MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 6

From a piece of cardboard, in the shape of a trapezium ABCD, and AB

$||$ CD and

$\angle$ BCD

$=90^o$ , quarter circle is removed. Given AB

$=$ BC

$=3.5$ cm and DE

$=2$ cm. Calculate the area of the remaining piece of the cardboard.(Take

$\pi$ to be

$\dfrac{22}{7}$ )

0%
$9.625cm^2$
0%
$6.125cm^2$
0%
$2.625cm^2$
0%
None of these

The area between the curves y=

$x^{2}$ and

$y=\frac{2}{1+x^{2}}$ is-

0%
$\pi -\frac{1}{3}$
0%
$\pi -2$
0%
$\pi -\frac{2}{3}$
0%
$\pi +\frac{2}{3}$

Explanation

we have,

$y={{x}^{2}}\,\,......\,\,\left( 1 \right)$

$y=\dfrac{2}{{{x}^{2}}+1}\,\,.......\,\,\left( 2 \right)$

Taking limit then,

By equation (1) and (2) to, and get,

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

Put x=1 and we get,

${{\left( 1 \right)}^{2}}=\dfrac{2}{{{\left( 1 \right)}^{2}}+1}$

$1=\dfrac{2}{2}$

$1=1$

Now, taking’

$x=-1$ and we get,

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

$1=1$

Then, the limit of this function of $1\,\,to\,\,-1$ .

The required area $=\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}$

$=\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}=2\int_{0}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}$

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

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

$=2\left[ 2\dfrac{\pi }{4}-2\times 0-\dfrac{1}{3} \right]$

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

Hence, this is the answer.

The area (in sq. units) of the region

$\left\{ {\left( {x,y} \right):{y^2} \ge 2x\,and\,{x^2} + {y^2} \le 4x,x \ge 0} \right\}$ is

0%
$\pi - \frac{4}{3}$
0%
$\pi - \frac{8}{3}$
0%
$\pi - \frac{{4\sqrt 2 }}{3}$
0%
$\frac{\pi }{2} - \frac{{2\sqrt 2 }}{3}$

The area of the region

$[(x, y) : x^{2} + y^{2} \leq 1 \leq x + y|$ is

0%
$\dfrac {\pi}{5}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi^{2}}{3}$
0%
$\dfrac {\pi}{4} - \dfrac {1}{2}$

The area of the region bounded by the x-axis and the curves

$y = \tan x\left( { - \frac{\pi }{3} \le x \le \frac{\pi }{3}} \right),and\,y = \cot x\left( {\frac{\pi }{6} \le x \le \frac{{3\pi }}{2}} \right)$ is

0%
$log 2$
0%
$2log2$
0%
$log\sqrt 2$
0%
$log\left( {\frac{3}{2}} \right)$

The area of the region bounded by the curves

$y=ex\log x$ and

$y=\dfrac{\log x}{ex}$ is

0%
$\dfrac{e^{2}-5}{4e}$
0%
$\dfrac{e^{2}+1}{2e}$
0%
$\dfrac{e^{2}}{2}$
0%
$None\ of\ these$

Area bounded by the curve

${x^2} = 4y$ and the straight line

$x=4y-2$ is

0%
$\frac{8}{9}$ sq. unit
0%
$\frac{9}{8}$ sq. unit
0%
$\frac{4}{3}$ sq. unit
0%
None of these

If the area enclosed between

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

$x=n{y}^{2}$ is

$\cfrac{1}{3}$ sq. units, then

$m,n$ can be roots of (where

$m,n$ are non zero real numbers)

0%
$2{x}^{2}+5x+1=0$
0%
$2{x}^{2}+3x-2=0$
0%
$2{x}^{2}+3x+2=0$
0%
$2{x}^{2}+5x-1=0$

