Limits, Continuity & Differentiability
Differential Equation from Functional Equation
nta_pyq_2025_apr
Grade 12
Question:
Let $f(x)$ be a real differentiable function such that $f(0) = 1$ and $f(x+y) = f(x)f'(y) + f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\displaystyle\sum_{n=1}^{100} \log_e f(n)$ is equal to:
Step-by-Step Solution
Key Concept: From the functional equation, derive a first-order ODE. Setting $x=y=0$: $f(0)=2f(0)f'(0)\Rightarrow f'(0)=1/2$. Setting $y=0$: $f(x)=f(x)/2+f'(x)/2\Rightarrow f'(x)/f(x)=1/2$.
$f'(0)=1/2$ (from $x=y=0$). $y=0$: $f(x)=\frac{1}{2}f(x)+f'(x)\Rightarrow f'(x)=\frac{1}{2}f(x)\Rightarrow f(x)=e^{x/2}$. $\sum_{n=1}^{100}\ln f(n)=\sum_{n=1}^{100}\frac{n}{2}=\frac{5050}{2}=2525$.
Correct Answer: 2525