Let f: R → R be a thrice differentiable odd function satisfying f′(x) ≥ 0, f^(′′)(x)=f(x), f(0)=0, f′(0)=3. Then 9 f(log ₑ 3) is equal to __________ .

Solution

Maths sir · 4 steps

Read f'' = f as f = Aeˣ + Be⁻ˣ, then let the two given conditions fix the constants.

1. Solve the equation. Try f = eᵐˣ in f'' - f = 0. That gives m² = 1, so m = ± 1 and f(x) = Aeˣ + Be⁻ˣ.
2. Use the conditions. f(0) = A + B = 0, so B = -A. Then f'(x) = Aeˣ + Ae⁻ˣ, and f'(0) = 2A = 3 gives A = 3/2.

   f(x) = 3/2(eˣ - e⁻ˣ)

3. Check the rest. This f is odd, and f'(x) = 3/2(eˣ + e⁻ˣ) > 0, so every given condition holds.
4. So. f(logₑ 3) = 3/2(3 - 1/3) = 3/2 × 8/3 = 4. Then 9f(logₑ 3) = 36.

The common slip: 72. Taking A = 3 instead of 3/2, by forgetting that f'(0) = 2A, doubles f and gives 72 instead of 36.

The idea to keep. For y'' = k² y, write y = Aeᵏˣ + Be⁻ᵏˣ straight away. Use y(0) and y'(0) to find A and B, and remember that e⁻ˣ gives a minus sign when you differentiate.

Answer from the NTA final answer key. Solution by PrepWiser.