5^x) * 5^x dx` is equal toA. `(5^(5^(x)))/((log5)^(3))+C`B. `5^(5^(5^(x)))(log5)^(3)+C`C. `(5^(5^(5^(x))))/((log5)^(3))+C`D. none of these

Answer» Correct Answer - c
Discussion
108.

`int(1)/(cosx-sinx)dx` is equal toA. `(1)/(sqrt(2))log|tan((x)/(2)+(3pi)/(8))|+C`B. `(1)/(sqrt(2))log|"cot"(x)/(2)|+C`C. `(1)/(sqrt(2))log|tan((x)/(2)-(3pi)/(8))|+C`D. `(1)/(sqrt(2))log|tan((x)/(2)-(pi)/(8))|+C`

Answer» Correct Answer - a
Discussion
109.

Evaluate:`int(e^tan^((-1)x))/((1+x^2))[(sec^(-1)sqrt(1+x^2)+cos^(-1)((1-x^2)/(1+x^2))]dx(x >0)dot`A. `I=e^(tan^(-1)x)(tan^(-1)x)+C`B. `I=e^(tan^(-1)x)(sec^(-1)sqrt(1+x^(2)))^(2)+C`C. `I=(1)/(2)e^(tan^(-1)x)(tan^(-1)x)^(2)+C`D. `I=e^(tan^(-1)x)("cosec"^(-1)sqrt(1+x^(2)))^(2)+C`

Answer» Correct Answer - b
Let `tan^(-1)x=thetaorx=tantheta` . Then,
`I=inte^(theta)(theta^(2)+2 theta)d theta=e^(theta)theta^(2)+C`
`rArrI=e^(tan^(-1)x)(tan^(-1)x)^(2)+C`
`rArrI=e^(tan^(-1)x)(sec^(-1)sqrt(1+x^(2)))^(2)+C`
Discussion
110.

Evaluate:`int(e^(2x)-2e^x)/(e^(2x)+1)dx`A. `log(e^(2x)+1)-tan^(-1)(e^(x))+C`B. `(1)/(2)log(e^(2x)+1)-tan^(-1)(e^(x))+C`C. `(1)/(2)log(e^(2x)+1)-2tan^(-1)(e^(x))+C`D. none of these

Answer» Correct Answer - c
Discussion
111.

Evaluate `inte^(3logx)(x^(4)+1)^(-1)dx`A. `log(x^(4)+1)+C`B. `(1)/(4)log(x^(4)+1)+C`C. `-log(x^(4)+1)`D. none of these

Answer» Correct Answer - b
Discussion
112.

Evaluate :`inta/(a^xb^x)dx`

Answer» `1/(a^xb^x)=1/(ab)^x`
`=(ab)^(-x)`
`=e^(loge(ab)*(-x))`
`int1/(a^xb^x)dx=inte^(loge(ab)*(-x)`
`=1/(-loge(ab))e^(loge(ab)^(-x)`
`=1/log_e(ab)*(ab)^(-x)+c`
`=(a^(-x)*b^(-x))/(log_e(ab))+c`.
Discussion
113.

If `intsin^(5)x cos^(4)x dx =A cos^(9) x +B cos^(7)x+C cos^(5)x+D`,then 9A +7B +5C=A. 1B. 0C. `-1`D. none of these

Answer» Correct Answer - b
`I=intsin^(5)xcos^(4)x dx`
`rArr I=intsin^(4)x cos^(4)x sinx dx`
`rArr I=- int (1-cos^(2)x)^(2)cos^(4) xd (cosx)`
`rArrI=- int(1-t^(2))^(2)t^(4)dt` , where t cos x
`rArrI=-int(t^(4)-2t^(6)+t^(8))dt`
`rArrI=-(t^(9))/(9)+(2)/(7)t^(7)-(1)/(5)t^(5)+D`
`rArrI=-(1)/(9)cos^(9)x+(2)/(7)cos^(7)x-(1)/(5)cos^(5)x+D`
`:. A=-(1)/(9),B =(2)/(7)andC=-(1)/(5)rArr9A+7B+5C=0`
Discussion
114.

