Question

Consider a region$R=\left\{(x,y)\in R^2:x^2\le y\le 2x\right\}$. If a line $y=\alpha$divides the area of region R into two equal parts, then which of the following is true?

• A. ${\alpha }^3-6{\alpha }^2+16=0$
• B. $3{\alpha }^2-8{\alpha }^{3/2}+8=0$
• C. ${\alpha }^3-6{\alpha }^{3/2}-16=0$
• D. $3{\alpha }^2-8\alpha +8=0$

Answer 1

The correct option is B

$3{\alpha }^2-8{\alpha }^{3/2}+8=0$

Explanation for the correct option:

To find the relation:

$y\ge x^2\Rightarrow$upper region of $y=x^2\left(x=\sqrt{y}\right)$

$y\le 2x\Rightarrow$lower region of $y=2x\left(x=\frac{y}{2}\right)$

According to the question.

Area of the region $OABCO=2\times$Area of the region $OACO$

$$\begin{gathered} \Rightarrow {\int }_0^4\left(\sqrt{y}-\frac{y}{2}\right)dy=2{\int }_0^{\alpha }\left(\sqrt{y}-\frac{y}{2}\right)dy \\ \Rightarrow {\left[\frac{2}{3}{y}^{3/2}-\frac{y^2}{4}\right]}_0^4=2{\left[\frac{2}{3}{y}^{3/2}-\frac{y^2}{4}\right]}_0^{\alpha } \\ \Rightarrow \frac{4}{3}=2\left[\frac{2}{3}{\alpha }^{3/2}-\frac{{\alpha }^2}{4}\right] \\ \Rightarrow 8=8{\alpha }^{3/2}-3{\alpha }^2 \\ \Rightarrow 3{\alpha }^2-8{\alpha }^{3/2}+8=0 \end{gathered}$$

Hence ,the correct option is (B).