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  • Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 9 maths textbook solution

Please solve RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 9 maths textbook solution

Answers (1)

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}

Posted by

Gurleen Kaur

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