Question

Let $f:R\to R$ be a function such that
$f\left(\frac{x+y}{3}\right)=\frac{f\left(x\right)+f\left(y\right)}{3},f\left(0\right)=0$ and $f^{\prime }\left(0\right)=3$ . Then

• A. $f(x)$ is a quadratic function
• B. $f(x)$ is continuous but not differentiable
• C. $f(x)$ is differentiable in $R$
• D. $f(x)$ is bounded in $R$

Answer 1

Answer: (C)

We have
$f\left(\frac{x+y}{3}\right)=\frac{f\left(x\right)+f\left(y\right)}{3},f\left(0\right)=0$ and $f^{\prime }\left(0\right)=3$
$f^{\prime }\left(x\right)=$ $\underset{h\to 0}{\lim }$$\frac{f\left(x+h\right)-f\left(x\right)}{h}$=$\underset{h\to 0}{\lim }$$\frac{f\left(\frac{3x+3h}{3}\right)-f\left(x\right)}{h}$
$=\underset{h\to 0}{\lim }$$\frac{\frac{f\left(3x\right)+f\left(3h\right)}{3}-\frac{f\left(3x\right)+f\left(0\right)}{3}}{h}$ $=\underset{h\to 0}{\lim }$ $\frac{f\left(3h\right)-f\left(0\right)}{3h}=3$
$\therefore f\left(x\right)=3x+c,\because f\left(0\right)=0\Rightarrow c=0$
$\therefore f\left(x\right)=3x$