Question

Let $S$ be the set of points where the function,$f\left(x\right)=|2-x|-3\left|\right|,x\in R$, is not differentiable. Then, the value of$\sum _{x\in S}f\left(f\right(x\left)\right)$ is equal to

Answer 1

Determine the value of $\sum _{x\in S}f\left(f\right(x\left)\right)$.

On checking the differentiability, we get

$f\left(x\right)=\left\{\begin{array}{l}1-xx<1\\ x-11\le x<3\\ 5-x3\le x<5\\ x-5x\ge 5\end{array}\right.$

Since, $f\left(x\right)=|2-|x-3\left|\right|$, so $f\left(x\right)$ is not differentiable at

$x=1,3,5$

$\therefore S=\left\{1,3,5\right\}$

Thus,

$$\begin{gathered} \sum _{x\in s}f\left(x\right)=f\left(f\right(1\left)\right)+f\left(f\right(3\left)\right)+f\left(f\right(5\left)\right) \\ =f\left(0\right)+f\left(2\right)+f\left(0\right) \\ =1+1+1 \\ =3 \end{gathered}$$

Therefore,$\sum _{x\in S}f\left(f\right(x\left)\right)=3$