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1.

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.

Answer» Correct Answer - 2
Here the mistake lies in the substitution `"tan"1/2x=t`, because `"tan"1/2x` is discontinuous at `x=pi` which is a point in the interval `[0,2pi]`.
Discussion
2.

The area bounded by the curves `y=cosx` and `y=sinx` between the ordinates `x=0` and `x=(3pi)/2` isA. `4sqrt(2) + 2`B. `4sqrt(2) +1`C. `4sqrt(2) + 1`D. `4sqrt(2) - 2`

Answer» Correct Answer - D
Discussion
3.

Obtain the area enclosed by region bounded by the curves `y = x ln x` and ` y = 2x-2x^(2)`.A. `7//6`B. `7//24`C. `12//7`D. `7//12`

Answer» Correct Answer - D
Discussion
4.

If `I_n=int_0^pi e^x(sinx)^n dx ,` then `(I_3)/(I_1)` is equal toA. `3//5`B. `1//5`C. `1`D. `2//5`

Answer» Correct Answer - A
`I_(3)=int_(0)^(pi)e^(x)(sinx)^(3)dx`
`=e^(x)(sinx)^(3)|_(0)^(pi)-3int_(0)^(pi)(sinx)^(2)cosx e^(x)dx`
`=0-3(sinx)^(2)cosx e^(x)|_(0)^(pi)+3int_(0)^(pi)(2sinx cos x cosx)`
`-sin x sin^(2)x)e^(x)dx`
`=0+6int_(0)^(pi)sin x cos^(2) xe^(2) dx-3 int_(0)^(pi) sin^(3) xe^(x) dx`
`=6int_(0)^(pi) sinx(1-sin^(2)x)e^(x)dx-3int_(0)^(pi)sin^(3)xe^(x)dx`
`=6int_(0)^(pi) sinxe^(x)dx-9int_(0)^(pi) sin^(3)x e^(x)dx=6I_(1)-9I_(3)`
or `10 I_(3)=6I_(1)`
or `(I_(3))/(I_(1))=3/5`
Discussion
5.

Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`

Answer» `I_(n)=int_(0)^(oo) (x^(2))^(n)x e^(-x^(2))dx`
Put `x^(2)=t` or `x dx=dt//2`
`:.I_(n)=1/2 int_(0)^(oo) t^(n)e^(-t)dt`
`=1/2[[-t^(n)e^(-t)]_(0)^(oo)+nint_(0)^(oo) t^(n-1)e^(-t)dt]`
`=1/2[0+n int_(0)^(oo) t^(n-1)e^(-t)dt]`
`=n/2int_(0)^(oo) t^(n-1)e^(-t)dt=nI_(n-1)`
or `I_(n-1)=(n-1)I_(n-2)`
or `I_(n)=n(n-1)(n-2).............1I_(0)`
`n!I_(0)=n! 1/2int_(0)^(oo) e^(-t)dt`
`=n! 1/2 [-e^(-t)]_(0)^(oo) =(n!)/2`
Discussion
6.

If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4

Answer» Correct Answer - A
Let `I=int_(0)^(3)x.f(x^(2))dx`
Put `x^(2)=t`
`rArr" "2x.dx=dt`
`rArr" "I=(1)/(2)int_(0)^(9)f(t)dt=(1)/(2).4=2`
Discussion
7.

`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1

Answer» Correct Answer - B
`underset(trarr0)(lim)int_(0)^(2pi)|(sin(x+t)-sinx)/(t)|dx`
`=int_(0)^(2pi)(underset(trarr0)(lim)|(2cos(x+(t)/(2))sin(t)/(2))/(t)|)dt`
`=int_(0)^(2pi)|cosx|dx=4`
Discussion
8.

Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.

Answer» Correct Answer - `((2sqrt(e^(pi)))/(e^(2)))^(m)`
Discussion
9.

`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`

Answer» Correct Answer - `pi/2`
Discussion
10.

`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`

Answer» Correct Answer - A
Discussion
11.

