MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 6

Let

$f\left(x\right)={x}^{2}-3x+2$ be a function, for all

$x\in R$ . On the basis of given information, answer the given question.

The area bounded by

$f\left(x\right),$ the

$x-$ axis and

$y-$ axis is,

0%
$\dfrac{1}{3}$ .sq.unit
0%
$\dfrac{2}{3}$ .sq.unit
0%
$\dfrac{3}{5}$ .sq.unit
0%
$\dfrac{5}{6}$ .sq.unit

Find the area of the region bounded by the curves

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

${x}^{2}=4ay$ .

0%
$\dfrac{16}{3}a^2\ sq.unit$
0%
$\dfrac{8}{3}a^2\ sq. unit$
0%
$\dfrac{6}{3}a^2\ sq. unit$
0%
None of these

Explanation

${y^2} = 4ax - \left( 1 \right)$

${x^2} = 4ay - \left( 2 \right)$

Solving equation 1 & 2

${y^2} = 4ax$

$\Rightarrow {\left( {\dfrac{{{x^2}}}{{4a}}} \right)^2} = 4ax$

$\Rightarrow \dfrac{{{x^4}}}{{16{a^2}}} = 4ax$

$\Rightarrow {x^4} = 64{a^3}x$

$\Rightarrow {x^4} - 64{a^3}x = 0$

$\Rightarrow x\left( {{x^3} - 64{a^3}} \right) = 0$

$\Rightarrow x = 0\,\,{x^3} = 64{a^3}\,\,\,x = 4a$

Area bounded

$= \int\limits_0^{4a} {\left( {\sqrt {4ax} - \frac{{{x^2}}}{{4a}}} \right)} dx$

$= \int\limits_0^{4a} {\left( {2\sqrt a \sqrt x - \frac{{{x^2}}}{{4a}}} \right)} dx$

$= 2\sqrt a \int\limits_0^{4a} {\sqrt x dx - \frac{1}{{4a}}\int\limits_0^{4a} {{x^2}dx} }$

$= {\left[ {2\sqrt a \times \frac{{2{x^{3/2}}}}{3}} \right]_0}^{4a} - \frac{1}{{4a}}{\left[ {\frac{{{x^3}}}{3}} \right]_0}^{4a}$

$= \dfrac{{4\sqrt a }}{3}\left[ {{{\left( {4a} \right)}^{3/2}}} \right] - \frac{1}{{12a}}\left[ {64{a^3}} \right]$

$= \dfrac{{4\sqrt a }}{3} \times 8a\sqrt a - \frac{{16}}{{3a}} \times {a^3}$

$= \dfrac{{32{a^2}}}{3} - \frac{{16{a^2}}}{3}$

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

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%
$\frac{\pi }{2} - \frac{{2\sqrt 2 }}{3}$

Let

$f$ and

$g$ be continuous function on

$a\le x\le b$ and set

$p\left(x\right)=$ max

$\left\{f\left(x\right),g\left(x\right)\right\}$ and

$q\left(x\right)=$ min

$\left\{f\left(x\right),g\left(x\right)\right\}$ , the area bounded by the curves

$y=p\left(x\right),y=q\left(x\right)$ and the ordinates

$x=a$ and

$x=b$ is given by

0%
$\int_{a}^{b}{\left(f\left(x\right)-g\left(x\right)\right)dx}$
0%
$\int_{a}^{b}{\left(p\left(x\right)-q\left(x\right)\right)dx}$
0%
$\int_{a}^{b}{\left|p\left(x\right)-q\left(x\right)\right|dx}$
0%
$\int_{a}^{b}{\left|f\left(x\right)-g\left(x\right)\right|dx}$

Explanation

Here,curve

$ABCD$ is max

$\left\{f\left(x\right),g\left(x\right)\right\}=p\left(x\right)$ and curve

$EBCF$ is min

$\left\{f\left(x\right),g\left(x\right)\right\}=q\left(x\right)$
Thus, the area can be determine by either

$\left(c\right)$ or

$\left(d\right)$
953274_1041218_ans_a4f983dc24e7457f9aee19880671ee05.png

Area bounded by the curve

$y=\ln{x}, y=0$ and

$x=3$ is

0%
$\left(\ln{9}-2\right)$ .sq.unit
0%
$\left(\ln{27}-2\right)$ .sq.unit
0%
$\ln\left(\dfrac{27}{{e}^{2}}\right)$ .sq.unit
0%
$>3$ .sq.unit

Explanation

Required Area

$=\int_{1}^{3}{\ln{x}dx}$

$=\left[x\ln{x}-x\right]_{1}^{3}$

$=\left(3\ln{3}-2\right)$

$=\ln{27}-2$ .sq.unit(option

$\left(b\right)$ is correct)

$=\ln{27}-\ln{{e}^{2}}$

$=\ln{\left(\dfrac{27}{{e}^{2}}\right)}$ .sq.unit(option

$\left(c\right)$ is correct)
Also,

$\ln{\left(\dfrac{27}{{e}^{2}}\right)}>3$

$\Rightarrow \ln{\left(\dfrac{27}{{e}^{2}}\right)}>3\ln{e}$

$\Rightarrow \ln{27}-\ln{{e}^{2}}>3\ln{e}$

$\Rightarrow \ln{27}-2\ln{e}>3\ln{e}$

$\Rightarrow \ln{27}>3\ln{e}+2\ln{e}$

$\Rightarrow \ln{27}>5\ln{e}$

$\Rightarrow \ln{27}>\ln{{e}^{5}}$ which is false.

