JEE Advanced 2021 — Paper 1 — Question 20

The region is defined by $0 \le x \le \frac{9}{4}$, $0 \le y \le 1$, $x \ge 3y$, and $x + y \ge 2$. Rewrite inequalities: $y \le \frac{x}{3}$ and $y \ge 2 - x$. Find intersection of $y = \frac{x}{3}$ and $y = 2 - x$: $\frac{x}{3} = 2 - x \Rightarrow x = \frac{3}{2}$, $y = \frac{1}{2}$. For $x \in [1.5, 2]$, the lower bound is $y = 2 - x$ and upper bound is $y = \frac{x}{3}$. For $x \in [2, 2.25]$, the lower bound is $y = 0$ and upper bound is $y = \frac{x}{3}$. Area $A = \int_{1.5}^{2} \left( \frac{x}{3} - (2 - x) \right) dx + \int_{2}^{2.25} \frac{x}{3} dx$. First integral: $\int_{1.5}^{2} \left( \frac{4x}{3} - 2 \right) dx = \left[ \frac{2x^2}{3} - 2x \right]_{1.5}^{2} = \left( \frac{8}{3} - 4 \right) - \left( \frac{4.5}{3} - 3 \right) = -\frac{4}{3} + \frac{3}{2} = \frac{1}{6}$. Second integral: $\int_{2}^{2.25} \frac{x}{3} dx = \left[ \frac{x^2}{6} \right]_{2}^{2.25} = \frac{81/16}{6} - \frac{4}{6} = \frac{81}{96} - \frac{64}{96} = \frac{17}{96}$.

Total area $= \frac{1}{6} + \frac{17}{96} = \frac{16}{96} + \frac{17}{96} = \frac{33}{96} = \frac{11}{32}$.

The answer is A.