MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 10

Find the area enclosed between the curves

$y^2- 2ye^{sin^{- 1}x} + x^2- 1 +[x] +e^{2sin^{- 1}x} = 0$

and line x = 0 and

$x=\frac{1}{2}$ is (where [.] denotes greatest integer function)

0%
$\dfrac{\sqrt{3}}{4}+\dfrac{\pi }{6}$
0%
$\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6}$
0%
$\dfrac{\sqrt{3}}{4}-\dfrac{\pi }{6}$
0%
$\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}$

Explanation

$\left (y-e^{sin^-1 x} \right )+x^2+[x]-1=0$

$\therefore x\in \left ( 0, 1/2 \right )$ ;[x] =0

$\Rightarrow y-e^{sin^- 1x} = \pm \sqrt{1- x^2}$

$\Rightarrow y=e^{sin^- 1x}\pm \sqrt{1- x^2}$

$\Rightarrow \overset {\frac{1}{2}}{\underset { 0 }{ \int } } y dx =\overset {\frac{1}{2}}{\underset { 0 }{ \int } } \left |e^{sin^- 1x}+ \sqrt{1+ x^2}- \left (e^{sin^- 1x}- \sqrt{1- x^2} \right ) \right |dx$

$\Rightarrow A = 2\overset {\frac{1}{2}}{\underset { 0 }{ \int } } \sqrt{1- x^2}dx$ ; Let

$x =sin\, \theta$ so

$dx =cos\, \theta \,d\, \theta$

$\Rightarrow A = \overset {\frac{\pi }{6}}{\underset { 0 }{ \int } } 2\,cos^2\,\theta\,d\, \theta=\left [ \theta +\dfrac{sin\, 2\, \theta }{2} \right ]_0^{\pi /6}$

$\Rightarrow A ={\pi /6}+\dfrac{\sqrt{3}}{4}$

$A_1$ is the area bounded by

$y= \cos x, y = \sin x$ &

$x=0$ and

$A_2$ the area bounded by

$y = \cos x , y = \sin x , y = 0$ in

$(0,\frac{\pi}{2})$ then

$\displaystyle \dfrac{A_1}{A_2}$ equals to :

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{\sqrt2}$
0%
$1$
0%
None of these

Explanation

$\displaystyle A_{1}=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x)dx =(\sin x +\cos x)_{0}^{\frac{\pi}{4}}$

$\displaystyle =\left ( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} -1\right ) =(\sqrt2-1)$

$\displaystyle A_{2}=\int^{0}_{\frac{\pi}{4}}\sin x dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x dx =\left [ -\cos x \right ]_{0}^{\frac{\pi}{4}}+(\sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

$\displaystyle =\left ( 1-\frac{1}{\sqrt{2}} \right )+\left ( 1-\frac{1}{\sqrt{2}} \right )=\left ( 2-\sqrt{2} \right )=\sqrt{2}(\sqrt{2}-1)$

Then,

$A_{1}:A_{2}=1:\sqrt{2}=\dfrac{1}{\sqrt{2}}$

The area bounded by the curves

$y = sin^{-1} |sin x|$ and

$y = (sin^{-1} | sin x|)^{2},$ where

$0\leq x\leq 2\pi ,$ is:

0%
$\dfrac{1}{3}+\dfrac{\pi ^{2}}{4}$ sq. units
0%
$\dfrac{1}{6}+\dfrac{\pi ^{3}}{8}$ sq. units
0%
$2$ sq. units
0%
$\displaystyle \frac{4}{3}+\pi ^{2}\frac{\pi -3}{6} sq.units$

Explanation

$y=sin^{-1}\left |sin x \right | =\begin{cases} x & 0\leq x< \pi /2 \\ \pi-x & \pi /2 \leq x< \pi \\ x-\pi & \pi \leq x< 3\pi /2 \\ 2\pi-x & 3\pi /2 \leq x< 2\pi \ \end{cases}$

$y=\left (sin^{-1}\left |sin x \right | \right )^{2}=\begin{cases} x^{2} & 0\leq x< \pi /2 \\ \left (\pi-x \right )^{2} & \pi /2 \leq x< \pi \\ \left (x-\pi \right )^{2} & \pi \leq x< 3\pi /2 \\ \left (2\pi-x \right )^{2} & 3\pi /2 \leq x< 2\pi \ \end{cases}$

Therefore required area

$\displaystyle=4\int_{0}^{1}\left (x-x^{2} \right )^{2}\mathrm{d} x+\int_{1}^{\frac{\pi }{2}}\left (x^{2}-x \right )\mathrm{d} x$

$\displaystyle =\frac{4}{3}+\pi ^{2}\frac{\pi -3}{6} sq.units$

$\left| z- (4 + 4i) \right| \geq 4$ , then area of the region bounded by the locii of

$z,\; iz,\; - z$ and

$-iz$ is:

0%
$4(4-\pi )$
0%
$16(4-\pi )$
0%
$16(\pi -1)$
0%
$4(\pi -1)$

Explanation

Locus of $|z-(4+4i)|=4$ is a circle with center at $(4,4)$ and radius $4$ is Complex plane.

Hence, locus of $|z-(4+4i)| \geq 4$ is all points either on or outside the circle with radius $4$ and center $(4,4)$ .

Similarly, locus of $|-z-(4+4i)| \geq 4$ is all points on or outside the circle with radius $4$ and center $(-4,-4)$ .

Locus of $|iz-(4+4i)| \geq 4$ is all points on or outside the circle with radius $4$ and center $(-4,4)$ .

Finally, locus of $|-iz-(4+4i)| \geq 4$ is all points on or outside the circle with radius $4$ and center $(4,-4)$ .

Hence, the area bounded by locus of all four will be the area enclosed by the four circles in argand plane as shown in the figure.
Area bounded $=$ area of shaded region
$= 64 - \pi r^{2}$
$=64-16\pi =16(4-\pi )$

If the area bounded by the curve

$y = f(x)$ , the coordinate axes and the line

$x = x_1$ is given by

$x_1e^{x_1}$ . Then

$f(x)$ equals

0%
$e^x$
0%
$xe^x$
0%
$xe^x-e^x$
0%
$xe^x+e_x$

If the area bounded by the curve

$|y|=sin^{-1}|x|$ and

$x=1$ is

$a(\pi+b)$ , then the value

$a-b$ is:

0%
1
0%
2
0%
3
0%
4

Explanation

From the graph, the area bounded is

$Area = \displaystyle 2\int_0^1 sin^{-1}x dx =\left[2x sin^{-1}x \right]_0^1+\int_0^1\frac{-2x}{\sqrt{1-x^2}}dx$

$=\pi + \left[ 2\sqrt{1-x^2}\right]_0^1=\pi-2$

$\therefore a=1$ and

$b=-2$

$\therefore a-b=3$

The parabola

$y^{2} = 4x$ and

$x^{2} = 4y$ divide the square region bounded by the lines

$x = 4, y = 4$ and the coordinate axes. If

$S_{1}, S_{2}, S_{3}$ are the areas of these parts numbered from top to bottom respectively, then

0%
$S_{1} : S_{2}\equiv 1 : 1$
0%
$S_{2} : S_{3}\equiv 1 : 2$
0%
$S_{1} : S_{3}\equiv 1 : 1$
0%
$S_{1} : (S_{1} + S_{2})\equiv 1 : 2$

Explanation

Area of region bounded by

${y}^{2}=4ax$ and

${y}^{2}=4by$ is

$\displaystyle\frac{16ab}{3}$ sq. units.