953460_1041488_ans_2715451c52af4b1fbdcb5ce12803351b.png

The area bounded by the curve

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

$x-axis$ is

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

Explanation

$A=2\int _{ 0 }^{ 1 }{ (x-1) }^{ 2 }dx\\ A={ [\cfrac { { (x-1) }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1 }$

$A=2[0-\cfrac{-1}{3}]$

$A=\cfrac{2}{3}\ sq\ units$

1037991_1048222_ans_959b27afd33249f2a7c2c8468d3d06ac.PNG

The area bounded by the curves

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

$\left|y\right|=2x$ is

$3$ sq unit.

0%
True
0%
False

Explanation

Both,

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

$\left|y\right|=2x$ are symmetrical about

$x-$ axis
Solving

${x}^{3}={\left(2x\right)}^{2}=4{x}^{2}$ we get

$\Rightarrow {x}^{3}-4{x}^{2}=0$

$\Rightarrow {x}^{2}\left(x-4\right)=0$

$\Rightarrow x=0,x=4$

$\therefore$ Required area

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

$=\left[{x}^{2}-\dfrac{2}{5}{x}^{\frac{5}{2}}\right]_{0}^{4}$

$=14-\dfrac{2}{5}\times 32$

$=\dfrac{16}{5}$ sq units
955281_1045209_ans_24bae7585ae647b18130b927437db5ad.png

Area bounded by the lines

$y=x, x=-1, x=2$ and x-axis is-

0%
$\dfrac{5}{2}sq.$ units
0%
$\dfrac{3}{2}sq$ units
0%
$\dfrac{1}{2}sq$ units
0%
None of these

Explanation

$Area=\dfrac{1}{2}\times2\times2+\dfrac{1}{2}\times1\times1$

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

$=\dfrac{5}{2}sq units$

1174800_1050111_ans_af5e98ca12e144e0aa33ebfa153a0a26.png

Find the area of the region

$\{(x, y):x^2+y^2\leq 4, x+y\geq 2\}$ .

0%
$\pi -2$
0%
$\pi -1$
0%
$2\pi -2$
0%
$4\pi -2$

Explanation

$A=$ area

$(OAB)-$ area of

$\triangle{OAB}\\=\cfrac{1}{4}(\pi. 2^2)-\cfrac{1}{2}\times 2\times 2\\=\pi-2$
1174689_1048847_ans_10dfdebe1ad74f98bb5c2b4ee83dd3b7.PNG

Area of the region bounded by the curve

$y={25}^{x}+16$ and curve

$y=b.{5}^{x}+4$ whose tangent at the point

$x=1,$ makes an angle

${\tan}^{-1}\left(40\log{5}\right)$ with the

$x-$ axis is:

0%
$2\log_{5}{\left(\dfrac{{e}^{4}}{27}\right)}$
0%
$4\log_{5}{\left(\dfrac{{e}^{4}}{27}\right)}$
0%
$3\log_{5}{\left(\dfrac{{e}^{4}}{27}\right)}$
0%
None of these

Explanation

For

$x=1,y=b.{5}^{x}+4=5b+4$ and

$\dfrac{dy}{dx}=b.{5}^{x}\log{5}$

$\Rightarrow 5b\log{5}=40\log{5}$

$\Rightarrow b=8$
The two curves intersect at points where

$8.{5}^{x}+4={25}^{x}+16$

$\Rightarrow {5}^{2x}-8.{5}^{x}+12=0$
On factorizing we get

$x=\log_{5}{2},x=\log_{5}{6}$
Hence the area of the given region

$=\int_{\log_{5}{2}}^{\log_{5}{6}}{\left[8.{5}^{x}-{25}^{x}-12\right]dx}$

$=\left[\dfrac{8.{5}^{x}}{\log_{e}{5}}-12x-\dfrac{{25}^{x}}{\log_{e}{25}}\right]_{\log_{5}{2}}^{\log_{5}{6}}$

$=\dfrac{-{5}^{\log_{5}{36}}}{\log_{e}{5}}+\dfrac{8.{5}^{x}}{\log_{e}{5}}-12\left(\log_{5}{6}-\log_{5}{2}\right)+\dfrac{{5}^{\log_{5}{4}}}{\log_{e}{25}}-\dfrac{16}{\log_{e}{5}}$
On simplification, we get

$=\dfrac{16}{\log_{e}{5}}-12\log_{e}{3}$

$=4\log_{5}{{e}^{4}}-4\log_{5}{27}$

$=4\log_{5}{\left(\dfrac{{e}^{4}}{27}\right)}$

A curve is such that the area of the region bounded by the coordinates axes, the curve and the ordinate of any point on it is equal to the cube of that ordinate the curve represent.

0%
a pair of straight lines
0%
a circle
0%
a parabola
0%
an ellipse

Explanation

Given that,

The curve and the coordinate of any point on its equal to the cube of

that ordinates.

Let (h, k) be point on it is equal to the cube of that ordinate.

Given the area of the region bounded by the co-ordinate axis in above figure.

We know that the area of rectangle is

$Area=length\times breath$

But Given that ,

Ordinate of point $=k$

$h\times k={{k}^{3}}\,$ (given function )

$h={{k}^{2}}$

Hence it represent the parabola.

hence option

$(C)$ is correct.