Explanation

$y=mx^{2}$

$y^{2}=m^{2}x^{4}$

$x=nm^{2}x^{1/3}$

$x = \left( \dfrac{1}{nm^{2}} \right)^{1/3}$

$y=m \dfrac{1}{(nm^{2})^{2/3}}$

$y= m\dfrac{1}{n^{2/3} m^{4/3}}$

$y=\dfrac{1}{(n^{2}m)^{1/3}}$

$(1/nm^{2})^{1/3}$

Area

$= \int_{0} \sqrt{\dfrac{x}{n}}- mx^{2}$

$=\left[ \right]^{(1/nm^{2})^{1/3}}$

$=\dfrac{2}{3\sqrt{n}} \left( \dfrac{1}{nm^{2}} \right)^{\dfrac{1}{3}. \dfrac{3}{2}}- \dfrac{m}{3} \left( \dfrac{1}{nm^{2}} \right)^{3 \times \dfrac{1}{3}}$

$=\dfrac{2}{3\sqrt{n}} \dfrac{1}{\sqrt{n}m}- \dfrac{m}{3} \dfrac{1}{nm^{2}}$

$=\dfrac{2}{3nm}- \dfrac{1}{3nm}$

$\dfrac{1}{3}= \dfrac{1}{3nm}$

$nm=1$

Product of roots

$=1$

In the given figure, a square

$OABC$ has been inscribed in the quadrant

$OPBQ$ . If

$OA=20cm$ then the area of the shaded region is

$\left[take\pi=3.14\right]$

0%
$214cm^{2}$
0%
$228cm^{2}$
0%
$222cm^{2}$
0%
$242cm^{2}$

Explanation

Area of shaded region

$=Area\ of \left[OPBQ\right]-Area\ of \left[OABC\right]$

$=\dfrac{1}{4}\pi {r}^{2}-{a}^{2}$

$\left[ \because OA=a=20cm\\ r=OB=OA\sqrt { 2 } =20\sqrt { 2 } cm \right]$

$=\dfrac{1}{4}\pi {\left(20\sqrt{2}\right)}^{2}-{20}^{2}$

$=200\pi-400=228\ {cm}^{2}$

The area bounded by

$y=x^2,x=y^2$ is

0%
$1$
0%
$\dfrac 16$
0%
$\dfrac 34$
0%
$None\ of\ these$

Explanation

$y=x^2\\y^2=x\implies y=\sqrt x$

The curves intersect at

$(0,0)$ and

$(1,1)$

Area between the curves is given by

$\displaystyle \int _0^1 \sqrt x-x^2dx\\\left.\dfrac 23 x^{3/2}+\dfrac{x^3}{3} \right|_0^1\\\dfrac 23+\dfrac13=1$

The area of the plane region bounded by the curve

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

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

0%
$-\dfrac{4}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{2}{3}$
0%
None of these

Explanation

The given curves are

$x+2y^2 = 0$ and

$x+3y^2 = 1$ .

The first curve is

$x+2y^2 = 0$ which implies

$2y^2 = -x$ and hence

$y^2 = -x/2$ , which is a parabola.

Second curve

$x+3y^2 = 1$ which is again a parabola as

$y^2 = -1/3 (x-1)$ .

Hence we are required to find the area between the two parabolas.

On solving the equations of the two parabolas, we get the points of intersection as

$(-2, 1)$ and

$(-2, -1)$ .

Hence we set

$x = -2$ and

$y = 1, -1$ .

Hence the required area is

$\int(-2y^2 -1 + 3y^2)dy$ , where the integral runs from 0 to 1.

$\int(y^2 -1) dy$ , where the integral runs from 0 to 1.

$(1/3 -1)$

$\dfrac{2}{3}$ .

Hence the area of the region bounded by the curves is

$\dfrac{4}{3}$ .

