Question

Let $f:R\to R$ be a function such that
$f(x+y)=f\left(x\right)+f\left(y\right),\forall x,y\in R$
If $f\left(x\right)$ is differentiable at $x=0$, then which one of the following is incorrect?

• A. $f\left(x\right)$ is continuous, $\forall x\in R$
• B. $f^{\prime }\left(x\right)$ is constant, $\forall x\in R$
• C. $f\left(x\right)$ is differentiable, $\forall x\in R$
• D. $f\left(x\right)$ is differentiable only in finite interval containing zero

Answer 1

The correct option is D $f\left(x\right)$ is differentiable only in finite interval containing zero

Let $f(x+y)=f\left(x\right)+f\left(y\right),\forall x,y\in R$
Put $x=0=y$
$\Rightarrow f\left(0\right)=f\left(0\right)+f\left(0\right)\Rightarrow f\left(0\right)=0$
Now $f^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}$
$f^{\prime \prime }\left(0\right)={lim}_{h\to 0}\frac{f\left(h\right)}{h}$
Now $f\left(x\right)={lim}_{h\to 0}\frac{f(x+h)f\left(x\right)}{h}={lim}_{h\to 0}\frac{f\left(x\right)+f\left(h\right)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f\left(h\right)}{h}=f^{\prime }\left(0\right)$
$\Rightarrow f\left(x\right)=x{f}^{\prime }\left(0\right)+C$
But $f\left(0\right)=0$
$\therefore$ $C=0$
Hence, $f\left(x\right)=x{f}^{\prime }\left(x\right),\forall x\in R$
Clearly, $f\left(x\right)$ is everywhere continuous and differentiable and $f^{\prime }\left(x\right)$ constant $\forall x\in R$