1001476_1054808_ans_6fd04181cbfb4633a6d8e77bc15df063.jpg

The area enclosed between the curves

$y = log ( x+ e) ; x = log_e \left(\dfrac{1}{y}\right)$ and x-axis is

0%
3
0%
1
0%
-2
0%
0.222

Explanation

$e^{y}=-\ln y+e$

$e^{y}+\ln y=e$

$\int_{0}^{0.642}{\log(x+e)}dx+\int_{0.642}^{\infty}{e^{-x}}dx$

$[x\log(x+e)-x]_{0}^{0.642}-[e^{-x}]_{0.642}^{\infty}$

$0.642\log(0.642+e)-0.642+e^{-0.642}$

$=0.222$

999151_1054812_ans_5a31fb6b6673446ebb1232fd3e123fa2.png

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 common area between the curve

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

$y^{2}=2x$ is

0%
$\dfrac{4}{3}+2\pi$
0%
$(2\sqrt2+\pi-1)$
0%
$(\sqrt2+\pi-1)$
0%
None of these

Explanation

$y^2=2x$

$y^2=8-x^2$

$\Rightarrow \ 8x^2=2x$

$x^2+2x=8$

$\Rightarrow \ (x+1)^2=9$

$\Rightarrow \ x=\pm 3-1$

$=2-4$

$x=2\ \Rightarrow \ y=\pm 2$

$A(2, 2),\ B(2, -2)$

$Ar=\displaystyle \int _{-2}^{2} (\sqrt {8-y^2} -y^2 /2) dy$

$=2\displaystyle \int _{0}^{2} (\sqrt {8-y^2} -y^2 /2) dy$

$=\left [\dfrac {y\sqrt {8-y^2}}{3} + \dfrac {8}{2}\sin^{-1} \left (\dfrac {x}{\sqrt 8} \right) -\dfrac {1}{2} y^3 /3\right]^2$

$=\left [\dfrac {2\sqrt {8-4}}{2}+4\sin^{-1} \left (\dfrac {2}{\sqrt 8} \right) - \dfrac {1}{2} \dfrac {8}{3} - 0-0 \right]$

$=2\left [2+4 \pi /4 -4/3 \right]$

$=2[\pi +2/3]=2\pi +4/3$

The area common to the circle

${x}^{2}+{y}^{2}=16{a}^{2}$ and the parabola

${y}^{2}=6ax$ is

0%
$4{ a }^{ 2 }\left( 8\pi -\sqrt { 3 } \right)$
0%
$\dfrac { 4{ a }^{ 2 }\left( 4\pi +\sqrt { 3 } \right) }{ 3 }$
0%
$\dfrac { 8{ a }^{ 2 }\left( 4\pi -\sqrt { 3 } \right) }{ 5 }$
0%
$none\ of\ these$

Explanation

${x^2} + {y^2} = 16{a^2} - \left( 1 \right)$

${y^2} = 6ax - \left( 2 \right)$

$solving\,eq\left( 1 \right)\left( 2 \right)$

${x^2} + 6ax = 16{a^2}$

${x^2} + 6ax - 16{a^2} = 0$

${x^2} + 8ax - 2ax - 16{a^2} = 0$

$x\left( {x + 8a} \right) - 2a\left( {x + 8a} \right) = 0$

$\left( {x + 8a} \right)\left( {x - 2a} \right) = 0$

$x = 2a, - 8a$

$Area\,bounded = 2\left[ {\int\limits_0^{2a} {{y_1}dx + \int\limits_{2a}^{4a} {{y_2}dx} } } \right]$

$= 2\left[ {\int\limits_0^{2a} {\sqrt {6a} \sqrt x dx + \int\limits_{2a}^{4a} {\sqrt {{{\left( {4a} \right)}^2} - {x^2}} } } dx} \right]$

$= 2\left[ {{{\left( {\sqrt {6a} \frac{{{x^{3/2}}}}{{3/2}}} \right)}_0}^{2a} + {{\left( {\frac{x}{2}\sqrt {16{a^2} - {x^2}} + \dfrac{{16{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{{4a}}} \right)}_{2a}}^{4a}} \right]$

$= 2\left[ {\dfrac{2}{3}\sqrt {6a} {{\left( {2a} \right)}^{3/2}} + \left( {8{a^2} \times \frac{\pi }{2} - \left( {a\sqrt {12{a^2}} + 8{a^2} \times \dfrac{\pi }{6}} \right)} \right)} \right]$

$= 2\left[ {\dfrac{{8\sqrt 3 }}{3}{a^2} + \dfrac{{8\pi {a^2}}}{3} - 2\sqrt 3 {a^2}} \right]$

$= 2\left[ {\dfrac{{2\sqrt 3 }}{3}{a^2} + \dfrac{{8\pi {a^2}}}{3}} \right]$

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

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

The area bounded by the curve :

$max$ {|x|,|y|}

$= 5$ is

0%
$10$
0%
$25$
0%
$100$
0%
$50$

Explanation

$max[|x|,|y|]=5$

$area=4\times5\times5$

$=100$

1157923_1053785_ans_685316fc858c426baa9e733c2d9af735.JPG

The area bounded by

$y = cos \, x , \, y = x + 1 , \, y = 0$ is

0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{5}{2}$

Explanation

Area

$=$ Area of

$\triangle{AOB}+$ Area of region

$AOC$

$=\dfrac{1}{2}\times 1\times 1+\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\cos{x}\,dx}$

$=\dfrac{1}{2}+\left[\sin{x}\right]_{0}^{\dfrac{\pi}{2}}$

$=\dfrac{1}{2}+\left[1-0\right]$

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

1458522_1058154_ans_aae9af27f4454639b6f63d92c099eedb.png

The area enclosed between the curve

$y^2 = x \, and \, y = |x|$ is

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

Explanation

$y=\mid x\mid$

$\left\{ x\quad ;\quad x\ge 0\\ -x\quad ;x<0 \right\}$

to find

$P$

$y^2=x$ and

$y=x$

$x^2=x\\x(x_1)=0$

$x=0,1\\P=1$

$A=\int _{ 0 }^{ 1 }{ \sqrt { x } -x } dx\\ A={ [\cfrac { 2{ x }^{ 3/2 } }{ 3 } -\cfrac { { x }^{ 2 } }{ 2 } ] }_{ 0 }^{ 1 }\\ A={ [\cfrac { 2 }{ 3 } -\cfrac { 1 }{ 2 } ] }\\ A=\cfrac { 1 }{ 6 }$