The maximum area of the triangle whose sides

$a, b \, and \, c$ satisfy

$0 \le a \le 1, \, 1 \le b \le 2$ and

$2 \le c \le 3$

0%
$1$
0%
$\frac {1}{2}$
0%
$2$
0%
$\frac {3}{2}$

The area enclosed between the curve

${y^2} = 4x$ and line

$y = x$ is

0%
$\frac {8}{3}$
0%
$\frac {4}{3}$
0%
$\frac {2}{3}$
0%
$\frac {1}{2}$

If the area anclosed by

$f\left( x \right) = \sin x + \cos x,y = a$ between two consecutive points of extremum is minimum, then the value of a is

0%
$0$
0%
$-1$
0%
$1$
0%
$2$

The area of the region bounded by the curve

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

$y = x$ is

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$

Area bounded by the curve

$y = \sin ^ { - 1 } x , y - a x i s$ and

$y = \cos ^ { - 1 } x$ is equal to

0%
$( 2 + \sqrt { 2 } )$ sq. unit
0%
$( 2 - \sqrt { 2 } )$ sq. unit
0%
$( 1 + \sqrt { 2 } )$ sq. unit
0%
$( \sqrt { 2 } - 1 )$ sq. unit

The area bounded by

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

$|x|+|y|=1$ is

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{2}{3}$
0%
$1$

Find the area enclosed between

$y=1+\left| x+1 \right| ,y=4$ .

0%
$9$ sq. units
0%
$25$ sq. units
0%
$18$ sq. units
0%
$21$ sq. units

Explanation

Given

$y=1+\left| x+1 \right|$

$=\begin{cases} -x,\quad \quad x<-1 \\ x+2,\quad \quad x>-1 \end{cases}$

Area

$=24-\left[ \int _{ -4 }^{ -1 }{ -x } dx+\int _{ -1 }^{ 2 }{ \left( x+2 \right) dx } \right]$

$==24-\left[ -\cfrac { 1 }{ 2 } (1-16)+{ \left[ \cfrac { { x }^{ 2 } }{ 2 } +2x \right] }_{ -1 }^{ 2 } \right] =24-\left[ \cfrac { 15 }{ 2 } +2+4-\cfrac { 1 }{ 2 } +2 \right]$

$=24-\left( 8+7 \right) =24-15=9$

1356376_1197837_ans_0cda998000e94ef8bb73dadcdc9c6e80.PNG

Area bounded by the curves

$y=x\sin x\ and x-axis$ between

$x=0\ and x=2\pi$ is

0%
$2\pi$
0%
$3\pi$
0%
$4\pi$
0%
None of these

The area of the figure bounded by the parabola

$(y-2)^2=x-1$ , the tangent to it at the point with the ordinate

$x=3$ , and the x-axis is

0%
$7$ sq. units
0%
$6$ sq. units
0%
$9$ sq. units
0%
None of these

Explanation

Given parabola is

$(y - 2)^2 = x - 1$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2(y-2)}$

When

$y = 3, x=2$

$\therefore \dfrac{dy}{dx} = \dfrac{1}{2}$ .

Tangent at

$(2,3)$ is

$y - 3 = \dfrac{1}{2}(x-2) \Rightarrow x-2y + 4 = 0$

$\therefore$ required area

$= \displaystyle \int _0^3 ((y-2)^2 + 1) dy - \int_0^3 (2y-4) dy$

$= \left| \dfrac{(y - 2)^3}{3} + y\right|_0^3 - |y^2 - 4y|_0^3$

$= \dfrac{1}{3} + 3 + \dfrac{8}{3} - (9 - 12 ) = 9$ sq. units.

1632632_1197023_ans_df9c172528ea4fb7a9d4da5d367e8505.png

The area bounded by the curve

$y=\cos x$ and

$y=\sin x$ between the ordinates

$x=0$ and

$x=\dfrac {3\pi}{2}$ is

0%
$4\sqrt {2}+2$
0%
$4\sqrt {2}-1$
0%
$4\sqrt {2}+1$
0%
$4\sqrt {2}-2$

Explanation

$\displaystyle \int_0^{3\pi/2}|\sin x-\cos x|dx----(1)$

$\displaystyle \int_0^{\pi / 4}(\cos x-\sin x)dx +\displaystyle \int_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx$

$\rightarrow \ \displaystyle \int_{5\pi /4}^{3\pi/2}(\cos x-\sin x)dx [\sin x+\cos x]_0^{\pi /4}$

