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(C) The function is given by $f(x) = |2 - |x - 3||$.First,we identify the points where $f(x)$ is not differentiable. The function $g(x) = |x - 3|$ is

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$3$

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(C) The function is given by $f(x) = |2 - |x - 3||$.First,we identify the points where $f(x)$ is not differentiable. The function $g(x) = |x - 3|$ is not differentiable at $x = 3$. The function $h(x) = 2 - |x - 3|$ is not differentiable at $x = 3$. The function $f(x) = |h(x)|$ is not differentiable where $h(x) = 0$ or where $h(x)$ is not differentiable.Setting $h(x) = 0$,we get $2 - |x - 3| = 0$,which implies $|x - 3| = 2$,so $x - 3 = 2$ or $x - 3 = -2$. Thus,$x = 5$ or $x = 1$.Therefore,the set of points $S$ where $f(x)$ is not differentiable is $S = \{1, 3, 5\}$.Now,we calculate $f(f(x))$ for each $x \in S$:For $x = 1$: $f(1) = |2 - |1 - 3|| = |2 - 2| = 0$. Then $f(f(1)) = f(0) = |2 - |0 - 3|| = |2 - 3| = 1$.For $x = 3$: $f(3) = |2 - |3 - 3|| = |2 - 0| = 2$. Then $f(f(3)) = f(2) = |2 - |2 - 3|| = |2 - 1| = 1$.For $x = 5$: $f(5) = |2 - |5 - 3|| = |2 - 2| = 0$. Then $f(f(5)) = f(0) = |2 - |0 - 3|| = |2 - 3| = 1$.Finally,$\sum_{x \in S} f(f(x)) = f(f(1)) + f(f(3)) + f(f(5)) = 1 + 1 + 1 = 3$.

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

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