Now

$\displaystyle{ S }_{ 1 }={ S }_{ 3 }=\int _{ 0 }^{ 4 }{ \frac { { x }^{ 2 } }{ 4 } dx } =\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }=\frac { 1 }{ 12 } \times 64=\frac { 16 }{ 3 }$ sq. unit

$\displaystyle\therefore { S }_{ 2 }+{ S }_{ 3 }=\int _{ 0 }^{ 4 }{ \sqrt { 4x } dx } =2{ \left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] }_{ 0 }^{ 4 }=\frac { 4 }{ 3 } \times 8=\frac { 32 }{ 3 }$ sq. unit

$\displaystyle\Rightarrow{ S }_{ 2 }=\frac { 16 }{ 3 }$ sq. unit

$\displaystyle\therefore { S }_{ 1 }:{ S }_{ 2 }:{ S }_{ 3 }=\frac { 16 }{ 3 } :\frac { 16 }{ 3 } :\frac { 16 }{ 3 } =1:1:1$

Area of the region bounded by the curve

$y = x^{2}$ and

$y = sec^{-1} [sin^{2}x]$ (where [ . ] denotes the greatest integer function) is

0%
$\frac{\pi}{3}\sqrt{\pi}$
0%
$\frac{2\pi\sqrt{\pi}}{3}$
0%
$\frac{4\pi\sqrt{\pi}}{3}$
0%
$\frac{6\pi\sqrt{\pi}}{3}$
0%
$\frac{3\pi\sqrt{\pi}}{2}$

Explanation

$y=\sec^{-1}[-\sin^{2}x]\Rightarrow \sec^{-1}(-1)$

$\Rightarrow y=\pi, x \neq n\pi$

$A=2\int_{0}^{\sqrt{\pi}}(\pi - x^{2})dx=\frac{4}{3}\pi\sqrt{\pi}$

The area bounded by

$y=\sec^ {-1}{x}, y= \text{cosec}^{-1}{x}$ and the line

$x-1=0$ is:

0%
$\ln(3+2\sqrt{2})-\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{2}+\ln(3+2\sqrt{2})$
0%
$\pi - \ln3$
0%
$\pi + \ln3$

Explanation

$\displaystyle A = \displaystyle {\int}_0^{\pi /4} \sec y dy + {\int}_{\pi /4}^{\pi /2} \text{cosec} y\ dy - 1\times \frac{\pi}{2}$

$\displaystyle=[\ln(\sec y + \tan y)]^{\pi /4}_0 + \left[\ln\tan\frac{y}{2}\right]^{\pi /2}_{\pi /4}-\frac{\pi}{2}$

$\displaystyle=[\ln(\sqrt{2}+1)-\ln1] + \ln \tan\frac{\pi}{4}- \ln\tan \frac{\pi}{8}-\frac{\pi}{2}$

$\displaystyle=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-\frac{\pi}{2}$

$= \ln\left(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)-\dfrac{\pi}{2}$

$\displaystyle=\ln(3+2\sqrt{2})-\frac{\pi}{2}$

Hence, option A is the correct answer.

The area enclosed by

$x^2 + y^2 = 4, y = x^2 + x + 1,$

$y=\left [ \sin^{2}\displaystyle \frac{x}{4}+\cos\displaystyle \frac{x}{4} \right ]$ and

$x$ -axis (where

$[.]$ denotes the greatest integer function) is:

0%
$\displaystyle \frac{2\pi}{3}+\sqrt{3}-\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{2\pi}{3}+2\sqrt{3}-\displaystyle \frac{1}{6}$
0%
$2\sqrt3-\frac{1}{3}$
0%
$\displaystyle \frac{\pi}{3}+\sqrt3$

Explanation

$\displaystyle \sin^{ 2 }\frac { x }{ 4 } +\cos\frac { x }{ 4 } =1-\cos ^{ 2 }{ \frac { x }{ 4 } + } \cos\frac { x }{ 4 }$

$\displaystyle =1-(\cos ^{ 2 }{ \frac { x }{ 4 } } - \cos\frac { x }{ 4 } )$

$\displaystyle=\frac { 5 }{ 4 } -\left( \cos\frac { x }{ 4 } -\frac { 1 }{ 2 } \right) ^{ { 2 } }$

Now, for

$-2\le x\le 2$

$\Rightarrow \displaystyle -\frac{1}{2}\le \frac{x}{4}\le \frac{1}{2}$

$\Rightarrow \cos^{2}\displaystyle \frac{x}{4}\leq \cos\displaystyle \frac{x}{4}$ for all

$x\in \left [-2,2 \right ]$

$\Rightarrow \cos^{ 2 }\dfrac { x }{ 4 } +\sin ^{ 2 }{ \dfrac { x }{ 4 } } \le \sin ^{ 2 }{ \dfrac { x }{ 4 } } +\cos\dfrac { x }{ 4 }, \ \forall \ x\in \left[ -2,2 \right]$

$\Rightarrow \left [\sin^{2}\displaystyle \frac{x}{4}+\cos\displaystyle \frac{x}{4} \right ]=1\ \forall\ \ x\in \left [-2,2\right ]$

Now from the figure, to find the enclosed area, we have to find point of intersections of

$y=1$ and the circle.

Solving both equations we get

$x=- \sqrt{3}$ and

$x = \sqrt{3}$

Also we have to find the points of intersection of

$y=1$ and

$y=x^2+x+1$ .

Solving both equations we get

$x=-1$ and

$x=0$

Now dividing the enclosed area into different intervals as per the points of intersections, we get required area

$=\displaystyle \int_{-2}^{-\sqrt{3}}\sqrt{4-x^{2}}\mathrm{d} x+\int_{-\sqrt{3}}^{-1} \mathrm {d} x+\int_{-1}^{0}(x^{2}+x+1)\mathrm{d} x+\int_{0}^{\sqrt{3}}\mathrm{d} x+\int_{\sqrt{3}}^{2}\sqrt{4-x^{2}}\mathrm{d} x$

$=\displaystyle \frac{2\pi }{3}+\sqrt{3}-\displaystyle \frac{1}{6}$

The area bounded by the function

$f(x)=x^{2}:R^{+}\rightarrow R^{+}$ and its inverse function is:

0%
$\dfrac{1}{2}$ sq.units
0%
$\dfrac{1}{3}$ sq.units
0%
$\dfrac{2}{3}$ sq.units
0%
$\dfrac{1}{6}$ sq.units

Explanation

Inverse of

$y = x^2$ is

$y = \sqrt x$ and required area

$= 2 \int_{o}^{1} (x - x^2) dx$

$\displaystyle = 2[{\frac {x^2}{2} - \frac {x^3}{3}}]_0^1 = 2[\frac {1}{2} - \frac {1}{3}] = \frac {1}{3}$ sq. units

Find the area of the region bounded by the curves

$y= log_{e}x$ ,

$y=\sin ^{4} \pi x$ ,

$x=0$

0%
$\displaystyle\ \frac{11}{8}sq.units$
0%
$\displaystyle\ \frac{9}{8}sq.units$
0%
$\displaystyle\ \frac{13}{8}sq.units$
0%
$\displaystyle\ \frac{15}{8}sq.units$

Explanation

$sin^{4}(\pi(x))$

$=(sin^{2}\pi(x))^{2}$

$=(\dfrac{1-cos2\pi(x)}{2})^{2}$

$=\dfrac{1}{4}[1+cos^{2}2\pi(x)-2cos2\pi(x)]$

$=\dfrac{1}{4}[1+\dfrac{1+cos4\pi(x)}{2}-2cos2\pi(x)]$

$=\dfrac{1}{4}[\dfrac{3+cos4\pi(x)}{2}-2cos2\pi(x)]$
Hence the required area

$=\int_{0} ^{1} \dfrac{1}{4}[\dfrac{3+cos4\pi(x)}{2}-2cos2\pi(x)]-ln(x).dx$

$=[\dfrac{3x}{8}+\dfrac{sin4\pi.x}{32\pi}-\dfrac{sin2\pi(x)}{4\pi}-x(ln(x)-1)]_{0} ^{1}$

$=\dfrac{3}{8}-(-1)$

$=\dfrac{11}{8}$ sq units.

State the following statement is True or False

The area bounded by the circle

$x^2 + y^2 = 1, x^2 + y^2 = 4$ and the pair of lines

$\sqrt{3} (x^2 + y^2) = 4xy$ , is equal to

$\dfrac{\pi}{2}$ . The statement is true or false.