$\rightarrow \ [\cos x-\sin x]_{\pi /4}^{5\pi /4}\rightarrow {\sin x+\cos x}_{5\pi /4}^{3\pi /4}$

$=\sqrt {2}-1+\sqrt {2}+\sqrt {2}+\sqrt {2}-1$

$4\sqrt {2}-2$

1353696_1204390_ans_5e40305cc335490e861a2734d42bd1f2.png

Let

$y=g(x)$ be the inverse of a bijective mapping

$f:R\rightarrow R\quad f(x)=3{ x }^{ 3 }+2x$ . The area bounded by graph of

$g(x)$ , the axis and the ordinate at

$x=5$ is

0%
$\displaystyle\frac { 5 }{ 4 }$
0%
$\displaystyle\frac { 7 }{ 4 }$
0%
$\displaystyle\frac { 9 }{ 4 }$
0%
$\displaystyle\frac { 13 }{ 4 }$

Explanation

Required Area

$=\displaystyle \int_{0}^{s}{f^{-1}(x)dx}$

$I=\displaystyle \int_{f^{-1}(0)}^{f^{-1}(s)}{f^{-1}(f(t)).f'(t)dt}$

put

$x=f(t)$

$f(0)=0$

$\displaystyle \int_{f^{-1}(0)}^{f^{-1}(s)}{(f'(t)dt}$

$f^{-1}(0)=0$

$\displaystyle \int_{0}^{1}{t(9t^2+2)dt}$

$f^{-1}(s)=1$

$\displaystyle \int_{0}^{1}{t(9t^3+2t)dt}$

$={ \left. \dfrac { 9+4 }{ 4 } +\dfrac { 2t^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }$

$=\dfrac{3}{4}+1=\dfrac{13}{4}\ Square\ Unit$

The area bounded by the curves

$|x|+|y|\ge1$ and

$x^2+y^2\le1$ is

0%
2 sq unit
0%
$\pi$ sq unit
0%
$(\pi-2)$ sq unit
0%
None of these

Explanation

Required area

$\pi \times 1^2-4\times \dfrac {1}{2}\times 1\times 1$

$= \pi -2$

$\therefore (\pi -2)sq \ unit$

1233116_1192357_ans_13dec5e6d3474d9aa9132ada15586b89.png

The area of the figure bounded by the parabola

$x = - 2 y ^ { 2 } \text { and } x = 1 - 3 y ^ { 2 }$ is ?

0%
$1 / 6$
0%
$2/3$
0%
$3/2$
0%
$4/3$

Explanation

$x = - 2{y^2} \to \left( i \right)\,\,\& \,\,x = 1 - 3{y^2} \to \left( {ii} \right)$

Equating equation (i) & (ii)

$\begin{array}{l} -2{ y^{ 2 } }=1-3{ y^{ 2 } } \\ -2{ y^{ 2 } }+3{ y^{ 2 } }=1 \\ { y^{ 2 } }=1 \\ y\pm 1 \end{array}$

So, area of region bounded by

$\begin{array}{l} A\Rightarrow 2\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-3{ y^{ 2 } } } \right) -\left( { -2{ y^{ 2 } } } \right) } \right] dy } \\ =2\int _{ 0 }^{ 1 }{ \left( { 1-{ y^{ 2 } } } \right) dy } \\ =2\left[ { y-\frac { { { y^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 }\Rightarrow 2\left[ { \left( { 1-\frac { 1 }{ 3 } } \right) -\left( { 0-\frac { 0 }{ 3 } } \right) } \right] \\ =2\left[ { \frac { 1 }{ 1 } -\frac { 1 }{ 3 } } \right] \Rightarrow 2\left[ { \frac { 2 }{ 3 } } \right] =\frac { 4 }{ 3 } \, \, sq.\, \, units \end{array}$

