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201.

The value of `int_-pi^pi cos^2x/[1+a^x].dx`,a>0 isA. `pi`B. `api`C. `(pi)/(2)`D. `2pi`

Answer» Correct Answer - C
Let `I=int_(-pi)^(pi)(cos^(2)x)/(1+a^(x))dx` . . . (i)
`=int_(pi)^(-pi)(cos^(2)(-x))/(1+a^(-x))d(-x)`
`rArrI=int_(-pi)^(pi)a^(x)(cos^(2)x)/(1+a)dx`
On adding Eqs . (i) and (ii) , we get
`2I=int_(-pi)^(pi)((1+a^(x))/(1+a^(x)))cos^(2)xdx`
`=int_(-pi)^(pi)cos^(2)xdx=2int_(0)^(pi)(1+cos2x)/(2)dx`
`=int_(0)^(pi)(1+cos2x)dx`
`=int_(0)^(pi)1dx+int_(0)^(pi)cos2xdx`
`=[x]_(0)^(pi)+2int_(0)^(pi//2)cos2xdx=pi+0`
`rArr2I=pirArrI=pi//2`
Discussion
202.

Let a and b be two positive real numbers. Then the value of `int_(a)^(b)(e^(x//a)-e^(b//x))/(x)dx` is

Answer» Correct Answer - A
`I=int_(a)^(b)(e^(x//a)-e^(b//a))/(x)dx`
Put `(x)/(a)=(b)/(y)`
`rArr" "I=int_(b)^(a)(e^((b)/(y))-e^((y)/(a)))/((ab)/(y))(-(ab)/(y^(2)))dy`
`" "=int_(a)^(b)(b^(b//x)-e^(x//a))/(x)dx=-I`
`rArr" "2I=0`
`rArr" "I=0`
Discussion
203.

`int_(1//3)^(3)(1)/(x)log_(e)(|(x+x^(2)-1)/(x-x^(2)+1)|)dx` is equal toA. `(8)/(3)`B. `-(8)/(3)`C. 0D. 3

Answer» Correct Answer - C
`I=int_(1//3)^(2)(1)/(x)log_(e)(|(x+x^(2)-1)/(x-x^(2)+1)|)dx`
Let `x=(1)/(t)rArr dx =-(1)/(t^(2))dt`
`rArr" "I=-int_(3)^(1//3)tlog_(e)(|((1)/(t)+(1)/(t^(2))-1)/((1)/(t)-(1)/(t^(2))+1)|)(1)/(t^(2))dt`
`" "=int_(1//3)^(3)(1)/(t)log_(e)(|(t-t^(2)+1)/(t+t^(2)-1)|)dt`
`" "=-int_(1//3)^(3)(1)/(x)log_(2)(|(x+x^(2)-1)/(x-x^(2)+1)|)dx`
`rArr" "I=-IrArr I=0`
Discussion
204.

The area (in sq. units) of the region described by `{(x,y) , y^(2) ge 2x` and `y ge 4x - 1}` isA. `7/32`B. `5/64`C. `15/64`D. `9/32`

Answer» Correct Answer - D
Discussion
205.

The area of the region described by `A = {(x,y) : x^(2)+y^(2) le 1` and `y^(2) le 1-x}` is :A. `pi/2-2/3`B. `pi/2+2/3`C. `pi/2+4/3`D. `pi/2+4/3`

Answer» Correct Answer - C
Discussion
206.

The integral `int_(1)^(e){((x)/(e))^(2x)-((e)/(x))^(x)} "log"_(e)x` dx is equal toA. `(3)/(2)-e-(1)/(2e^(2))`B. `-(1)/(2)+(1)/(e)-(1)/(2e^(2))`C. `(1)/(2)-e-(1)/(e^(2))`D. `(3)/(2)-(1)/(e)-(1)/(2e^(2))`

Answer» Correct Answer - A
Let `I=int_(1)^(e){((x)/(e))^(2x)-((e)/x)^(x)} "log"_(e) x dx`
Now , put `((x)/(e))^(x)=t rArrx log _(e)((x)/(e))= log t`
`rArrx (log_(e)x-log_(e))=logt`
`rArr[x((1)/(x))+(log_(e)x-log_(e)e)]dx=(1)/(t)dt`
`rArr(1+log_(e)x-1)dx=(1)/(t)dtrArr(log_(e)x)dx=(1)/(t)dt`
Also , upper limit `x=erArrt=1` and lower limit `x=1rArrt=(1)/(e)`
`I=int_(1//e)^(1)(t^(2)-(1)/(t))*(1)/(t)dtrArrI=int_(1//e)^(1)(t-t^(2)) dt`
`I=[((t^(2))/(2)+(1)/(t))]_(1/(e))^(1)={((1)/(2)+1)-((1)/(2e^(2))+e)}=(3)/(2)-e-(1)/(2e^(2))`
Discussion
207.