1037543_1062027_ans_0cbd72ae173241bb87f63a337d36e3fb.png

The value of

$\int_{-1}^{1}{f\left(x\right)dx},$ is

0%
$\dfrac{2}{15\left(k+1\right)}\left(23-10k\right)$
0%
$\dfrac{2}{15\left(k+1\right)}\left(23+10k\right)$
0%
$\dfrac{2}{15\left(k+1\right)}\left(10k-17\right)$
0%
$\dfrac{2}{15\left(k+1\right)}\left(10k+17\right)$

Explanation

$\int_{0}^{t}{\left(f\left(x\right)-\left({x}^{4}-4{x}^{2}\right)\right)dx}=k\int_{0}^{t}{\left(2{x}^{2}-{x}^{3}-f\left(x\right)\right)dx}$
Differentiating w.r.t

$t$ we get

$f\left(t\right)-{t}^{4}+4{t}^{2}=k\left(2{t}^{2}-{t}^{3}-f\left(t\right)\right)$

$\Rightarrow \left(1+k\right)f\left(t\right)={t}^{4}-4{t}^{2}+k\left(2{t}^{2}-{t}^{3}\right)$

$\therefore f\left(t\right)=\dfrac{{t}^{4}-4{t}^{2}+k\left(2{t}^{2}-{t}^{3}\right)}{k+1}$

$\int_{-1}^{1}{f\left(x\right)dx}=\int_{-1}^{1}{\dfrac{{x}^{4}-4{x}^{2}+k\left(2{x}^{2}-{x}^{3}\right)}{k+1}dx}$

$=\int_{-1}^{1}{\dfrac{{x}^{4}-k{x}^{3}+\left(2k-4\right){x}^{2}}{k+1}dx}$
We know that

$\int_{-a}^{a}{f\left(x\right)dx}=0$ if

$f\left(x\right)=-f\left(x\right)$
Thus,

$\int_{-1}^{1}{{x}^{3}dx}=0$ .Thus the above integral becomes,

$\int_{-1}^{1}{\dfrac{{x}^{4}+\left(2k-4\right){x}^{2}}{k+1}dx}$
Using the concept

$\int_{-a}^{a}{f\left(x\right)dx}=2\int_{0}^{a}{f\left(x\right)dx}$ we have

$=\dfrac{2}{k+1}\left[\dfrac{{x}^{5}}{5}\right]_{0}^{1}+\dfrac{2k-4}{k+1}\times 2\left[\dfrac{{x}^{3}}{3}\right]_{0}^{1}$

$=\dfrac{2}{k+1}\times \dfrac{1}{5}+\dfrac{4k-8}{k+1}\dfrac{1}{3}$

$=\dfrac{6+20k-40}{15\left(k+1\right)}$

$=\dfrac{20k-34}{15\left(k+1\right)}=\dfrac{2\left(10k-17\right)}{15\left(k+1\right)}$

Let

$P(x, y)$ be a moving point in the

$x-y$ plane such that

$[x].[y]=2$ , where [.] denotes the greatest integer function, then area of the region containing the points

$P(x, y)$ is equal to:

0%
$1$ sq. units
0%
$2$ sq. units
0%
$4$ sq. units
0%
None of these

Explanation

$[x][y]=2$

Hence the area

$=4\times1$

$=4$

1380987_1066436_ans_7e7a042c507c42f3b0f001eae1109582.png

The area bounded by the curve f(x) = x + sin x and its inverse function between the ordinates

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

0%
$4 \pi$
0%
$8 \pi$
0%
$4$
0%
$8$

Explanation

R.E.F image

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

So area between

$f(x)$ &

$f^{-1} (x)$ is double

the area between

$f(x) \& \,y = x$

Area (A) =

$\displaystyle\int_0^{2\pi} (x + \sin x)^{dx} \, \int_0^{2 \pi} x dx$

$= \displaystyle \int_0^{2 \pi} \sin x \, dx$

$= \displaystyle\int^{\pi}_0 \, \sin x dx - \int_{\pi}^{2 \pi} \sin x dx$

$= [-\cos x]_0^{\pi} - [-\cos x ]_{\pi}^{2 \pi}$

$= 2 + 2 = 4$

Actual Area =

$2 \times A = 8$

Option D is correct.

1048988_1071233_ans_1d74c34f5eb74a71a0dcaa735aabaceb.png

$\theta \le x\le \pi$ ; then the area bounded by the curve

$y=x$ and

$y=x+\sin x$ is

0%
$2$
0%
$4$
0%
$2\pi$
0%
$4\pi$

Explanation

$A=\int _{ 0 }^{ \pi }x+\sin{x}-xdx$

$A=\int _{ 0 }^{ \pi }\sin{x}dx$

$A={ [-\cos x] }_{ 0 }^{ \pi}$

$A=-cos \pi -cos 0$

$a=1-(-1)=2$

1038267_1062433_ans_0917c68d77aa4e35b43aa48923d885e9.png

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

Consider the functions

$f(x)$ and

$g(x)$ , both defined from

$R \rightarrow R$ and are defined as

$f(x)=2x-x^{2}$ and

$g(x)=x^{n}$ where

$n \in N$ . If the area between

$f(x)$ and

$g(x)$ in first quadrant is

$1/2$ then

$n$ is not a divisor of :