1213656_1213322_ans_c8f70f105c6b41f6b1d5bee087fcfb69.PNG

The area bounded by parabola

$y^{2}=x$ , straight line

$y=4$ and

$y-axis$ is-

0%
$\dfrac{16}{3}$
0%
$7\sqrt{2}$
0%
$\dfrac{32}{3}$
0%
$\dfrac{64}{3}$

Explanation

R.E.F image

Taking a rectangle of width

dy and length as x-coordinate

of parabola, Area would be,

$A = ^{4}\int _{0}xdy = ^{4}\int _{0}y^{2}dy =\left [\dfrac{y^{3}}{3}\right]_{4}^{0}$

$= \dfrac{64}{3}\Rightarrow (D)$

1382178_1214416_ans_512003a7976742f7b4167de612c427fd.JPG

The angle between the curves

$y=sin x$ and

$y=cos x$ is :

0%
$tan^{-1}(2\sqrt{2})$
0%
$tan^{-1}(3\sqrt{2})$
0%
$tan^{-1}(3\sqrt{3})$
0%
$tan^{-1}(5\sqrt{2})$

The area bounded by

$\dfrac { | x | } { a } + \dfrac { | y | } { b } = 1$ where

$a > 0$ and

$b > 0$ is

0%
$\dfrac { 1 } { 2 } a b$
0%
$ab$
0%
$2ab$
0%
$4ab$

Explanation

$\dfrac{|x|}{a}+\dfrac{|y|}{b}=1$

$a > o,b>o$

We should consider only 1st quadrant

are of

$\triangle =\dfrac{1}{2}\times a\times b$

$=\dfrac{1}{2}ab$ .

1330751_1209606_ans_3e4bc054d9f44fd3b263c05d4c12742f.png

Area bounded by

$y=x^2$ and line

$y=x$

0%
2/3
0%
1/6
0%
1/3
0%
1/4

The area bounded by the curve

$y=(x+1)^2,y=(x-1)^2$ and the line

$y=0$ is

0%
$\dfrac{1}{6}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$

Explanation

R.E.F image

$\rightarrow y = (x+1)^{2}$ is obtained

by shifting origin to (-1,0)

$x^{2} = y,$ for

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

Similarly (1,0)

As graph is symmetric

about y-axis, area A

would be,

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

$= 2\int_{0}^{1} x^{2}-2x+1dx = 2\left[x\dfrac{3}{3}-x^{2}+x\right]_{0}^{1}$

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

The area (in

$sq\ unit$ ) of the region

$\left\{(x,y):y^{2}\ge 2x\ and\ x^{2}+y^{2} \le 4x,x \ge 0,y\ \ge 0\right\}$ is:-

0%
$\dfrac {\pi}{2}+\dfrac {2\sqrt {2}}{3}$
0%
$\pi-\dfrac {4}{3}$
0%
$\pi-\dfrac {8}{3}$
0%
$\pi-\dfrac {4\sqrt {2}}{3}$

Area common to the cutve

$y=\sqrt {9-x^{2}}$ and

$x^{2}-y^{2}=6x$ is:

0%
$\dfrac {\pi +\sqrt {3}}{4}$
0%
$\dfrac {\pi -\sqrt {3}}{4}$
0%
$3\left(\pi +\dfrac {\sqrt {3}}{4}\right)$
0%
$None\ of\ these$

If the point

$(\lambda, \lambda +1)$ lies inside the region bounded by the curve

$x=\sqrt{25-y^2}$ and y-axis, then

$\lambda$ belongs to the interval.

0%
$(-1, 3)$
0%
$(-4, 3)$
0%
$(-\infty, -4)\cup (3, \infty)$
0%
None of these

The area enclosed by the curves

$y=x^{2},y=x^{3},x=0$ and

$x=p$ , where

$p > 1$ , is

$\dfrac{1}{6}$ . then

$p$ equals

0%
$8/3$
0%
$16/3$
0%
$4/3$
0%
$2$

Area of the region bounded by the curve

$y=e^x$ and lines

$x=0$ and

$y=e$ is?