If `int_0^3 (3ax^2+2bx+c)dx=int_1^3 (3ax^2+2bx+c)dx` where `a,b,c` are constants then `a+b+c=`A. `a+b+c=3`B. `a+b+c=1`C. `a+b+c=0`D. `a+b+c=2`

Answer» Correct Answer - C
`int_(0)^(3)(3ax^(2)+2bx+c)dx=int_(1)^(3)(3ax^(2)+2bx+c)dx`
`rArr" "int_(0)^(1)(3ax^(2)+2bx+c)dx+int_(1)^(3)(3ax^(2)+2bx+c)dx`
`" "=int_(1)^(3)(3ax^(2)+2bx+c)`
`rArr" "int_(0)^(1)(3ax^(2)+2bx+c)dx=0`
`rArr" "[(3ax^(3))/(3)+(2bx^(2))/(2)+cx]_(0)^(1)=0`
`rArr" "a+b+c=0`
Discussion
12.

`lim_(nrarroo) sum_(k=1)^(n)(k^(1//a{n^(a-(1)/(a))+k^(a-(1)/(a))}))/(n^(a+1))` is equal toA. 1B. 2C. 43467D. 4

Answer» Correct Answer - A
`underset(nrarroo)(lim)sum_(k=1)^(n)(k^(1//a){n^(a-(1)/(a))+k^(a-(1)/(a))})/(n^(a+1))`
`" "=underset(nrarroo)(lim)sum_(k=1)^(n)(1)/(n).{((k)/(n))^(1//a)+((k)/(n))^(a)}`
`" "=int_(0)^(1)(x^(1//a)+x^(a))dx`
`" "={(x^((1//a)+1))/((1)/(a)+1)+(x^(a+1))/(a+1)}_(0)^(1)`
`" "=(a)/(a+1)+(1)/(a+1)=1`
Discussion
13.

Prove that:`y=int_(1/8)^(sin^2x)sin^(-1)sqrt(t)dt+int_(1/8)^(cos^2x)cos^(-1),w h e r e0lt=xlt=pi/2,`is the equation of a straight line parallel to the x-axis. Find theequation.

Answer» Here we have to prove that `y=` constant or derivative of `y` w.r.t `x` is zero. ltbr. `y=int_(1/8)^(sin^(2)x) sin^(-1)sqrt(5)dt, dt+int_(1/8)^(cos^(2)x) cos^(-1)sqrt(5)dt`…………..1
`(dy)/(dx)=sin^(-1)sqrt(sin^(2)x)2sinx cosx`
`+cos^(-1)sqrt(cos^(2)x)(-2cosxsinx)`
`=2x sin x cos x-2x sin x cos x`
`=0` for all `x`
Therefore the curve in equation 1 is a straight line parallel to the `x`-axis.
Now, since `y` is constant it is independent of `x`. So let us select `x=pi//4`.Then
`y=int_(1//8)^(1//2)sin^(-1)sqrt(5)dt+int_(1//8)^(1//2)cos^(-1)sqrt(t)dt`
`=int_(1//8)^(1//2)(sin^(-1)sqrt(5)+cos^(-1)sqrt(t))dt`
`=int_(1//8)^(1//2)pi//2dt`
`=(pi)/2[1/2-1/8]`
`=(3pi)/16`
Therefore, equation of the line is `y=(3pi)/16`
Discussion
14.

Let `f: Rvec(0,1)`be a continuous function. Then, which of the following function (s) has(have) the value zero at some point in the interval (0,1)?`e^x-int_0^xf(t)sintdt`(b) `f(x)+int_0^(pi/2)f(t)sintdt``x-int_0^(pi/2-x)f(t)costdt`(d)`x^9-f(x)`A. `e^(x) - underset(0)overset(x)intf(t) sin t dt`B. `f(x) + underset(0)overset(pi/2) intf(t) sin t dt`C. `x - underset(0)overset(pi/2-x)int cos t dt`D. `x^(9) - f(x)`

Answer» Correct Answer - CD
Discussion
15.

