Consider the region $R=\{(x, y) ∈ R^2 : x^2 <y ≤x\}$. If a line $y = α$ divides the area of region R into two equal parts, then which of the following is true? |
$3α^2-8α+8=0$ $α^3-6α^{3/2} -16=0$ $α^3-6α^2+16=0$ $3α^2-8α^{3/2}+8=0$ |
$3α^2-8α^{3/2}+8=0$ |
It is given that $y = α$ divides the shaded region into two equal parts. $∴\int\limits_0^4(x_2-x_1)dy=2\int\limits_0^α(x_2-x_1)dy$ $⇒\int\limits_0^4\left(\sqrt{y}-\frac{y}{2}\right)dy=2\int\limits_0^α\left(\sqrt{y}-\frac{y}{2}\right)dy$ $⇒\left[\frac{2}{3}y^{3/2}-\frac{1}{4}y^2\right]_0^4=2\left[\frac{2}{3}y^{3/2}-\frac{1}{4}y^2\right]_0^α$ $⇒\frac{4}{3}=2\left(\frac{2}{3}α^{3/2}-\frac{1}{4}α^2\right)$ $⇒\frac{2}{3}=\frac{2}{3}α^{3/2}-\frac{1}{4}α^2$ $⇒1=α^{3/2}-\frac{3}{8}α^2$ $⇒8α^{3/2}-3α^2-8=0⇒3α^2-8α^{3/2}+8=0$ |