Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x+y)=f(x)+f(y)$, as $f(x)$ is differentiable at $x=0$. $$\begin{array}{rl}\Rightarrow & f^{\prime }(0)=k\\ \text{ Now, }{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{f(x)+f(h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{f(h)}{h}\left[\frac{0}{0}\text{ from }\right]\end{array}$$ Given, $f(x+y)=f(x)+f(y),\forall x,y$ $$\begin{array}{rlrl}& \therefore & f(0)& =f(0)+f(0),\\ & \text{ when }& x& =y=0\Rightarrow f(0)=0\end{array}$$ Using L'Hospital's rule, $$\underset{h\to 0}{\lim }\frac{f^{\prime }(h)}{1}=f^{\prime }(0)=k$$ $\Rightarrow f^{\prime }(x)=k$, on integrating both sides, $f(x)=kx+C$, as $f(0)=0\Rightarrow C=0$ So, $f(x)=kx$ $\therefore f(x)$ is continuous for all $x\in R$ and $f^{\prime }(x)=k$, i.e. constant for all $x\in R$. Hence, both (b) and (c) are correct. Solutions (Q. Nos. 12-13) $$\left(\begin{array}{l}3\mathrm{W}\\ 2\mathrm{R}\end{array}\right)(\underset{u_1}{\underbrace{1\mathrm{W}}})\}\text{ Initial }$$ Head appears. $\left(\begin{array}{l}2\mathrm{W}\\ 2\mathrm{R}\end{array}\right)\underset{u_1}{\mathrm{wW}}(\underset{u_2}{\underbrace{2\mathrm{W}}})$ or ${\left(\begin{array}{l}3\mathrm{W}\\ 1\mathrm{R}\end{array}\right)\underset{u_1}{\stackrel{1\mathrm{R}}{u_1}}\left(\begin{array}{c}1\mathrm{W}\\ 1\mathrm{R}\end{array}\right))}_{u_2}^{u_2}$