JEE Main 2024 Mathematics question
Let the area of the region {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ min{x² + 2, 2x + 2}} be A. Then 12A is equal to .
Answer: 164
Solution
min(x² + 2, 2x + 2) = x² + 2 for 0 ≤ x ≤ 2 and 2x + 2 for 2 ≤ x ≤ 3. So A = ∫₀²(x² + 2)dx + ½(6 + 8)(1) = 20/3 + 7 = 41/3, and 12A = 164.