NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.2

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Chapter 8 Ex.8.2 Question 1

Find the area of the circle $$4{x^2} + 4{y^2} = 9$$ which is interior to the parabola $${x^2} = 4y$$.

Solution

Solving $$4{x^2} + 4{y^2} = 9$$ and $${x^2} = 4y$$, point of intersection $$B\left( {\sqrt 2 ,\frac{1}{2}} \right)$$ and $$D\left( { - \sqrt 2 ,\frac{1}{2}} \right)$$.

Required area is symmetrical about y-axis.

$$ar\left( {OBCDO} \right) = 2 \times ar\left( {OBCO} \right)$$

Draw BM perpendicular to OA

Coordinates of M are $$\left( {\sqrt 2 ,0} \right)$$

\begin{align}ar\left( {OBCO} \right)&= ar\left( {OMBCO} \right)--ar\left( {OMBO} \right)\\ &= \int_0^{\sqrt 2 } {\sqrt {\frac{{\left( {9 - 4{x^2}} \right)}}{4}} dx - \int_0^{\sqrt 2 } {\frac{{{x^2}}}{4}} } dx\\ &= \frac{1}{2}\int_0^{\sqrt 2 } {\sqrt {9 - 4{x^2}} } dx - \frac{1}{4}\int_0^{\sqrt 2 } {{x^2}dx} \\ &= \frac{1}{4}\left[ {x\sqrt {9 - 4{x^2}} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2x}}{3}} \right]_0^{\sqrt 2 } - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_0^{\sqrt 2 }\\& = \frac{1}{4}\left[ {\sqrt 2 \sqrt {9 - 8} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right] - \frac{1}{{12}}{\left( {\sqrt 2 } \right)^3}\\ &= \frac{{\sqrt 2 }}{4} + \frac{9}{8}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3} - \frac{{\sqrt 2 }}{6}\\ &= \frac{{\sqrt 2 }}{{12}} + \frac{9}{8}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\\ &= \frac{1}{4}\left( {\frac{{\sqrt 2 }}{3} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right)\end{align}

Required area OBCDO $$= \left( {2 \times \frac{1}{4}\left[ {\frac{{\sqrt 2 }}{3} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right]} \right) = \left[ {\frac{{\sqrt 2 }}{6} + \frac{9}{4}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right]$$ units.

Chapter 8 Ex.8.2 Question 2

Find the area bounded by curves $${\left( {x - 1} \right)^2} + {y^2} = 1$$ and $${x^2} + {y^2} = 1$$.

Solution

Solving $${\left( {x - 1} \right)^2} + {y^2} = 1$$ and $${x^2} + {y^2} = 1$$, point of intersection $$A\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)$$ and $$B\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)$$

Required area is symmetrical about x-axis.

$$ar\left( {OBCAO} \right) = 2 \times ar\left( {OCAO} \right)$$

Join AB, intersects OC at M

AM is perpendicular to OC

Coordinates of $$M\left( {\frac{1}{2},0} \right)$$

\begin{align}ar\left( {OCAO} \right) &= ar\left( {OMAO} \right) + ar\left( {MCAM} \right)\\& = \left[ {\int_0^{\frac{1}{2}} {\sqrt {1 - {{\left( {x - 1} \right)}^2}} dx} + \int_{\frac{1}{2}}^1 {\sqrt {1 - {x^2}} dx} } \right]\\ &= \left[ {\frac{{x - 1}}{2}\sqrt {1 - {{\left( {x - 1} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {x - 1} \right)} \right]_0^{\frac{1}{2}} + \left[ {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x} \right]_{\frac{1}{2}}^1\\ &= \left[ { - \frac{1}{4}\sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2} - 1} \right) - \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right] \\& \qquad+ \left[ { + \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right) - \frac{1}{4}\sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} - \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{8} + \frac{1}{2}\left( { - \frac{\pi }{6}} \right) - \frac{1}{2}\left( { - \frac{\pi }{2}} \right)} \right] + \left[ {\frac{1}{2}\left( {\frac{\pi }{2}} \right) - \frac{{\sqrt 3 }}{8} - \frac{1}{2}\left( {\frac{\pi }{6}} \right)} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{4} - \frac{\pi }{{12}} + \frac{\pi }{4} + \frac{\pi }{4} - \frac{\pi }{{12}}} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{4} - \frac{\pi }{6} + \frac{\pi }{2}} \right]\\& = \left[ {\frac{{2\pi }}{6} - \frac{{\sqrt 3 }}{4}} \right]\end{align}

Required Area OBCAO is $$\left( {2 \times \left[ {\frac{{2\pi }}{6} - \frac{{\sqrt 3 }}{4}} \right]} \right) = \left[ {\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right]$$ units.

Chapter 8 Ex.8.2 Question 3

Find the area of the region bounded by the curves $$y = {x^2} + 2,y = x,x = 0$$ and $$x = 3$$

Solution

\begin{align}ar\left( {OCBAO{\rm{ }}} \right)& = ar\left( {ODBAO} \right)--ar\left( {ODCO} \right)\\ &= \int_0^3 {\left( {{x^2} + 2} \right)dx} - \int_0^3 {xdx} \\& = \left[ {\frac{{{x^3}}}{3} + 2x} \right]_0^3 - \left[ {\frac{{{x^2}}}{2}} \right]_0^3\\& = \left[ {9 + 6} \right] - \left[ {\frac{9}{2}} \right]\\ &= 15 - \frac{9}{2}\\ &= \frac{{21}}{2}\end{align}

Chapter 8 Ex.8.2 Question 4

Using integration finds the area of the region bounded by the triangle whose vertices are $$\left( {-1,0} \right),\left( {1,3} \right)$$ and $$\left( {3,2} \right)$$.

