The area of the region $\{(x,y): 0 \leq x \leq \frac{9}{4}, \quad 0 \leq y \leq 1, \quad x \geq 3y, \quad x+y \geq 2\}$ is

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Area Under Curves

The area of the region

{(x, y) : 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3 y, x + y \geq 2}

(A) $\frac{11}{32}$ ✓ Correct
(B) $\frac{35}{96}$
(C) $\frac{37}{96}$
(D) $\frac{13}{32}$

Answer: (A)

Solution

The region is bounded by x=0, x=9/4, y=0, y=1, x=3y, and x+y=2. The vertices of the region are P(3/2,1/2), Q(2,0), R(9/4,0), and S(9/4,3/4). Using the shoelace formula for the area of a quadrilateral with these vertices: Area = (1/2)|det₁ + det₂ + det₃ + det₄| where each determinant is formed by consecutive vertices. Computing: Area = (1/2)|(0-1) + (0-0) + (27/16-0) + (9/8-9/8)| = (1/2)|−1 + 27/16| = (1/2) · 11/16 = 11/32.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option

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