MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 11

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

Find the area bounded by

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

$y = x$ .

0%
$\displaystyle \frac{1}{8}sq.units$
0%
$\displaystyle \frac{1}{4}sq.units$
0%
$\displaystyle \frac{1}{12}sq.units$
0%
$\displaystyle \frac{1}{6}sq.units$

The area of the figure bounded by

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

$x-y-1= 0$ is:

0%
$2/3$
0%
$4/3$
0%
$8/3$
0%
$11/3$

Explanation

From figure required area

$\displaystyle -\int _{ 0 }^{ -\frac { 1 }{ 2 } }{ \left( \sqrt { 2x+1 } -\left( 1-x \right) \right) dx } +\int _{ -\frac { 1 }{ 2 } }^{ 4 }{ \left( \sqrt { 2x+1 } -\left( 1-x \right) \right) dx }$

$\displaystyle ={ \left[ \frac { 1 }{ 2 } \frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ -\frac { 1 }{ 2 } }+{ \left[ \frac { 1 }{ 2 } \frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ -\frac { 1 }{ 2 } }^{ 4 }=\frac { 11 }{ 3 }$

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

$1:1:1$

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

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

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

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

The curve

$f(x) = \displaystyle Ax^{2}+Bx+C$ passes through the point (1, 3) and line

$4x + y = 8$ is tangent to it at the point (2, 0). The area enclosed by

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

0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{16}{3}$
0%
$\dfrac{32}{3}$

Explanation

Given curve is

$\displaystyle y=f(x)=Ax^{2}+Bx+C.........(i)$

It passes through

$(1, 3)$

$\displaystyle \therefore 3=A+B+C..........(ii)$

Point

$(2, 0)$ also lie on the curve

$(i)$

$\displaystyle \therefore 0=4A+2B+C.........(iii)$

$\displaystyle -C=4A+2B$
from eq (ii)

$\displaystyle 3=A+B-4A-2B$

$\displaystyle B=3(1-A)........(v)$

Slope of tangent is

$-4$

$\displaystyle \therefore -4=4A+B.........(iv)$

$\displaystyle \therefore$ from

$(ii), (iii)$ &

$(iv)$ and

$(v)$ we get

$A = -1, B = 0, C = 4$

Thus, the required curve is

$\displaystyle y=-x^{2}+4$
Hence required area

$=$ area of

$\displaystyle \Delta OAB-\int_{0}^{2}\left ( -x^{2}+4 \right )dx=8-\left(-\frac{8}{3}+8\right)=\frac{8}{3}$

For which of the following values of m is the area of the region bounded by the curve

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

$y = mx$ equals to 9/2 ?

0%
- 4
0%
- 2
0%
2
0%
4

Explanation

The two curves meet at

$\displaystyle mx=x-x^{2}\:or\:x^{2}=x\left ( 1-m \right )\therefore x=0,1-m$

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

$\displaystyle \frac{9}{2}$ if m < 1or

$\displaystyle \left ( 1-m \right )^{3}\left [ \frac{1}{2}-\frac{1}{3} \right ]=\frac{9}{2}$ or

$\displaystyle \left ( 1-m \right )^{3}=27$

$\displaystyle \therefore m =-2$
But if m > 1 then 1 - m is negative then

$\displaystyle \left [ \left ( 1-m \right )\frac{x^{2}}{2}-\frac{x^{}3}{3} \right ]^{0}_{1-m}=\frac{9}{2}-\left ( 1-m \right )^{3}\left ( \frac{1}{2}-\frac{1}{3} \right )=\frac{9}{2}$

$\displaystyle \therefore -\left ( 1-m \right )^{3}=-27$ or

$\displaystyle 1-m=-3\therefore m=4$

The area bounded between the curve

$\displaystyle y = \tan x$ ; tangent drawn to it at

$\displaystyle x=\frac { \pi }{ 4 }$ and

$\displaystyle y\ge 0$ is

0%
$\displaystyle \frac { 1 }{ 4 } \left( { \log }_{ e }4-1 \right)$
0%
$\displaystyle \frac { 1 }{ 2 } \left( {\log }_{ e }4-1 \right)$
0%
$\displaystyle \frac { 1 }{ 2 } \left( { \log }_{ e }4+1 \right)$
0%
$\displaystyle \frac { 1 }{ 4 } \left( {\log }_{ e }4+1 \right)$