0%
$e-1$
0%
$1$
0%
$2$
0%
$e^2-1$

The area (in sq units) of the region

$\{ (x,y):{ y }^{ 2 }\ge 2x$ and

${ x }^{ 2 }+{ y }^{ 2 }\le 4x ,\chi \ge 0, Y\ge 0\}$

0%
$\pi -\frac { 4 }{ 3 }$
0%
$\pi -\frac { 8 }{ 3 }$
0%
$\pi -\frac { 4\sqrt { 2 } }{ 3 }$
0%
$-\frac { 2\sqrt { 2 } }{ 3 }$

The area bounded by the curve

$xy^{2}=1$ and the lines

$x=1$ ,

$x=2$ is

0%
$4\left( {\sqrt 2 - 1} \right)$
0%
$4\left( {\sqrt 2 + 1} \right)$
0%
$2\left( {\sqrt 2 - 1} \right)$
0%
$2\left( {\sqrt 2 + 1} \right)$

The area in square units bounded by the curves

$y = x ^ { 3 } , y = x ^ { 2 }$ and the ordinates

$x = 1 , x = 2$ is

0%
$\frac { 17 } { 12 }$
0%
$\frac { 12 } { 13 }$
0%
$\frac { 2 } { 7 }$
0%
$\frac { 7 } { 2 }$

Explanation

$\int _{ 1 }^{ 2 }{ \left( { x }^{ 3 }{ -x }^{ 2 } \right) dx } \\ ={ \left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 1 }^{ 2 }\\ =\left[ 4-\frac { 8 }{ 3 } \right] -\left[ \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right] \\ =\left[ \frac { 4 }{ 3 } \right] -\left[ -\frac { 1 }{ 12 } \right] \\ =\frac { 4 }{ 3 } +\frac { 1 }{ 12 } \\ =\frac { 17 }{ 12 }$
1213504_1228356_ans_f126c23106ac4bde8bb41f5ff77026d7.jpg

1213504_1228356_ans_f126c23106ac4bde8bb41f5ff77026d7.jpg

The area bounded by the curves

$y=\sqrt{-x}$ and

$x=-\sqrt{-y}$ , where

$x,y\le0$ , is equal to

0%
$\dfrac{2}{3}sq. unit$
0%
$\dfrac{1}{3}sq. unit$
0%
$\dfrac{1}{2}sq. unit$
0%
$Cannot\ be\ determined$

Explanation

Given the curve

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

$x=-\sqrt{-y}$

Now, the area bound by the both curve

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

$=\displaystyle \int_{-1}^{0}{-x^2 dx}+\displaystyle \int_{-1}^{0}{\sqrt{-x}dx}$

change

$=\displaystyle \int_{-1}^{0}{\sqrt{x} dx}-\displaystyle \int_{-1}^{0}{x^2 dx}$

$=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{3}\right]_0^1$

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

$=\dfrac{1}{3}\ sq.\ unit$

Hence, this is the answer.

Let

$T$ be the triangle with vertices

$\left (0,0\right), \left (0,{c}^{2}\right)\ and \left (c,{c}^{2}\right)$ and let

$R$ be the region between

$y=cx$ and

$y={x}^{2}\ where c>0$ then

0%
Area $\left( R \right) =\dfrac { { c }^{ 3 } }{ 6 }$
0%
Area of $R=\dfrac {e^{x}}3{3}$
0%
$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=3$
0%
$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=\dfrac {3}{2}$

Let

$f(x)=minimum (x+1,\sqrt{1-x})$ for all

$x \le 1$ . Then the area bounded by

$y=f(x)$ and the x-axis is

0%
$\dfrac{7}{3}$ sq. units
0%
$\dfrac{1}{6}$ sq. units
0%
$\dfrac{11}{6}$ sq. units
0%
$\dfrac{7}{6}$ sq. units

Tangents are drawn from a point

$P$ to a parabola

$y^{2}=4ax$ . The area enclosed by the tangents and the corresponding chord of contact is

