The area of the region {(x,y):0≤x≤49,0≤y≤1,x≥3y,x+y≥2} is
(A)3211✓ Correct
(B)9635
(C)9637
(D)3213
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