If $f(x)$ is differentiable for all $x \in \mathbb{R}$ and satisfies the relation

$x=\mathop {\lim }\limits_{n \to \infty }\frac{\left[1^2(f(x))^x\right]+\left[2^2(f(x))^x\right]+\ldots+\left[n^2(f(x))^x\right]}{n^3}$ where [.] denotes the greatest integer function, then $f^{\prime}(x)=$

Let $f(x)=|1-2 x|$, then

A function $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfies $f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)+f(0)}{3}$ for all $x, y \in \mathbb{R}$. If the function ' $f$ ' is differentiable at $x=0$, then $f$ is