Let \(f: R \rightarrow R\) be a function such that \(f(x+y)=f(x)+f(y), \forall x, y \in R\). If \(f(x)\) is differentiable at \(x=0\), then

\(f(x+y)=f(x)+f(y)\), as \(f(x)\) is differentiable at \(x=0\).
\[
\begin{aligned}
\Rightarrow & f^{\prime}(0)=k \\
\text { Now, } f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(h)}{h} \quad\left[\frac{0}{0} \text { from }\right]
\end{aligned}
\]
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 \(\left.\left(\begin{array}{l}3 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right) \underset{u_1}{\stackrel{1 \mathrm{R}}{u_1}}\left(\begin{array}{c}1 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right)\right)_{u_2}^{u_2}\)