Maths NCERT Exemplar Solutions Class 12th Chapter Eight: Overview, Questions, Preparation

Sol:

$\begin{array}{l}Wehave,y^2=9x,y=3x\\ Solvingthetwoequations,wehave\\ {\left(3x\right)}^2=9x\\ \Rightarrow 9x^2-9x=0\Rightarrow 9x\left(x-1\right)=0\\ \therefore x=0,1\\ Areaoftheshadedregion\\ =ar\left(regionOAB\right)-ar\left(\Delta OAB\right)\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{1}dx=}{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{9x}dx}-{\displaystyle \underset{0}{\overset{1}{\int }}3xdx}\\ =3{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{x}dx}-3{\displaystyle \underset{0}{\overset{1}{\int }}xdx=3\times \frac{2}{3}}{\left[x^{3/2}\right]}_0^1-3{\left[\frac{x^2}{2}\right]}_0^1\\ =2\left[{\left(1\right)}^{3/2}-0\right]-\frac{3}{2}\left[{\left(1\right)}^2-0\right]=2\left(1\right)-\frac{3}{2}\left(1\right)=2-\frac{3}{2}=\frac{1}{2}sq.units\\ Hence,therequiredarea=\frac{1}{2}sq.units.\end{array}$

2. Find the area of the region bounded by the parabolas $y^2=2px$  and $x^2=2py$ .

Sol:

$\begin{array}{l}Wearegiventhat{x}^2=2py\dots \left(i\right)\\ and{y}^2=2px\dots \left(ii\right)\\ Fromeqn.\left(i\right)wegety=\frac{x^2}{2p}\\ Puttingthetheofyineqn.\left(ii\right),wehave\\ {\left(\frac{x^2}{2p}\right)}^2=2px\Rightarrow \frac{x^4}{4p^2}=2px\\ \Rightarrow x^4=8p^{3}x\Rightarrow x^4-8p^{3}x=0\\ \Rightarrow x\left(x^3-8p^3\right)=0\\ \therefore x=0,2p\\ Requiredarea=Areaoftheregion\left(OCBA-ODBA\right)\\ ={\displaystyle \underset{0}{\overset{2p}{\int }}\sqrt[]{2px}dx-}{\displaystyle \underset{0}{\overset{2p}{\int }}\frac{x^2}{2p}dx}\\ =\sqrt[]{2p}{\displaystyle \underset{0}{\overset{2p}{\int }}\sqrt[]{x}dx-}\frac{1}{2p}{\displaystyle \underset{0}{\overset{2p}{\int }}{x}^{2}dx}=\sqrt[]{2p}.\frac{2}{3}{\left[x^{3/2}\right]}_0^{2p}-\frac{1}{2p}.\frac{1}{3}{\left[x^3\right]}_0^{2p}\\ =\frac{2\sqrt[]{2}}{3}\left[{\left(2p\right)}^{3/2}-0\right]-\frac{1}{6p}\left[{\left(2p\right)}^3-0\right]\\ =\frac{2\sqrt[]{2}}{3}\sqrt[]{p}.2\sqrt[]{2}{p}^{\frac{3}{2}}-\frac{1}{6p}.8p^3\\ =\frac{8}{3}.p^2-\frac{8}{6}{p}^2=\frac{8}{6}{p}^2=\frac{4}{3}{p}^{2}sq.units\\ Hence,therequiredarea=\frac{4}{3}{p}^{2}sq.units.\end{array}$

1. Find the area of the region bounded by the curves $y^2=2x$ and $x^2+y^2=4x$ .

Sol.