0%
True
0%
False

Explanation

The angle

$\theta$ between the lines represented by

$\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$ is given by

$\theta = \displaystyle tan^{-1} \frac{\sqrt{h^2 - ab}}{|a+b|}$

$= tan^{-1}\displaystyle \frac{2\sqrt{2^2 - 3}}{\sqrt{3} + \sqrt{3}} = \frac{1}{\sqrt{3}}$
Gives

$\displaystyle \theta = \frac{\pi}{6}$
Hence, the shaded area

$=\displaystyle \frac{\pi/6}{2 \pi} \times \pi (2^2 - 1^2) = \frac{\pi}{4}$

Find the area bounded by the curves

$\displaystyle\ y=\sqrt{1-x^{2}}$ and

$\displaystyle\ y=x^{3}-x$ . Also find the ratio in which the y-axis divide this area

0%
$\displaystyle\ \frac{\pi }{2}$ , $\displaystyle\ \frac{\pi -1}{\pi +1}$
0%
$\displaystyle\ \frac{\pi }{4}$ , $\displaystyle\ \frac{\pi -1}{\pi +1}$
0%
$\displaystyle\ \frac{\pi }{2}$ , $\displaystyle\ \frac{\pi +1}{\pi -1}$
0%
None of these

Explanation

The two curves are

$y=\sqrt { 1-{ x }^{ 2 } }$ ....(1)

and

$y={x}^{3}-x$ ...(2)

The point of intersection are

$P\left(-1,0\right);Q\left(1,0\right)$

Consider

$y=\sqrt { 1-{ x }^{ 2 } }$

On squaring both sides, we get

${x}^{2}+{y}^{2}=1$

But

$y=\sqrt { 1-{ x }^{ 2 } } \ge0$ by the definition of square root which is a semi-circle with center

$\left(0,0\right)$ and radius

$1$ and above X-axis.

Consider

$y={ x }^{ 3 }-x=x\left( x-1 \right) \left( x+1 \right)$

Now for

$x\le-1,0\le x\le1;y\le0$

and for

$-1\le x\le0,x\ge1;y\ge0$

Taking into account the oddness of the function and the intervals of constant sign.

(We can construct its graph by finding the maxima and minima at

$\displaystyle x=\pm\frac{1}{\sqrt{3}}$

Thus the required Area

$={A}_{1}+{A}_{2}$

where

$\displaystyle { A }_{ 1 }=\int _{ -1 }^{ 0 }{ \left[ \sqrt { 1-{ x }^{ 2 } } -{ x }^{ 3 }+x \right] dx } =\frac { \pi }{ 4 } -\frac { 1 }{ 4 }$

and

$\displaystyle { A }_{ 2 }=\int _{ 0 }^{ 1 }{ \left[ \sqrt { 1-{ x }^{ 2 } } -{ x }^{ 3 }+x \right] dx } =\frac { \pi }{ 4 } +\frac { 1 }{ 4 }$

$\Rightarrow$ Required area

$\displaystyle=\frac{\pi}{2}$ and required ratio

$\displaystyle =\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \pi -1 }{ \pi +1 }$

Find the area of the region enclosed between the two circles

$\displaystyle\ x^{2}+y^{2}=1$ &

$(x-1)^{2}+y^{2}=1$

0%
$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{2}$ sq.units
0%
$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{2}$ sq.units
0%
$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{4}$ sq.units
0%
$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{4}$ sq.units

Explanation

The circles intersect at

$x=0.5$
Hence the required area will be

$=\displaystyle 2\int_{\frac{1}{2}} ^{1} \sqrt{1-x^{2}}-\sqrt{1-(x-1)^{2}}.dx$

$=2\times\dfrac{1}{2}\left[\sin^{-1}(x)+x\sqrt{1-x^{2}}-\sin^{-1}(x-1)-(x-1)\sqrt{1-(x-1)^{2}}\right]_{\frac{1}{2}} ^{1}$

$=\left[\sin^{-1}(x)+x\sqrt{1-x^{2}}-\sin^{-1}(x-1)-(x-1)\sqrt{1-(x-1)^{2}}\right]_{\frac{1}{2}} ^{1}$

$=\dfrac{\pi}{2}-\left(\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{4}+\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{4}\right)$

$=\dfrac{\pi}{2}-\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{2}$

$=\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{2}$ sq units.

A polynomial function f(x) satisfies the condition

$f(x+1) =f(x)+2x+1$ . Find

$f(x)$ if

$f(0)=1$ . Find also the equations of the pair of tangents from the origin on the curve

$y=f(x)$ and compute the area enclosed by the curve and the pair of tangents.

0%
$f(x)=x^2+1;$ , $y=\pm 2x;$ , $\displaystyle A=\frac{2}{3}$ sq.units
0%
$f(x)=x^2-1;$ , $y=\pm 2x;$ , $\displaystyle A=\frac{2}{3}$ sq.units
0%
$f(x)=x^2+1;$ , $y=\pm 2x;$ , $\displaystyle A=\frac{3}{2}$ sq.units
0%
$f(x)=x^2-1;$ , $y=\pm 2x;$ , $\displaystyle A=\frac{3}{2}$ sq.units

Explanation

Let

$f(x)=x^{2}+1$
Hence

$f(0)=1$ and

$f(x+1)$

$=(x+1)^{2}+1$

$=x^{2}+2x+2$

$=(x^{2}+1)+2x+1$

$=f(x)+2x+1$ .
Therefore

$f(x)=x^{2}+1$ satisfies all the conditions required.
Let the tangents be

$y=mx$
Hence

$f'(x)_{x=h}=m$
Hence

$2x=m$

$x=\dfrac{m}{2}$ ...(i)
Now

$mx=x^{2}+1$

$x^{2}-mx+1=0$

$x=\dfrac{m\pm\sqrt{m^{2}-4}}{2}$

$=\dfrac{m}{2}$ ... from i.
Hence

$\sqrt{m^{2}-4}=0$
Or

$m=\pm2$ .
Hence the equations of the tangents are

$y=\pm2x$
The required area will be

$=|2\int_{0}^{1} x^{2}+1-2x.dx|$ or

$|2\int _{0}^{-1} x^{2}+1+2x.dx|$
Hence

$=|2\int_{0}^{1} x^{2}+1-2x.dx|$

$=2\int_{0} ^{1}(x-1)^{2}.dx$

$=2.|[\dfrac{(x-1)^{3}}{3}]|_{0} ^{1}$

$=\dfrac{2}{3}$ sq units.

Find the area enclosed the curves :

$y=ex\log { x }$ and

$\displaystyle y=\frac { \log { x } }{ ex }$ where

$\log { e } =1$

0%
$\displaystyle\frac { { e }^{ 2 }-5 }{ 4e }$
0%
$\displaystyle\frac { { e }^{ 2 }+5 }{ 4e }$
0%
$\displaystyle\frac { { e }^{ 2 }-3 }{ 2e }$
0%
$\displaystyle\frac { { e }^{ 2 }+3 }{ 2e }$

Explanation

Both the curves are defined for

$x>0$
Both are positive when

$x>1$ and negative when

$0<x<1$
We know

$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x } \right) \rightarrow -\infty }$
Hence,

$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x } \right) \rightarrow -\infty }$ .Thus y-axis is asymptote of second curve.
And

$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ ex\log { x } } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \left( 0 \right) \times \infty form \right]$

$\displaystyle =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\log { x } }{ \frac { 1 }{ x } } } \left( -\frac { \infty }{ \infty } form \right)$

$\displaystyle =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\left( \frac { 1 }{ x } \right) }{ \left( -\frac { 1 }{ { x }^{ 2 } } \right) } } =0$ (using L'hopital rule)
Thus, the first curve starts from

$(0,0)$ but does not include

$(0,0)$
Now, the given curves intersect, therefore

$\displaystyle ex\log { x } =\frac { \log { x } }{ ex } \Rightarrow \left( { e }^{ 2 }{ x }^{ 2 }-1 \right) \log { x } =0$

$\displaystyle \Rightarrow x=1,\frac { 1 }{ e } \quad \quad \quad \quad \left( since\quad x>0 \right)$
Therefore required area

$\displaystyle =\int _{ \frac { 1 }{ e } }^{ 1 }{ \left( \frac { \left( \log { x } \right) }{ ex } -ex\log { x } \right) } dx$

$\displaystyle=\frac { 1 }{ e } { \left[ \frac { { \left( \log { x } \right) }^{ 2 } }{ 2 } \right] }_{ \frac { 1 }{ e } }^{ 1 }-e{ \left[ \frac { { x }^{ 2 } }{ 4 } \left( 2\log { x } -1 \right) \right] }_{ \frac { 1 }{ e } }^{ 1 }$

$\displaystyle =\frac { { e }^{ 2 }-5 }{ 4e }$

The area included between the curve

${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ and

$\displaystyle \sqrt { \left| x \right| } +\sqrt { \left| y \right| } =\sqrt { a } \left( a>0 \right)$ is:

0%
$\displaystyle \left( \pi +\frac { 2 }{ 3 } \right) { a }^{ 2 }$
0%
$\displaystyle \left( \pi -\frac { 2 }{ 3 } \right) { a }^{ 2 }$
0%
$\displaystyle \frac { 2 }{ 3 } { a }^{ 2 }$
0%
$\displaystyle \frac { 2\pi }{ 3 } { a }^{ 2 }$

Explanation

The graphs

$\displaystyle \left| x \right| +\left| y \right| =a$ and

$\displaystyle { \left| x \right| }^{ 2 }+{ \left| y \right| }^{ 2 }={ a }^{ 2 }$ are as shown in the figure

From the figure, it is clear that when powers of

$|x|$ and

$|y|$ both are reduced to half the straight lines get stretched inside.

