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  • Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 7

Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 7

Answers (1)

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}

Posted by

Gurleen Kaur

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