JEE Advanced 2018 — Paper 2 — Question 38

Given functional equation: $f(x+y)=f(x)f'(y)+f'(x)f(y)$ for all $x,y$. Set $y=0$: $f(x)=f(x)f'(0)+f'(x)f(0)$. With $f(0)=1$, we get $f'(x)=f(x)(1-f'(0))$. Set $x=0$: $f(y)=f(0)f'(y)+f'(0)f(y)$ ⇒ $f'(y)=f(y)(1-f'(0))$, same relation. Evaluate at $y=0$: $f'(0)=f(0)(1-f'(0)) = 1-f'(0)$ ⇒ $2f'(0)=1$ ⇒ $f'(0)=\frac12$. Then $f'(x)=f(x)\left(1-\frac12\right)=\frac12 f(x)$. This is $\frac{df}{dx}=\frac12 f$. Solve: $\int \frac{df}{f} = \int \frac12 dx$ ⇒ $\ln f(x) = \frac{x}{2}+C$. Using $f(0)=1$ gives $C=0$, so $f(x)=e^{x/2}$. Hence $\log_e f(4) = \ln e^{4/2} = \ln e^2 = 2$.