If `fa n dg`are continuous function on `[0,a]`satisfying `f(x)=f(a-x)a n dg(x)(a-x)=2,`then show that`int_0^af(x)g(x)dx=int_0^af(x)dxdot`

Answer» Correct Answer - NA
`int_(0)^(a)f(x)g(x)dx`
`=int_(0)^(a)f(a-x)g(a-x)dx`
`=int_(0)^(a)f(x).{2-g(x)}dx`
`=2int_(0)^(a)f(x)dx-int_(0)^(a)f(x)g(x)dx`
or `2int_(0)^(a)f(x)f(x)dx=2int_(0)^(a)f(x)dx`
or `int_(0)^(a)f(x)g(x)dx=int_(0)^(a)f(x)dx`
Discussion
16.

Evaluate : (i) `int_(-oo)^(oo) (dx)/(x^(2)+2x+2)` , (ii) `int_(sqrt(2))^(oo)(dx)/(xsqrt(x^(2)-1))`, (iii) `int_(0)^(4)(x^(2))/(1+x)dx` (iv) `int_(0)^(pi//2) sqrt(costheta)sin^(3)theta`

Answer» Correct Answer - (i) `pi` , (ii) `pi/4` , (iii) `4 + ln 5` , (iv) `8/21`
Discussion
17.

Evaluate : (i) `int_(0)^(oo) (dx)/(e^(x)+e^(-x))`, (ii) `int_(0)^(1) (x)/(1+sqrt(x))dx` , (iii) `int_(0)^(pi//2) (sinxcosx)/(cos^(2)x+3cosx+2)` (iv) `int_(0)^(pi/2) (sin2theta d theta)/(sin^(4)theta+cos^(4)theta)`, (v) `int_(0)^(pi//4) (sinx+cosx)/(9+16 sin 2x)dx`

Answer» Correct Answer - (i) `pi/4` , (ii) `5/3 - 2 ln 2` , (iii) `ln (9/8)` , (iv) `pi/2` , (v) `1/20 ln 3`
Discussion
18.

The value of `int_1^e((tan^(-1)x)/x+(logx)/(1+x^2))dxi s``tane`(b) `tan^(-1)e``tan^(-1)(1/e)`(d) none of theseA. `tane`B. `tan^(-1)e`C. `tan^(-1)(1//e)`D. none of these

Answer» Correct Answer - B
`int_(1)^(e)((tan^(-1)x)/x+(logx)/(1+x^(2)))dx`
`=int_(1)^(e)(tan^(-1)x)/x dx+int_(1)^(e)(logx)/(1+x^(2)) dx`
`=int_(1)^(e)(tan^(-1)x)/x dx+(logx tan^(-1)x)_(1)^(e)-int_(1)^(e)(tan^(-1)x)/x dx`
`=tan^(-1)e`
Discussion
19.

`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt` The range of `f(x)` isA. `[-(sqrt(3))/2,(sqrt(3))/2]`B. `[-(sqrt(5))/3,(sqrt(5))/3]`C. `[-(sqrt(5))/2,(sqrt(5))/2]`D. none of these

Answer» Correct Answer - B
`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
20.

If `int_(0)^(x^(2)(1+x))f(t)dt = x` then find `f(2)`

Answer» Correct Answer - `1/5`
Discussion
21.

`int_(pi//2)^(0)sin^(11)xdx`

Answer» Correct Answer - `128/693`
Discussion
22.

Evaluate `int_(0)^(pi)(xdx)/(1+cosalphasinx),0ltalphaltpi`.

Answer» Correct Answer - `(alphapi)/(sin alpha)`
Let `I= int _(0)^(pi) (x)/(1+ cos alpha sin x)dx` . . . (i)
`rArr I = int _(0)^(pi) ((pi-x))/(1+cos alpha sin (pi- x))dx`
` rArr I= int _(0)^(pi) ((pi-x))/(1+ cos alpha sin a )dx` . . . (ii)
On adding Eqs . (i) and (ii) , we get
`rArr 2 I = pi int _(0)^(pi) ("sec "^(2)(x)/(2)dx)/((1+ " tan" ^(2)(x)/(2))+2 cos alpha " tan " (x)/(2))`
Put `"tan " (x)/ (2)= t rArr "sec" ^(2)(x)/(2) dx = 2 dt`
`rArr 2 I = 2 pi int _(0)^(oo) (dt)/((t + cos alpha )^(2)+sin ^(2)alpha)`
`I= (pi)/(sinalpha) [ tan ^(-1) ((t- coas alpha)/( sin alpha))]_(0)^(oo)`
` = (pi)/(sin alpha)[ tan^(-1)(oo)- tan ^(-1) (cot alpha0]`
` = (pi)/(sin alpha)((pi)/(2)-((pi)/(2)- alpha))=(alphapi)/( sinalpha)`
`:. I= (alpha pi)/( sin alpha)`
Discussion
23.

