Let $f: R \rightarrow R$ be a function such that $f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}$  for all  $x, y \in R$. f(0) = 0 and f'(0) = 3. Then, f(x) is

differentiable in R

We have,

$f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}$  for all  $x, y \in R$

$\Rightarrow f\left(\frac{x}{3}\right)=\frac{f(x)+f(0)}{3}$  for all  $x \in R$    [Putting y = 0]

$\Rightarrow 3 f\left(\frac{x}{3}\right)=f(x)$  for all  $x \in R$         [∵ f(0) = 0]

Now,

$f'(x) =\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\Rightarrow f'(x) =\lim\limits_{h \rightarrow 0} \frac{f\left(\frac{3 x+3 h}{3}\right)-f(x)}{h}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{\frac{f(3 x)+f(3 h)}{3}-f(x)}{h}$            $\left[∵ f\left(\frac{x+y}{3}\right) =\frac{f(x)+f(y)}{3}\right]$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(3 x)+f(3 h)-3 f(x)}{3 h}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{3 f(x)+f(3 h)-3 f(x)}{3 h}$               $\begin{bmatrix}  ∵ 3f\left(\frac{x}{3}\right)=f(x) \\ \Rightarrow 3f(x) = f(3x)\end{bmatrix}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(3 h)}{3 h}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(3 h)-f(0)}{3 h}$                [∵  f(0) = 0]

$\Rightarrow f'(x)=f'(0)$                [∵  f'(0) = 3]

$f'(x)=3$

$\Rightarrow f(x)=3 x+C$

But, f(0) = 0. Therefore, C = 0