$\begin{array}{l}Equationsofthecurvearegivenby\\ x^2+y^2=4x\dots \left(i\right)\\ and{y}^2=2x\dots \left(ii\right)\\ \Rightarrow x^2-4x+y^2=0\\ \Rightarrow x^2-4x+4-4+y^2=0\\ {\left(x-2\right)}^2+y^2=4\\ Clearlyitistheequationofacirclehavingitscentre\left(2,0\right)andradius2.\\ Solving{x}^2+y^2=4xand{y}^2=2x\\ x^2+2x=4x\\ \Rightarrow x^2+2x-4x=0\\ \Rightarrow x^2-2x=0\\ \Rightarrow x\left(x-2\right)=0\\ \therefore x=0,2\\ Areaoftherequiredregion=2\left[{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{4-{\left(x-2\right)}^2}dx-}{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x}dx}\right]\\ \left[\because Parabolaandcirclebotharesymmetricalaboutx-axis\right]\\ =2{\left[\frac{x-2}{2}\sqrt[]{4-{\left(x-2\right)}^2}+\frac{4}{2}si{n}^{-1}\frac{x-2}{2}\right]}_0^2-2.\sqrt[]{2}.\frac{2}{3}{\left[x^{3/2}\right]}_0^2\\ =2\left[\left(0+0\right)-\left(0+2si{n}^{-1}\left(-1\right)\right)\right]-\frac{4\sqrt[]{2}}{3}\left[{\left(2\right)}^{3/2}-0\right]\\ =2\times 2.\left(-\frac{\pi }{2}\right)-\frac{4\sqrt[]{2}}{3}.2\sqrt[]{2}=2\pi -\frac{16}{3}=2\left(\pi -\frac{8}{3}\right)sq.units\\ Hence,therequiredarea=2\left(\pi -\frac{8}{3}\right)sq.units.\end{array}$

2. Find the area bounded by the curve $y=sinx$ between $x=0$ and $x=2\pi$ .

Sol. 

$\begin{array}{l}Requiredarea={\displaystyle \underset{0}{\overset{\pi }{\int }}sinxdx}+{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|sinx\right|dx}\\ ={\left[cosx\right]}_0^{\pi }+{\left|\left(-cosx\right)\right|}_0^{2\pi }\\ =-\left[cos\pi -cos0\right]+\left[cos2\pi -cos\pi \right]\\ =-\left[-1-1\right]+\left[1+1\right]=2+2=4sq.units\\ Hence,therequiredarea=4sq.units.\end{array}$

1. The area of the region bounded by the y-axis, $y=cosx$  and $y=sinx$ , $0\le x\le \frac{\pi }{2}$ is:

(A) $\sqrt[]{2}$ sq. units

(B) $\sqrt[]{2}$ +1sq. units

(C) $\sqrt[]{2}-1$  sq. units

(D) $2\sqrt[]{2}-1$ sq. units

Sol.

$\begin{array}{l}Giventhaty-axis,y=cosx,y=sinx,0\le x\le \frac{\pi }{2}\\ Requiredarea={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}cosx}dx-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}sinx}dx\\ ={\left[sinx\right]}_0^{\frac{\pi }{4}}-{\left[-cosx\right]}_0^{\frac{\pi }{4}}\\ =\left[sin\frac{\pi }{4}-sin0\right]+\left[cos\frac{\pi }{4}-cos0\right]\\ =\left[\frac{1}{\sqrt[]{2}}-0+\frac{1}{\sqrt[]{2}}-1\right]=\frac{2}{\sqrt[]{2}}-1=\left(\sqrt[]{2}-1\right)sq.units\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

2. The area of the region bounded by the curve $x^2=4y$  and the straight line $x=4y-2$ is:

(A) $\frac{3}{8}$ sq. units

(B) $\frac{5}{8}$ sq. units

(C) $\frac{7}{8}$ sq. units

(D) $\frac{9}{8}$ sq. units

Sol.

$\begin{array}{l}Giventhat:Theequationofparabolais{x}^2=4y\dots \left(i\right)\\ andequationofstraightlinex=4y-2\dots \left(ii\right)\\ Solvingeqn\left(i\right)and\left(ii\right)weget\\ y=\frac{x^2}{4}\Rightarrow x=4\left(\frac{x^2}{4}\right)-2\\ \Rightarrow x=x^2-2\Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x-2\right)\left(x+1\right)=0\therefore x=-1,x=2\\ Requiredarea={\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x+2}{4}}dx-{\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x^2}{4}}dx\\ =\frac{1}{4}{\left[\frac{x^2}{2}+2x\right]}_{-1}^2-\frac{1}{4}.\frac{1}{3}{\left[x^3\right]}_{-1}^2\\ =\frac{1}{4}\left[\left(\frac{4}{2}+2\right)-\left(\frac{1}{2}-2\right)\right]-\frac{1}{12}\left[8+1\right]\\ =\frac{1}{4}\left[6+\frac{3}{2}\right]-\frac{1}{12}\left[9\right]=\frac{1}{4}\times \frac{15}{2}-\frac{3}{4}=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}sq.units\\ Hence,thecorrectoptionis\left(d\right).\end{array}$