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

• A

  $f(x)$ is differentiable only in a finite interval containing zero
• Correct answer
  B

  $f(x)$ is continuous, $\forall x \in R$
• Correct answer
  C

  $f^{\prime}(x)$ is constant, $\forall x \in R$
• D

  $f(x)$ is differentiable except at finitely many points

Solution

$f(x+y)=f(x)+f(y)$, 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 $\left.\left(\begin{array}{l}3 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right) \underset{u_1}{\stackrel{1 \mathrm{R}}{u_1}}\left(\begin{array}{c}1 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right)\right)_{u_2}^{u_2}$

Hence, the correct answer is options: 'B', 'C'.