$4a^{2}$ . Then point

$P$ satisfies

0%
$y^{2}=4ax$
0%
$y^{2}=2a(x+a)$
0%
$y^{2}=4a(x-a)$
0%
$y^{2}=4a(x+a)$

The area of the figure bounded by the curves

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

$x - y - 1 = 0$ is

0%
$\dfrac {2}{3}$
0%
$\dfrac {4}{3}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {16}{3}$

Area (in

$sq.$ unit) of region bounded by

$y=2\cos x,\ y=3\tan x$ and

$y-$ axis is

0%
$1+3ln \left(\dfrac {2}{\sqrt {3}}\right)$
0%
$1+\dfrac {3}{2}ln3-3ln2$
0%
$1+\dfrac {3}{2}ln3-ln2$
0%
$ln \left(\dfrac {3}{2}\right)$

Explanation

Given that,

$y=2\cos x, y=3\tan x$

$2\cos x=3\tan x$

$2\cos x=3\dfrac{\sin x}{\cos x}$

$2\cos^2x=3\sin x$

$2(1-\sin^2x)=3\sin x$

$2-2\sin^2x=3\sin x$

$2\sin^2x+3\sin x-2=0$

On solving,

$x=\dfrac{\pi}{6}$ and y-axis

So,

$x=0$

then,

$\displaystyle\int^{\pi/6}_0(2\cos x-3\tan x)dx$

$=2\displaystyle\int^{\pi/6}_0\cos xdx-3\displaystyle\int^{\pi/6}_0\tan xdx$

$=2[\sin x]^{\pi/6}_0-3[log \sec x]^{\pi/6}_0$

$=2\left[\sin\dfrac{\pi}{6}-\sin\theta\right]-3\left[log\sec\dfrac{\pi}{6}-log\sec 0\right]$

$=2\left[\dfrac{1}{2}-0\right]-3\left[log \dfrac{2}{\sqrt{3}}-log 1\right]$

$=\dfrac{2}{2}-0-3\left[log 2-log\sqrt{3}-0\right]$

$=1-3[log 2-log \sqrt{3}]$

$=1-3log 2+3log \sqrt{3}$

$=1-3log 2+3log (3)^{\dfrac{1}{2}}$

$=1-3log 2+\dfrac{3}{2}log 3$

$=1+\dfrac{3}{2}log 3-3log 2$ .

1196171_1242318_ans_a3d76ad1fa244cbcb063009e87c036d5.png

Let

$f\left( x \right)$ be a non-negative continuous function such that the area bounded by the curve

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

$x=\cfrac { \pi }{ 4 }$ ,

$x=\beta >\cfrac { \pi }{ 4 }$ is

$\left( \beta \sin { \beta } +\cfrac { \pi }{ 4 } \cos { \beta } +\sqrt { 2 } \beta -\cfrac { \pi }{ 2 } \right)$ . Then

$f\left( \cfrac { \pi }{ 2 } \right)$ is

0%
$\left( 1-\dfrac { \pi }{ 4 } -\sqrt { 2 } \right)$
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$\left( 1-\dfrac { \pi }{ 4 } +\sqrt { 2 } \right)$
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$\left( \cfrac { \pi }{ 4 } +\sqrt { 2 } -1 \right)$
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$\left( \cfrac { \pi }{ 4 } -\sqrt { 2 } +1 \right)$

Explanation

Given that,

$\displaystyle\int^{\beta}_{\pi/4}f(x)dx=\beta \sin\beta +\dfrac{\pi}{4}\cos\beta +\sqrt{2}\beta -\dfrac{\pi}{2}$

Differentiating w.r.t x and we get

$f(\beta)=\beta \cos\beta +\sin\beta -\dfrac{\pi}{4}\sin\beta +\sqrt{2}-0$

$f\left(\dfrac{\pi}{2}\right)=\left(1-\dfrac{\pi}{4}\right)\sin\dfrac{\pi}{2}+\sqrt{2}$

$=\left(1-\dfrac{\pi}{4}\right)\times 1+\sqrt{2}$

$=1-\dfrac{\pi}{4}+\sqrt{2}$

Hence, this is the answer.

