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…
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
- $f(x)$ is differentiable only in a finite interval containing zero
- $f(x)$ is continuous, $\forall x \in R$
- $f^{\prime}(x)$ is constant, $\forall x \in R$
- $f(x)$ 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
Asked in: JEE Advanced 2011 (Paper 1)
Practice more Continuity and Differentiability questions on Aicharya