0%
$12$
0%
$15$
0%
$20$
0%
$30$

Explanation

$Area=\int\limits_{a}^{b}{\left( upper\,curve \right)-\left( lower\,curve \right)dx}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( f\left( x \right)-g\left( x \right) \right)dx}$

$\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( 2x-{{x}^{2}}-{{x}^{n}} \right)dx}$

$\,\,\,\,\,\,\,\,\,=\left[ x-\frac{{{x}^{3}}}{3}-\frac{{{x}^{n+1}}}{n+1} \right]_{0}^{1}$

$\,\,\,\,\frac{1}{2}=1-\frac{1}{3}-\frac{1}{n+1}$

$\Rightarrow \frac{1}{2}=\frac{2}{3}-\frac{1}{n+1}$

$\Rightarrow \frac{1}{n+1}=\frac{2}{3}-\frac{1}{2}$

$\Rightarrow \frac{1}{n+1}=\frac{1}{6}$

$\Rightarrow n+1=6$

$n=5$

$Hence\,n\,is\,not\,a\,divisor\,of \ 12$

The area bounded by the curves

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

$y-$ axes in first quadrant is:

0%
$\sqrt {2}-1$
0%
$\sqrt {2}$
0%
$\sqrt {2}+1$
0%
None of the above

Explanation

The area bounded by the curves

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

$y-$ axis in the first quadrant is,

$A =\displaystyle \int_{0}^{\pi/4} (\cos x - \sin x) dx$

$= [\sin x + \cos x]_{0}^{\pi/4}$

$= \left(\sin \dfrac{\pi}{4} + \cos\dfrac{ \pi}{4}\right) - (\sin 0 + \cos 0)$

$= \sqrt{2} - 1$

The area bounded by

$y=x^2, y=[x+1], x \leq 1$ and the y-axis is, where

$[.]$ is greatest integer function

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$1$
0%
$\dfrac{7}{3}$

Explanation

Since,

$0\leq x \leq 1$

Therefore,

$y=[x+1]=1$

Hence, required area =

$\int_{0}^{1} \sqrt{y} \ dy$

$=\dfrac{2}{3}[y^{3/2}]_{0}^{1}$

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

1494610_1082655_ans_6bdfd1da3fc74601a306bf1bf85becb7.PNG

The area between the curves y = tanx, y = cotx and x - axis in the interval

$[0,\pi / 2]$ is

0%
$log 2$
0%
$log 3$
0%
$log \sqrt{2}$
0%
None of these

Explanation

$A=2\int _{ 0 }^{ \pi /4 }{\tan{x} } dx$

$A=2{ [-\log \cos { x } ] }_{ 0 }^{ \pi /4 }$

$A= 2[-\log \cos { \pi/4 } ]$

$-[-\log \cos { 0 } ]$

$A=-2\log \sqrt {1/2}$

$A=\log 2$

1039575_1074194_ans_fb0b65c4b075491fadf8bca275bcef43.PNG

The are included between the curves

$y^2 = 4ax \,$ and

$\, x^2 = 4 ay$ is ____ sq units.

0%
$\dfrac{16a^2}{3}$
0%
$\dfrac{8a^2}{3}$
0%
$\dfrac{4a^2}{3}$
0%
$\dfrac{5a^2}{3}$

Explanation

We have,

$y^2 = 4ax$ --------------------------- (1)

$x^2 = 4ay$ ---------------------------- (2)

(1) and (2) intersects hence

$x = \dfrac{y^2}{4a}$ (a > 0)

$\Rightarrow (\dfrac{y^2}{4a})^2 = 4ay$

$\Rightarrow y^4 = 64a^3y$

$\Rightarrow y^4 – 64a^3y = 0$

$\Rightarrow y[y^3 – (4a)^{3}] = 0$

$\Rightarrow y = 0, 4a$

When y = 0, x = 0 and when y = 4a, x = 4a.

The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).

Area

$=\int_{b}^{a}\left [ y_1-y_2 \right ]dx$

$=\int_{0}^{4a}\left [ \sqrt{4ax}-\dfrac{x^2}{4a} \right ]dx$

$\left [ \dfrac{4\sqrt{a}}{3}(x)^{\dfrac{3}{2}} - \dfrac{x^3}{12a}\right ]_0^{4a}$

$=\dfrac{32a^2}{3}-\dfrac{16a^2}{3}$

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

The area of the region enclosed by the curves

$y=x$ ,

$x=e$ ,

$y=\dfrac{1}{x}$ and the positive

$x-axis$ is .

0%
$\dfrac{3}{2}$ square units
0%
$\dfrac{5}{2}$ square units
0%
$\dfrac{1}{2}$ square units
0%
$1$ square units

Explanation

$A=\int _{ 0 }^{ 1 }{ x } dx+\int _{ 0 }^{ e }{ \cfrac { 1 }{ x } } dx\\ A={ [\cfrac { { x }^{ 2 } }{ 2 } ] }_{ 0 }^{ 1 }+{ [\ln { x } ] }_{ 1 }^{ e }\\ A=\cfrac { 1 }{ 2 } +\ln { e } -\ln { 1 } \\ A=\cfrac { 1 }{ 2 } +1-0\\ A=\cfrac { 3 }{ 2 }$
1034229_1093100_ans_8c06fef9b20b45998d42f7124ffcc9b2.PNG