Thus, the required area

$=4$

$[$ shaded area in the first quadrant

$]$

$\displaystyle =4\left[ \frac { \pi { a }^{ 2 } }{ 4 } -\int _{ 0 }^{ a }{ \left( \sqrt { a } -\sqrt { x } \right) ^{ 2 }dx } \right] =\left( \pi -\frac { 2 }{ 3 } \right) { a }^{ 2 }$

$[$ In the first quadrant

$x,y>0,$ therefore,

$\displaystyle \sqrt { \left| x \right| } +\sqrt { \left| y \right| } =\sqrt { a } \Rightarrow \sqrt { x } +\sqrt { y } =\sqrt { a } \Rightarrow y={ \left( \sqrt { a } -\sqrt { x } \right) }^{ 2 }]$

Sketch the region bounded by the curves

$\displaystyle y=x^{2}$ &

$\displaystyle\ y= 2/(1+x^{2})$ . Find the area:

0%
$\displaystyle\ \pi -\frac{2}{3}$
0%
$\displaystyle\ \pi -\frac{1}{3}$
0%
$\displaystyle\ \pi -\frac{5}{3}$
0%
$\displaystyle\ \pi -\frac{7}{3}$

Explanation

$y=x^{2}$ and

$y=\dfrac{2}{1+x^{2}}$
Therefore

$x^{2}=\dfrac{2}{1+x^{2}}$

$x^{4}+x^{2}-2=0$

$(x^{2}+2)(x^{2}-1)=0$
Now x is real.
Therefore

$x^{2}+2=0$ is not possible.
Hence

$x=\pm1$ .
Hence the required area is

$=\int_{-1} ^{1} x^{2}-\dfrac{2}{1+x^{2}}$

$=2\int _{0} ^{1} x^{2}-\dfrac{2}{1+x^{2}}$

$=2[\dfrac{x^{3}}{3}-2tan^{-1}(x)]_{0} ^{1}$

$=2[\dfrac{1}{3}-2.\dfrac{\pi}{4}]$

$=|\dfrac{2}{3}-\pi|$

$=\pi-\dfrac{2}{3}$ sq.units.

The ratio in which the area bounded by the curves

$y^{2}=4x$ and

$x^{2}=4y$ is divided by the line

$x=1$ is

0%
$64 : 49$
0%
$15 : 34$
0%
$15 : 49$
0%
none of these

Explanation

For the point of intersection.

$\dfrac{x^{4}}{16}=4x$

$x=0$ and

$x=4$ ...(i)
Now the area bounded by the curve between x=0 and x=1

$=\int_{0} ^{1} 2\sqrt{x}-\dfrac{x^{2}}{4}$

$=[\dfrac{4x\sqrt{x}}{3}-\dfrac{x^{3}}{12}]_{0}^{1}$

$=\dfrac{4}{3}-\dfrac{1}{12}$

$=\dfrac{15}{12}$ ...(i)

The area bounded by the curve between x=1 and x=4 is

$=\int_{1} ^{4}2\sqrt{x}-\dfrac{x^{2}}{4}$

$=[\dfrac{4x\sqrt{x}}{3}-\dfrac{x^{3}}{12}]_{1}^{4}$

$=\dfrac{32}{3}-\dfrac{64}{12}-\dfrac{15}{12}$

$=\dfrac{128-64-15}{12}$

$=\dfrac{128-79}{12}$

$=\dfrac{49}{12}$ ...(ii)
Hence the required ratio

$=\dfrac{i}{ii}$

$=\dfrac{15}{49}$ .

If f(x) be an increasing function defined on [a, b] then
max {f(t) such that

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x) & min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a),
min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x).
On the basis of above information answer the following questions.

$\int_{0}^{\pi }max\left \{ \sin x, \cos x \right \}dx$ is equal to

0%
$\sqrt{2}-1$
0%
$\displaystyle 1-\frac{1}{\sqrt{2}}$
0%
$\displaystyle 1+\frac{1}{\sqrt{2}}$
0%
None of these

Explanation

First plot the graph of

$y=\sin x$ ,

$y=cos x$ by a dotted curve & find their point of intersections.
Now we find any two consecutive points of intersections. In between these points either

$\sin x> \cos x$ or

$\cos x> \sin x$ then in order to get max

$\left ( \sin x, \cos x \right )$ we take those segments for which one function is greater than the other function.
In the figure, A is the point of intersection of

$\sin x$ and

$\cos x$ .

$\therefore$

$\displaystyle A\left ( \frac{\pi }{4}, \frac{1}{\sqrt{2}} \right )$

$\therefore$ max

$\left ( \sin x, \cos x \right )$

$\forall 0\leq x\leq \pi$

$\therefore$

$\displaystyle =\begin{cases} \cos x & \forall 0\leq x\leq \dfrac{\pi }{4} \\ \sin x & \forall \dfrac{\pi }{4}\leq x\leq \pi \end{cases}$

$\therefore$

$\int_{0}^{\pi }max \left ( \sin x\cos x \right )dx=\int_{0}^{\pi /4}\cos xdx+\int_{\pi /4}^{\pi }\sin xdx =\sqrt{2}+1$

The ratio of the area's bounded by the curves

$\displaystyle y^{2}=12x$ and

$\displaystyle x^{2}=12y$ is divided by the line

$\displaystyle x=3$ is

0%
15 : 49
0%
9 : 15
0%
7 : 15
0%
7 : 5

Explanation

For the point of intersection

$2\sqrt{3x}-\dfrac{x^{2}}{12}=0$
Or

$\sqrt{x}[24\sqrt{3}-x\sqrt{x}]=0$

$x=0$ and

$x=12$ .
Hence

$\int _{0} ^{3} (2\sqrt{3x}-\dfrac{x^{2}}{12}).dx$

$=[\dfrac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\dfrac{x^{3}}{36}]_{0} ^{3}$

$=\dfrac{45}{4}$ ...(i)

And

$\int _{3} ^{12} (2\sqrt{3x}-\dfrac{x^{2}}{12}).dx$

$=[\dfrac{4\sqrt{3}}{3}.x^{\frac{3}{2}}-\dfrac{x^{3}}{36}]_{3} ^{12}$

$=\dfrac{147}{4}$ ...(ii)
Hence
Ratio:

$\dfrac{i}{ii}$

$=\dfrac{45}{147}$

$=\dfrac{15}{49}$ .

The function

$\displaystyle f\left ( x \right )=\max \left \{ x^{2},\left ( 1-x \right )^{2},2x\left ( 1-x \right ) \forall 0\leq x\leq 1\right \}$ then area of the region bounded by the curve

$\displaystyle y=f\left ( x \right )$ , x-axis and

$\displaystyle x=0,x=1$ is equals,

0%
$\displaystyle \frac{27}{17}$
0%
$\displaystyle \frac{9}{17}$
0%
$\displaystyle \frac{18}{17}$
0%
None of these

Explanation

Solving

$y={ x }^{ 2 }$ and

$y=2x\left( 1-x \right)$ we get

$\displaystyle x=0,\frac { 2 }{ 3 }$

And solving

$y=1-{ x }^{ 2 }$ and

$y=2x\left( 1-x \right)$ , we get

$\displaystyle x=1,\frac { 1 }{ 3 }$

Therefore
The required area

$\displaystyle A=\int _{ 0 }^{ 1 }{ f\left( x \right) dx } =\int _{ 0 }^{ \frac { 1 }{ 3 } }{ { \left( 1-x \right) }^{ 2 } } dx+\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } }{ 2x } \left( 1-x \right) dx\quad +\int _{ \frac { 2 }{ 3 } }^{ 1 }{ { x }^{ 2 }dx }$

