JEE Main 2024 Mathematics question
Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and h′(0) = h′(1) = 2. If g(x) = h(eˣ, then g′(0) is equal to:
Answer: (D)
Solution
g′(x) = h′(eˣ)eˣ + h(eˣ′(x). At 0: h′(1) × 1 × e⁰ + h(1) × e⁰ × h′(0) = 2 + 2 = 4.