Q34 / 75 · JEE Main 2025 (22 Jan Shift 1) · Mathematics

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

  • Correct answer
    A
    $2525$
  • B
    $5220$
  • C
    $2384$
  • D
    $2406$

Solution

$\because f(x+y)=f(x) \cdot f(y)+f(x) \cdot f(y), \forall x, y \in R$ ....(i) And $f(0)=1$ ....(ii) Now replace $x$ by zero and $y$ by zero we get $\begin{aligned} & f(0)=f(0) f(0)+f(0) f(0) \\ & 1=f(0)+f(0) \\ & \therefore \quad f^{\prime}(0)=\frac{1}{2} ...(iii) \end{aligned}$ Now replace $y$ by zero in equation (i), we get $f(x)=\frac{1}{2} f(x)+f^{\prime}(x)$ or, $\frac{1}{2} f(x)=f^{\prime}(x)$ then $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{2}$ hence $\ln |f(x)|=\frac{x}{2}+c$ Put $x=0$, we get $c=0$ $\therefore \quad \ln |f(x)|=\frac{x}{2}$ Then $\sum_{n=1}^{100} \ln (f(\eta))=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots+\frac{100}{2}\right)$ $=\frac{5050}{2}=2525$

Hence, the correct answer is option: 'A'.

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