Show that`int_0^(pi/2)sqrt((sin2theta))sinthetadtheta=pi/4`

Answer» Let `I=int_(0)^(pi//2)sqrt((sin 2theta))sin theta d theta `……………1
`:.I=int_(0)^(pi//2)sqrt(sin(2(pi/2)-theta))sin(pi/2-theta)d theta`
`=int_(0)^(pi//2)sqrt((sin 2theta))cos theta d theta` …………..2
Adding 1 and 2 we get
`2I=int_(0)^(pi//2)sqrt((sin 2theta))(sin theta+costheta)d theta`
or `I=1/2int_(0)^(pi//2)sqrt(1-(sin theta -cos theta)^(2))(sin theta+cos theta) d theta`
`=1/2int_(-1)^(1)sqrt(1-t^(2))dt`[ Let `sin theta -cos theta=t`]
`=1/2[1/2 t sqrt((1-t^(2)))+1/2sin^(-1)t]_(-1)^(1)=(pi)/4`
Discussion
24.

Prove that`pi/6

Answer» Correct Answer - NA
Since `0ltx^(3)ltx^(2)`, we have (for `0ltxlt1`)
`x^(2)lt x^(2)+x^(3)lt2x^(2)`
or `-2x^(2)lt-x^(2)-x^(3)lt-x^(2)`
or `4-2x^(2)lt4-x^(2)-x^(3)lt4-x^(2)`
or `sqrt(4-2x^(2))ltsqrt(4-x^(2)-x^(3))ltsqrt(4-x^(2))`
or `1/(sqrt(4-x^(2)))lt 1/(sqrt(4-x^(2)-x^(3)))lt 1/(sqrt(4-2x^(2)))`
or `int_(0)^(1)1/(sqrt(4-x^(2)))dx lt int_(0)^(1)1/(sqrt(4-x^(2)-x^(3)))dx lt int_(0)^(1)1/(sqrt(4-2x^(2)))dx`
or `sin^(-1)(x/2)]_(0)^(1)lt int_(0)^(1) (dx)/(sqrt(4-x^(2)-x^(3))) lt 1/(sqrt(2))"sin"^(-1)x/(sqrt(2))]_(0)^(1)`
or `(pi)/6 lt int_(0)^(1)(dx)/(sqrt(4-x^(2)-x^(3)))lt (pi)/(4sqrt(2))`
Discussion
25.

Evaluate:`int_0^(pi/4)(tan^(-1)((2cos^2theta)/(2-sin2theta)))sec^2thetadthetadot`

Answer» Given integral is `int_(0)^(pi//4)"tan"^(-1)((2cos^(2)theta)/(2-sin2theta))sec^(2) theta d theta`
`=int_(0)^(pi//4)"tan"^(-1)(1/(sec^(2)theta-tan theta))sec^(2) theta d theta`
`=int_(0)^(pi//4)"tan"^(-1)(1/(1+tan^(2)theta-tan theta))sec^(2) theta d theta`
Put `tan theta=t`. Then `sec^(2)theta d theta=dt`
The given integral reduces to
`int_(0)^(1)"tan"^(-1)(1/(1+t^(2)-1))dt=int_(0)^(1)"tan"^(-1)((t-(t-1))/(1+t(t-1)))dt`
`=int_(0)^(1)"tan"^(-1)t dt -int_(0)^(1)tan^(-1)(t-1)dt`
`=int_(0)^(1) tan^(-1)tdt-int_(0)^(1)tan^(-1)((1-t)-1)dt`
`=2int_(0)^(1)1tan^(-1)t dt`
`=2[t tan^(-1)t]_(0)^(1)-2int_(0)^(1)t/(1+t^(2))dt`
(Integrating by parts)
`(pi)/2-[In(1+t^(2))]_(0)^(1)=(pi)/2-In2`.
Discussion
26.