$\displaystyle ={ \left( -\frac { 1 }{ 3 } { \left( 1-x \right) }^{ 3 } \right) }_{ 0 }^{ \frac { 1 }{ 3 } }+{ \left( { x }^{ 2 }-\frac { 2{ x }^{ 3 } }{ 3 } \right) }_{ \frac { 2 }{ 3 } }^{ \frac { 2 }{ 3 } }+{ \left( \frac { { x }^{ 3 } }{ 3 } \right) }_{ \frac { 2 }{ 3 } }^{ 1 }$

$\displaystyle =\frac { 19 }{ 81 } +\frac { 13 }{ 81 } +\frac { 19 }{ 81 } =\frac { 17 }{ 27 }$

Compute the area of the curvilinear triangle bounded by the y-axis & the curve,

$\displaystyle\ y=\tan x$ &

$\displaystyle\ y=(2/3) \cos x$

0%
$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$
0%
$\displaystyle\ \frac{1}{3}- ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$
0%
$\displaystyle\ \frac{2}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$
0%
$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{1}}{2} \right ]sq.units$

Explanation

For the point of intersection

$tan(x)=\dfrac{2}{3}cos(x)$
Or

$3tan(x)=2cos(x)$

$3sin(x)=2cos^{2}(x)$
Or

$3sin(x)=2-2sin^{2}(x)$
Or

$2sin^{2}(x)+3sin(x)-2=0$
Hence

$sin(x)=\dfrac{-3\pm\sqrt{9+16}}{4}$

$=\dfrac{-3\pm5}{4}$
Hence

$sin(x)=-2$ ...(not possible) and

$sin(x)=\dfrac{1}{2}$
Hence

$x=\dfrac{\pi}{6}$ ,

$x=\dfrac{5\pi}{6}$ .
Hence the required area will be

$=|\int_{0} ^{\frac{\pi}{6}} tan(x)-\dfrac{2}{3}cos(x)|$

$=ln|sec(x)|-\dfrac{2}{3}sin(x)|_{0} ^{\frac{\pi}{6}}$

$=|ln(\dfrac{2}{\sqrt{3}})-\dfrac{1}{3}|$

$=\dfrac{1}{3}-ln(\dfrac{2}{\sqrt{3}})$

$=\dfrac{1}{3}+ln|\dfrac{\sqrt{3}}{2}|$ sq units.

The area bounded by

$\displaystyle x=a\cos ^{3}\theta,y=a\sin ^{3}\theta$ is:

0%
$\displaystyle \frac{3\pi a^{2}}{16}$
0%
$\displaystyle \frac{3\pi a^{2}}{8}$
0%
$\displaystyle \frac{3\pi a^{2}}{32}$
0%
$\displaystyle 3\pi a^{2}$

Explanation

Eliminating

$\theta$ , we have

$\displaystyle { x }^{ \dfrac { 2 }{ 3 } }+{ y }^{ \dfrac { 2 }{ 3 } }={ a }^{ \dfrac { 2 }{ 3 } }\\ x=0\Rightarrow y=\pm a\\ y=0\Rightarrow x=\pm a$
Symmetric about both axis
Required area

$\displaystyle =4\int _{ 0 }^{ a }{ y } dx=4\int _{ \dfrac { \pi }{ 2 } }^{ 0 }{ y } \dfrac { dx }{ dt } dt$

$\displaystyle y=a\sin ^{ 3 }{ t } ,x=a\cot ^{ 3 }{ t } \Rightarrow \frac { dx }{ dt } =-3a\cos ^{ 3 }{ t } \sin { t }$
Area

$\displaystyle =4\int _{ \dfrac { \pi }{ 2 } }^{ 0 }{ a } \sin ^{ 3 }{ t } .\left( -3a\cos ^{ 2 }{ t } \sin { t } \right) dt$

$\displaystyle =12{ a }^{ 2 }\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sin ^{ 4 }{ t } } \cos ^{ 2 }{ t } dt$

$\displaystyle =\dfrac { { 12a }^{ 2 }\dfrac { 5 }{ 2 } !\dfrac { 3 }{ 2 } ! }{ 2\dfrac { 8 }{ 2 } ! } =\dfrac { { 6a }^{ 2 }\times \dfrac { 3 }{ 2 } \times \dfrac { 1 }{ 2 } \times \sqrt { \pi } .\dfrac { 1 }{ 2 } \sqrt { \pi } }{ 3\times 2 }$

$\displaystyle =\dfrac { 3 }{ 8 } { \pi a }^{ 2 }$

The area lying in the first quadrant inside the circle

${ x }^{ 2 }+{ y }^{ 2 }=12$ and bounded by the parabolas

${ y }^{ 2 }=4x,{ x }^{ 2 }=4y$ is:

0%
$\displaystyle 2\left( \frac { \sqrt { 2 } }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 } } \right)$
0%
$\displaystyle 4\left( \frac { \sqrt { 2 } }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 } } \right)$
0%
$\displaystyle \left( \frac { \sqrt { 2 } }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 } } \right)$
0%
none of these

Explanation

The required area

$\displaystyle =\int _{ 0 }^{ 2 }{ 2\sqrt { x } dx } +\int _{ 2 }^{ 2\sqrt { 2 } }{ \sqrt { 12-{ x }^{ 2 } } dx } -\int _{ 0 }^{ 2\sqrt { 2 } }{ \frac { { x }^{ 2 } }{ 4 } dx }$

$\displaystyle ={ \left[ \frac { 4{ x }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 2 }+{ \left[ \frac { x }{ 2 } \sqrt { 12-{ x }^{ 2 } } +\frac { 12 }{ 2 } \sin ^{ -1 }{ \frac { x }{ 2\sqrt { 3 } } } \right] }_{ 2 }^{ 2\sqrt { 3 } }-\frac { 1 }{ 4 } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2\sqrt { 2 } }$

$\displaystyle =4\left( \frac { \sqrt { 2 } }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 } } \right)$

If f(x) be an increasing function defined on [a, b] then
max {f(t) such that

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x) & min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a),
min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x).
On the basis of above information answer the following questions.

Let

$\displaystyle f\left ( x \right )=min \left \{ 1, 1-\cos x, 2\sin x \right \}$ then

$\displaystyle \int_{0}^{\pi}f\left ( x \right )dx$ is

0%
$\displaystyle \frac{\pi }{3}+1-\sqrt{3}$
0%
$\displaystyle \frac{2\pi }{3}-1+\sqrt{3}$
0%
$\displaystyle \frac{5\pi }{6}+1-\sqrt{3}$
0%
$\displaystyle \frac{\pi }{6}+1-\sqrt{3}$

Explanation

min

$\left \{ 1, 1-\cos x, 2\sin x\forall 0\leq x\leq \pi \right \}$

$\displaystyle =\begin{cases} 1-\cos x & 0\leq x\leq \dfrac{\pi }{2} \\ 1 & \dfrac{\pi }{2}\leq x\leq \dfrac{5\pi }{6} \\ 2\sin x & \dfrac{5\pi }{6}\leq x\leq \pi \end{cases}$

$\therefore$

$\displaystyle \int_{0}^{\pi }f\left ( x \right )dx=\int_{0}^{\pi /2}\left ( 1-\cos x \right )dx+\int_{\pi /2}^{5\pi /6}1dx+\int_{5\pi /6}^{\pi }2\sin xdx$

$=\left [ x-\sin x \right ]_{0}^{\pi /2}+\left [ x \right ]_{\pi /2}^{5\pi /6}-2\left [ \cos x \right ]_{5\pi /6}^{\pi }$

$\displaystyle =\frac{\pi }{2}-1+\frac{5\pi }{6}-\frac{\pi }{2}-2\left [ -1+\frac{\sqrt{3}}{2} \right ]=\frac{5\pi }{6}+1-\sqrt{3}$

The area of the plane region bounded by the curves

$x+2y^{2}=0$ and

$x+3y^{2}=1$ is

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{5}{3}$

Explanation

Solving the equation

$x+2y^{2}=0$ &

$3y^{2}+x=1$
we get

$A(-2, 1)$ &

$B(-2, -1)$
Required Area

$\displaystyle =2\left [ \int_{0}^{1}xdy-\int_{0}^{1}xdy \right ]$

$\displaystyle =2\int_{0}^{1}\left ( 1-3y^{2}+2y^{2} \right )dy$

$\displaystyle =2\int_{0}^{1}\left ( 1-y^{2} \right )dy$

$\displaystyle =2\left [ y-\frac{y^{3}}{3} \right ]_{0}^{1}=2\left ( 1-\frac{1}{3} \right )=\frac{4}{3}$ sq.units

