Question

Let the area of the region $\left\{(x, y):|2 x-1| \leq y \leq\left|x^2-x\right|, 0 \leq x \leq 1\right\}$ be $A$ Then $(6 A +11)^2$ is equal to ____

Answer 1

Correct Answer: 125

$y\ge |2x-1|,y\le \left|x^2-x\right|$
Both curve are symmetric about $x=\frac{1}{2}$ Hence
$A=2\underset{\frac{3-\sqrt{5}}{2}}{\overset{\frac{1}{2}}{\int }}\left(\left(x-x^2\right)-(1-2x)\right)dx$
$A=2\underset{\frac{3-\sqrt{5}}{2}}{\overset{\frac{1}{2}}{\int }}\left(-x^2+3x-1\right)dx=2{\left(\frac{-x^3}{3}+\frac{3}{2}{x}^2-x\right)}_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}$
On solving $6A+11=5\sqrt{5}$
$(6A+11{)}^2=125$
So , the correct answer is 125.