Explanation

$y=\tan X$

now slope(m) of the line is

$\dfrac{dy}{dx}_{(x=\dfrac{\pi}{4})}=\sec^{2}X=2$

now in graph of

$\tan X$ when

$x=\dfrac{\pi}{4},y=1$

so equation of line is

$(y-y_1)=m(x-x_1)$

so equation of line is

$y=2x+1-\dfrac{\pi}{2}$

now at

$y=0, x=\dfrac{\pi}{4}-\dfrac{1}{2}$

now

$A=\int_{0}^{\tfrac{\pi}{4}}(\tan X)-\int_{(\tfrac{\pi}{4}-\tfrac{1}{2})}^{\tfrac{\pi}{4}}(y=2x+1-\dfrac{\pi}{2})$

$A=\ln(\sec X)-[x^2-\dfrac{\pi x}{2}+x]$

now putting upper and lower limt,and on simplified the term, we will get

$A=\dfrac{(\ln 4-1)}{4}$

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

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

$\Rightarrow a=32, c=3$

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

Suppose

$g(x) = 2x+1$ and

$h(x) = \displaystyle 4x^{2}+4x+5$ and

$h(x) = (fog)(x)$ The area enclosed by the graph of the function

$y = f(x)$ and the pair of tangents drawn to it from the origin is

0%
8/3
0%
16/3
0%
32/3
0%
none

Explanation

$g(x) = 2x+1$ and

$h(x) = \displaystyle 4x^{2}+4x+5 = (2x+1)^2 + 4 = g(x)^2 + 4$

$h(x) = (fog)(x)$

$f(x) = x^2+4$
Let the point where tangent touches the parabola be (x,y)=

$(x, x^2+4)$
Slope =

$2x$
Equation of tangent

$\cfrac{x^2+4}{x} = \cfrac{2x}{1}$

$x = \pm 2$

$y = 8$
Area enclosed

$= 2(\cfrac{1}{2} 4 \times 8 - \int _4 ^8 \sqrt{y-4}dy)$

$= \cfrac{16}{3}$

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,

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

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

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

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

$= \cfrac{1}{3}$

The area of the figure bounded by

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

$x-y-1=0$ is

0%
$\displaystyle \frac { 16 }{ 3 }$
0%
$\displaystyle \frac { 8 }{ 3 }$
0%
$\displaystyle \frac { 4 }{ 3 }$
0%
None of these

Explanation

Solving

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

$x-y-1=0$

we get

$\displaystyle \frac { { y }^{ 2 }-1 }{ 2 } =y+1\Rightarrow { y }^{ 2 }-2y-3=0$

$\Rightarrow \left( y-3 \right) \left( y+1 \right) =0\Rightarrow y=3,-1$

Area of strip is

$\displaystyle =\int _{ -1 }^{ 3 }{ \left( y+1-\frac { { y }^{ 2 }-1 }{ 2 } \right) dy } ={ \left[ \frac { { y }^{ 2 } }{ 2 } -\frac { { y }^{ 3 } }{ 6 } +\frac { 3 }{ 2 } y \right] }_{ -1 }^{ 3 }$

$\displaystyle =\left( \frac { 9 }{ 2 } -\frac { 1 }{ 2 } \right) -\left( \frac { 27 }{ 6 } -\frac { -1 }{ 6 } \right) +\frac { 3 }{ 2 } \left[ 3-\left( -1 \right) \right] =\frac { 16 }{ 3 } \quad$

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,

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

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

The smaller area enclosed by

$y=f(x)$ , where

$f(x)$ is polynomial of least degree satisfying

$\displaystyle{ \left[ \lim _{ x\rightarrow 0 }{ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } } \right] }^{ \tfrac { 1 }{ x } }=e$ and the circle

$x^2+y^2=2$ above the

$x-$ axis is

0%
$\displaystyle\frac { \pi }{ 2 } +\frac { 3 }{ 5 }$
0%
$\displaystyle\frac { \pi }{ 2 } -\frac { 3 }{ 5 }$
0%
$\dfrac { \pi }{ 2 } -\dfrac { 6 }{ 5 }$
0%
None of these

Explanation

Since

$\displaystyle\lim _{ x\rightarrow 0 }{ { \left[ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } \right] }^{ \tfrac { 1 }{ x } } }$ exists, so

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { f\left( x \right) }{ { x }^{ 3 } } } =0$

$\therefore f(x)={ a }_{ 4 }{ x }^{ 4 }+{ a }_{ 5 }{ x }^{ 5 }+...+{ a }_{ n }{ x }^{ n },a_n\neq 0,n\ge 4$