Find the area bounded by the curves

$\displaystyle x = y^{2}$ and

$\displaystyle x = 3-2y^{2}$

0%
$2$ sq. units
0%
$4$ sq. units
0%
$6$ sq. units
0%
$8$ sq. units

Explanation

The two curves represent parabolas with vertices at

$\left( 0,0 \right)$ and

$\left( 3,0 \right)$
They intersect at

$\left( 1,1 \right)$ and

$\left( 1,-1 \right)$
So the required area is
area of OPMQO

$= 2$ ( area of OPMO)

$\displaystyle =2\left( \int _{ 0 }^{ 1 }{ \sqrt { x } } dx+\int _{ 1 }^{ 3 }{ \sqrt { \frac { 3-x }{ 2 } } dx } \right)$

$\displaystyle =2\left( \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } } \right) _{ 0 }^{ 1 }{ - }\left[ 2\frac { 1 }{ \sqrt { 2 } } .\frac { 2 }{ 3 } { \left( 3-x \right) }^{ \frac { 3 }{ 2 } } \right] _{ 1 }^{ 3 }$

$\displaystyle =2\left[ \frac { 2 }{ 3 } -\left( 0-\frac { 1 }{ \sqrt { 2 } } \frac { 2 }{ 3 } { 2 }^{ \frac { 3 }{ 2 } } \right) \right] =2\left( \frac { 2 }{ 3 } +\frac { 4 }{ 3 } \right) =4$

The area bounded by the curves

$y=\log x$ ,

$y=\log \left | x \right |$ ,

$y=\left | \log x \right |$ and

$y=\left | \log \left | x \right | \right |$

0%
4 sq. units
0%
6 sq. units
0%
10 sq. units
0%
None of these

Explanation

Required Area

$=2\int_{0}^{1}\left | \log \left | x \right | \right |dx$

$\displaystyle =2\left [ \left ( x\left | \log \left | x \right | \right | \right ) \right ]_{0}^{1}-\int_{0}^{1}\left ( -\frac{1}{x} \right )xdx$

$=2\left [ \left ( 1-0 \right )+\left ( x \right )_{0}^{1} \right ]=4$ units

Let y=f(x) be the given curve and

$x=a$ ,

$x=b$ be two ordinates then area bounded by the curve

$y=f(x)$ , the axis of x between the ordinates

$x=a$ &

$x=b$ , is given by definite integral

$\int_{a}^{b}ydx$ or

$\int_{a}^{b}f\left ( x \right )dx$ and the area bounded by the curve

$x=f(y)$ , the axis of y & two abscissae

$y=c$ &

$y=d$ is given by

$\int_{c}^{d}xdy$ or

$\int_{c}^{d}f\left ( x \right )dy$ . Again if we consider two curves

$y=f(x)$ ,

$y=g(x)$ where

$f\left ( x \right )\geq g\left ( x \right )$ in the interval [a, b] where

$x=a$ &

$x=b$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by

$\int_{a}^{b}\left [ f\left ( x \right )-g\left ( x \right ) \right ]dx$
On the basis of above information answer the following questions.

The area bounded by parabolas

$y=x^{2}+2x+1$ &

$y=x^{2}-2x+1$ and the line

$\displaystyle y=\frac{1}{4}$ is equal to

0%
$\displaystyle \frac{2}{3}$ square unit
0%
$\displaystyle \frac{1}{3}$ square unit
0%
$\displaystyle \frac{3}{2}$ square unit
0%
$\displaystyle \frac{1}{2}$ square unit

Explanation

Given curve is

$y=x^{2}+2x+1=\left ( x+1 \right )^{2}$ (i)
is upward parabola with vertex at (-1, 0) meet y-axis at (0, 1) and the curve

$y=x^{2}-2x+1=\left ( x-1 \right )^{2}$ (ii)
is also upward parabola with vertex at (1, 0) meet, y-axis at (0, 1)
Also

$\displaystyle y=\frac{1}{2}$ (iii)
A line parallel to x-axis meeting(i) at

$\displaystyle \left ( -\frac{1}{2}, \frac{1}{4} \right )$ ,

$\displaystyle \left ( -\frac{3}{2}, \frac{1}{4} \right )$ & meeting (ii) at

$\displaystyle \left ( \frac{3}{2}, \frac{1}{4} \right )$ ,

$\displaystyle \left ( \frac{1}{2}, \frac{1}{4} \right )$
Required area is the shaded region given by 2 Area (LTNL) (by symmetry)

$\displaystyle =2\int_{0}^{1/2}\left \{ \left ( x-1 \right )^{2}-\frac{1}{4} \right \}dx=2\left [ \frac{\left ( x-1 \right )^{3}}{3}-\frac{x}{4} \right ]_{0}^{1/2}$

$\displaystyle =2\left [ \left ( -\frac{1}{24}-\frac{1}{8} \right )+\left ( \frac{1}{3} \right ) \right ]=2\left [ -\frac{1}{6}+\frac{1}{3} \right ]=\frac{1}{3}$ square unit

If f(x) be an increasing function defined on [a, b] then
max {f(t) such that

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x) & min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(a),
min {f(t),

$a\leq t\leq x$ ,

$a\leq x\leq b$ }=f(x).
On the basis of above information answer the following questions.

$\int_{0}^{\pi /2}min\left \{ \sin x, \cos x \right \}dx$ equals

0%
$2\left ( \sqrt{3}-1 \right )$
0%
$\sqrt{2}\left ( \sqrt{2}-1 \right )$
0%
$\left ( \sqrt{3}-1 \right )$
0%
$\sqrt{2}\left ( \sqrt{2}+1 \right )$

Explanation

min

$\left \{ \sin x, \cos x \right \}$

$\displaystyle \forall 0\leq x\leq \frac{\pi }{2}$ (from the graph of

$y=\sin x$ ,

$y=\cos x$

$\displaystyle \forall 0\leq x\leq \frac{\pi }{2}$ )

$\displaystyle =\begin{cases} \sin x & 0\leq x\leq \frac{\pi }{4} \\ \cos x & \frac{\pi }{4}\leq x\leq \frac{\pi }{2} \end{cases}$

$\therefore$ Required area

$\displaystyle =\int_{0}^{\pi /4}\sin xdx+\int_{\pi /4}^{\pi /2}\cos xdx$

$\displaystyle =\left ( -\cos x \right )_{0}^{\pi /4}+\left ( \sin x \right )_{\pi /4}^{\pi /2}=\left ( -\frac{1}{\sqrt{2}}+1 \right )+\left ( 1-\frac{1}{\sqrt{2}} \right )$

$=2-\sqrt{2}=\sqrt{2}\left ( \sqrt{2}-1 \right )$

The parabolas

$y^{2}=4x$ and

$x^{2}=4y$ divide the square region bounded by the lines

$x=4, y=4$ and the coordinate axes. If

$S_{1}$ ,

$S_{2}$ ,

$S_{3}$ are respectively the areas of these parts numbered from top to bottom;

$S_{1}: S_{2}: S_{3}$ is

0%
$1: 2: 3$
0%
$1: 2: 1$
0%
$1: 1: 1$
0%
$2: 1: 2$

Explanation

Total area

$=4\times 4=16$ sq. units
Area of

$\displaystyle S_{3}=\int_{0}^{4}\frac{x^{2}}{4}dx=\frac{16}{3}=S_{1}$

$\therefore$

$\displaystyle S_{2}=16-\frac{16}{3}\times 2=\frac{16}{3}$

$\therefore$

$S_{1}:S_{2}:S_{3}$ is

$1:1:1$

The area bounded by

${ y }^{ 2 }+8x=16$ and

${ y }^{ 2 }-24x=48$ is

$\displaystyle \frac { a\sqrt { 6 } }{ c }$ , then

$a+c=$

0%
$30$
0%
$32$
0%
$35$
0%
None

Explanation

The curve

${ y }^{ 2 }+8x=16$ and

${ y }^{ 2 }-24x=48$ cuts at

$x=-1$

${ y }^{ 2 }=24\Rightarrow y=\pm \sqrt { 24 }$

Required area

$\displaystyle =\int _{ -\sqrt { 24 } }^{ \sqrt { 24 } }{ \left( { x }_{ 1 }-{ x }_{ 2 } \right) dy } =\int _{ -\sqrt { 24 } }^{ \sqrt { 24 } }{ { \left( \left( 2-\frac { { y }^{ 2 } }{ 8 } \right) -\left( \frac { { y }^{ 2 } }{ 24 } -2 \right) \right)}dy }$

