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?

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

Answer 1

Answer: (D)

* $y \geq x^{2} \Rightarrow $ upper region of
$y=x^{2}$ $y \leq 2 x \Rightarrow $ lower region of $y=2 x$
According to ques, area of $OABC =2$ area of $OAC$
$\Rightarrow \int\limits_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d y=2 \int\limits_{0}^{a}\left(\sqrt{y}-\frac{y}{2}\right) \cdot d y$
$\Rightarrow \frac{4}{3}=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{1}{4} . \alpha^{2}\right]$
${3 \alpha^{2}-8 \alpha^{3 / 2}+8=0}$