Since,

$f(x)$ is of least degree

$\Rightarrow f(x)={ a }_{ 4 }{ x }^{ 4 }$
The graph of

$y=x^4$ and

$x^2+y^2=2$ are shown in the figure

$\therefore$ The required area

$\displaystyle=2\int _{ 0 }^{ 1 }{ \left( \sqrt { 2-{ x }^{ 2 } } -{ x }^{ 4 } \right) } dx=\frac { \pi }{ 2 } +\frac { 3 }{ 5 }$

The line

$3x + 2y =13$ divides the area enclosed by the curve

$\displaystyle 9x^{2}+4y^{2}-18x-16y-11=0$ in two parts Find the ratio of the larger area to the smaller area

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

Explanation

$9{ x }^{ 2 }+4{ y }^{ 2 }-18x-16y-11=0\Rightarrow 9\left( { x }^{ 2 }-2x \right) +4\left( { y }^{ 2 }-4y \right) =11\\ \Rightarrow 9\left( { \left( x-1 \right) }^{ 2 }-1 \right) +4\left( { \left( y-2 \right) }^{ 2 }-4 \right) =11\Rightarrow 9{ \left( x-1 \right) }^{ 2 }+4{ \left( y-2 \right) }^{ 2 }=36$

$\displaystyle \\ \Rightarrow \dfrac { { \left( x-1 \right) }^{ 2 } }{ 4 } +\dfrac { { \left( y-2 \right) }^{ 2 } }{ 9 } =1$
Let

$x-1=X$ and

$y-2=Y$

$\displaystyle \Rightarrow \frac { { X }^{ 2 } }{ 4 } +\frac { { Y }^{ 2 } }{ 9 } =1$
Hence

$3x+2y=13$

$\displaystyle \Rightarrow 3\left( X+1 \right) +2\left( Y-2 \right) =13\Rightarrow 3X+2Y=6\Rightarrow \frac { X }{ 2 } +\frac { Y }{ 3 } =1$
Therefore area of triangle

$\displaystyle OPQ=\frac { 1 }{ 2 } \times 2\times 3=3$
Also area of ellipse

$=\pi$ (semimajor axes)(semi minor axis)

$=\pi .3.2=6\pi$

$\displaystyle { A }_{ 1 }=\frac { 6\pi }{ 4 } -$ area of

$\displaystyle \triangle OPQ=\frac { 3\pi }{ 2 } -3$

$\displaystyle { A }_{ 2 }=3\left( \frac { 6\pi }{ 4 } \right) +$ area of

$\displaystyle \triangle OPQ=\frac { 9\pi }{ 2 } +3$
Hence

$\displaystyle \dfrac { { A }_{ 2 } }{ { A }_{ 1 } } =\dfrac { \dfrac { 9\pi }{ 2 } +3 }{ \dfrac { 3\pi }{ 2 } -3 } =\dfrac { 3\pi +2 }{ \pi -2 }$

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

$a=9$ .

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

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

$=2-\sqrt { 2 }$

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

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

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

$=2$

Find the value(s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line

$\displaystyle y=\frac{a^{2}-ax}{1+a^{4}}$ & the parabola

$\displaystyle y=\frac{x^{2}+2ax+3a^{2}}{1+a^{4}}$ is the greatest

0%
$\displaystyle a=2^{1/4}$
0%
$\displaystyle a=5^{1/4}$
0%
$\displaystyle a=7^{1/4}$
0%
$\displaystyle a=3^{1/4}$

Explanation

$\displaystyle A=\int_{-a}^{-2a}\frac{a^{2}-ax-\left ( x^{2}+2ax+3a^{2} \right )}{1+a^{4}}dx$

$\displaystyle =\frac{3}{2}\frac{a^{3}}{1+a^{4}}$ Now f(a)

$\displaystyle =\frac{3}{2}\frac{a^{3}}{1+a^{4}}\Rightarrow f'(a)=0\Rightarrow \left ( 1+a^{4} \right )3a^{2}-a^{3}4a^{3}=0$

$\displaystyle \Rightarrow a_{min}=0,\:\:a_{min}=3^{1/4}$

Find the positive value of 'a' for the which the parabola

$\displaystyle y=x^{2}+1$ bisects the area of the rectangle with vertices

$(0, 0), (a, 0), (0, \displaystyle a^{2}+1)$ and

$(a, \displaystyle a^{2}+1)$

0%
$\displaystyle \sqrt 2$
0%
$\displaystyle \sqrt 3$
0%
$\displaystyle \sqrt 5$
0%
$\displaystyle \sqrt 7$

