RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 03:18 PM IST

It is quite famous that the solutions of RD Sharma books are an efficient path for any student to follow to ace the subject of mathematics. The Class 12 RD Sharma chapter 20 exercise MCQ solution is one of the examples for it. The solution is constructive in every way out and is also exam-oriented, simple, and essential. The RD Sharma class 12th exercise MCQ can be used for self-practice and evaluation of marks.

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RD Sharma Class 12 Solutions Chapter20 MCQ Areas Of Bounded Region - Other Exercise
Areas of Bounded Regions Excercise:MCQ
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter20 MCQ Areas Of Bounded Region - Other Exercise

Areas of Bounded Regions Excercise:MCQ

Areas of Bounded Region exercise multiple choice question 1

Answer:
1(b)
Hint:
Make the graph of 2^kx and then apply integration
Given:
$y=2^{kx} \text { and } x=0, \quad x=2 \text { is }\frac{3}{log_{e}2}$
Explanation:

Area of the bounded region
$=\int_{0}^{2}2^{kx}dx$
$\begin{aligned} &\frac{3}{log_{e}2}=\int_{0}^{2}2^{kx}dx \\ &\frac{3}{log_{e}2}=\left [ \frac{2^{kx}}{log_{e}2} \right ]_{0}^{2} \\ &3=\left [ 2^{kx} \right ]_{0}^{2} \\ &3=(2^{k})^{2}-(2^{k})^{0} \\ &3=2^{2k}-1 \end{aligned}$ $\begin{aligned} &\left [ \because \int a^{x}dx=\frac{a^{x}}{log_{e}a} \right ] \\ &\left [ \because a^{0}=1 \right ] \end{aligned}$
$\begin{aligned} &4=2^{2k} \\ &2^{2}=2^{2k} \\ &2k=2 \\ &k=1 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 2

Answer:
(c)
Hint:
Find the bounded region between the parabolas
Given:
y² = 4x and x² = 4y
Explanation:

Intersection point is
$\begin{aligned} &y^{2}=4x \\ &=4(2\sqrt{y}) \qquad \qquad \qquad [\because x^{2}=4y] \\ &y^{2}=8\sqrt{y} \\ &y^{4}=64y \\ &y^{4}-64y=0 \\ &y(y^{3}-64)=0 \\ &y=0, \: \: y^{3}=64,\: \: y=4 \\ &\therefore x=4 \end{aligned}$
Area between the parabolas
$\begin{aligned} &=\int_{0}^{4}\left ( 2\sqrt{x}-\frac{x^{2}}{4} \right )dx \\ &=\frac{1}{4}\int_{0}^{4}(8\sqrt{x}-x^{2})dx \\ &=\frac{1}{4}\left [ \frac{8x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{3} \right ]_{0}^{4} \\ &=\frac{1}{4}\left [ \frac{16}{3}(4)^{\frac{3}{2}}-\frac{4^{3}}{3}-0+0 \right ] \\ &=\frac{1}{4}\left [ \frac{16}{3}(2^{2})^{\frac{3}{2}}-\frac{64}{3} \right ] \\ &=\frac{1}{4}\left [ \frac{128}{3}-\frac{64}{3} \right ]\\ &=\frac{16}{3}sq.units \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 3

Answer:
(b)
Hint:
Construct the graph
Given:
y=log_ex and x - axis, the straight line x = e
Explanation:

Area bounded by the curve, x - axis and x = e
$\begin{aligned} &=\int_{1}^{e}log_{e}x dx \\ &=[x.log_{e}x]_{1}^{e}-\int_{1}^{e}x.\frac{1}{x}dx \\ &=e\, log_{e}e-log_{e}1-\int_{1}^{e}dx \\ &=e(1)-0-[x]_{1}^{e} \qquad \qquad \qquad [\text { Integration by parts }]\\ &=e-[e-1] \\ &=e-e+1 \\ &=1sq.units \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 4

Answer:
(b)
Hint:
Find the region between parabola and line
Given:
$y=2-x^{2},\quad x+y=0$
Explanation:

To find the point of intersection of
$y=2-x^{2} \text { and } x+y=0$
$\begin{aligned} &x=-y \\ &y=2-x^{2} \\ &-x=2-x^{2} \\ &x^{2}-x-2=0 \\ &x^{2}-2x+x-2=0 \\ &x(x-2)+1(x-2)=0 \\ &(x+1)(x-2)=0 \\ &x=-1,2 \end{aligned}$
Area of the region ABCD,
$\begin{aligned} &A=\int_{-1}^{2}(2-x^{2}-(-x))dx \\ &=\int_{-1}^{2}(2-x^{2}+x)dx \\ &=\left [ 2x-\frac{x^{3}}{3}+\frac{x^{2}}{2} \right ]_{1}^{2} \\ &=\left [ 2(2)-\frac{(2)^{3}}{3} +\frac{(2)^{2}}{2}\right ]-\left [ 2(-1)-\frac{(-1)^{3}}{3} +\frac{(-1)^{2}}{2} \right ] \\ &=\left [ 4-\frac{8}{3}+2 \right ]-\left [ -2+\frac{1}{3}+\frac{1}{2} \right ] \\ &=6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2} \\ &=5-\frac{1}{2} \\ &=\frac{9}{2}sq.units \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 5

Answer:
$\frac{32}{3}(b)$
Hint:
Find the region between parabola and y - axis
Given:
x = 4 - y² and y-axis (x=0)
Explanation:

Area bounded by region
$\begin{aligned} &=\int_{-2}^{2}(4-y^{2})dy \\ &=\left [ 4y-\frac{y^{3}}{3} \right ]_{-2}^{2} \\ &=8-\frac{8}{3}+8-\frac{8}{3} \\ &=16-\frac{16}{3} \\ &=\frac{32}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 6

