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  • Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 39

Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 39

Answers (1)

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}

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Gurleen Kaur

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