Explanation

Area of the rectangle outside the parabola

$= \int _0 ^{a} (x^2+1)dx$

$= \cfrac{a^3}{3} + a$
Given it is equal to half the area of rectangle

$\cfrac{a(a^2+1)}{2} = \cfrac{a^3}{3} + a$
On solving

$a = \sqrt{3}$

A figure is bounded by the curves

$\displaystyle y=\left | \sqrt{2}\sin \frac{\pi x}{4} \right |$

$y = 0, x = 2$ &

$x = 4$ . At what angles to the positive

$x$ -axis straight lines must be drawn through

$(4, 0)$ , so that these lines divide the figure into three parts of the same size

0%
$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi },\pi +\tan^{-1} \frac{4\sqrt{2}}{3\pi }$
0%
$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{2\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{2\pi }$
0%
$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{3\pi }$
0%
$\displaystyle \pi +\tan ^{-1}\frac{2\sqrt{2}}{2\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{2\pi }$

Explanation

Let equation of line is

$y = mx - 4m$

$\displaystyle A=\int_{2}^{4}\sqrt{2}\sin \frac{\pi }{4}x\ dx=\left [ -\sqrt{2}\frac{4}{\pi }\cos \frac{\pi x}{4} \right ]^{4}_{2}=\frac{4\sqrt{2}}{\pi }....(i)$

Also area of

$\displaystyle \Delta ABC=\frac{1}{2}.2.\left ( -2m_{1} \right )=-2m_{1}...(ii)$ from (i) and (ii)

$\displaystyle -2m_{1}=\frac{4\sqrt{2}}{3\pi }\Rightarrow m_{1}=\frac{-2\sqrt{2}}{3\pi }$

$\displaystyle \Rightarrow \tan \left ( \pi -\theta _{1} \right )=\frac{-2\sqrt{2}}{3\pi }\Rightarrow \pi -\theta _{1}=\tan ^{-1}\frac{2\sqrt{2}}{3\pi }$

$\displaystyle \Rightarrow \theta _{1}=\pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi }$ or

$\displaystyle \frac{1}{2}.(2)\left ( -2m_{2} \right )=\frac{8\sqrt{2}}{3\pi }$

$\displaystyle \Rightarrow m_{2}=\frac{-4\sqrt{2}}{3\pi }\Rightarrow \tan \left ( \pi -\theta _{2} \right )=\frac{-4\sqrt{2}}{3\pi }$

$\displaystyle \Rightarrow \theta _{2}=\pi -\tan ^{-1}\frac{4\sqrt{2}}{3\pi }$

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

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

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%
$\displaystyle f(x)=x^{2}+1;y=\pm x; A=\frac{2}{3}sq. units$
0%
$\displaystyle f(x)=x^{2}+1;y=\pm x; A=\frac{4}{3}sq. units$
0%
$\displaystyle f(x)=x^{2}+1;y=\pm 2x; A=\frac{2}{3}sq. units$
0%
$\displaystyle f(x)=x^{2}+1;y=\pm 2x; A=\frac{4}{3}sq. units$

Explanation

$\displaystyle f\left ( x+1 \right )=f(x)+2x+1$

$\displaystyle \Rightarrow f''\left ( x+1 \right )=f''(x)\:\:\:\:\forall \times \in R$ Let

$f"(x) = a \displaystyle \Rightarrow$

$f'(x) = ax + b \displaystyle \Rightarrow f(x)=\frac{ax^{2}}{2}+bx+c\Rightarrow c=1$

$\displaystyle \left [ \because f(0)=1 \right ]$
Now

$f(x + 1) - f(x) = 2x + 1 \displaystyle \Rightarrow \left [ \frac{a}{2}\left ( x+1 \right )^{2}+b\left ( x+1 \right )+c \right ]-\left [ \frac{ax^{2}}{2}+bx+c \right ] =2x+1$

$\displaystyle \Rightarrow ax+\frac{a}{2}+b=2x+1$ on comparing we get a = 2
or

$\displaystyle \frac{a}{2}+b=1\Rightarrow b=0$

$\displaystyle \therefore f(x)=x^{2}+1.........(i)$
Now let equation of tangent be

$y = mx ....... (ii)$
From (i) and (ii), we get

$\displaystyle x^{2}-mx+1=0\Rightarrow m=\pm 2$

$\displaystyle \therefore tangents\:\:\: are\:\:\:y=2x \:or\:\:y=-2x$

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

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.

