Let f: R arrow R be a function such that f(x+y)=f(x)+f(y), x, y ∈ R...

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