Let be differentiable for all . If and for , then
- (A)✓ Correct
- (B)
- (C)
- (D)
Answer: (A)
Solution
By the Mean Value Theorem on there is a point with . Since throughout the interval, , so . With , this gives .
JEE Main · 2005 · 08 May 2005 · Offline · Single correct option