Answer:
(a)
Hint:
Find
$A_{n}>A_{n+1}\: or\: A_{n}<A_{n+1}$
Given:
$y=(tan\, x)^{n}, x=0, y=0 \: and\: x=\frac{\pi }{4}$
Explanation:

$\begin{aligned} &A_{n}=\int_{0}^{\frac{\pi }{4}}(tan\, x)^{n}dx \\ &A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \qquad\qquad\dots(i) \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n}\, xdx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: tan^{2}dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x-1)dx \\ &A_{n}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: sec^{2}x dx-\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, xdx \\ &A_{n}+A_{n-2}=\int_{0}^{\frac{\pi }{4}}tan^{n-2}\, x\: (sec^{2}x)dx \end{aligned}$
Let,
$\begin{aligned} &t=tan\: x \\ &dt=sec^{2}xdx\\ &x=0\: \: x=\frac{\pi }{4}\\ &t=0\: \: t=1\\ &A_{n}+A_{n-2}=\int_{0}^{1}t^{n-2}dt\\ &=\left [ \frac{t^{n-1}}{n-1} \right ]_{0}^{1}\\ &=\frac{1}{n-1} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 7

Answer:
(c)
Hint:
Integration
Given:
$x^{2}+y^{2}-6x-4y+12\leq 0$
$y\leq x\: and\: x\leq \frac{5}{2}$
Explanation:
$\begin{aligned} &x^{2}+y^{2}-6x-4y+12\leq 0\\ &x^{2}-6x+9-9+y^{2}-4y+4-4+12\leq 0\\ &(x-3)^{2}+(y-2)^{2}-1\leq 0\\ &(x-3)^{2}+(y-2)^{2}\leq 1\\ &(y-2)^{2}\leq 1-(x-3)^{2}\\ &(y-2)\leq\sqrt{1-(x-3)^{2}}\\ &y\leq2\pm \sqrt{1-(x-3)^{2}}\\ &\text{ As }x\leq \frac{5}{2},y=2- \sqrt{1-(x-3)^{2}} \end{aligned}$
Area,
$\begin{aligned} &=\int_{2}^{\frac{5}{2}}xdx-\int_{2}^{\frac{5}{2}}\left ( 2-\sqrt{1-(x-3)^{2}} \right )dx\\ &=\left [ \frac{x^{2}}{2} \right ]_{2}^{\frac{5}{2}}-\left [ 2x-\frac{(x-3)}{2}\sqrt{1-(x-3)^{2}}+\frac{1}{2}sin^{-1}(x-3) \right ]_{2}^{\frac{5}{2}}\\ &=\frac{25}{8}-2-\left [ 2\left ( \frac{1}{2} \right )-\frac{1}{2}\left ( \frac{-1}{2} \right )\times \left ( \frac{\sqrt{3}}{2} \right )+\frac{1}{2}sin^{-1}\left ( \frac{-1}{2} \right )-sin^{-1}(-1)\right ] \end{aligned}$
$\begin{aligned} &=\frac{9}{8}-1+\frac{\sqrt{3}}{8}+\frac{1}{2}\left ( \frac{-\pi }{6}+\frac{\pi }{2} \right )\\ &=\frac{1}{8}+\frac{\sqrt{3}}{8}-\frac{\pi }{6}\\ &=\frac{-\pi }{6}+\frac{1+\sqrt{3}}{8} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 8

Answer:
(a)
Hint:
Integration
Given:
$y=log_{e}(x+e),x=log_{e}\left ( \frac{1}{y} \right ),x-axis$
Explanation:
$\begin{aligned} &x-axis(y=0)\\ &0=log_{e}(x+e)\\ &x+e=e^{0}\\ &x+e=1\\ &x=1-e\\ &x=log_{e}\frac{1}{0}\\ &x=\infty \end{aligned}$
Area,
$\begin{aligned} &=\int_{1-e}^{0}\left [ log(x+e) \right ]dx+\int_{0}^{\infty }\left [ e^{-x} \right ]dx\\ &=\left [ (x+e)log(x+e) \right ]_{1-e}^{0}-\left [ \frac{(x+e)}{x+e} \right ]_{1-e}^{0}\lim_{k\rightarrow \infty }\left [ -e^{-x} \right ]_{0}^{k}\\ &=e[1-1]-[-1]+\lim_{k\rightarrow \infty }\left [ -e^{-x} \right ]_{0}^{k}\\ &=1+(1-0)\\ &=2 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 9

Answer:
(c)
Hint:
Integration
Given:
(y-2)² = x-1, the tangent to it at the point with ordinate 3 and x-axis.
Explanation:
(y-2)² = x-1
Tangent,
$\begin{aligned} &2(y-2)\frac{\mathrm{d} y}{\mathrm{d} x}=1\\ &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2(y-2)} \end{aligned}$
At y = 3
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \end{aligned}$
Therefore, the equation of tangent at (2,3)
$\begin{aligned} &y-3=\frac{1}{2}(x-2)\\ &x-2y+4=0\\ &x=2y-4 \end{aligned}$
The area of bounded region
$\begin{aligned} &=\int_{0}^{3}\left [ (y-2)^{2}+1-(2y-4) \right ]dy\\ &=\int_{0}^{3}\left [ y^{2}-6y+9 \right ]dy\\ &=\left [ \frac{y^{3}}{3}-\frac{6y^{2}}{2}+3y \right ]_{0}^{3}\\ &=\frac{3^{3}}{3}\\ &=9 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 10