The value of `int_(0)^(oo)(logx)/(a^(2)+x^(2))dx` isA. `(2piloga)/(a)`B. `(pi log a)/(2a)`C. `pi loga`D. 0

Answer» Correct Answer - B
`I=int_(0)^(oo)(logx)/(a^(2)+x^(2))dx`
Put `x=(a^(2))/(y),dx=-(a^(2))/(y^(2))dy`
`therefore" "I=int_(oo)^(0)(log((a^(2))/(y)))/(a^(2)+(a^(4))/(y^(2)))((-a^(2))/(y^(2)))dy`
`" "=int_(oo)^(0)((loga^(2)-logy))/(a^(2)+y^(2))(-dy)`
`" "=log(a^(2))int_(0)^(oo)(dy)/(a^(2)+y^(2))-int_(0)^(oo)(logy)/(a^(2)+y^(2))dy`
`" "log(a^(2))(1)/(a)(tan^(-1)((y)/(a)))_(0)^(oo)-I`
`rArr" "2I=(2loga)/(a).(pi)/(2)`
`rArr" "I=(piloga)/(2a)`
Discussion
208.

The value of the integral `int_(e^(-1))^(e^2)|((log)_e x)/x|dxi s``3/2`(b) `5/2`(c) 3(d) 5A. `3//2`B. `5//2`C. 3D. 5

Answer» Correct Answer - D
`int_(e^(-1))^(e^(2))|(log_(e)x)/(x)|dx=int_(e^(-1))^(1)|(log_(e)x)/(x)|dx-int_(1)^(e^(2))|(log_(e)x)/(x)|dx`
`["since , 1 is turning point for"|(log_(e)x)/(x)|" for + ve and - ve values"]`
`=-int_(e^(-1))^(1)(log_(e)x)/(x)dx+int_(1)^(e^(2))|(log_(e)x)/(x)|dx`
`=-(1)/(2)[(log_(e)x)^(2)]_(e^(-1))^(1)+(1)/(2)[(log_(e)x)^(2)]_(1)^(e^(2))`
`=-(1)/(2){0-(-1)^(2)}+(1)/(2)(2^(2)-0)=(5)/(2)`
Discussion
209.

The integral `int_(2)^(4) (logx^(3))/(logx^(2)+log(36-12x+x^(2)))dx` is equal toA. 2B. 4C. 1D. 6

Answer» Correct Answer - C
Discussion
210.

`int_(0)^(4)(|x-1|+|x-3|)dx`

Answer» Correct Answer - 10
Discussion
211.

`int_(-2)^(4) f[x]dx`, where `[x]` is integral part of x.

Answer» Correct Answer - 3
Discussion
212.

`int_(0)^(9)[sqrt(t)]dt`.

Answer» Correct Answer - 13
Discussion
213.

`int_(0)^(pi)(x)/(1+sinx)dx`.

Answer» Correct Answer - `pi`
Discussion
214.

Evaluate the following `int_(-1)^(1)|x|dx`.

Answer» Correct Answer - 1
Discussion
215.

`int_(-pi//4)^(pi//2)(secxdx)/(1+2^(x))dx`.

Answer» Correct Answer - `ln(sqrt(2)+1)`
Discussion
216.

The value of `int_(1)^(3) (|x-2|+[x])dx` is `([x]` stands for greatest integer less than or equal to x)A. 7B. 5C. 4D. 3

Answer» Correct Answer - A
Discussion
217.

A farmer `F_1`has a land in the shape of a triangle withvertices at `P(0, 0), Q(1, 1)`and `R(2, 0)`. From this land, a neighbouring farmer `F_2`takes away the region which lies between the side `P Q`and a curve of the form `y=x^n (n >1)`. If the area of the region taken away by thefarmer `F_2`is exactly 30% of the area of ` P Q R`, then the value of `n`is _______.

Answer» Correct Answer - 4
Discussion
218.

Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b` If `f(x)` is an odd function, thenA. `phi(x)` is also an odd functionB. `phi(x)` is an even functionC. `phi(x)` is neither an even nor an odd functionD. for `phi(x)` to be an even function, it must satisfy `int_(0)^(a)f(x)dx=0`

Answer» Correct Answer - B
`f(x)` is an odd function. Thus, `f(x)=-f(-x)`
`phi(-x)=int_(0)^(-x)f(t)dt`
Put `t=-y`
`:. phi(-x)=int_(-a)^(x)f(-t)(-dt)`
`=int_(-a)^(x)f(t)dt=int_(-a)^(a)f(t)dt+int_(a)^(x)f(t)dt`
`=0+int_(a)^(x)f(t)dt=phi(x)`.
Discussion
219.

