If the area of the region {(x,y):1+x2≤y≤min{x+7,11−3x}} is A , then 3 A is equal to
(A)50✓ Correct
(B)49
(C)46
(D)47
Answer: (A)
Solution
The two upper boundaries meet where x+7=11−3x, i.e. x=1. The parabola y=1+x2 meets y=x+7 at x=−2 (the relevant point) and meets y=11−3x at x=2. Splitting at x=1, A=∫−21(x+6−x2)dx+∫12(10−3x−x2)dx=227+619=350. Therefore 3A=50.
JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option