Let `I_(n)=inttan^(n)xdx , n gt 1`. `I_(4)+I_(6)=atan^(5)x+bx^(5)+C` , where C is a constant of integration , then the ordered pair ( a , b) is equal toA. `((1)/(5),-1)`B. `(-(1)/(5),0)`C. `(-(1)/(5),1)`D. `((1)/(5),0)`

Answer» Correct Answer - d
We have , `:. I_(4)+I_(6)=inttan^(4)dx+inttan^(6)x dx`
`rArr I_(4)+I_(6)=int(tan^(4)x+tan^(6)x)dx`
`rArrI_(4)+I_(6)=inttan^(4)x(1+tan^(2)x)dx`
`rArr I_(4)+I_(6)=inttan^(4)xsec^(2)xdx =inttan^(4)xd (tanx)`
`rArrI_(4)+I_(6)=(1)/(5)tan^(5)x+C`
But, `I_(4)+I_(6)=atan^(5)x+bx^(5)+C`
`:. a=(1)/(5),b=0`
Hence , `( a ,b)=((1)/(5),0)`
Discussion
115.

If `int(1)/(1-sin ^(4)x)dx=(1)/(2) tanx+Atan^(-1){f(x)}+C`, thenA. `A=(1)/(2sqrt(2))andf(x) =sqrt(2)tanx`B. `A=sqrt(2)andf(x) =sqrt(2)tanx`C. `A=-sqrt(2)andf(x) =sqrt(2)tanx`D. none of these

Answer» Correct Answer - a
Let `I=int(1)/(1-sin^(4)x)dx=int(1)/((1-sin^(2)x)(1+sin^(2)x))dx`
`rArrI=int(sec^(2)x)/(1+sin^(2)x)dx=int(1+tan^(2)x)/(1+2tan^(2)x)d(tanx)`
`rArrI=(1)/(2)int(tan^(2)x+(1)/(2)+(1)/(2))/(tan^(2)x+(1)/(2))d(tanx)`
`rArrI=(1)/(2)int(1+(1)/(2)*(1)/(tan^(2)x+(1)/(2)))d(tanx)`
`rArrI=(1)/(2)[tan x +(sqrt(2))/(2)tan^(-1)(sqrt(2)tanx)]+C`
`rArrI=(1)/(2)tan x +(sqrt(2))/(4)tan^(-1)(sqrt(2)tanx)+C`
Hence , `A=(sqrt(2))/(4)=(1)/(2sqrt(2))andf(x)=sqrt(2)tanx`.
Discussion
116.

If `int(sin^(4)x)/(cos^(8)x)dx=atan^(7)x +btan^(5)x+C` , thenA. 7a = 5bB. 5a = 7bC. 7a + 5b = 0D. 5a + 7b = 0

Answer» Correct Answer - a
We have `I=int(sin^(4)x)/(cos^(8)x)dx`
`rArr I=int(tan^(4)x)/(cos^(4)x)dx` " " `["Dividing" N^(r) and^(r) "by" cos^(4)x]`
`rArr I=inttan^(4)x(1+tan^(2)x)dx (tanx)`
`rArrI=int(tan^(4)x+tan^(6)x)d (tanx)=(tan^(5)x)/(5)+(tan^(7)x)/(7)+C`
`:. a=(1)/(7)and b =(1)/(5)rArr7a=5b`
Discussion
117.

If `I=int(sin2x)/((3+4cosx)^(3))dx` , then I=A. `(3cosx+8)/((3+4cosx)^(2))+C`B. `(3+8cosx)/(16(3+4cosx)^(2))+C`C. `(3+cosx)/((3+4cosx)^(2))+C`D. `(3-8 cosx)/(16(3+4cosx)^(2))+C`

Answer» Correct Answer - b
We have, `I=2int(cosxsinx)/((3+4cos)^(3))dx`
`rArrI=-(1)/(2)int(cosx)/((3+4cosx)^3)d(3+4cosx)`
`rArrI=-(1)/(2)int(t-3)/(4t^(3))dt " where " t=3+4cosx`
`rArrI=(1)/(8)((1)/(t)-(3)/(2t^(2)))+C`
`rArrI=(2t-3)/(16t^(2))+C=(8 cosx+3)/(16(3+4cosx)^(2))+C`
Discussion
118.

If `int(1)/((x^(2)+1)(x^(2)+4))dx=Atan^(-1)x+B" tan"^(-1)(x)/(2)+C` , thenA. `A=1//3,B=-2//3`B. `A=-1//3,B=2//3`C. `A=-1//3,B=1//3`D. `A=1//3,B=-1//6`

Answer» Correct Answer - a
Discussion
119.

If `int(dx)/(5-4cosx)=Atan^(-1)(Btanx//2)+C`, thenA. `A=1,B=1//3`B. `A=2//3,B=1//3`C. `A=-1,B=1//3`D. `A=1//3, B=2//3`

Answer» Correct Answer - b
Discussion
120.

