Question

Let $f : R \rightarrow R$ 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 $f '\left(0\right) = 3$ . 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 $f '\left(0\right) = 3$
$f '\left(x\right)=$ $\displaystyle \lim_{h \to 0}$$\frac{f \left(x+h\right)-f \left(x\right)}{h}$=$\displaystyle \lim_{h \to 0}$$\frac{f \left(\frac{3x+3h}{3}\right)-f \left(x\right)}{h}$
$=\displaystyle \lim_{h \to 0}$$\frac{\frac{f \left(3x\right)+f \left(3h\right)}{3}-\frac{f \left(3x\right)+ f \left(0\right)}{3}}{h}$ $=\displaystyle \lim_{h \to 0}$ $\frac{f \left(3h\right)-f \left(0\right)}{3h}=3$
$\therefore f\left(x\right) = 3x + c, \because f \left(0\right) =0 \Rightarrow c=0$
$\therefore f \left(x\right)=3x$