Evaluating integrals dependent on a parameter: Differentiate I with respect to the parameter within the sign an integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as a function of the parameter then integrate then result of get I. Constant of integration can be computed by giving some arbitrary values to the parameter and the corresponding value of I. The value of `(dI)/(da)` when `I=int_(0)^(pi//2) log((1+asinx)/(1-asinx)) (dx)/(sinx)` (where `|a|lt1`) isA. `(pi)/(sqrt(1-a^(2)))`B. `-pisqrt(1-a^(2))`C. `sqrt(1-a^(2))`D. `(sqrt(1-a^(2)))/(pi)`

Answer» Correct Answer - A
Let `I(a)=int_(0)^(pi//2) log((1+a sin x)/(1-a sinx))(dx)/(sinx)`
`(dI)/(da)=int_(0)^(pi//2) (2sinx)/(1-a^(2)sin^(2)x)(dx)/(x sin x)`
`=int_(0)^(pi//2)(2sec^(2)xdx)/(1+tan^(2)x-a^(2)tan^(2)x)`
`=int_(0)^(pi//2) (2sec^(2)xdx)/(1+tan^(2)x-a^(2)tan^(2)x)`
`=int_(0)^(oo) (2dt)/(1+(1-a^(2))t^(2))` (put `tan x=t`)
`=2/(sqrt(1-a^(2)))[tan^(-1)(tsqrt(1-a^(2)))]_(0)^(oo) =(pi)/(sqrt(1-a^(2))`
`:. I=pisin^(-1)a` [as `I(0)=0`]
Discussion
220.

Evaluating integrals dependent on a parameter: Differentiate I with respect to the parameter within the sign an integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as a function of the parameter then integrate then result of get I. Constant of integration can be computed by giving some arbitrary values to the parameter and the corresponding value of I. If `int_(0)^(pi)(dx)/((a-cosx))=(pi)/(sqrt(a^(2)-1))`, then the value of `(dx)/((sqrt(10)-cosx))` isA. `(pi)/81`B. `(7pi)/162`C. `(7pi)/81`D. none of these

Answer» Correct Answer - C
`int_(0)^(pi)(dx)/((a-cosx))=(pi)/(sqrt(a^(2)-1))`
Differentiating both sides with respect to `a` we get
`-int_(0)^(pi) (dx)/((a-cosx)^(2))=(-pia)/((a^(2)-1)^(3//2))`
Again differentiating with respect to `a`, we get
`2int_(0)^(pi) (dx)/((a-cosx)^(3))-(pi(1+2a^(2)))/((a^(2)-1)^(5//2))`
Putting `a=sqrt(10)`, we get `int_(0)^(pi) (dx)/((sqrt(10)-cosx)^(3))=(7pi)/81`
Discussion
221.

Evaluating integrals dependent on a parameter: Differentiate I with respect to the parameter with in the sign an integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as a function of the parameter then integrate then result of get I. Constant of integration can be computed by giving some arbitrary values to the parameter and the corresponding value of I. The value of `int_(0)^(1)(x^(a)-1)/(logx)dx` isA. `log(a-1)`B. `log(a+1)`C. `alog(a+1)`D. none of these

Answer» Correct Answer - B
Let `I(a)=int_(0)^(1)(x^(a)-1)/(logx)dx`……………..1
Differentiating w.r.t `a` keeping `x` is constant, we get
`(dI(a))/(da)=int_(0)^(1)((x^(a)-1)/(logx))dx`
`=int_(0)^(1)(x^(a)logx)/(logx)dx`
`=int_(0)^(1)x^(a)dx`
`=(x^(a+1))/(a+1)|_(0)^(1)=1/((a+1))`
Integrating both sides w.r.t `a` we get
`I(a)=log(a+1)+c`
For `a=0,1(0)=log1+c` [from equation 1]
`0=0+c`
`:. I=log(a+1)`
Discussion
222.

Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b` If `m,n` are even integers and `p,q epsilon R`, then `int_(p+n alpha)^(q+n alpha)g(t)dt` is equal toA. `int_(p)^(q)g(x)dx`B. `(n-m)int_(0)^(alpha)g(x)dx`C. `int_(p)^(alpha)g(x)dx+(n-m)int_(0)^(alpha)g(2x)dx`D. `int_(p)^(q)g(x)dx+((n-m))/2int_(0)^(2alpha)g(x)dx`

Answer» Correct Answer - D
`int_(p+m alpha)^(q+n alpha) g(t)dt=int_(p+m alpha)^(p)g(x)dx+int_(p)^(q)g(x)dx+int_(q)^(q+n alpha) g(x)dx`
`=-m/2 int_(0)^(2 alpha) g(x)dx+int_(p)^(q)g(x)dx+n/2 int_(0)^(2alpha) g(x)dx`
`=int_(p)^(q)g(x)dx+((n-m)/2)int_(0)^(2alpha) g(x)dx`
Discussion
223.

Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b` If `f(x)` is an even function, thenA. `phi(x)` is also an even functionB. `phi(x)` is an odd functionC. if `f(a-x)=-f(x)`, then `phi(x)` is an even functionD. if `f(a-x)=-f(x)` then `phi(x)` is an odd function

Answer» Correct Answer - D
If `f(x)` is an even function, then
`phi(-x)=-int_(-a)^(x)f(t)dt`
`=-int_(-a)^(a)f(t)dt-int_(a)^(x)f(t)dt`
`=-2int_(0)^(a)f(t)dt-int_(a)^(x)f(t)dt` [as `f(x)` is an even function]
Now `int_(0)^(a)f(t)dt=int_(0)^(a)f(a-t)dt`
`-int_(0)^(a)-f(t)dt` [using `f(a-x)=-f(x))`]
or `int_(0)^(a)f(t)dt=0`
or `phi(-x)=-int_(a)^(x)f(t)dt=-phi(x)`
Thus `phi(x)` is an odd function.
Discussion
224.

The value of `int_0^1(x^4(1-x)^4)/(1+x^2) dx` isA. `(22)/(7)-pi`B. `(2)/(105)`C. 0D. `(71)/(15)-(3pi)/(2)`

Answer» Correct Answer - A
Let `I=int_(0)^(1)(x^(4)(1-x)^(4))/(1+x^(2))dx = int_(0)^(1)((x^(4)-1)(1-x)^(4)+(1-x)^(4))/((1+x^(2)))dx`
`=int_(0)^(1)(x^(2)-1)(1-x)^(4)dx+int_(0)^(1)((1+x^(2)-2x)^(2))/((1+x^(2)))dx`
`=int_(0)^(1){(x^(2)-1)(1-x)^(4)+(1+x^(2))-4x+(4x^(2))/((1+x^(2)))}dx`
`=int_(0)^(1){(x^(2)-1)(1-x)^(4)+(1+x^(2))-4x+4-(4)/(1+x^(2)))dx`
`=int_(0)^(1)(x^(6)-4x^(5)+5x^(4)-4x^(2)+4-(4)/(1+x^(2)))dx`
`=[(x^(7))/(7)-(4x^(6))/(6)+(5x^(5))/(5)-(4x^(3))/(3)+4x-4 tan^(-1)x]_(0)^(1)`
`=(1)/(7)-(4)/(6)+(5)/(5)-(4)/(3)+4-4((pi)/(4)-0)=(22)/(7)-pi`
Discussion
225.

The valueof the integral `int_0^(1/2)(1+sqrt(3))/(((x+1)^2(1-x)^6)^(1/4))dx`is ______.

Answer» Correct Answer - 2
`I=int_(0)^(1//2)(1+sqrt(3))/(((x+1)^(2)(1-x)^(6))^(1//4))dx`
`=int_(0)^(1//2) ((1+sqrt(3))dx)/((1+x)^(2)[((1-x)^(6))/((1+x)^(6))]^(1//4)`
Put `(1-x)/(1+x)=t`
`implies(-2dx)/((1+x)^(2))=dt`
`:.I=int_(1)^(1//3)((1+sqrt(3))dt)/(-2t^(3//2))=(-(1+sqrt(3)))/2xx|(-2)/(sqrt(t))|_(1)^(1//3)`
`=(1+sqrt(3))(sqrt(3)-1)=2`
Discussion
226.

`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt` The value of `int_(0)^(pi//2) f(x)dx` isA. `1`B. `-2`C. `-1`D. `2`