Evaluate:`int(secx)/(log(secx+tanx)dx`

Answer» `d/dx[log(secx+tanx)]=1/(sex+tanx)(secxtanx+sec^2x)`
`=(secx+tanx)/(secx+tanx)`
`=secx`
`intsecx/(log(secx+tanx))dx=int(d[log(secx+tanx)])/log(secx+tanx)`
`=log[log(secx+tanx)]+c`.
Discussion
121.

`int(1)/(tanx+cotx+secx+"cosec "x)dx` is equal toA. `(1)/(2)(sinx+cosx+x)+C`B. `(1)/(2)(sinx-cosx-x)+C`C. `(1)/(2)(cosx-xsinx)+C`D. none of these

Answer» Correct Answer - d
Let `=int(1)/(tanx+cotx+secx+"cosec " x)dx` Then ,
`I=int(sinxcosx)/(1+sinx+cosx)dx`
`rArrI=(1)/(2)int((1+2sinx cosx-1))/(1+sinx+cosx)dx=(1)/(2)int((sinx+cosx)^(2)-1)/(1+sinx+cosx)`
`rArr I=(1)/(2)int((sinx+cosx+1)(sinx+cosx-1))/(1+sinx+cosx)dx`
`rArr I=(1)/(2)int(sinx+cosx-1)dx`
`rArrI=(1)/(2)(-cosx+sinx-1)+C`
Discussion
122.

The integral `int(sec^2x)/((secx+tanx)^(9/2))dx`equals (for some arbitrary constant `K)dot``-1/((secx+tanx)^((11)/2)){1/(11)-1/7(secx+tanx)^2}+K``1/((secx+tanx)^(1/(11))){1/(11)-1/7(secx+tanx)^2}+K``-1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K``1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K`A. `-(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`B. `(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`C. `-(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`D. `(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`

Answer» Correct Answer - c
We have , `I=int(sec^(2)x)/((secx+tanx)^(9//2))dx`
Let `secx-tanx=1//t and, secx(secx+tanx)dx=dt`.
`thereforesecx dx=(1)/(t)dtand, secx=(1)/(2)(t+(1)/(t))`
`thereforeI=(1)/(2)int((1)/(t)(t+(1)/(t)))/(t^(9//2))dt=(1)/(2)int(1)/(t^(9//2))+(1)/(t^(13//2))dt`
`rArr I=-(1)/(7t^(7//2))-(1)/(11t^(11//2))+K`
`rArrI=-(1)/(t^(11//2)){(t^(2))/(7)+(1)/(11)}+K`
`rArr I=-(1)/((secx+tanx)^(11//2)){(1)/(11)+(1)/(7)(secx+tanx)^(2)}+K`
Discussion
123.

The value of the integral `int(xsin x^(2)e^(secx^(2)))/(cos^(2)x^(2))dx` , isA. `(1)/(2)e^(secx^(2))+C`B. `(1)/(2)e^(sinx^(2))+C`C. `(1)/(2)sinx^(2)e^(cos^(2)x^(2))+C`D. none of these

Answer» Correct Answer - a
Discussion
124.

`int secx/(sqrt(sin(2x+alpha)+sinalpha))dx`A. `sqrt(secalpha(tanx+tanalpha))+C`B. `sqrt(2 secalpha(tanx-tanalpha))+C`C. `sqrt(2 secalpha(tanalpha-tanalpha))+C`D. none of these

Answer» Correct Answer - a
Let `I=int(secx)/(sqrt(2x+alpha+sinalpha))dx`
`rArr I=int(secx)/(sqrt(2 sin(x+alpha)cosx))dx`
`rArrI=(1)/(sqrt(2))int(sec^(2)x)/(sqrt(tanx cos alpha+sinalpha))dx`
`rArrI=(1)/(sqrt(2)cosalpha)int(1)/(sqrt(tanx cosalpha+sinalpha))d(tanx cosalpha+sinalpha)`
`rArrI=(1)/(sqrt(2)cos alpha)xx2sqrt(tan x cosalpha+sinalpha)+C`
`rArr I=sqrt(2sec^(2)alpha(tan xcosalpha+sinalpha))+C`
`rArrI=sqrt(2secalpha(tanx+tanalpha))+C`.
Discussion
125.

`int{1+2tanx(tanx+secx)}^(1//2)dx` is equal toA. `logsec x (secx -tanx)+C`B. `log"cosec" (secx+tanx)+C`C. `logsecx(secx+tanx+C)`D. `log(secx+tanx)+C`

Answer» Correct Answer - c
Discussion