Let `C_1 and C_2`, be the graph of the functions `y= x^2 and y= 2x, 0

Answer» Correct Answer - `f(x) = x^(3) - x^(2)`
Discussion
27.

The function `f`and `g`are positive and continuous. If `f`is increasing and `g`is decreasing, then`int_0^1f(x)[g(x)-g(1-x)]dx`is always non-positiveis always non-negativecan take positive and negative valuesnone of theseA. is always non-positiveB. is always non-negativeC. can take positive and negative valuesD. none of these

Answer» Correct Answer - A
`I=int_(0)^(1)f(x)[g(x)-g(1-x)]dx`
`=-int_(0)^(1)f(1-x)[g(x)-g(1-x)]dx`
or `2I=int_(0)^(1)[f(x)-f(1-x)][g(x)-g(1-x)]dx`
`=2int_(0)^(1//2)[f(x)-f(1-x)].[g(x)-g(1-x)]dxle0`
Discussion
28.

If `I=int_(-20pi)^(20pi)|sinx|[sinx]dx` (where [.] denotes the greatest integer function) then the value of `I` isA. `-40`B. `40`C. `20`D. `-20`

Answer» Correct Answer - A
`int_(-20pi)^(20pi)|sinx|[sinx]dx=int_(0)^(20pi)|sinx|([sinx]+[-sinx])dx`
`=-20int_(0)^(pi)(sinx)dx`
`=-20(=cosx)_(0)^(pi)`
`=20(2)`
`=-40`
Discussion
29.

`int_(0)^(pi)(x tanx)/(secx+cosx)dx` isA. `(pi^(2))/4`B. `(pi^(2))/2`C. `(3pi^(2))/2`D. `(pi^(2))/3`

Answer» Correct Answer - A
Let `I=int_(0)^(pi)(x tan x)/(sec x +cosx)`……………1
`=int_(0)^(pi)((pi-x)tan(pi-x))/(sec(pi-x)+cos(pi-x))`
`=int_(0)^(pi)((pi-x)tanx)/(secx+cosx) dx`…………….2
Adding 1 and 2 gives
`2I=pi int_(0)^(pi) (tanx)/(secx+cosx)dx`
`=pi int_(0)^(pi)((sinx)/(cosx))/(1/(cosx)+cosx)dx=pi int_(0)^(pi) (sinx)/(1+cos^(2)x) dx`
Put `cosx=z`.Therefore `-sinxdx=dz`
When `x=0, z=1`, when `x=pi,z=-1`
`:.2I=pi int_(1)^(-1)(-dz)/(1+z^(2))=pi int_(-1)^(1)(dz)/(1+z^(2))`
`=pi|tan^(-1)z|_(-1)^(1)`
`=pi[tan^(-1)1-tan^(-1)(-1)]`
`=pi((pi)/4+(pi)/4)=(2pi^(2))/4`
or `I=(pi^(2))/4`
Discussion
30.

Consider the polynomial f`(x)= 1+2x+3x^2+4x^3`. Let s be the sum of all distinct real roots of `f(x)`and let `t= |s|`.A. `(-(1)/(4),0)`B. `(-11, -(3)/(4))`C. `(-(3)/(4), - (1)/(2))`D. `(0,(1)/(4))`

Answer» Correct Answer - C
Discussion
31.

Consider the polynomial f`(x)= 1+2x+3x^2+4x^3`. Let s be the sum of all distinct real roots of `f(x)`and let `t= |s|`.A. `(3/4,3)`B. `(21/64,11/16)`C. `(9,10)`D. `(0,(21)/(64))`

Answer» Correct Answer - A
Discussion
32.

Consider the polynomial f`(x)= 1+2x+3x^2+4x^3`. Let s be the sum of all distinct real roots of `f(x)`and let `t= |s|`.A. increasing in `(-t, 1/4)` and decreasing in `(-(1)/(4),t)`B. decreasing in `(-t, -1/4)` and increasing in `(-(1)/(4),t)`C. increasing in `(-t,t)`D. decreasing in `(-t,t)`

Answer» Correct Answer - B
Discussion
33.

