JEE Main 2025 — 2 April, Morning Shift — Question 42

Given $f''(x) = f(x)$, the general solution is $f(x) = A e^x + B e^{-x}$. Using $f(0) = 0$: $A + B = 0 \Rightarrow B = -A$, so $f(x) = A(e^x - e^{-x})$. Using $f'(0) = 3$: $f'(x) = A(e^x + e^{-x})$, so $f'(0) = A(1+1) = 2A = 3 \Rightarrow A = \frac{3}{2}$. Thus $f(x) = \frac{3}{2}(e^x - e^{-x})$. Now $f(\ln 3) = \frac{3}{2}(e^{\ln 3} - e^{-\ln 3}) = \frac{3}{2}\left(3 - \frac{1}{3}\right) = \frac{3}{2} \cdot \frac{8}{3} = 4$. Therefore $9 f(\ln 3) = 9 \times 4 = 36$.