Solution

BL and CM are perpendicular to x-axis.

$$ar\left( {\Delta ACB} \right) = ar\left( {ALBA} \right) + ar\left( {BLMCB} \right) - ar\left( {AMCA} \right)$$

Equation of AB is

\begin{align}y - 0 &= \frac{{3 - 0}}{{1 + 1}}\left( {x + 1} \right)\\y &= \frac{3}{2}\left( {x + 1} \right)\end{align}

\begin{align}ar\left( {ALBA} \right) &= \int_{ - 1}^1 {\frac{3}{2}\left( {x + 1} \right)} dx\\& = \frac{3}{2}\left[ {\frac{{{x^2}}}{2} + x} \right]_{ - 1}^1\\ &= \frac{3}{2}\left[ {\frac{1}{2} + 1 - \frac{1}{2} + 1} \right]\\& = 3\end{align}

Equation of BC is

\begin{align}y - 3 &= \frac{{2 - 3}}{{3 - 1}}\left( {x - 1} \right)\\y& = \frac{1}{2}\left( { - x + 7} \right)\end{align}

\begin{align}ar\left( {BLMCB} \right) &= \int_1^3 {\frac{1}{2}\left( { - x + 7} \right)} dx\\ &= \frac{1}{2}\left[ { - \frac{{{x^2}}}{2} + 7x} \right]_1^3\\ &= \frac{1}{2}\left[ { - \frac{9}{2} + 21 + \frac{1}{2} - 7} \right]\\ &= 5\end{align}

Equation of AC is

\begin{align}y - 0 &= \frac{{2 - 0}}{{3 + 1}}\left( {x + 1} \right)\\y& = \frac{1}{2}\left( {x + 1} \right)\end{align}

\begin{align}ar\left( {AMCA} \right) &= \frac{1}{2}\int_{ - 1}^3 {\left( {x + 1} \right)} dx\\ &= \frac{1}{2}\left[ {\frac{{{x^2}}}{2} + x} \right]_{ - 1}^3\\ &= \frac{1}{2}\left[ {\frac{9}{2} + 3 - \frac{1}{2} + 1} \right]\\ &= 4\end{align}

Therefore, $$ar\left( {\Delta ABC} \right) = \left( {3 + 5 - 4} \right) = 4 \,\rm{units}$$

Chapter 8 Ex.8.2 Question 5

Using integration find the area of the triangular region whose sides have the equations $$y = 2x + 1,y = 3x + 1$$ and $$x = 4$$.

Solution

Vertices of triangle are $$A\left( {0,1} \right),B\left( {4,13} \right)$$ and $$C\left( {4,9} \right)$$.

\begin{align}ar\left( {\Delta ACB} \right)& = ar\left( {OLBAO} \right) - ar\left( {OLCAO} \right)\\ &= \int_0^4 {\left( {3x + 1} \right)dx} - \int_0^4 {\left( {2x + 1} \right)dx} \\ &= \left[ {\frac{{3{x^2}}}{2} + x} \right]_0^4 - \left[ {\frac{{2{x^2}}}{2} + x} \right]_0^4\\& = \left( {24 + 4} \right) - \left( {16 + 4} \right)\\ &= 28 - 20\\ &= 8\end{align}

Chapter 8 Ex.8.2 Question 6

Smaller area enclosed by the circle $${x^2} + {y^2} = 4$$ and the line $$x + y = 2$$ is

(A) $${\rm{2}}\left( {\pi - 2} \right)$$

(B) $$\pi - 2$$

(C) $$2\pi - 1$$

(D) $${\rm{2}}\left( {\pi + 2} \right)$$

Solution

\begin{align}ar\left( {ACBA} \right)& = ar\left( {OACBO} \right) - ar\left( {\Delta OAB} \right)\\ &= \int_0^2 {\sqrt {4 - {x^2}} dx} - \int_0^2 {\left( {2 - x} \right)dx} \\& = \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\frac{x}{2}} \right]_0^2 - \left[ {2x - \frac{{{x^2}}}{2}} \right]_0^2\\& = \left[ {2 \times \frac{\pi }{2}} \right] - \left[ {4 - 2} \right]\\ &= \left( {\pi - 2} \right)\end{align}

Chapter 8 Ex.8.2 Question 7

Area lying between the curve $${y^2} = 4x$$ and $$y = 2x$$ is

(A) $$\frac{{\rm{2}}}{3}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{4}$$

(D) $$\frac{3}{4}$$

Solution

Points of intersection of curve $${y^2} = 4x$$ and $$y = 2x$$ are $${\rm{O}}\left( {0,0} \right)$$ and$$A\left( {1,2} \right)$$.

Draw AC perpendicular to x-axis.

Coordinates of C are (1,0)

\begin{align}ar\left( {OBAO} \right) &= ar\left( {\Delta OCA} \right) - ar\left( {OCABO} \right)\\& = \int_0^1 {2xdx} - \int_0^1 {2\sqrt x } dx\\ &= 2\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - 2\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1\\& = \left| {1 - \frac{4}{3}} \right|\\& = \left| { - \frac{1}{3}} \right|\\ &= \frac{1}{3}\end{align}

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