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Let $f (x) = | | x - 1 | - 1 |$ then all the number of points where $f (x)$ is not differentiable is (are)

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detailed solution

Correct option is C

$\begin{array}{l} f (x) = \{\begin{cases} | x - 1 | - 1 & if | x - 1 | \geq 1 \\ 1 - | x - 1 | & if | x - 1 | < 1 \end{cases} \\ = \{\begin{array}{cc} - x & x \leq 0 \\ x & 0 < x \leq 1 \\ 2 - x & 1 < x < 2 \\ x - 2 & x \geq 2 \end{array} \end{array}$

$f$ is not differentiable at $0, 1, 2$ as these are corner points on the graph.

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detailed solution

Correct answer is 3

$\begin{array}{l} f (x) = \{\begin{cases} | x - 1 | - 1 & if | x - 1 | \geq 1 \\ 1 - | x - 1 | & if | x - 1 | < 1 \end{cases} \\ = \{\begin{array}{cc} - x & x \leq 0 \\ x & 0 < x \leq 1 \\ 2 - x & 1 < x < 2 \\ x - 2 & x \geq 2 \end{array} \end{array}$

$f$ is not differentiable at $0, 1, 2$ as these are corner points on the graph.

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