1034229_1093100_ans_8c06fef9b20b45998d42f7124ffcc9b2.PNG

The area bounded by the curve

$y=cos ax$ in one are of the curve is where

$a=4n+1,n\in integer$

0%
$2a$
0%
$1/a$
0%
$2/a$
0%
$2{a^2}$

Explanation

Area

$= \left| {\int\limits_0^{\dfrac{\pi }{2}} {\cos \,ax\,dx} } \right|$

$= {\left[ {\dfrac{{\sin \,\,ax\,\,}}{a}} \right]_0}^{\dfrac{\pi }{2}}$

$= \dfrac{1}{a} - 0$

$= \dfrac{1}{a}$

Area enclosed between the curves

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

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

0%
$\dfrac{{3\pi - 14}}{3}$ sq.units
0%
$\dfrac{{\pi - 8}}{3}$ sq.units
0%
$\dfrac{{2\pi - 8}}{3}$ sq.units
0%
None of these

Explanation

We have,

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

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

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

$y=1-{{x}^{2}}$ ……. (2)

So, area of first quadrant

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

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

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

We know that

$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{1}{2}a\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

Therefore,

$=\left[ \dfrac{1}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}\left( x \right) \right]_{0}^{1}-\left( x-\dfrac{{{x}^{3}}}{3} \right)_{0}^{1}$

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

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

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

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

$=\dfrac{\pi }{4}-\dfrac{7}{6}$

$=\dfrac{3\pi -14}{12}$

Therefore,

Area of all quadrants

$=4\times \left( \dfrac{3\pi-14}{12} \right)$

$=\dfrac{3\pi-14}{3}$

Hence, this is the answer.

1012160_1093002_ans_8b1bc5e61d864c4380cac15d55ff4a93.jpg

The area bounded by the curves

$y=xe, y=-xe$ and the line

$x=1$ is-

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

Explanation

Consider the given curves.

$y=xe$

$y=-xe$

$x=1$

Therefore, area bounded by these curves

$=\int_{0}^{1}{\left( xe-\left( -xe \right) \right)}dx$

$=2e\int_{0}^{1}{x}dx$

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

$=e\left[ {{x}^{2}} \right]_{0}^{1}$

$=e\left[ {{1}^{2}}-0 \right]$

$=e$

Hence, this is the answer.

1043588_1088501_ans_7ce9c2bfe26a4fb59d3be92e8c06c21d.jpg

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%
$None\ of\ these$

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%
$\frac {3}{2}$

Area bounded by curve

$y = k \sin \,x$ between

$x = \pi$ and

$x = 2\pi$ , is

0%
$2k$ sq. unit
0%
$0$
0%
$\dfrac{k^2}{2}$ sq. unit
0%
None of these

Explanation

$y=k\sin x, x=[\pi,2\pi]$

Area required = Area BCD

$=\int_{2\pi}^{\pi}k\sin dx$

$=k\int_{2\pi}^{\pi}\sin xdx$

$k[=\cos x]^{2\pi}_\pi$

$=-k[\cos 2\pi-\cos \pi]$

$=-k[1-(-1)]$

$-2k$

since area cannot be negative

$\therefore Area =2ksq.unit$

2115421_1105483_ans_84548591cdb44ff3b72f7d43d2330ceb.png

Suppose that

$F(\alpha)$ denotes the area of the region bounded by

$x=0$ ,

$x=2$ ,

$y^2=4x$ and

$y=|\alpha x-1|+|\alpha x-2|+\alpha x$ , where

$\alpha \in \{0, 1\}$ . Then the value of

$F(\alpha)+\dfrac{8\sqrt{2}}{3}$ , when

$\alpha =0$ , is

0%
$4$
0%
$5$
0%
$6$
0%
$9$

The area bounded by the curve

$y=sin(x-[x]),y=sin1,\,x=1$ and the x-axis is

0%
$sin1$
0%
$1-sin1$
0%
$1+sin1$
0%
$1-\cos1$

The area of the region bounded by the curves

$y=x^2$ and

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

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

Explanation

$y=x^2$ and

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

meeting point

$x^2 \left(1+x^2\right)=2$

$x=1$ and

$x=-1$ are roots

$\Rightarrow \left(x^2-1\right)$ is factor

$\qquad \quad\left( { x }^{ 2 }+2 \right) \\ \left. { x }^{ 2 }-1 \right) \overline { { x }^{ 4 }+{ x }^{ 2 }-2 } \\ \quad \quad \quad \ { x }^{ 4 }-{ x }^{ 2 }\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 2{ x }^{ 2 }-2 \qquad \qquad \qquad x^2+2=0\\ \qquad \quad 2{ x }^{ 2 }-2 \qquad \qquad \qquad x^2=-2\ not\ positive\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 0\quad$

$\displaystyle \therefore area=\int _{ -1 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx }$

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

$\displaystyle =-\left[ \int _{ -1 }^{ 1 }{ { x }^{ 2 }dx } -\int _{ -1 }^{ 1 }{ \dfrac { 2 }{ 1+{ x }^{ 2 } } dx } \right]$

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

$\displaystyle =-\left[ \left[ \dfrac { 1 }{ 3 } -\left( -1/3 \right) \right] -2\left[ \pi /4-\left( -\pi /4 \right) \right] \right]$

$\displaystyle =-\left[ \dfrac { 2 }{ 3 } -2\left[ \pi /4+\pi /4 \right] \right]$

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

The area of the region bounded by

$\left| arg\left( z+1 \right) \right| \le \frac { \pi }{ 3 }$ and

$\left|z+1 \right| \le \frac { \pi }{ 4 }$ is given by

0%
$\dfrac{4\pi}{3}$
0%
$\dfrac{16\pi}{3}$
0%
$\dfrac{2\pi}{3}$
0%
$\dfrac{20\pi}{3}$

Area common to the curve

$y^2 = 16x$ and

$y = 2x$ , is :

0%
$\dfrac{16}{3}$ sq. units
0%
$\dfrac{17}{3}$ sq. units
0%
$\dfrac{19}{3}$ sq. units
0%
$\dfrac{20}{3}$ sq. units