Answer:
(a)
Hint:
Integration
Given:
$y=sin\, x, x=0\: \: and\: \: x=\pi ,x-axis$
Explanation:
$\begin{aligned} &\int_{0}^{\pi }sin\, xdx\\ &=[-cos\, x]_{0}^{\pi }\\ &=-cos\, \pi +cos\, 0\\ &=1+1\\ &=2 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 11

Answer:
(b)
Hint:
Integration
Given:
$y^{2}=4ax, x^{2}=4ay$
Explanation:
$\begin{aligned} &\left ( \frac{x^{2}}{4a} \right )^{2}=4ax\\ &\frac{x^{4}}{16a^{2}}=4ax\\ &x^{4}-64a^{3}x=0\\ &x(x^{3}-64a^{3})=0\\ &x=0 \quad x=4a \end{aligned}$
Area,
$\begin{aligned} &=\int_{0}^{4a}\left [ 2\sqrt{ax}-\frac{x^{2}}{4a} \right ]dx\\ &=\left [ 2\sqrt{a}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{3}}{12a} \right ]_{0}^{4a}\\ &=\frac{4}{3}\sqrt{a}(4a)^{\frac{3}{2}}-\frac{(4a)^{3}}{12a}-0\\ &=\frac{4}{3}\sqrt{a}2a^{\frac{3}{2}}-\frac{16a^{2}}{3}\\ &=\frac{32a^{2}}{3}-\frac{16a^{2}}{3}\\ &=\frac{16a^{2}}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 12

Answer:
(b)
Hint:
Integration
Given:
y = x⁴ - 2x³ + x²+ 3 , x - axis and minima of y
Explanation:
Finding the minima of y
$\begin{aligned} &y = x^{4} - 2x^{3} + x^{2} + 3 \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=4x^{3}-6x^{2}+2x \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12x^{2}-12x+2 \\ \end{aligned}$
Now,
$\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &4x^{3}-6x^{2}+2x=0\\ &2x(2x^{2}-3x+1)=0\\ &x=0 \qquad 2x^{2}-3x+1=0\\ & \qquad \qquad (2x-1)(x-1)=0\\ &x=0,\: 1,\: \frac{1}{2} \end{aligned}$
At x = 0
$\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2> 0 \end{aligned}$
At x = 1
$\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12-12+2 =2> 0 \end{aligned}$
At
$\begin{aligned} &x=\frac{1}{2} \end{aligned}$ $\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=-1 < 0 \end{aligned}$

So,f(x) has minimum value at

$\begin{aligned} &x=\frac{1}{2} \end{aligned}$

The minimum value is y = 1

$\begin{aligned} &1=x^{4}-2x^{3}+x^{2}+3\\ &x^{4}-2x^{3}+x^{2}+2=0\\ &x=1 \end{aligned}$

$\begin{aligned} &\int_{0}^{1} (x^{4}-2x^{3}+x^{2}+3)dx\\ &=\left [ \frac{x^{5}}{5}-2\left ( \frac{x^{4}}{x} \right )+\frac{x^{3}}{3}+3x \right ]_{0}^{1}\\ &=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=0\\ &=\frac{91}{30} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 13

Answer:
(b)
Hint:
Integration
Given:
$y^{2}=4ax, \text { latus rectum }=a , \text{ x-axis }(y=0)$
Explanation:

y = 0
x = 0
$\begin{aligned} &=\int_{0}^{a}2\sqrt{ax}dx\\ &=2\sqrt{a}\int_{0}^{a}\sqrt{x}dx\\ &\text { Area }=2\sqrt{a}\left [ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right ]_{0}^{a}\\ &=\frac{4}{3}\sqrt{a}\left [ a \right ]^{\frac{3}{2}}\\ &=\frac{4a^{2}}{3} \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 14

Answer:
(c)
Hint:
Integration
Given:
$[(x,y):x^{2}+y^{2}\leq 1\leq x+y]$
Explanation:

$\begin{aligned} &x^{2}+y^{2}=1\\ &y^{2}=1-x^{2}\\ &y=\sqrt{1-x^{2}},\: 0< x< 1\\ &x+y=1\\ &y=1-x \end{aligned}$
Area
$\begin{aligned} &=\int_{0}^{1}(\sqrt{1-x^{2}})-(1-x)dx\\ &=\left [ \frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}sin^{-1}\frac{x}{2}-x+\frac{x^{2}}{2} \right ]_{0}^{1}\\ &=\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right )-1+\frac{1}{2}\\ &=\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right )-\frac{1}{2}\\ &=\frac{1}{2}\left ( \frac{\pi }{6}-1 \right ) \end{aligned}$

Areas of Bounded Regions exercise 5 multiple choice question 15

Answer:
(c)
Hint:
Integration
Given:
$y=2x^{2},\: \: y=x^{2}+4$
Explanation:
$\begin{aligned} &2x^{2}=x^{2}+4\\ &x^{2}=4\\ &x=\pm 2\\ &\text { Area }=\int_{-2}^{2}(2x^{2}+4-2x^{2})dx\\ &=\int_{-2}^{2}(-x^{2}+4)dx \end{aligned}$
$\begin{aligned} &=\left [ \frac{-x^{3}}{3}+4x \right ]_{-2}^{2}\\ &=\frac{-8}{3}+8-\frac{8}{3}+8\\ &=16-\frac{16}{3} \\ &=\frac{48-16}{3}\\ &=\frac{32}{3}\text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 16