Let the straight line x = b divide the area enclosed by `y = (1-x)^(2), y = 0` and `x = 0` into two parts `R_(1) (0 le x le b)` and `R_(2) (b le x le 1)` such that `R_(1) - R_(2) = 1/4`. Then b equalsA. `3/4`B. `1/2`C. `1/3`D. `1/4`

Answer» Correct Answer - B
Discussion
34.

Let `S`be the area of the region enclosed by `y=e^-x^2,y=0,x=0,a n dx=1.`Then`Sgeq1/e`(b) `Sgeq1=1/e``Slt=1/4(1+1/(sqrt(e)))`(d) `Slt=1/(sqrt(2))+1/(sqrt(e))(1-1/(sqrt(2)))`A. `S ge 1/e`B. `S ge 1 - 1/e`C. `S le 1/4 (1+(1)/(sqrt(e )))`D. `S le 1/(sqrt(2)) + 1/(sqrt(e ))(1-1/(sqrt(2)))`

Answer» Correct Answer - A::B::D
Discussion
35.

The area of the region enclosed by the curvesv `y=x,x=e,y=1/x` and the positive X-axis, isA. `1/2` square unitsB. 1 square unitsC. `3/2` square unitsD. `5/2` square units

Answer» Correct Answer - C
Discussion
36.

Find area bounded by the curve `y = lnx + tan^(-1)x` and x-axis between ordiantes `x = 1` and `x = 2`.

Answer» `y = ln x + tan^(-1)x`
Domain `x gt 0 , (dy)/(dx) = 1/x + (1)/(1+x^(2)) gt 0`
`y` is increasing and `x = 1, y = (pi)/(4) rArr y` is positive in `[1,2]`
y in increasing and `x = 1, y = (pi)/(4) rArr y` is positive in `[1,2]`
`:.` Required area `= underset(1)overset(2)int(lnx+tan^(-1)) dx = [xlnx - x + x tan^(-1)- 1/2ln(1+x^(2))]_(1)^(2)`
`= 2ln2-2+2tan^(-1)2-1/2 ln5-0+1-tan^(-1)1/2 ln 2`
`= 5/2 ln 2- 1/2 ln 5+2 tan^(-1) - (pi)/(4) - 1`
Discussion
37.

Prove that `int_0^1sqrt((1+x)(1+x^3))dx`cannot exceed `sqrt((15)/8)`.

Answer» `int_(0)^(1)sqrt((1+x)(1+x^(3)))dx le sqrt((int_(0)^(1)(1+x)dx)(int_(0)^(1)(1+x^(3))dx))`
`=sqrt((x+(x^(2))/2)_(0)^(1)(x+(x^(4))/4)_(0)^(1))`
`=sqrt((3/2)(5/4))`
`=sqrt(15/8)`
Discussion
38.

Evaluate : `int_(0)^(a)x^(5//2)sqrt(a-x)dx`

Answer» `{:("Put" x = a sin^(2)theta,rArr,dx=2asinthetacostheta d theta),("Lower limit" : x = 0, rArr, theta = 0),("Upper limit" x = a , rArr, theta = pi/2):}`
`underset(0)overset(pi)intx^(5//2)sqrt(a-x)dx=underset(0)overset(pi/2)int2a^(4)sin^(6)theta d theta = 2a^(4) xx (pi)/(2).((5.31)(1))/(8.64.2) = (5pia^(4))/(128)`
Discussion
39.

Estimate the absolute value of the integral `int_(10)^(19)(sinx)/(1+x^8)dx`

Answer» Since `|sinx|le1,` then
`|(sinx)/(1+x^(8))|le1/(|1+x^(8)|)`………………1
But `10lexle19`. So `1+x^(8)gtx^(8)ge10^(8)`
or `1/(1+x^(8))lt1/(x^(8))le1/(10^(8))`
or `1/(|1+x^(8)|)le1/(10^(8))`…………..2
From 1 and 2 we get `|(sinx)/(1+x^(8))|le10^(-8)`
Then `int_(10)^(19)(sinx)/(1+x^(8))dx le(19-10)xx10^(-8)`
`=9xx10^(-8)=(10-1)xx10^(8)`
`=10^(-7)-10^(-8)lt10^(-7)`
Hence `|int_(10)^(19)(sinxdx)/(1+x^(8))|lt10^(-7)`.
Discussion
40.