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

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

The area of the region bounded by the parabola

$y={x}^{2}-4x+5$ and the straight line

$y=x+1$ is

0%
$\cfrac{1}{2}$
0%
$2$
0%
$3$
0%
$\cfrac{9}{2}$

Explanation

Given equation of parabola is

$y={x}^{2}-4x+5$

$\Rightarrow$

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

$\Rightarrow$

${(x-2)}^{2}=(y-1)....(i)$
and equation of line is

$y=x+1$

$\Rightarrow$

$x-y=-1.....)(ii)$

On putting the value of

$(y-1)$ from eqs

$(ii)$ in

$(i)$ we get

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

$\Rightarrow$

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

$\Rightarrow$

${x}^{2}-5x+4=0$

$\Rightarrow$

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

$\Rightarrow$

$x(x-4)-1(x-4)=0$

$\Rightarrow$

$(x-1)(x-4)=0$ or

$x=1,4$
then from is

$(ii)$

$y=2,5$

$\therefore$ Required area

$=\displaystyle \int _{ 1 }^{ 4 }{ \left( -{ x }^{ 2 }+5x-4 \right) dx } ={ \left[ \cfrac { -{ x }^{ 3 } }{ 3 } +\cfrac { 5{ x }^{ 2 } }{ 2 } -4x \right] }_{ 1 }^{ 4 }$

$=\left[ -\cfrac { 64 }{ 3 } +40-16+\cfrac { 1 }{ 3 } -\cfrac { 5 }{ 2 } +4 \right]$

$=\left(-21-\cfrac { 5 }{ 2 } +28\right)=\cfrac { 9 }{ 2 }$

Area bounded by

$\displaystyle y=1-\ell { n }^{ 2 }x,x-$ axis which is common with

$\displaystyle x\ge 1$ , is -

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

Explanation

$\displaystyle A=\int _{ 1 }^{ e }{ \left( 1-\ell { n }^{ 2 }x \right) dx } =\left( e-1 \right) -\int _{ 1 }^{ e }{ 1.\ell { n }^{ 2 }xdx }$

$\displaystyle =e-1-\left[ [x\ell { n }^{ 2 }x{ ] }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ x.2\ell nx.\frac { 1 }{ x } dx } \right]$

$\displaystyle =e-1-\left[ e-0-2.[x\ell nx-x{ ] }_{ 1 }^{ e } \right]$

$\displaystyle =e-1-e+2\left[ e-e-0+1 \right]$

$=1$

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

$A=A^*-A^@$

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

The area of the region bounded by the curves

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

$y^{2} = 2x$ is

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

Explanation

Given curves,

$x^{2} + y^{2} = 8$ .....(i)
and

$y^{2} = 2x$ .....(ii)

On solving Eqs. (i) and (ii), we get

$x^{2} + 2x - 8 = 0$

$x^{2} + 4x - 2x - 8 = 0$

$x (x - 4) - 2 (x + 4) = 0$

$(x- 2) (x +4) = 0$

$\therefore x = 2$ and

$y = \pm 2$

$\therefore$ Required area

$= 2[\text {Area of OAP} + \text {Area of PAB}]$

$= 2\left [\displaystyle\int_{0}^{2} \sqrt {2x} dx + \int_{2}^{2\sqrt {2}} \sqrt {8 - x^{2}} dx \right ]$

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

$= 2\left [\dfrac {2\sqrt {2}}{3} (2^{3/2}) + 4\times \dfrac {\pi}{2} - 2 - 4\times \dfrac {\pi}{4}\right ]$

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

$= 2\left [\dfrac {8}{3} - 2 + \pi \right ] + 2\left (\dfrac {2}{3} + \pi \right ) = 2\pi + \dfrac {4}{3}$

Area common to the curves

$5x^2 = 0$ and

$2x^2 + 9 = 0$ is equal to

0%
$12 \sqrt 3$
0%
$6 \sqrt3$
0%
$36$
0%
$18$

Explanation

Given curves

$y=5x^2$

$\Rightarrow x^2=\dfrac{y}{5}$ ....(1)
which is a parabola opening upward having vertex at (0,0)

Other curve is

$2x^2+9=y$ .....(2)

$\Rightarrow x^2=\dfrac{y-9}{2}$
which is a parabola having vertex at (0,9).