Answer» Correct Answer - C
`f(x)=sinx+sin int_(-pi//2)^(pi//2) f(t)dt+cosx int_(-pi//2)^(pi//2) tf(t)dt`
`=sinx (1+int_(-pi//2)^(pi//2) f(t)dt)+cosx int_(-pi//2)^(pi//2) tf(t)dt`
`=Asinx+Bcosx`
Thus, `A=1+int_(-pi//2)^(pi//2) f(t)dt`
`=1+int_(-pi//2)^(pi//2) (Asint+Bcost)dt`
`=1+2Bint_(0)^(pi//2) cost dt`
`=A=1+2B` ............1
`B=int_(-pi//2)^(pi//2) tf(t)dt`
`=int_(-pi//2)^(pi//2)t(Asint+Bcost)dt`
`=2Aint_(0)^(pi//2) t sin tdt`
`=2A[-tcost+sint]_(0)^(pi//2)`
`:.B=2A`
From equations 1 and 2 we get
`A=-1//3, B=-2//3`
`:.f(x)=-1/3(sinx+2cosx)`
Thus, the range fo `f(x)` is `[-(sqrt(5))/3,(sqrt(5))/3]`
`f(x)=-1/2(sinx+2cosx)`
`=-(sqrt(5))/3 sin (x+tan^(-1)2)`
`=-(sqrt(5))/3cos(x-"tan"^(-1)1/2)`
`f(x)` in invertible if `-(pi)/2 le x=tan^(-1)2le(pi)/2`
or `-(pi)/2-tan^(-1)2 le xle (pi)/2 -tan^(-1) 2`
or `0lex-"tan"^(-1)1/2 le pi`
or `"tan"^(-1)1/2 le x le pi+"tan"^(-1)1/2`
or `pi le x-"tan"^(-1)1/2le 2pi`
or `x epsilon [pi+cot^(-1)2, 2pi +cot^(-1) 2]`
`int_(0)^(pi//2) f(x)dx=-1/3int_(0)^(pi//2) (sinx+2cosx)dx`
`=-1/3[-cosx+2sinx]_(0)^(pi//2) =-1`
Discussion
227.

Let `u=int_0^oo (dx)/(x^4+7x^2+1` and `v=int_0^x (x^2dx)/(x^4+7x^2+1)` thenA. `pi//3`B. `pi//6`C. `pi//12`D. `pi//9`

Answer» Correct Answer - B
`u=int_(0)^(oo) (dx)/(x^(4)+7x^(2)+1)` and `v=int_(0)^(oo) (x^(2)dx)/(x^(4)+7x^(2)+1)`
`:. u+v=int_(0)^(oo) (1+x^(2))/(x^(4)+7x^(2)+1)dx`
`=int_(0)^(oo) (1/(x^(2))+1)/((x-1/x)^(2)+9)dx`
`=1/3["tan"^(-1)((x-1/x)/3)]_(0)^(oo)`
`=1/3[pi//2+pi//2]=pi//3`
`:.u+v=pi//3`
Now `u-v=int_(0)^(oo) (1-x^(2))/(x^(4)+7x^(2)+1)dx`
Let `x=1/t` or `x=-(dt)/(t^(2))`
`:.u-v=int_(oo)^(0)(1-1/(t^(2)))/(1/(t^(4))+7/(t^(2))+1)(1-1/(t^(2)))dt`
`=-int_(0)^(oo)(1-t^(2))/(t^(4)+7t^(2)+1) dt`
`=-(u-v)`
`:. u-v=0`
From 1 and 2, we get `u=v=pi//6`
Discussion
228.

The values of `a`for which the integral `int_0^2|x-a|dxgeq1`is satisfied are`(2,oo)`(b) `(-oo,0)``(0,2)`(d) none of theseA. `[2,oo)`B. `(-oo,0]`C. `(0,2)`D. none of these

Answer» Correct Answer - A::B::C
For `ale0` given equation becomes
`int_(0)^(2)(x-a)dxge1` or `alt1/2`
So` ale0`
For `0 lt a lt 2`
`int_(0)^(2)|x-a|dxge1` or `int_(0)^(a)(a-x)dx+int_(a)^(2)(x-a)dxge1`
or `(a^(2))/2+2-2a+(a^(2))/2ge1`
or `a^(2)-2a+1ge0` or `(a-1)^(2)ge0`
So `0lt a lt 2`
For `a ge 2`.
`int_(0)^(2)|x-a|dxge1`
or `int_(0)^(2)(a-x)dxge1`
or `2a-2ge1` or `age3/2`
So `age2`
Discussion
229.

`Iff(2-x)=f(2+x)a n df(4-x)=f(4+x)`for all `xa n df(x)`is a function for which `int_0^2f(x)dx=5,t h e nint_0^(50)f(x)dx`is equal to125 (b) `int_(-4)^(46)f(x)dx``int_1^(51)f(x)dx`(d) `int_2^(52)f(x)dx`A. `125`B. `int_(-4)^(46)f(x)dt`C. `int_(1)^(51)f(x)dx`D. `int_(2)^(52)f(x)dx`