Answer:
(d)
Hint:
Integration
Given:
$y=x^{2}+1,\: \: x+y=3$
Explanation:
Intersection points are
$\begin{aligned} &x+y=3\\ &x+x^{2}+1=3\\ &x^{2}+x-2=0\\ &x^{2}+2x-x-2=0\\ &x(x+2)-1(x+2)=0\\ &x=1,\: \: x=-2 \end{aligned}$
Area
$\begin{aligned} &=\int_{-2}^{1}\left [ (3-x)-(x^{2}+1) \right ]dx\\ &=\int_{-2}^{1}\left [ 2-x-x^{2} \right ]dx\\ &=\left [ 2x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \right ]_{-2}^{1}\\ &=2-\frac{1}{2}-\frac{1}{3}-\left [ -4-\frac{4}{2}+\frac{8}{3} \right ]\\ &=\frac{3}{2}-\frac{1}{3}+6-\frac{8}{3}\\ &=\frac{9}{2} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 17

Answer:
(b)
Hint:
Integration
Given:
$\text { The ratio of } y=cos\, x \text { and } y=cos2x \text { from } x=0 \text { to }x=\frac{\pi }{3}$
Explanation:
$\begin{aligned} &\text {Area of }cos\, x=A_{1}=\int_{0}^{\frac{\pi }{3}}cos\, xdx\\ &=\left [ sin\, x \right ]_{0}^{\frac{\pi }{3}}\\ &=sin\frac{\pi }{2}-sin\, 0\\ &=\frac{\sqrt{3}}{2} \end{aligned}$
$\begin{aligned} &\text {Area of }cos\, 2x=A_{2}=\int_{0}^{\frac{\pi }{3}}cos\, 2xdx\\ &=sin\frac{2\pi }{3}-sin\, 0\\ &=\frac{1}{2}sin\left ( \pi -\frac{\pi }{3} \right )-0\\ &=sin\frac{\pi }{3}\\ &=\frac{\frac{\sqrt{3}}{2}}{2}\\ &=\frac{\sqrt{3}}{4} \end{aligned}$
$\begin{aligned} &A_{1}:A_{2}=\frac{\left ( \frac{\sqrt{3}}{2} \right )}{\left ( \frac{\sqrt{3}}{4} \right )}=2:1 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 18

Answer:
(d)
Hint:
Integration
Given:
$y=cos\, x,\: x-axis,\: x=0,\: x=2\pi$
Explanation:

$\begin{aligned} &\int_{0}^{\frac{\pi }{2}}cos\, x+\int_{\frac{\pi }{2}}^{\frac{3\pi }{2}}(-cos\, x)dx+\int_{\frac{3\pi }{2}}^{2\pi }(cos\, x)dx\\ &=[sin\, x]_{0}^{\frac{\pi }{2}}+[-sin\, x]_{\frac{\pi }{2}}^{\frac{3\pi }{2}}+[sin\, x]_{\frac{3\pi }{2}}^{2\pi }\\ &=sin\frac{\pi }{2}-sin\, 0-\left [ sin\frac{3\pi }{2}-sin\frac{\pi }{2} \right ]+sin2\pi -sin\frac{3\pi }{2}\\ &=1+1+1+1\\ &=4 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 19

Answer:
(a)
Hint:
Integration
Given:
$y^{2}=x,\: 2y=x$
Explanation:
Intersection point
$\begin{aligned} &y^{2}=x\\ &y^{2}=2y\\ &y^{2}-2y=0\\ &y(y-2)=0\\ &y=0,\: \: y=2\\ &\Rightarrow x=4 \end{aligned}$
Area
$\begin{aligned} &=\int_{0}^{4}\left ( \sqrt{x}-\frac{x}{2} \right )dx\\ &=\left [ \frac{2}{3}x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{2}}{4} \right ]_{0}^{4}\\ &=\frac{2}{3}\times 8-4\\ &=\frac{4}{3} \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 20

Answer:
(c)
Hint:
Integration
Given:
$y=4x-x^{2} \text { and } x-axis (y=0)$
Explanation:
$\begin{aligned} &y=0\\ &0=4x-x^{2}\\ &x^{2}-4x=0\\ &x(x-4)=0\\ &x=0,\: x=4 \end{aligned}$
Area
$\begin{aligned} &=\int_{0}^{4}(4x-x^{2})dx\\ &=\left [ \frac{4x^{2}}{2}-\frac{x^{3}}{3} \right ]_{0}^{4}\\ &=2(4)^{2}-\frac{(4)^{3}}{3}-0+0\\ &=32-\frac{64}{3}\\ &=\frac{96-64}{3}\\ &=\frac{32}{3}\text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 21

Answer:
(b)
Hint:
Integration
Given:
$y^{2}(2a-x)=x^{3} \text { and } x=2a \text { above }x-axis$
Explanation:
$\begin{aligned} &x=2a\\ &y^{2}(2a-x)=x^{3}\\ &y^{2}=\frac{x^{3}}{2a-x}\Rightarrow y=\sqrt{\frac{x^{3}}{2a-x}}\\ \end{aligned}$
Let
$\begin{aligned} &x=2a\: sin^{2}\theta \\ &dx=4a\: sin\: \theta\: cos\: \theta \end{aligned}$
Area
$\begin{aligned} &=\int_{0}^{2a}\sqrt{\frac{x^{3}}{2a-x}}dx\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{(2a)^{3}(sin^{2}\theta )^{3}}{2a-2asin^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [W\! hen\: x=0, \theta=0\: and\: x=2a,\theta =\frac{\pi }{2} ]\\ &=\int_{0}^{\frac{\pi }{2}}\sqrt{\frac{8a^{3}sin^{6}\theta }{2a\, cos^{2}\theta }}(4a\, sin\, \theta \: cos\, \theta\: d\theta ) \qquad \qquad [\because 1-sin^{2}\theta =cos^{2}\theta ] \end{aligned}$
$\begin{aligned} &=8a^{2}\int_{0}^{\frac{\pi }{2}}\sqrt{sin^{6}\theta }(sin\, \theta d\theta )\\ &=8a^{2}\int_{0}^{\frac{\pi }{2}}sin^{4}\theta d\theta \qquad \qquad [\because \int sin^{4}xdx=\frac{3x}{8}-\frac{1}{4}sin2x+\frac{1}{32}sin4x+c]\\ &=8a^{2}\left [ \frac{3\theta }{8}-\frac{1}{4}sin2\theta + \frac{1}{32}sin4\theta \right ]_{0}^{\frac{\pi }{2}}\\ &=8a^{2}\left [ \frac{3\left ( \frac{\pi }{2} \right )}{8}-\frac{1}{4}sin\pi +\frac{1}{32}sin2\pi \right ]-0\\ &=8a^{2}\times \frac{3\left ( \frac{\pi }{2} \right )}{8}\\ &=\frac{3}{2}\pi a^{2} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 22