1196176_1242263_ans_c6c3c5a1067a4dcda008f7510650fc50.png

The area of (in sq. units ) of the region described by

$A={(x,y): x^2+y^2 \leq 1\ and\ y^2 \leq 1-x }$

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

The area of the region bounded by the curves

$y=|x-1|$ and

$y=3-|x|$ is?

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$2$ sq. units
0%
$3$ sq. units
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$4$ sq. units
0%
$6$ sq. units

The area bounded by

$y=|x-1|$ ,

$y=0$ and

$|x|=2$ is?

0%
$4$
0%
$5$
0%
$3$
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$10$

Explanation

$\textbf{Step -1: Plotting the graph.}$

$\text{Given equations, }$

$y=|x-1|$ ,

$y=0 \text {and } |x|=2$

$\text{Blue lines indicates lines formed by } |x| = 2$

$\text{Red line indicates line formed by } y = |x-1|$

$\text{Green lines indicates line formed by } |y = 0$

$\text{A, B, C, D, E are the points formed by the intersection of given equations with x and y-axis}$

$\text{and also among themselves}$

$\textbf{Step -2: Finding the area . }$

$\text{Area of }\Delta ABC= \dfrac12AB\times BC$

$=\dfrac12(3\times3 )= \dfrac92$

$\text{Area of }\Delta CDE = \dfrac12CE\times DE$

$=\dfrac12(1\times1)=\dfrac12$

$\text{Total area }=\dfrac92+\dfrac12$

$=5$

$\textbf{Thus, the area under the graph is }\mathbf{5 unit^2 .}$

Area of the region containing all points

$(x, y)$ satisfying

$0\le y\le \sqrt{4-x^{2}}, y \le x^{2}+x+1$ and

$y=\left[\sin^{2}\dfrac{\pi}{4}+\cos\dfrac{x}{4}\right]$ is equal to ( where [.] denotes the greatest integer function ).

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$\dfrac{4\pi-1}{6}+\sqrt{3}$
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$\dfrac{4\pi+1}{6}+\sqrt{3}$
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$\dfrac{2\pi-1}{6}+\sqrt{3}$
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$\dfrac{2\pi+1}{6}+\sqrt{3}$

The area bounded by

$x^2=4ay$ and

$y=2a$ is?

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$\dfrac{16\sqrt{2}a^2}{3}$
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$\dfrac{16a^2}{3}$
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$\dfrac{8a^2}{3}$
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$\dfrac{8\sqrt{2}a^2}{3}$

Explanation

REF.Image

$x^{2}=4ay, y=2a$

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

$=2A_{2}$

$= 2\int_{0}^{2\sqrt{2}a}(2a-\dfrac{x^{2}}{4a})dx$

$= 2[2ax-\dfrac{x^{3}}{4a\times 3}]_{0}^{2\sqrt{2}a}$

$= 2[4\sqrt{2}a^{2}-\dfrac{4a^{2}\sqrt{2}}{3}]$

$= \dfrac{16}{3}\sqrt{2}a^{2}$

1194874_1249801_ans_0d2cfd0650c24782ad9b68d0b232688b.jpg

The area enclosed between the curves

${y^2} = x$ and

$y = |x|\;is$ -

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$\dfrac {2}{3}$
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$1$
0%
$\dfrac {1}{6}$
0%
$\dfrac {1}{3}$

Explanation

POint of intersection P

$y^2=x\\x^2=x\\x=1\quad and\quad 0$

Area enclosed

$=\int_0^1{\sqrt x-x}\\ \Rightarrow\cfrac{2}{3}x^{2/3}-\cfrac{x^2}{2}]^1_0\\ \Rightarrow\cfrac{2}{3}-\cfrac{1}{2}\\ \Rightarrow\cfrac{4-3}{6}=\cfrac{1}{6}$

1210430_1244484_ans_8007b410710247088874f0c08edb0a18.png

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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