Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\because f(x+y)=f(x)\cdot f(y)+f(x)\cdot f(y),\forall x,y\in R$ ....(i) And $f(0)=1$ ....(ii) Now replace $x$ by zero and $y$ by zero we get $\begin{array}{rl}& f(0)=f(0)f(0)+f(0)f(0)\\ & 1=f(0)+f(0)\\ & \therefore f^{\prime }(0)=\frac{1}{2}...(iii)\end{array}$ 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 $c=0$ $\therefore \ln |f(x)|=\frac{x}{2}$ Then ${\sum }_{n=1}^{100}\ln (f(\eta ))=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\dots +\frac{100}{2}\right)$ $=\frac{5050}{2}=2525$