Answer:
(b)
Hint:
Integration
Given:
$x^{2}=4y,\: x=2 \text { and } x-axis (y=0)$
Explanation:
$\begin{aligned} &y=0\\ &x^{2}=4(0)\\ &x=0\\ &\text { Area }=\int_{0}^{2}\frac{x^{2}}{4}dx\\ &=\left [ \frac{x^{3}}{12} \right ]_{0}^{2}\\ &=\frac{8}{12}\\ &=\frac{2}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 23

Answer:
(c)
Hint:
Integration
Given:
$\text { Area }=(b-1)sin(3b+4)\\ x=1,\: x=b$
Explanation:
$\begin{aligned} &\text { Area }=\int_{1}^{b}f(x)dx\\ &(b-1)sin(3b+4)=\int_{1}^{b}f(x)dx \end{aligned}$
Differentiate both sides with respect to b,
$\begin{aligned} &f(b)=3(b-1)cos(3b+4)+sin(3b+4)\\ &\text { Put } b=x\\ &f(x)=3(x-1)cos(3x+4)+sin(3x+4) \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 24

Answer:
(a)
Hint:
Integration
Given:
$y^{2}=8x,x^{2}=8y$
Explanation:
$\begin{aligned} &y^{2}=8x\\ &x=\frac{y^{2}}{8}\\ &x^{2}=8y\\ &\left ( \frac{y^{2}}{8} \right )^{2}=8y\\ &\left ( \frac{y^{4}}{64} \right )=8y\\ &y^{4}-512y=0\\ &y(y^{3}-512)=0\\ &y=0,\: \: y^{3}=512\\ &y^{3}=8^{3}\\ &y=8 \end{aligned}$
$\begin{aligned} &\text { Area }=\int_{0}^{8}\left ( \sqrt{8x}-\frac{x^{2}}{8} \right )dx\\ &=\sqrt{8x}\int_{0}^{8}\sqrt{x}dx-\frac{1}{8}\int_{0}^{8}x^{2}dx\\ &=\sqrt{8}\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{8}-\frac{1}{8}\left [ \frac{x^{3}}{3} \right ]_{0}^{8}\\ &=\frac{2}{3}\sqrt{8}\left [ 8^{\frac{3}{2}}-0 \right ]-\frac{1}{24}\left [ 8^{3} -0\right ]\\ &=\frac{2\times 8^{2}}{3}-\frac{8^{2}}{3}\\ &=\frac{64}{3} \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 25

Answer:
(c)
Hint:
Integration
Given:
$\begin{aligned} &y^{2}=8x, x-axis(y=0) \text { and latus rectum }\\ &y^{2}=4\times 2x,\: \text { latus rectum }=2\\ &\text {Area }=2\int_{0}^{2}\sqrt{8x}dx\\ &=4\sqrt{2}\int_{0}^{2}\sqrt{x}dx=4\sqrt{2}\left [ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right ]_{0}^{2}\\ &=4\sqrt{2}\times \frac{2}{3}\left ( 2\frac{3}{2}-0 \right )\\ &=\frac{8\sqrt{2}}{3}\times 2\sqrt{2}\\ &=\frac{32}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 26

Answer:
(d)
Hint:
Integration
Given:
$y=x^{3},\: x-axis,\: x=-2,\: x=1$
Explanation:

Area OAB
$\begin{aligned} &=\int_{0}^{1}x^{3}dx\\ &=\left [ \frac{x^{4}}{4} \right ]_{0}^{1}\\ &=\frac{1}{4} \end{aligned}$
Area OCD
$\begin{aligned} &=\int_{-2}^{0}x^{3}dx\\ &=\left [ \frac{x}{4} \right ]_{-2}^{0}\\ &=-4 \end{aligned}$
As area is always positive
=4
Total area
$\begin{aligned} &=4+\frac{1}{4}\\ &=\frac{17}{4} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 27

Answer:
(c)
Hint:
Integration
Given:
$y=x\left | x \right |,\: x=-1,\: x=1$
Explanation:

$y=x\left | x \right |=\left\{\begin{matrix} x^{2} &x> 0 \\ -x^{2} & x< 0 \end{matrix}\right.$
Area of bounded region,
$=\int_{-1}^{1}x\left | x \right |dx$
Area AOB,
$\begin{aligned} &=\int_{-1}^{0}-x^{2}dx\\ &=\left [ \frac{-x^{3}}{3} \right ]_{-1}^{0}\\ &=\frac{-1}{3} \end{aligned}$
Area is always positive,
$\begin{aligned} &=-\left ( \frac{-1}{3} \right )=\frac{1}{3} \end{aligned}$
Area OCD,
$\begin{aligned} &=\int_{0}^{1}x^{2}dx\\ &=\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ &=\frac{1}{3} \end{aligned}$
Total area
$\begin{aligned} &=\frac{1}{3}+\frac{1}{3}\\ &=\frac{2}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 28