Explanation

$y^2=16x,y=2x$

$\Rightarrow 4\sqrt{x}=2x$

$\sqrt{x}=2\Rightarrow x=4$

Area $=\int_{0}^{4}(\sqrt{16x}-2x)dx$

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

$=\left [ \dfrac{4x^{\frac{3}{2}}}{\frac{3}{2}}-x^2 \right ]_0^4$

$=\dfrac{8}{3}[8-0]-[16-0]$

$=\dfrac{64}{3}-16$

$=\dfrac{16}{3}$ sq.units

The curves

$y = x^{2} - 1, y = 8x - x^{2} - 9$ at

0%
Intersect at right angles at $(2, 3)$
0%
Touch each other at $(2, 3)$
0%
Do not intersect at $(2, 3)$
0%
Intersect at an angle $\dfrac {\pi}{3}$

Explanation

$y={ x }^{ 2 }-1=8x-{ x }^{ 2 }-9\\ 2{ x }^{ 2 }-8x+8=0\\ { x }^{ 2 }-4x+4=0\\ { (x-2) }^{ 2 }=0\\ x=2\\ y={ x }^{ 2 }-1={ (2) }^{ 2 }-1=3$

So, curves intersect at

$(2,3)$

Now,

${ m }_{ 1 }=\cfrac { dy }{ dx }$ of

$({ x }^{ 2 }-1)$ at

$(2,3)$

${ m }_{ 1 }=2x=2(2)=4$

${ m }_{ 2 }=\cfrac { dy }{ dx }$ of

$(8x-{ x }^{ 2 }-9)$ at

$(2,3)$

${ m }_{ 2 }=8-2x=8-2(2)=4$

Since

${ m }_{ 1 }={ m }_{ 2 }$

we can say that curves touch each other at

$(2,3)$ .

The area bounded by the curves

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

$x=1$ and

$x=\beta$ is

$(\beta -1)\sin(3\beta +4)$ . Then

$f(x)$ is

0%
$(x-1)\cos(3x+4)$
0%
$\sin(3x+4)$
0%
$\sin(3x+4)+3(x-1)\cos(3x+4)$
0%
$\sin(3x+4)+x$

Two vertices of a rectangle are on the positive x-axis. The other two vertices lie on the lines

$y=4x$ and

$y=-5x+6$ . Then the maximum area of the rectangle is?

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

Explanation

Let the two vertex lying be

$(a,0)$ and

$(b,0)$ where

$a<b$

Since

$y=4x$ is containing coordinate corresponding to

$(a,0)$ ; we get

Point to be

$(a,4a)$

and the other with

$y=-5x+6$ , we get point as

$(b,-5b+6)$

Now as

$b-a=-5b+6-4a$ (Length)

$a+2d=2$

Area of Rectangle

$=$ Length

$\times$ breadth

$=(b-a)(4a)$

Area

$=(b-a)(4a)=(\cfrac { 2-a }{ 2 } -a)(4a)=(2-3a)(2a)$

Area

$=4a-6{ a }^{ 2 }$

For Maxima/Minima,

$\cfrac { d(Area) }{ da } =0\quad \Rightarrow 4-12a=0\\ a=\cfrac { 1 }{ 3 }$

Area

$=(4a-{ 6a }^{ 2 })=\cfrac { 4 }{ 3 } -\cfrac { 6 }{ 9 } =\cfrac { 2 }{ 3 }$ sq. unit

The area bounded by the curves

$x^2=4ay$ and

$y^2=4ax$ is,

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

The area of the region bounded by the curve

${a^4}{y^2} = \left( {2a - x} \right){x^5}$ is to that curve whose radius is

$a$ , is given by the ration.

0%
$5:4$
0%
$5 : 8$
0%
$2 : 3$
0%
$3 : 2$

Explanation

${{a}^{4}}{{y}^{2}}=\left( 2a-x \right){{x}^{5}}$

$cut\text{ off x-axis when y=0}$

$\left( 2a-x \right){{x}^{5}}=0$

$A_{1}^{{}}=\int\limits_{0}^{2a}{\dfrac{\sqrt{\left( 2a-x \right)}{{x}^{5/2}}}{{{a}^{2}}}}dx$

$put\,x=\,2a{{\sin }^{2}}\theta$

$dx=4a\sin \theta \cos \theta d\theta$

$\therefore A_{1}^{{}}=\int\limits_{0}^{\pi /2}{\dfrac{\sqrt{2a}\cos \theta {{\left( 2a \right)}^{5/2}}{{\sin }^{5}}\theta \times 4a\sin \theta \cos \theta }{{{a}^{2}}}}d\theta$

$\,\,\,\,\,\,\,\,\,=32{{a}^{2}}\int\limits_{0}^{\pi /2}{{{\sin }^{6}}\theta {{\cos }^{2}}\theta d\theta }$

$\,\,\,\,\,\,\,=32{{a}^{2}}\dfrac{\left( 5.3.1 \right)\left( 1 \right)}{8.6.4.2}\dfrac{\pi }{2}\,\,\,\,\left[ by\,Walli'sformula \right]$