Answer» Correct Answer - A::B::D
`f(2-x)=f(2+x),f(4-x)=f(4+x)`
or `f(4+x)=f(4-x)=f(2+2-x)=f(2-(2-x))=f(x)`
Thus, the period of `f(x)` is 4.
`int_(0)^(50)f(x)dx=int_(0)^(48)f(x)dx+int_(48)^(50)f(x)dx`
`=12 int_(0)^(4)f(x)dx+int_(0)^(2)f(x)dx`
[In second integral, replacing `x` by `x+48` and then using `f(x)=f(x+48)`]
`=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4-x)dx)+5`
`=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4+x)dx)+5`
`=24int_(0)^(2)f(x)dx+5=125`
`int_(-4)^(46)f(x)dx=int_(-4)^(-2)f(x)dx+int_(-2)^(-2+48)f(x)dx`
`=int_(0)^(2)f(x+4)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx`
`=125`
also `int_(2)^(52)f(x)dx=int_(2)^(4)f(x)dx+int_(4)^(4+48)f(x)dx`
`=int_(0)^(2)f(4-x)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(4+x)dx+24int_(0)^(2)f(x)dx`
`=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx`
`=125`
`int_(1)^(51)f(x)dx=int_(1)^(3)f(x)dx+int_(3)^(3+48)f(x)dx`
`=int_(1)^(3)f(x)dx+12int_(0)^(4)f(x)dx`
`=int_(0)^(2)f(x+1)dx+24int_(0)^(2)f(x)dx`
`!=125`
Discussion
230.

If `I_(n)=int_(0)^(pi//4) tan^(n)x dx, (ngt1` is an integer ), thenA. `I_(n)+I_(n-2)=1/(n+1)`B. `I_(n)+I_(n-2)=1/(n-1)`C. `I_(2)+I_(4),I_(6),…….` are in H.P.D. `1/(2(n+1))ltI_(n)lt1/(2(n-1))`

Answer» Correct Answer - B::C::D
`I_(n)=int_(0)^(pi//4)tan^(n)xdx`
`=int_(0)^(pi//4)tan^(n-2)xtan^(2)xdx`
`=int_(0)^(pi//4)sec^(2)x tan^(n-2)x dx-int_(0)^(pi//4) tan^(n-2) xdx`
`=int_(0)^(1)t^(n-2)dt=I_(n-2)`, where `t=tanx`
`:.I_(n)+I_(n-2)=((t^(n-1))/(n-1))_(0)^(1)=1/(n-1)`
Thus `I_(2)+I_(4)+I_(4)+I_(6)..............` are in H.P.
For `0lt xlt pi//4` we have `0 lt tan^(n-2)x`
So `0ltI_(n)ltI_(n-2)` or `I_(n)+I_(n+2)lt 2I_(n)ltI_(n)+I_(n-2)`
or `1/(n+1)lt 2I_(n) lt 1/(n-1)` or `1/(2(n+1)) lt I_(n) lt 1/(2(n-1))`
Discussion
231.

If `I_(1)=int_(0)^(1)(dx)/(e^(x)(1+x))` and `I_(2)=int_(0)^(pi//4)(e^(tan^(7)theta)sintheta)/((2-tan^(2)theta)cos^(3)theta d theta`,then find the value of `(l_(1))/(l_(2))`.

Answer» Correct Answer - `2//e`
`I_(2)int_(0)^(pi//4)(e^(tan^(2)theta).tan theta)/((2-tan^(2)theta))sec^(2) d theta`
Put `tan^(2)theta=t`
`:.2 tan theta sec^(2) d theta =dt`
`:.I_(2)=1/2int_(0)^(1)(e^(t)dt)/((2-t))`
`=1/2 int_(0)^(1)(e^(1-t)dt)/(1+t)`
`=e/2int_(0)^(1)(dt)/(e^(t)(t+1))=e/2.I_(1)`
`:. (I_(1))/(I_(2))=2/e`
Discussion
232.

If `f(x)` in inegrable over `[1,2]` then `int_(1)^(2) f(x) dx` is equal to :A. `underset(nrarroo)"lim"(1)/(n)underset(r=1)overset(n)sumf(r/n)`B. `underset(nrarroo)"lim"(1)/(n)underset(r=n+1)overset(2n)sumf(r/n)`C. `underset(nrarroo)"lim"(1)/(n)underset(r=1)overset(n)sumf((r+n)/n)`D. `underset(nrarroo)"lim"1/n underset(r=1)overset(2n)sumf(r/n)`

Answer» Correct Answer - B::C
Discussion
233.

