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 ∈ R.

Question

Let \(f(\mathrm{x})\) be a real differentiable function such that f(0) = 1 and \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f^{\prime}(\mathrm{y})+f^{\prime}(\mathrm{x}) f(\mathrm{y})\) for all \(x, y \in \mathbf{R}.\) Then \(\sum_{\mathrm{n}=1}^{100} \log _{\mathrm{e}} f(\mathrm{n})\) is equal to :

(1) 2384

(2) 2525

(3) 5220

(4) 2406

Answer 1

Correct option is (2) 2525  

\(f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(x)\)

Put = x = y = 0 

\(f(0)=f(0) f^{\prime}(0)+f^{\prime}(0) f(0)\)

\(\mathrm{f}^{\prime}(0)=\frac{1}{2}\)

Put y = 0 

\(f(x)=f(x) f^{\prime}(0)+f^{\prime}(x) f(0)\)

\(f(x)=\frac{1}{2} f(x)+f^{\prime}(x)\)

\(f^{\prime}(x)=\frac{f(x)}{2}\)

\(\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2}\)

\(\Rightarrow \ell ny =\frac{x}{2}+c\)  

\(\because \mathrm{f}(0)=1 \Rightarrow \mathrm{C}=0\)  

\(\ell ny =\frac{\pi}{2} \Rightarrow f(x)=e^{x / 2}\)

\(\ell n f(n)=\frac{n}{2}\)   

\(\sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2}\)   

=2525