Now, solving eqn (1) and (2), we get

$5x^2=2x^2+9$

$\Rightarrow 3x^2=9$

$\Rightarrow x=\pm\sqrt{3}$

$\Rightarrow y=15$
Point of intersection of the curves is

$(-\sqrt{3},15)$ and

$(\sqrt{3},15)$

Required area

$=2(ar OAB)$

$=2 \int ^\sqrt 3 _0 (y_1-y_2)dx$

$= 2 \int ^\sqrt 3 _0 \{ (2x^2 + 9) - 5 x^2\} dx$

$2 \int ^\sqrt 3 _ 0 (9-3x^2) dx$

$= 2 \left( 9x - 3 \displaystyle \frac{x^3}{3}\right) ^\sqrt3 _0$

$= 2(9 \sqrt 3 - 3 \sqrt 3 ) = 12 \sqrt 3$ sq. units

The area of the region, bounded by the curves

$y = \sin^{-1} x + x (1 - x)$ and

$y = \sin^{-1} x - x (1 - x)$ in the first quadrant, is

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

Explanation

$\sin^{-1} x$ is defined, if

$-1 \leq x \leq 1$
In first quadrant

$0\leq x\leq 1$ and

$x(1 - x) \geq 0$

$\therefore y = \sin^{-1} x + x (1 - x)$ ...... (i)
Lies above

$y = \sin^{-1} x - x (1 - x)$ ...... (ii)
On solving, we get

$2x (1 - x) = 0$

$\Rightarrow x = 0, 1$

$\therefore$ Required area

$=\displaystyle \int_{0}^{1} (y_{1} - y_{2})dx$

$=\displaystyle \int_{0}^{1} [\left \{\sin^{-1} x + x(1 - x)\right \} - \left \{\sin^{-1} x - x(1 - x) \right \}]dx$

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

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

The area of the region enclosed between by the

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

${y^2} = 6x$ .

0%
$\dfrac{2}{3} (\sqrt 3 + 4\pi)$ sq. units
0%
$\dfrac{4}{3} (\sqrt 3 + 4\pi)$ sq. units
0%
$\dfrac{2}{3} (\sqrt 3 + 8\pi)$ sq. units
0%
$\dfrac{4}{3} (\sqrt 3 + 8\pi)$ sq. units

Explanation

Point of intersection of the parabola and the circle is obtained by solving the equations:

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

${y}^{2}=6x$

$\Rightarrow\,{x}^{2}+6x-16=0$

$\Rightarrow\,{x}^{2}+8x-2x-16=0$

$\Rightarrow\,x\left(x+8\right)-2\left(x+8\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+8\right)=0$

$\Rightarrow\,x=2,\,x=-8$

$\therefore\,x=2$ is the only possible solution(from the fig.)

$\therefore\,$ when

$x=2,\,y=\pm\sqrt{6\times 2}=\pm\,2\sqrt{3}$

$\therefore\,B\left(2,2\sqrt{3}\right)$ and

${B}^{\prime}\left(2,-2\sqrt{3}\right)$ are the points of intersection of parabola and the circle.

Required area

$=area\,of\,OBA{B}^{\prime}O=2\,area\,of \,OBAO$

$=2\left[area\,of\,OBDO+area\,of\,DBAD\right]$

$=2\left[\displaystyle\int_{0}^{2}{\sqrt{6x}dx}+\displaystyle\int_{2}^{4}{\sqrt{16-{x}^{2}}dx}\right]$

$=2\left[\sqrt{6}\left[\dfrac{{x}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{2}+\left[\dfrac{1}{2}x\sqrt{16-{x}^{2}}+\dfrac{1}{2}\times 16{\sin}^{-1}{\dfrac{x}{4}}\right]_{2}^{4}\right]$

$=2\left[\left[\sqrt{6}\times\dfrac{2}{3}\times{2}^{\frac{3}{2}}-0\right]+\left[\dfrac{1}{2}\times\,4\sqrt{16-16}+\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{4}{4}}-\dfrac{1}{2}\times\,2\sqrt{16-4}-\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{1}{2}}\right]\right]$