The intercepts on x-axis made by tangents to thecurve, `y=int_0^x|t|dt , x in R ,`which are parallel to the line `y""=""2x`, are equalto(1) `+-2`(2) `+-3`(3) `+-4`(4) `+-1`A. `+-1`B. `+-2`C. `+-3`D. `+-4`

Answer» Correct Answer - A
`y=int_(0)^(x)|t|dt`
Case I: If `xgt0`
`y=int_(0)^(x)tdt=[(t^(2))/2]_(0)^(x)=(x^(2))/2implies(dy)/(dx)=x=2` (given)
`implies x=2` and `y=2`
`:.` eqution of tangent is `(y-2)=2(x-2)`
or `y-2x+2=0`
Hence `x` intercept `=`
Case II: `xlt0`
`-y=int_(0)^(pi)-tdt=[(-t^(2))/2]_(0)^(x)=(-x^(2))/2`
`:.(dy)/(dt)=-x=2`
`:. x=-2,:.y=-2`,
`:.` equation iof tangent is `y+2=2(x+2)`
or `2x-y+2=0`
`:.x` intencept `=-1`
Discussion
234.

Let `f(x) = int_(x)^(x+(pi)/(3))|sin theta|d theta(x in [0,pi])`A. `f(x)` is stricity increasing in this intervalB. `f(x)` is differentiable in this intervalC. Range of `f(x)` is `[2-sqrt(3)-1]`D. `f(x)` has a maxima at `x = pi/3`

Answer» Correct Answer - B::C::D
Discussion
235.

Given an even function f defined and integrable everywhere and periodic with period 2.Let `g(x)=int_0^x f(t) dt and g(1)=A.`A. `g(2n) = 0` for every integer nB. `g(x)` is an even functionC. `g(x)` and `f(x)` have the same periodD. `g(x)` is an odd function

Answer» Correct Answer - A::B::C
Discussion
236.

If `I = int_(0)^(2x)sin^(2)xdx`, thenA. `I = 2 underset(0)overset(pi)intsin^(2)xdx`B. `I = 4 underset(0)overset(pi//2)intsin^(2)xdx`C. `I = underset(0)overset(2pi)intcos^(2)xdx`D. `I = 8 underset(0)overset(pi//4)intsin^(2)xdx`

Answer» Correct Answer - A::B::C
Discussion
237.

Evaluate : (i) `int_(0)^(1)sin^(-1)xdx` , (ii) `int_(1)^(2)(lnx)/(x^(2))dx`, (iii) `int_(0)^(1)x^(2)sin^(-1)xdx`.

Answer» Correct Answer - (i) `ln (sqrt(3))` , (ii) `7//6` , (iii) `3/8`
Discussion
238.

`int_(0)^(1) x^(5)sin^(-1)xdx`

Answer» Correct Answer - `(pi)/(14)-(16)/(245)`
Discussion
239.

`int_(0)^(9) x(a^(2)-x^(2))^(7/2)dx`

Answer» Correct Answer - `(a^(9))/(9)`
Discussion
240.

` lim_(nrarr0) sum_(r=1)^(n) ((r^(3))/(r^(4)+n^(4)))` equals to :A. `ln 2`B. `1/2 ln 2`C. `1/3 ln 2`D. `1/4 ln 2`

Answer» Correct Answer - D
Discussion
241.

`int_(0)^(2) sqrt(2-x)dx`.

Answer» Correct Answer - `(pi)/(2)`
Discussion
242.

The value of `lim_(n rarroo) sum_(r=1)^(n)(1)/(sin{((n+r)pi)/(4n)}).(pi)/(n) ` is equal toA. `2ln(sqrt2-1)`B. `4 ln (sqrt2-1)`C. `4 ln (sqrt2+1)`D. `ln sqrt2`

Answer» Correct Answer - C
`S=underset(nrarroo)(lim)sum_(r=1)^(n)(pi)/(n).(1)/(sin{(pi)/(4)+(pir)/(4n)})`
`" "=pi int_(0)^(1)(dx)/(sin((pi)/(4)x+(pi)/(4)))=4log_(e)(sqrt2-1)`
Discussion
243.

`lim_(nrarroo) {1/n+(n^(2))/((n+1)^(3))+(n^(2))/((n+2)^(3))+"......"+1/(8n)}`

Answer» Correct Answer - `3/8`
Discussion
244.

`lim_(nrarroo) [(1)/(n)+(n^(2))/((n+1)^(3))+(n^(2))/((n+2)^(3))+...+(1)/(8n)]` is equal toA. `(3)/(8)`B. `(1)/(4)`C. `(1)/(8)`D. None of these

Answer» Correct Answer - A
`underset(nrarroo)(lim)[(1)/(n)+(n^(2))/((n+1)^(3))+(n^(2))/((n+2)^(3))+...+(1)/(8n)]`
`" "=underset(nrarroo)(lim)[(n^(2))/((n+0)^(3))+(n^(2))/((n+2)^(3))+(n^(2))/((n+2)^(3))+...+(n^(2))/((n+n)^(3))]`
`" "=underset(nrarroo)(lim)sum_(r=0)^(n)(n^(2))/((n+r)^(3))`
`" "=underset(nrarroo)(lim)sum_(r=0)^(n)(1)/(n)(1)/((1+(r)/(n)))`
`" "=int_(0)^(1)(dx)/((1+x^(3)))=[-(1)/(2(1+x)^(2))]_(0)^(1)`
`" "=-(1)/(2)((1)/(4)-1)=(3)/(8)`
Discussion
245.

