JEE Advanced 2022 Paper 2, Mathematics Section 1 Q8: Area as Definite Integral

By symmetry the region is symmetric about the $y$-axis; compute the area on $[0,3/4]$ and double.

Find the intersection of $f$ and $g$ on $[0,3/4]$: $x^2+\frac{5}{12}=2-\frac{8x}{3}$, i.e. $12x^2+32x-19=0$. Solving the quadratic gives $x=\frac{1}{2}$ as the positive root.

On $[0,1/2]$, the parabola $f$ lies below $g$, so $\min \{f,g\}=f$. On $[1/2,3/4]$, the line $g$ lies below $f$, so $\min \{f,g\}=g$.

Area on the right half: ${\int }_0^{1/2}\left(x^2+\frac{5}{12}\right)dx+\frac{1}{2}\cdot \left(\frac{3}{4}-\frac{1}{2}\right)\cdot g(1/2)={\left[\frac{x^3}{3}+\frac{5x}{12}\right]}_0^{1/2}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{2}{3}=\frac{1}{24}+\frac{5}{24}+\frac{1}{12}=\frac{1}{4}+\frac{1}{12}=\frac{1}{3}$.

Doubling: $\alpha =\frac{2}{3}$, hence $9\alpha =6$.