Question

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

• A. 3.0
• B. 3.00
• C. 3

Answer 1

Answer: (A), (B), (C)

There will be three points $x=1,3,5$ at which $f\left(x\right)$ is non-differentiable.
So, $f\left(f\right(1\left)\right)+f\left(f\right(3\left)\right)+f\left(f\right(5\left)\right)$
$=f\left(0\right)+f\left(2\right)+f\left(0\right)$
$=1+1+1$
$=3$