RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

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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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Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive.

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

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Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

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Data Analyst

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Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

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5 Jobs Available

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

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A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

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Welding Engineer

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QA Manager

4 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

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An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.