$\displaystyle =4{ \left[ y \right] }_{ -\sqrt { 24 } }^{ \sqrt { 24 } }-\frac { 1 }{ 18 } { \left[ { y }^{ 3 } \right] }_{ -\sqrt { 24 } }^{ \sqrt { 24 } }$

$=8\sqrt { 24 } -\dfrac { 2 }{ 18 } (24\sqrt { 24 })$

$\displaystyle =\frac { 16\sqrt { 24 } }{ 3 } =\frac { 32\sqrt { 6 } }{ 3 }$

Area is given as

$\dfrac{a\sqrt{6}}{c}$

$\Rightarrow a=32, c=3$

$\therefore a+c=32+3=35$

The area enclosed between the curves

$y=x^3$ and

$y=\sqrt{x}$ is (in square units)

0%
$\displaystyle\frac{5}{3}$
0%
$\displaystyle\frac{5}{4}$
0%
$\displaystyle\frac{5}{12}$
0%
$\displaystyle\frac{12}{5}$

Explanation

$y=\sqrt { x }$ or

${ y }^{ 2 }=x\left( y\ge 0 \right)$ and

$y={ x }^{ 3 }.$

We get points of intersection

$(0,0)$ and

$(1,1)$

$\therefore$ Required area

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 3 } \right) } dx$

$\displaystyle =\left[ \dfrac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\dfrac { { x }^{ 4 } }{ 4 } \right] _{ 0 }^{ 1 }=\dfrac{2}{3}-\dfrac{1}{4}=\dfrac { 5 }{ 12 }$ sq. unit.

Find the area bounded by

$\displaystyle y = \cos ^{-1}x,y=\sin ^{-1}x$ and

$y-$ axis

0%
$\displaystyle \left ( 2-\sqrt{2} \right )$ sq. units
0%
$\displaystyle \left ( \sqrt{2}-{2} \right )$ sq. units
0%
$2 \sqrt{2}$ sq. units
0%
$\sqrt {2}$ sq. units

Explanation

$y=\cos ^{ -1 }{ x }$ and

$y=\sin ^{ -1 }{ x }$ intersect at

$P\equiv\displaystyle \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { \pi }{ 4 } \right)$
Hence, required area is,

$\displaystyle \int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \left( \cos ^{ -1 }{ x } -\sin ^{ -1 }{ x } \right) dx } =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \cos ^{ -1 }{ x } dx } -\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \sin ^{ -1 }{ x } dx }$

$\displaystyle ={ \left[ x\cos ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } -{ \left[ x\sin ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$

$\displaystyle ={ \left[ \dfrac1{\sqrt2}\cos ^{ -1 }{\dfrac1{\sqrt2} } \right] } -{ \left[ \dfrac1{\sqrt2}\sin ^{ -1 }{ \dfrac1{\sqrt2} } \right] }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ 2x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$

$\displaystyle =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ 2x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$

Assuming

$1-x^2 = t^2$

Differentiating both sides, we get

$-2xdx = 2tdt$

Substituting in the integral, we get

$\displaystyle =\int _{ \frac { 1 }{ \sqrt { 2 } } }^12\ dt$

$=2-\sqrt { 2 }$

Consider two curves

$\displaystyle C_{1}:y=\frac{1}{x}$ and

$\displaystyle C_{2}$ :

$y = \displaystyle lnx$ on the xy plane Let

$\displaystyle D_{1}$ denotes the region surrounded by

$\displaystyle C_{1}$ ,

$\displaystyle C_{2}$ and the line

$x = 1$ and

$\displaystyle D_{2}$ denotes the region surrounded by

$\displaystyle C_{1}$ ,

$\displaystyle C_{2}$ and the line

$x = a$ If

$\displaystyle D_{1}$ =

$\displaystyle D_{2}$ then the value of 'a':

0%
$\displaystyle \frac{e}{2}$
0%
$e$
0%
$e-1$
0%
$2(e-1)$

Explanation

Let the x point of intersection of the two curves be (k)

$D1 = \int _1 ^k (\cfrac{1}{x} - \ln \ \ x)dx$

$D2 = \int _k ^a (- \cfrac{1}{x} \ln \ \ x)dx$

$D1 +D2 =0$

$\int _1 ^a (\cfrac{1}{x}dx = \int _1 ^a (\ln \ \ x)dx$

$\ln \ \ a = a \ \ \ln \ \ a - a +1$

$(\ln \ \ a -1)(1-a) = 0$

$a =e$

Suppose

$y = f(x)$ and

$y = g(x)$ are two functions whose graphs intersect at three points

$(0, 4), (2, 2)$ and

$(4, 0)$ with

$f(x) > g(x)$ for

$0 < x < 2$ and

$f(x) < g(x)$ for

$2 < x < 4$ . if

$\displaystyle \int_{0}^{4}\left ( f(x)-g(x) \right )dx=10$ and

$\displaystyle \int_{2}^{4}\left ( g(x)-f(x) \right )dx=5$ , the area between two curves for

$0 < x < 2$ , is:

0%
5
0%
10
0%
15
0%
20

Explanation

$f(x) > g(x)$ for

$0 < x < 2$ and

$f(x) < g(x)$ for

$2 < x < 4$ .
if

$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$ and

$\displaystyle \int_{2}^{4}\left [ g(x)-f(x)) \right ]dx=5$ ,

$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$

$\displaystyle \int_{0}^{2} f(x)-g(x) dx + \int_{2}^{4} f(x)-g(x) dx =10$

$\displaystyle \int_{0}^{2} ( f(x)-g(x)) dx + \int_{2}^{4} ( g(x)-f(x)) dx =10$

$\displaystyle \int_{0}^{2} ( f(x)-g(x)) dx =15$

The area bounded by the curves

$\displaystyle y=-\sqrt{-x}$ and

$\displaystyle x=-\sqrt{-y}$ were

$\displaystyle x,y\leq 0$

0%
Can not be determined
0%
is 1/3
0%
is 2/3
0%
is same as that of the figure by the curves $\displaystyle y=\sqrt{-x};x\leq 0$ and $\displaystyle x=\sqrt{-y};y\leq 0$

Explanation

$y_1=-\sqrt{-x}$

$y_2= -x^2$
Area bound is given by,

$\displaystyle = \int _{-1} ^0 (y_1+y_2)dx$

$\displaystyle = \int _{-1} ^0 (-\sqrt{-x}+x^2)dx$

$\displaystyle = \left[\frac{-x^{3/2}}{3/2}+\dfrac {x^3}{3}\right]_{-1}^0$

$=(0-0)-\left(\dfrac{-2}{3}+\dfrac{1}{3}\right)$

$= \cfrac{1}{3}$

Area of the region enclosed between the curves

$\displaystyle x=y^{2}-1$ and

$\displaystyle x = \left | y \right |\sqrt{1-y^{2}}$ is

0%
$1$
0%
$4/3$
0%
$2/3$
0%
$2$

Explanation

Required area is,

$\displaystyle A=2\int_{0}^{1}\left [ y\sqrt{1-y^{2}}-\left ( y^{2}-1 \right ) \right ]dy$

$\displaystyle =\left(\frac{-2}{3}\left ( 1-y^{2} \right )^{3/2}\right)_{0}^{1}-\left ( \frac{2y^{3}}{3}-2y \right )^{1}_{0}=2$

Find the area of the region bounded by the curves

$\displaystyle x=\frac{1}{2},x=2,y=logx$ and

$y=2^{x}$

0%
$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$
0%
$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$
0%
$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$
0%
$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$

Explanation

Therefore the required area

$\displaystyle =\int _{ \dfrac { 1 }{ 2 } }^{ 2 }{ \left( { 2 }^{ x }-\log { x } \right) dx }$

$\displaystyle ={ \left[ \dfrac { { 2 }^{ x } }{ \log { 2 } } -\left( x\log { x } -x \right) \right] }_{ \dfrac { 1 }{ 2 } }^{ 2 }$

$\displaystyle =\frac { 4-\sqrt { 2 } }{ \log 2 } -\frac { 5 }{ 2 } \log 2+\frac { 3 }{ 2 }$

Area bounded by

$\displaystyle y=2x-{ x }^{ 2 }$ &

$\displaystyle (x-1{ ) }^{ 2 }+{ y }^{ 2 }=1$ in first quadrant, is:

0%
$\displaystyle \frac { \pi }{ 2 } -\frac { 4 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } -\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } +\frac { 4 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } +\frac { 2 }{ 3 }$

Explanation

Area of circle is

$\pi r^2$

But here only half circle is in

$1^{st}$ quadrant

Then,

$A^*=\dfrac{\pi r^2}{2}$ .......

