Mathematics - Continuity - Differentiability Question with Solution | TestHub

Let $f(x+y)=f(x)+f(y),\forall x,y\in R$ Put $x=0=y$ $\Rightarrow f(0)=f(0)+f(0)$ $\Rightarrow f(0)=0$ Now, $f^{\prime }(0)={\lim }_{h\to 0}\frac{f(0+h)-f(0)}{h}$ $f^{\prime \prime }(0)={\lim }_{h\to 0}\frac{f(h)}{h}$ Now, $f(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ $={\lim }_{h\to 0}\frac{f(x)+f(h)-f(x)}{h}$ $$\begin{array}{l}\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})={\lim }_{\mathrm{h}\to 0}\frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}={\mathrm{f}}^{\prime }(0)\\ \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{xf}(0)+\mathrm{C}\end{array}$$ But $f(0)=0$ $$\therefore \mathrm{C}=0$$ Hence, $f(x)=xf(0),\forall x\in R$ Clearly, $\mathrm{f}(\mathrm{x})$ is everywhere continuous and differentiable and ${\mathrm{f}}^{\prime }(\mathrm{x})$ is constant. $\forall \mathrm{x}\in \mathrm{R}$