JEE Advanced 2018 Paper 2Question 1
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function with $f(0)=1$ and satisfying the equation $$f(x+y)=f(x)f'(y)+f'(x)f(y)\text{ for all }x,y\in\mathbb{R}.$$ Then, the value of $\log_e(f(4))$ is _____.
Differential Equation
Answer$2$
Solution
Concept. Put $x=y=0$ to find $f'(0)$, then $y=0$ gives a differential equation. Formulas used. (1) $f'=kf,\ f(0)=1\Rightarrow f=e^{kx}$Step 1 - $f'(0)$. With $x=y=0$: $f(0)=2f(0)f'(0)$, so $f'(0)=\frac12$. Step 2 - Put $y=0$. $f(x)=f(x)\cdot\frac12+f'(x)\cdot1$, so $f'=\frac f2$. Step 3 - Solve. By (1), $f(x)=e^{x/2}$. Step 4 - Value. $\ln f(4)=2$. Answer: 2