Question

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

• A. $\frac{f\left(x\right)}{x}$ is differentiable in R
• B. $f\left(x\right)$ is continuous but not differentiable in R
• C. $f\left(x\right)$ is continuous in R
• D. None of these

Answer 1

The correct option is C $f\left(x\right)$ is continuous in R

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