`int_(0)^(pi//4)(2sec^(2)x+x^(3)+2)dx`

Answer» Correct Answer - `(pi^(4))/(1024) + (pi)/2+2`
Discussion
41.

`int_(0)^(pi/3)(x)/(1+secx)dx`

Answer» Correct Answer - `(pi)/(18) - (pi)/(3sqrt(3))+2ln(2/sqrt(3))`
Discussion
42.

Examples: Find the area of the region bounded by the curve `y^2 = 2y - x` and the y-axis.

Answer» Correct Answer - `4//3` sq units
Discussion
43.

Evaluate `Lt_(n to oo)[(1)/(1+n)+(1)/(2+n)+(1)/(3+n)+"...."+(1)/(10n)] `

Answer» `underset(nrarr0)(Lt)[(1)/(1+n)+(1)/(2+n)+(1)/(3+n)+"......"+(1)/(10n)] = underset(nrarroo)("Lt")underset(r=1)overset(9n)sum(1)/(r+n)`
`= underset(nrarroo)(Lt)underset(r=1)overset(9n)sum(1)/(n) (1)/((r/n)+1) = underset(0)overset(9)int(dx)/(x+1)=[ln(x+1)]_(0)^(9)=ln10`
Discussion
44.

Find the area enclosed betweent the curve `y = x^(2)+3, y = 0, x = - 1, x = 2`.

Answer» Correct Answer - `51/4` sq. unit
Discussion
45.

Evaluate: `lim_(nrarr0) (((2n)!)/(n!n^(n)))^(1/n)`

Answer» Let `y= underset(n rarroo)("lim") (((2n)!)/(n!n^(n)))^(1/n)rArr lny=underset(nrarroo)(lim)1/nln(((2n)!)/(n!n^(n)))`
`= underset(nrarr0)(lim)1/nln((2n(n-1)(2n-2)"......"(n+1))/(n^(n)))`
`= underset(nrarroo)(lim)underset(r=1)overset(n)sum(1)/(n)[ln(1+r//n)]=underset(0)overset(1)intln(1+x)dx = (xln(1+x))_(0)^(1)-underset(0)overset(1)int(x)/(1+x)dx`
`= (xln(1+x))_(0)^(1)-(x-ln(1+x))_(0)^(1)=ln2-(1-ln2)=ln4//erArr y = 4//e`
Discussion
46.

Draw a graph of the function `f(x) = cos^(-1) (4x^(3)-3x), x in [-1,1]` and find the ara enclosed between the graph of the function and the x-axis varies from 0 to 1.

Answer» Correct Answer - `3(sqrt(3)-1)` sq. units
Discussion
47.

Find the area of the region which is inside the parabola `y = - x^(2) + 6x - 5`, out the side the parabola `y = - x^(2) + 4x - 3` and left of the stragiht line `y = 3x-15`.

Answer» Correct Answer - `73/6`
Discussion
48.

Find the area bounded by the curves `x = y^(2)` and `x = 3-2y^(2)`.

Answer» Correct Answer - `4` sq. units
Discussion
49.

Find the area bounded by the curve `y = x^(2) - 2x + 5`, the tangent to it at the point `(2,5)` and the axes of the coordinates.

Answer» Correct Answer - `8//3` sq. units
Discussion
50.

Let ABC be a triangle with vertices `A -=(6,,2sqrt3+1))),B-=(4,2)and C-=(8,2)`. Let R be the region consisting of all those points P inside `DeltaABC` which satisfyd`(P, BC) >= max{d(P,AB);d(P,AC)}`, where d(P, L) denotes the distance of the point from the line L, then

Answer» Correct Answer - `(4sqrt(3))/(3)`
Discussion