Consider a region R = { {( x , y ) ∈ R2 :
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?

  1. $\alpha^{3}-6 \alpha^{2}+16=0$

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

  3. $\alpha^{3}-6 \alpha^{3 / 2}-16=0$

  4.  $3 \alpha^{2}-8 \alpha^{3 / 2}+8=0$


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

$\Rightarrow \int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d y=2 \int_{0}^{a}\left(\sqrt{y}-\frac{y}{2}\right) \cdot d y$

$\Rightarrow \frac{4}{3}=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{1}{4} \cdot \alpha^{2}\right]$

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

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