Answer:
(b)
Hint:
Integration
Given:
$y=cos\, x, y=sin\, x,0\leq x\leq \frac{\pi }{2}$
Explanation:

Intersection point,
$\begin{aligned} &cos\, x=sin\, x\\ &-tan\, x=1\\ &x=\frac{\pi }{4}, 0\leq x\leq \frac{\pi }{2} \end{aligned}$
Area AOB
$\begin{aligned} &=\int_{0}^{\frac{\pi }{4}}(cos\, x-sin\, \, x)dx\\ &=\left [ sin\, x+cos\, x \right ]_{0}^{\frac{\pi }{4}}\\ &=sin\frac{\pi }{4}-sin\, 0+cos\frac{\pi }{4}-cos\, 0\\ &=\frac{1}{\sqrt{2}}+0+\frac{1}{\sqrt{2}}-1\\ &=-1+\frac{2}{\sqrt{2}}\\ &=-1+\sqrt{2}\\ &=\sqrt{2}-1 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 29

Answer:
(c)
Hint:
Integration
Given:
$x^{2}+y^{2}=16,y^{2}=6x$
Explanation:

Area OABC
$\begin{aligned} &=2 \text { Area }O\! AB\\ &=\left [ \text { Area }O\! AD+\text { Area }AB\! D \right ] \qquad \qquad \dots(i) \end{aligned}$
Intersection points are
$\begin{aligned} &x^{2}+y^{2}=16 \\ &x^{2}+6x=16 \qquad \qquad \qquad \left [ \because y^{2}=6x \right ]\\ &x^{2}+6x-16=0\\ &x^{2}+8x-2x-16=0\\ &x(x+8)-2(x+8)=0\\ &x-2=0 \qquad x+8=0\\ &x=2, x=-8 \end{aligned}$
Not possible in first and third quadrant
x=2
From (i)
Area OABC,
$\begin{aligned} &=2\left [ \int_{0}^{2}\sqrt{6x}dx+\int_{2}^{4}\sqrt{16-x^{2}}dx \right ]\\ &=2\left [ \sqrt{6}\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{2}+2\left [ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}\frac{x}{4} \right ]_{2}^{4}\\ &=\frac{4}{3}\sqrt{6}\left [ (2)^{\frac{3}{2}-1} \right ]+2\left [ \frac{4}{2}\sqrt{16-(4)^{2}}+8sin^{-1}\frac{4}{4} \right ]-\left [ \frac{2}{2}\sqrt{16-2^{2}}+8sin^{-1}\frac{2}{4} \right ] \end{aligned}$
$\begin{aligned} &=\frac{4\sqrt{6}}{3}\left ( 2\sqrt{2} \right )+2\left [ 0+2\times \frac{\pi }{2}-\sqrt{12}-8sin^{-1}\frac{1}{2} \right ]\\ &=\frac{16\sqrt{3}}{3}+2\left [ 4\pi -2\sqrt{3}-8\frac{\pi }{6} \right ] \\ &=\frac{16\sqrt{3}}{3}+8\pi -4\sqrt{3}-\frac{8\pi }{3}\\ &=\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\\ &=\frac{4}{3}\left ( 4\pi +\sqrt{3} \right ) \text { sq.units } \end{aligned}$
Area of circle
$\begin{aligned} &=\pi r^{2}\\ &=\pi (4)^{2}\\ &=16\pi \end{aligned}$
Exterior area
$\begin{aligned} &=16\pi -\frac{4}{3}\left ( 4\pi +\sqrt{3} \right )\\ &=16\pi -\frac{16}{3}\pi -\frac{4}{\sqrt{3}}\\ &=\frac{4}{3}(8\pi -\sqrt{3}) \text{sq.units} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 30

Answer:
(b)
Hint:
Integration
Given:
x² + y² = 4 and x + y = 2
Explanation:

x² + y² = 4
r = 2
Area OAB,
$\begin{aligned} &=\int_{0}^{2}\sqrt{4-x^{2}}-(2-x)dx\\ &=\int_{0}^{2}\sqrt{4-x^{2}}-2+xdx\\ &=\int_{0}^{2}\sqrt{4-x^{2}}dx+\int_{0}^{2}(x-2)dx\\ &=\left [ \frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2}sin^{-1}\frac{x}{2} \right ]_{0}^{2}+\left [ \frac{x^{2}}{2}-2x \right ]_{0}^{2}\\ &=\frac{2}{2}\sqrt{4-4}+\frac{4}{2}sin^{-1}\frac{2}{2}-0+2sin^{-1}0+\frac{4}{2}-4-0+0\\ &=2sin^{-1}1+2-4\\ &=2\times \frac{\pi }{2}-2\\ &=\pi -2 \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 31

Answer:
(b)
Hint:
Integration
Given:
y²= 4x and y = 2x
Explanation:

$\begin{aligned} &y=2x\\ &y^{2}=4a\Rightarrow (2x)^{2}=4x\\ &4x^{2}=4x\\ &4x^{2}-4x=0\\ &4x(x-1)=0\\ &x=0,x=1 \end{aligned}$
Area OAB
$\begin{aligned} &=\int_{0}^{1}\left ( 2\sqrt{3}-2x \right )dx\\ &=\left [ \frac{2x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2x^{2}}{2} \right ]_{0}^{1}\\ &=\left [ \frac{4}{3}x^{\frac{3}{2}}-x^{2} \right ]_{0}^{1}\\ &=\frac{4}{3}(1)^{\frac{3}{2}}-(1)^{2}-0+0\\ &=\frac{4}{3}-1\\ &=\frac{1}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 32

Answer:
(a)
Hint:
Integration
Given:
x² + y² = 4 and x = 0, x = 2
Explanation:

$x^{2}+y^{2}=4\Rightarrow y=\sqrt{4-x^{2}}$
Area OAB,
$\begin{aligned} &=\int_{0}^{2}\sqrt{4-x^{2}}dx\\ &=\left [ \frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2}sin^{-1}\frac{x}{2} \right ]_{0}^{2}\\ &=\frac{2}{2}\sqrt{4-4}+2sin^{-1}\frac{2}{2}-0-2sin^{-1}0\\ &=2sin^{-1}1\\ &=2\left ( \frac{\pi }{2} \right )\\ &=\pi \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 33

Answer:
(b)
Hint:
Integration
Given:
y²=4x, y-axis (x = 0) and y=3
Explanation:

$\begin{aligned} &x=0\\ &y^{2}=4(0)=0\\ &y=0 \text { and } y=3\\ &9=4x\\ &x=\frac{9}{4} \end{aligned}$
Area CAB
$\begin{aligned} &=\int_{0}^{\frac{9}{4}}(3-2\sqrt{x})dx\\ &=\left [ 3x-\frac{2x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{\frac{9}{4}}\\ &=\left [ 3x-\frac{4}{3}x^{\frac{3}{2}} \right ]_{0}^{\frac{9}{4}}\\ &=3\left ( \frac{9}{4} \right )-\frac{4}{3}\left ( \frac{9}{4} \right )^{\frac{3}{2}}-0+0\\ &=\frac{27}{4}-\frac{4}{3}\left ( \left ( \frac{3}{2} \right )^{2} \right )^{\frac{3}{2}}\\ &=\frac{27}{4}-\frac{4}{3}\times \frac{27}{8}\\ &=\frac{27}{4}-\frac{9}{2}\\ &=\frac{9}{4} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 34

Answer:
(d)
Hint:
Area of the circle = 4 X (Area of circle in first quadrant)
Given:
x² + y²= 2
Explanation:

$\begin{aligned} &x^{2}+y^{2}=2\\ &y^{2}=2-x^{2} \qquad \qquad \qquad \left [ r^{2}=2\Rightarrow r=\sqrt{2} \right ]\\ &y=\sqrt{2-x^{2}} \end{aligned}$
Area of the circle = 4 X (Area of OAB)
$\begin{aligned} &=4\int_{0}^{\sqrt{2}}\sqrt{2-x^{2}}dx\\ &=4\left [ \frac{x}{2}\sqrt{2-x^{2}}+\frac{2}{2}sin^{-1}\frac{x}{\sqrt{2}} \right ]_{0}^{\sqrt{2}}\\ &=4\left [ \frac{\sqrt{2}}{2}\sqrt{2-(\sqrt{2})^{2}}+1sin^{-1}\frac{\sqrt{2}}{\sqrt{2}}-\frac{0}{2}\sqrt{2-0}-sin^{-1}0 \right ]\\ &=4\left [ sin^{-1}1 \right ]\\ &=4\left ( \frac{\pi }{2} \right )\\ &=2\pi \text{ sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 35

Answer:
(b)
Hint:
Area in all quadrants are equal. So, we do 4 X Area of one quadrant
Given:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Explanation:

Area of the ellipse
$\begin{aligned} &=4\times \text { Area of OAB }\\ &=4\times \int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}dx \qquad \qquad \qquad \dots (i)\\ &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ &\frac{y^{2}}{b^{2}}=1-\frac{x^{2}}{a^{2}}\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}}\\ &y=\frac{b}{a}\sqrt{a^{2}-x^{2}} \end{aligned}$
From (i)
$\begin{aligned} &=\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx\\ &=\frac{4b}{a}\left [ \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a} \right ]_{0}^{a}\\ &=\frac{4b}{a} \left [ \frac{a}{2}\sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{a}{a}-\frac{0}{2}\sqrt{a^{2}-0}-\frac{a^{2}}{2}sin^{-1}\frac{0}{a} \right ]\\ &=\frac{4b}{a} \left [ 0+\frac{a^{2}}{2}sin^{-1}1-0-0 \right ]\\ &=\frac{4b}{a} \left [ \frac{a^{2}}{2}\left ( \frac{\pi }{2} \right ) \right ]\\ &=\pi ab \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 36

Answer:
(b)
Hint:
Find intersection point then area
Given:
y = x² and y = 16
Explanation:

$\begin{aligned} &y=16\\ &16=x^{2}\\ &x=\pm 4 \end{aligned}$
Area of the region
$\begin{aligned} &=\int_{-4}^{4}16-x^{2}dx\\ &=\left [ 16x-\frac{x^{2}}{3} \right ]_{-4}^{4}\\ &=16(4+4)-\left [ \frac{(4)^{3}}{3}-\frac{(-4)^{3}}{3} \right ]\\ &=128-\left [ \frac{2\times 4\times 4\times 4}{3} \right ]\\ &=128-\frac{128}{3}\\ &=\frac{256}{3} \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 37

Answer:
(b)
Hint:
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c$
Given:
In first quadrant, x- axis (y = 0), y = x, x² + y² = 32
Explanation:

