QJEEM25Ma6535: JEE Main 2025 Mathematics - Class 12 JEEM25Ma6535 Solution: Let be a real differentiable function such that and for all .…

$\begin{aligned} & f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(x) \\ & \text { Put }=x=y=0 \\ & f(0)=f(0) f^{\prime}(0)+f^{\prime}(0) f(0) \\ & f^{\prime}(0)=\frac{1}{2} \\ & \text { Put } y=0 \\ & f(x)=f(x) f^{\prime}(0)+f^{\prime}(x) f(0) \\ & f(x)=\frac{1}{2} f(x)+f^{\prime}(x) \\ & f^{\prime}(x)=\frac{f(x)}{2}\end{aligned}$

$\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2}$

$\Rightarrow \ln y=\frac{x}{2}+c$

$\begin{aligned} & \because \mathrm{f}(0)=1 \Rightarrow \mathrm{C}=0 \\ & \ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2} \\ & \ln \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2} \\ & \sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2} \\ & =2525\end{aligned}$