$( r=1)$

$\Rightarrow A^{*}=\dfrac{\pi}{2} ......(1)$

Now area bounded by parabola

$y=2x-x^2$ is

$A^@=\displaystyle \int_{0}^{2}(2x-x^2)$

$A=\dfrac{4}{3} ......(2)$

Now area bounded by both curves is

$A=A^*-A^@$

$\Rightarrow A=\dfrac{\pi}{2}-\dfrac{4}{3}$

In what ratio does the x-axis divide the area of the region bounded by the parabolas

$\displaystyle y=4x-x^{2}$ and

$\displaystyle y=x^{2}-x$ ?

0%
4 : 121
0%
4 : 144
0%
4 : 169
0%
4 : 100

For what value of 'a' is the area of the figure bounded by

$\displaystyle y=\frac{1}{x}, y=\frac{1}{2x-1}$

$x = 2$ &

$x = a$ equal to

$\displaystyle ln\frac{4}{\sqrt{5}}$ ?

0%
$\displaystyle a=4\:$
0%
$\displaystyle a=8\:$
0%
$\displaystyle a=4\: or \frac{2}{5}\left ( 6-\sqrt{21} \right )$
0%
none of these

Explanation

The required area is

$=\int_{2} ^{a} \dfrac{1}{x}-\dfrac{1}{2x-1}.dx$

$=[lnx-\dfrac{ln(2x-1)}{2}]_{2} ^{a}$

$=[ln\dfrac{x}{\sqrt{2x-1}}]_{2} ^{a}$

$=-ln\dfrac{(2)}{\sqrt{3}}+ln\dfrac{a}{\sqrt{2a-1}}$

$=ln\dfrac{4}{\sqrt{5}}$

Hence

$ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{4}{\sqrt{5}}+ln\dfrac{(2)}{\sqrt{3}}$

$\Rightarrow ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{8}{\sqrt{15}}$
Hence

$2a-1=15$
Hence

$a=8$ .
Therefore a=8 is one solution.
Now

$\dfrac{a^{2}}{2a-1}=\dfrac{64}{15}$

$\Rightarrow 15a^{2}=128a-64$

$\Rightarrow 15a^{2}-128a+64=0$

Hence

$a=8$ and

$a=\dfrac{8}{\sqrt{15}}$

If the area enclosed by the parabolas

$\displaystyle y=a-x^{2}$ and

$\displaystyle y=x^{2}$ is

$\displaystyle 18\sqrt {2}$ sq. units Find the value of 'a'

0%
$a = -9$
0%
$a= 6$
0%
$a =9$
0%
$a=-6$

Explanation

For the point of intersection

$x^{2}=a-x^{2}$

$2x^{2}=a$

$x=\pm\sqrt{\dfrac{a}{2}}$
Hence the required area will be

$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} x^{2}-(a-x^{2}).dx$

$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$

$=2\int_{0} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$

$=2[\dfrac{2x^{3}}{3}-ax]_{0} ^{\sqrt{\frac{a}{2}}}$

$=2[\dfrac{2.a\sqrt{a}}{6.\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$

$=2[\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$

$=18\sqrt{2}$

$|\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}|=9\sqrt{2}$

$\dfrac{2a\sqrt{a}}{3\sqrt{2}}=9\sqrt{2}$

$a\sqrt{a}=27$

$a^{\frac{3}{2}}=3^{3}$

$a=3^{2}$
Hence

$a=9$ .

Area enclosed between the curves

${ y }^{ 2 }=x$ and

${ x }^{ 2 }=y$ is equal to

0%
$\displaystyle 2\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) dx }$
0%
$\displaystyle \frac { 1 }{ 3 }$
0%
area of region $\left\{ \left( x,y \right) :{ x }^{ 2 }\le y\le \left| x \right| \right\}$
0%
$\displaystyle \frac { 2 }{ 3 }$

Explanation

The

$\displaystyle { y }^{ 2 }=x$ and

${ x }^{ 2 }=y$ intersect each other at point

$(0,0)$ and

$(1,1)$ as shown below.

$\therefore$ required area is given by

$\displaystyle A=\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) } dx=2{ A }_{ 2 }$

Where

${ A }_{ 2 }=$ area between

$y=x$ and

$y={x}^{2}$ between

$x=0$ and

$x=1=\left\{ \left( x,y \right) :{ x }^{ 2 }\le y\le \left| x \right| \right\}$

$\displaystyle =2\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx=2\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }$

$\displaystyle =2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =\frac { 1 }{ 3 }$ square units.

Find the area enclosed between the curves

$\displaystyle y=\log_{e}\left ( x+e \right ), x=\log_{e}\left ( 1/y \right )$ & the x-axis

0%
1 sq. units
0%
2 sq. units
0%
3 sq. units
0%
4 sq. units

Explanation

Area enclosed

$= \int _{1-e} ^0 ln(x+e) dx + \int _{0} ^{\infty} e^{-x} dx$

$= (xln(x+e) -x)| _{1-e} ^0 -e^{-x} | _{0} ^{\infty}$

$=2$

Let

$f(x)$ be a continuous function given by

$\displaystyle f\left ( x \right )=2x$ for

$\displaystyle \left | x \right |\leq 1$ for

$\displaystyle f\left ( x \right )=x^{2}+ax+b$ for

$\displaystyle \left | x \right |> 1$ . Find the area of the region in the third quadrant bounded by the curves

$\displaystyle x=-2y^{2}$ and

$y = f(x)$ lying on the left of the line

$8x + 1 = 0$

0%
$\dfrac{235}{192} ; a = 2 ; b = - 1$
0%
$\dfrac{235}{192} ; a = -1 ; b = 2$
0%
$\dfrac{257}{192} ; a = -1 ; b = 2$
0%
$\dfrac{257}{192} ; a = 2 ; b = - 1$

Let

$\displaystyle C_{1}$ &

$\displaystyle C_{2}$ be two curves passing through the origin as shown in the figure A curve C is said to "bisect the area" the region between

$\displaystyle C_{1}$ &

$\displaystyle C_{1}$ if for each point P of C the two shaded regions A & B shown in the figure have equal areas Determine the upper curve

$\displaystyle C_{2}$ given that the bisecting curve C has the equation

$\displaystyle y=x^{2}$ & that the lower curve

$\displaystyle C_{1}$ has the equation

$\displaystyle y=x^{2}/2$

0%
$\displaystyle \left ( 16/9 \right )x^{2}$
0%
$\displaystyle \left ( 25/9 \right )x^{2}$
0%
$\displaystyle \left ( 25/16 \right )x^{2}$
0%
$\displaystyle \left ( 9/25 \right )x^{2}$

Explanation

According to question

$\displaystyle \int_{0}^{a^{2}}\left ( -f^{-1}\left ( y \right )+\sqrt{y} \right )\:\:dy=\int_{0}^{a}\left ( x^{2}-\frac{x^{2}}{2} \right )dx$

$\displaystyle \Rightarrow \left [ f^{-1}(a^{2})-a \right ]2a=-\frac{a^{2}}{2}\Rightarrow f^{-1}(a^{2})=\frac{3a}{4}\Rightarrow f\left ( \frac{3a}{4} \right )-a^{2}$
or

$\displaystyle f(x)=\frac {16}{9}x^{2}$

Find the area bounded by

$y = x + sinx$ and its inverse between

$x = 0$ and

$x = \displaystyle 2\pi$

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

graph of

$x+\sin x$ and its inverse is shown in the figure which is symmetric with

$y=x$

let

$f(x)=x+\sin x$

from the figure it is clear that

$A=\int _{ 0 }^{ 2\pi }{ \left( f\left( x \right) -f^{ -1 }\left( x \right) \right) dx } =4\int _{ 0 }^{ \pi }{ \left( f\left( x \right) -x \right) dx } =4\int _{ 0 }^{ \pi }{ \left( x+\sin { x } -x \right) dx } =8$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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