Question:
Consider a region $R=\left\{(x, y) \in R^{2}: x^{2} \leq y \leq 2 x\right\}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts, then
which of the following is true?
Correct Option: , 4
Solution:
$* \mathrm{y} \geq \mathrm{x}^{2} \Rightarrow$ upper region of $\mathrm{y}=\mathrm{x}^{2}$
$\mathrm{y} \leq 2 \mathrm{x} \Rightarrow$ lower region of $\mathrm{y}=2 \mathrm{x}$
According to ques, area of $\mathrm{OABC}=2$ area of OAC
$\Rightarrow \int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d y=2 \int_{0}^{a}\left(\sqrt{y}-\frac{y}{2}\right) \cdot d y$
$\Rightarrow \frac{4}{3}=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{1}{4} \cdot \alpha^{2}\right]$
$\Rightarrow 3 \alpha^{2}-8 \alpha^{3 / 2}+8=0$