JEE Advanced 2021 · Paper 1, Mathematics question 2

Slice horizontally. For each $y$, $x$ runs from $\max(3y,\,2-y)$ to $\frac94$. The two lower bounds cross at $y=\frac12$, and the strip vanishes when $3y=\frac94$, i.e. $y=\frac34$.

The two pieces.

• $0\le y\le\frac12$: $\displaystyle\int_0^{1/2}\left(\frac94-2+y\right)dy=\frac18+\frac18=\frac14$.
• $\frac12\le y\le\frac34$: $\displaystyle\int_{1/2}^{3/4}\left(\frac94-3y\right)dy=\frac9{16}-\frac32\cdot\frac5{16}=\frac3{32}$.

Total. $\frac14+\frac3{32}=\frac{11}{32}$. (A)