$\,\,\,\,\,\,=\dfrac{5\pi {{a}^{2}}}{8}$

$Area\,of\,circle\,A_{2}^{{}}=\pi {{a}^{2}}$

$\dfrac{A_{1}^{{}}}{A_{2}^{{}}}=\dfrac{5}{8}$

$A_{1}^{{}}:A_{2}^{{}}=5:8$

The area enclosed between the curves

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

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

$\\ (a>0)$ is

$1sq.unit$ . then

$a=$

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

Explanation

$y=a(ay^2)^2\\y=a^3y^4\\y(y^3a^3-1)=0$

$y=0,1/a$

$x=ay^2\\x=a/a^2$

$x=1/a$

$\\ A=[\int _{ 0 }^{ 1/a }{ \sqrt { \cfrac { x }{ a } } -a{ x }^{ 2 } } dx\\ A={ [\cfrac { 2{ x }^{ 3/2 } }{ 3\sqrt { a } } -\cfrac { a{ x }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1/a }\\ A=\cfrac { 2{ (1/a) }^{ 3/2 } }{ 3\sqrt { a } } -\cfrac { a{ (1/a) }^{ 3 } }{ 3 } \\ A=\cfrac { 2 }{ 3{ a }^{ 2 } } -\cfrac { 1 }{ 3{ a }^{ 2 } } \\ A=\cfrac { 1 }{ 3{ a }^{ 2 } }$

given

$A=1\\\cfrac{1}{3a^2}=1\\3a^2=1\\a^2=1/3\\a=\sqrt{1/3}$

If area bounded by

$f(x)=x^{\frac{1}{3}}(x-1)$

$x-$ axis is A then find the value of

$28A$ .

0%
$5$
0%
$6$
0%
$7$
0%
$9$

Explanation

We have,

$f(x)=y=x^{1/3}(x-1)$

Since,

$x-axis$ means

$y=0$

Therefore,

$f(x)=y=x^{1/3}(x-1)=0$

$x=0,1$

Area bounded by this function is

$A$ .

$A=\left|\displaystyle \int_{0}^{1}x^{1/3}(x-1)dx\right|$

$A=\left|\displaystyle \int_{0}^{1}(x^{4/3}-x^{1/3})dx\right|$

$A=\left|\displaystyle \left(\dfrac{x^{7/3}}{7/3}-\dfrac{x^{4/3}}{4/3}\right)_{0}^{1}\right|$

$A=\left|\displaystyle \left(\dfrac{3}{7}(1)-\dfrac{3}{4}(1)\right)\right|$

$A=\left|\displaystyle \left(\dfrac{12-21}{28}\right)\right|$

$A=\left|\displaystyle \left(\dfrac{-9}{28}\right)\right|$

$A=\dfrac{9}{28}$

Therefore,

$28A=28\times \dfrac{9}{28}=9$

Hence, this is the answer.

The area of the region bounded by the curves

$y=|x-2|$ and

$y=4-|x| is-$

0%
2
0%
4
0%
5
0%
6

Explanation

From the attached fig.,

$ABCD$ is the region bounded by the curves

$y = \left| x - 2 \right|$ and

$y = 4 - \left| x \right|$ .

Since

$ABCD$ is a rectangle, therefore

$ar{\left( ABCD \right)} =$ area of rectangle

$= AB \times BC$

Now,

$AB = \sqrt{{\left( -1 - 0 \right)}^{2} + {\left( 4 - 3 \right)}^{2}} = \sqrt{2}$

$BC = \sqrt{{\left( 3 - 0 \right)}^{2} + {\left( 1 - 4 \right)}^{2}} = 3 \sqrt{2}$

$\therefore ar{\left( ABCD \right)} = \sqrt{2} \times 3 \sqrt{2} = 6 \text{ sq. units}$

Hence the area of region bounded by the curves

$y = \left| x - 2 \right|$ and

$y = 4 - \left| x \right|$ is

$6$ units.

1180816_1152613_ans_1a96503685664d09a7a9f918b11596d6.png

The area enclosed between the curves

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

$x={ ay }^{ 2 }$

$(a>0)$ is

$1\ sq.unit$ . then

$a=$

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

Explanation

$\Rightarrow x = a(ax^2)^2$

$\Rightarrow x = 0, \dfrac{1}{a} \Rightarrow y = 0, \dfrac{1}{a}$

$\therefore$ Point of intersection are

$(0,0)$ and

$\left(\dfrac{1}{a} , \dfrac{1}{a} \right)$ .

Thus, required area

$OABCO =$ Area of curve

$OCBDO -$ area of curve

$OABDO$

$\Rightarrow \displaystyle \int_0^{1/a} \left(\sqrt{\dfrac{x}{a}} - ax^2 \right) dx = 1$ (given)

$\Rightarrow \left[\dfrac{1}{\sqrt{a}} . \dfrac{x^{3/2}}{3/2} - \dfrac{ax^3}{3} \right]_0^{1/a} = 1$

$\Rightarrow \dfrac{2}{3a^2} - \dfrac{1}{3a^2} = 1$

$\Rightarrow a^2 = \dfrac{1}{3} \Rightarrow a = \dfrac{1}{\sqrt{3}} \, \, \, (\because a > 0)$

Hence, this is the answer.

1291424_1140623_ans_b2e5868857c4434997b5304d3cd4a64c.png

The area of the region bounded by

$y=(x-4)^2, y=16-x^2$ and the x axis,is

0%
$16$
0%
$32$
0%
$\dfrac{64}{3}$
0%
$64$

Explanation

$y_1=x^2+16-8x$

$y_2=16-x^2$

$16-x^2=x^2=16-8x$

$\Rightarrow 2x^2-8x=0$

$x^2-4=0$

$x=0,4$

$\overset { 4 }{ \underset { 0 }{ \int } } (y_1-y_2)dx$

$\Rightarrow { }{ \underset { 0 }{ \int } } (2x^2-8x)dx$

$\Rightarrow { \left[ \dfrac { { 2x }^{ 3 } }{ 3 } -{ 4x }^{ 2 } \right] }_{ 0 }^{ 4 }$

$\left| \dfrac { 128 }{ 3 } -64 \right| =\dfrac { 64 }{ 3 }\\$

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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