$=2\left[\dfrac{8\sqrt{3}}{3}+\dfrac{8\pi}{2}-2\sqrt{3}-\dfrac{8\pi}{6}\right]$

$=2\left[\dfrac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)\right]$

$=2\left[\dfrac{2\sqrt{3}}{3}+8\times\dfrac{2\pi}{6}\right]$

$=\dfrac{4\sqrt{3}}{3}+\dfrac{16\pi}{3}$

1495741_1051360_ans_c9349aa70c524f45a5352689e333bc59.png

The area of the region enclosed between parabola

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

$y=mx$ is

$\cfrac{1}{48}$ . Then, the value of

$m$ is

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

Explanation

Equation of parabola is

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

$y=mx$
For intersection point of both curves put

$x={y}^{2}$ , we get

$y=m{y}^{2}\Rightarrow$

$y(my-1)=0$

$\Rightarrow$

$y=0$ or

$y=\cfrac{1}{m}$
Then,

$x=0$ or

$x=\cfrac{1}{{m}^{2}}$

$\therefore$ Intersection points are

$(0,0)$ and

$P(\cfrac{1}{{m}^{2}},\cfrac{1}{m})$

$\therefore$ Required area

$=\displaystyle \int _{ 0 }^{ 1/m }{ \left| \left( \cfrac { y }{ m } -{ y }^{ 2 } \right) \right| dy } =\left| { \left[ \cfrac { { y }^{ 2 } }{ 2m } -\cfrac { { y }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1/m } \right| \quad$

$=\left| \cfrac { 1 }{ 2{ m }^{ 3 } } -\cfrac { 1 }{ 3{ m }^{ 3 } } \right| =\left| \cfrac { 1 }{ 6{ m }^{ 3 } } \right| =\cfrac { 1 }{ 48 }$

$\Rightarrow \cfrac { 1 }{ 6{ m }^{ 3 } } =\pm \cfrac { 1 }{ 48 } \Rightarrow { m }^{ 3 }=\pm 8$
Now if

${m}^{3}=8$

${ m }^{ 3 }=8\quad \Rightarrow { m }^{ 3 }={ \left( 2 \right) }^{ 3 }\Rightarrow m=2$
If

${m}^{3}=-8$

$\Rightarrow { m }^{ 3 }=-8$

$\Rightarrow { m }^{ 3 }={ -\left( 2 \right) }^{ 3 }$

$\Rightarrow m=-2$

The area of the region bounded by the curves,

$y^{2} = 8x$ and

$y = x$ is

0%
$\dfrac {64}{3}$
0%
$\dfrac {32}{3}$
0%
$\dfrac {16}{3}$
0%
$\dfrac {8}{3}$

Explanation

Given curves,

$y^2=8x \rightarrow$ (i)

and

$y=x \rightarrow$ (ii)

On solving Eqs. (i) and (ii), we get

$x^2-8x=0$

$x(x-8)=0$

and

$y=0,8$

$\therefore$ Required area (OPA)

$=\displaystyle \int_0^8(\sqrt{8x}-x)dx$

$=\left[2\sqrt2. \dfrac23.x^{3/2}-\dfrac{x^2}2\right]_0^8$

$=\dfrac{4\sqrt2}3.(8)^{3/2}-\dfrac{8^2}2$

$=\dfrac{32}3$

The area of the region described by

$\left \{(x, y)/ x^{2} + y^{2} \leq 1\ and\ y^{2} \leq 1 - x\right \}$ is

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

The area (in square units) of the region bounded by the curves

$x=y^2$ and

$x=3-2y^2$ is

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

Explanation

Required area

$\displaystyle =2\int_0^1(3-2y^2-y^2)dy=6\int_0^1(1-y^2)dy=6\left(y-\dfrac{y^3}{3}\right)_0^1=4$

Since given curves are symmetrical about x-axis

The area (in square units) bounded by the curves

$x\, =\, -2y^2$ and

$x\, =\, 1-3y^2$ is

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

Explanation

$x=-2y^2$

$x=1-3y^2$

$-2y^2=1-3y^2\Rightarrow y^2=1\Rightarrow y=\pm 1$

Area between the curves

$= A_1-A_2$

$\displaystyle =\int^1_{-1}(1-3y^2+2y^2)dy$

$=\int^1_{-1}(1-y^2)dy$

$=[y-\dfrac{y^3}{3}]^1_{-1}=\dfrac{4}{3}$

The area (in square units) of the region bounded by

$x=-1, x=2, y=x^2+1$ and

$y=2x-2$ is

0%
10
0%
7
0%
8
0%
9

Explanation

$A_1=ar (ABC)$

$=\frac{1}{2}\times |2||4|$

$=4$ units

$A_2=ar(CDE)$

$=\frac{1}{2}\times 1\times 2$

$A_2=1$ units

Area of region between parabola and x-axis

$A_3=\int_{-1}^2(x^2+1)dx=\left[\dfrac{x^3}{3}+x\right]_{-1}^{2}$

Area required

$=A_3+A_1-A_2$

$=6+4-1$

$=9$ units

Area of region

$\displaystyle \left\{ \left( x,y \right) \in { R }^{ 2 }:y\ge \sqrt { \left| x+3 \right| } ,5y\le x+9\le 15 \right\}$ is equal to