Area bounded by the region consisting of points `(x,y)` satisfying `y le sqrt(2-x^(2)), y^(2)ge x, sqrt(y)ge -x` isA. `pi/2`B. `pi`C. `2pi`D. `pi//4`

Answer» Correct Answer - A
Discussion
246.

The value of `int_(pi/2)^( pi/2) (sin^(2)x)/(1+2^(x)) dx` is :A. `4pi`B. `(pi)/(4)`C. `pi/8`D. `pi/2`

Answer» Correct Answer - B
Discussion
247.

Let `g(x) = cosx^(2), f(x) = sqrt(x)`, and `alpha, beta (alpha lt beta)` be the roots of the quadratic equation `18x^(2) -9pix + pi^(2) = 0`. Then the area (in sq. units) bounded by the curve `y = (gof) (x)` and the lines `x = alpha,x = beta` and `y = 0` isA. `1/2(sqrt(3) - sqrt(2))`B. `1/2(sqrt(3)-1)`C. `1/2(sqrt(3)-1)`D. `1/2(sqrt(3)+1)`

Answer» Correct Answer - C
Discussion
248.

`int_(2-a)^(2+a)f(x)dxi se q u a lto[w h e r ef(2-alpha)=f(2+alpha)AAalpha in R]`2`int_2^(2+a)f(x)dx`(b) `2int_0^af(x)dx``2int_2^2f(x)dx`(d) none of theseA. `2int_(2)^(2+a)f(x)dx`B. `2int_(0)^(a)f(x)dx`C. `2int_(2)^(2)f(x)dx`D. none of these

Answer» Correct Answer - A
`f(2-alpha)=f(2+alpha)`
Thus, function is symmetric about the line `x=2`
`int_(2-a)^(2+a)f(x)dx=2int_(2)^(2+a)f(x)dx`
Discussion
249.

If `f(x)`is continuous for all real values of `x ,`then`sum_(r=1)^nf(r-1+x)dxi se q u a lto``int_0^nf(x)dx`(b) `int_0^1f(x)dx``nint_0^1f(x)dx`(d) `(n-1)int_0^1f(x)dx`A. `int_(0)^(n)f(x)dx`B. `int_(0)^(1)f(x)dx`C. `nint_(0)^(1)f(x)dx`D. `(n-1)int_(0)^(1)f(x)dx`

Answer» Correct Answer - A
`sum_(r=1)^(n)int_(0)^(1)f(r-1+x)dx`
`=int_(0)^(1)f(x)dx+int_(0)^(1)f(1+x)dx+int_(0)^(1)f(2+x)dx+`……………
`+ int_(0)^(1) f(n-1+x)dx`
`= int_(0)^(1) f(x) dx+int_(1)^(2)f(x)dx+int_(2)^(3) f(x)dx+int(r-1)^(2)f(x)dx+`………………
`+int_(n-1)^(n)f(x)dx`.
`=int_(0)^(n)f(x)dx`.
Discussion
250.

`int_(5/2)^5(sqrt((25-x^2)^3))/(x^4)dxi se q u a lto``pi/6`(b) `(2pi)/3``(5pi)/6`(d) `pi/3`A. `(pi)/6`B. `(2pi)/3`C. `(5pi)/6`D. `(pi)/3`

Answer» Correct Answer - D
`I=int_(5//2) ^(5)(sqrt((25-x^(2))^(3)))/(x^(4)) dx`
Let `x=5 sin theta`
`:.dx=5 cos theta d theta`
`:. I=int_(pi//6)^(pi//2)(sqrt((25-25sin^(2) theta)^(3)))/(5^(4)sin^(4) theta)`
`=int_(pi//6)^(pi//2) (5^(3)cos ^(3) theta. 5 cos theta)/(5^(4)sin^(4) theta) d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta (cosec^(2) theta-1)d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta cosec^(2) theta d theta -int_(pi//6)^(pi//2) cot^(2) theta d theta`
`=int_(pi//6)^(pi//2) cot^(2) theta cosec^(2) theta d theta -int_(pi//6)^(pi//2) (cosec^(2) theta -1)d theta`
`=[-(cot^(3) theta)/3+cot theta + theta]_(pi//6)^(pi//2)`
`=-0+0+(pi)/2-(-(3sqrt(3))/3+sqrt(3)+(pi)/6)=(pi)/3`
Discussion