Intersection point,
$\begin{aligned} &y = x\\ &x^{2}+x^{2}=32\\ &2x^{2}=32\\ &x^{2}=16\\ &x=\pm 4 \end{aligned}$
x = -4 is not possible as we have to find in first quadrant and in first quadrant x > 0
$\begin{aligned} &x=4\\ &y^{2}+x^{2}=32\\ &x^{2}=32\\ &x=4\sqrt{2} \end{aligned}$
Now, y = x
Therefore, y = 4
Area of ABCD = Area of ABD + Area of BCD
$\begin{aligned} &=\int_{0}^{4}xdx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}dx\\ &=\left [ \frac{x^{2}}{2} \right ]_{0}^{4}+\left [ \frac{x}{2}\sqrt{32-x^{2}}+\frac{1}{2}(32)sin^{-1}\frac{x}{4\sqrt{2}} \right ]_{4}^{4\sqrt{2}}\\ &=8+\frac{4\sqrt{2}}{2}\sqrt{32(4\sqrt{2})^{2}}+\frac{1}{2}(32)sin^{-1}\frac{4\sqrt{2}}{4\sqrt{2}}-\frac{4}{2}\sqrt{32-(4)^{2}}-16sin-1\frac{4}{4\sqrt{2}}\\ &=8+0+16sin^{-1}1-2\sqrt{32-16}-16sin^{-1}\frac{1}{\sqrt{2}} \end{aligned}$
$\begin{aligned} &=8+16\left ( \frac{\pi }{2} \right )-2(4)-16 \left ( \frac{\pi }{4} \right )\\ &=8+8\pi -8-4\pi \\ &=4\pi \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 38

Answer:
(c)
Hint:
Put the limit y = -1,1
Given:
x = 2y + 3, y = 1, y = -1
Explanation:
Area of the bounded region
$\begin{aligned} &=\int_{-1}^{1}(2y+3)dy\\ &=\left [ \frac{2y^{2}}{2}+3y \right ]_{-1}^{1}\\ &=\left [ y^{2}+3y \right ]_{-1}^{1}\\ &=1+3-(-1)^{2}-3(-1)\\ &=4-1+3\\ &=6 \text { sq. units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 39

Answer:
(d)
Hint:
Find the intersection point
Given:
$\begin{aligned} &x^{2}=4y\\ &x=4y-2 \qquad \qquad \qquad \dots (i) \end{aligned}$
Explanation:
Intersection points are
$\begin{aligned} &x^{2}=4y\\ &(4y-2)^{2}=4y\\ &16y^{2}+4-16y=4y\\ &16y^{2}-20y+4=0\\ &4y^{2}-5y+1=0\\ &4y^{2}-4y-y+1=0\\ &4y(y-1)-1(y-1)=0\\ &(y-1)(4y-1)=0\\ &y=1, y=\frac{1}{4} \\ &y=1\Rightarrow x=2\\ &y=\frac{1}{4}\Rightarrow x=4\left ( \frac{1}{4}\right )-2=-1 \qquad \text {[From (i)]}\end{aligned}$
Area
$\begin{aligned} &=\int_{-1}^{2}\left [ \left ( \frac{x+2}{4} \right )-\frac{x^{2}}{4} \right ]dx\\ &=\left [ \frac{x^{2}}{8}+\frac{2x}{4}-\frac{x^{2}}{12} \right ]_{-1}^{2}\\ &=\frac{1}{4}\left [ \left ( 2+4-\frac{8}{3} \right )-\left ( \frac{1}{2}-2+\frac{1}{3} \right ) \right ]\\ &=\frac{1}{4}\left [ \frac{10}{3}-\left ( \frac{-7}{6} \right ) \right ]\\ &=\frac{1}{4}\left ( \frac{9}{2} \right )\\ &=\frac{9}{8} \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 40

Answer:
(a)
Hint:
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\left ( \frac{x}{a} \right )+c$
Given:
$y= \sqrt{16-x^{2}}, x-axis(y=0)$
Explanation:
$\begin{aligned} &y=0\\ &0=\sqrt{16-x^{2}}\\ &16-x^{2}=0\\ &x^{2}=16\\ &x=\pm 4\\ \end{aligned}$
Area bounded by the curve
$\begin{aligned} &=\int_{-4}^{4}\sqrt{16-x^{2}}dx\\ &=\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{4^{2}}{2}sin^{-1}\frac{x}{4} \right ]_{-4}^{4}\\ &=\left [ \frac{4}{2}\sqrt{16-4^{2}}+\frac{4^{2}}{2}sin^{-1}\frac{4}{4} \right ]-\left [ \frac{-4}{2}\sqrt{16-(-4)^{2}}+\frac{4^{2}}{2}sin^{-1}\left ( \frac{-4}{4} \right ) \right ] \qquad \qquad \left [ \because sin^{-1}1=\frac{\pi }{2} \right ]\\ &=0+8sin^{-1}1-0+8sin^{-1}1\\ &=8\times 2\times \frac{\pi }{2}\\ &=8\pi \text { sq.units } \end{aligned}$

Areas of Bounded Regions exercise multiple choice question 41

Answer:
(a)
Hint:
Find the graph between y² = x and 2y = x
Given:
y² = x and 2y = x
Explanation:

Intersection point,
$\begin{aligned} &y=\frac{x}{2}\\ &y^{2}=x\\ &\left ( \frac{x}{2} \right )^{2}=x\Rightarrow \frac{x^{2}}{4}=x\\ &x^{2}-4x=0\\ &x(x-4)=0\\ &x=0,x=4 \end{aligned}$
Area of the bounded region ABC
$\begin{aligned} &=\int_{0}^{4}\left ( \sqrt{x}-\frac{x}{2} \right )dx\\ &=\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{2}}{4} \right ]_{0}^{4}\\ &=\frac{2}{3}(4)^\frac{3}{2}-\frac{(4)^{2}}{4}\\ &=\frac{4}{3} \text { sq.units } \end{aligned}$

The RD Sharma class 12 solution of Area of bounded region exercise MCQ is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise MCQ consists of a total of 41 questions that covers up all the essential concepts of the chapter that is mentioned below:

Algorithm to find Area using vertical and horizontal stripes
Method to find the Area between two curves
The Area between two curves using vertical and horizontal stripes
Area of the region
Area of the region bounded by the curve
Area of the region bounded by a parabola

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