0%
$\displaystyle \frac { 1 }{ 6 }$
0%
$\displaystyle \frac { 4 }{ 3 }$
0%
$\displaystyle \frac { 3 }{ 2 }$
0%
$\displaystyle \frac { 5 }{ 3 }$

Explanation

Area

$ABE$ (under parabola)

$=\displaystyle \int^{-3}_{-4}\sqrt{-x-3}dx=\frac{2}{3}$

Area

$BCD$ (under parabola)

$=\displaystyle \int^{1}_{-3}\sqrt{x+3}dx=\frac{16}{3}$
Area of trapezium

$ACDE$

$=\dfrac{1}{2}(1+2)5=\dfrac{15}{2}$
Required area

$= \dfrac{15}{2}-\dfrac{16}{3}-\dfrac{2}{3}=\dfrac{3}{2}$

Area bounded by curve

$y=x^2$ and

$y=2-x^2$ is ?

0%
$\dfrac{8}{3}$ sq units
0%
$\dfrac{3}{8}$ sq units
0%
$\dfrac{3}{2}$ sq units
0%
None of these

Explanation

$y=x^2$ and

$y=2-x^2$

now both he curve intersect each other ,

from both the equation

$x^2=2-x^2$

$x=+1,-1$

now area bounded by both curve is

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

because both are even function hence

$A =2[\displaystyle \int_{0}^{1}(2-x^2)-\displaystyle \int_{0}{1}(x^2)$ ]

$A=2[(2x-\dfrac{x^3}{3})-(\dfrac{x^3}{3})]$

now on putting upper and lower value of limit ,we will get

$A=\dfrac{8}{3}$

If the area bounded by the curves

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

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

$(a>0)$ is

$1$ sq.units, then the value of

$a$ is

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

Explanation

Points of intersection of

$y=ax^2$ and

$x=ay^2$ are

$(0,0)$ and

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

Hence,

$\displaystyle \int _0 ^{\tfrac1a} \left (\sqrt {\dfrac {x}{a}}-ax^2\right)dx=1$

$\Rightarrow \displaystyle\cfrac{2x^{\tfrac32}}{3\sqrt a} \big|_0^{\tfrac1a} - \cfrac{ax^3}3\bigr|_0^{\tfrac1a} =1$

$\Rightarrow \cfrac{2}{3a^2} - \cfrac1{3a^2} =1$

$\Rightarrow \cfrac{1}{3a^2} =1$

$\Rightarrow a=\dfrac {1}{\sqrt3}$ ....As

$(a>0)$

The ratio of the areas of two regions of the curve

$C_1 : 4x^2 + \pi^2y^2 = 4\pi^2$ divided by the curve

$C_2 : y = -sgn \left(x - \dfrac{\pi}{2}\right) \cos x$ (where sgn(x) denotes signum function) is -

0%
$\dfrac{\pi^2 +4}{\pi^2-2\sqrt{2}}$
0%
$\dfrac{\pi^2-2}{\pi^2+2}$
0%
$\dfrac{\pi^2+6}{\pi^2+3\sqrt{3}}$
0%
$\dfrac{\pi^2+1}{\pi^2-\sqrt{2}}$

$z$ is not purely real then area bounded by curves

$lm\left(z+\dfrac{1}{z}\right) = 0$ and

$|z-1| = 2$ is (in square units)-

0%
$4\pi$
0%
$3\pi$
0%
$2\pi$
0%
$\pi$

The graphs of

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

$g(x)=cx^{3}(c>0)$ intersect at the points

$(0, 0)$ and

$(\dfrac{1}{c}, \dfrac{1}{c^{2}})$ . If the region which lies between these graphs and over the interval

$[0, \dfrac{1}{c}]$ has the area equal to

$(\dfrac{2}{3})sq.\ units$ , then the value of

$c$ is :

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

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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