MCQ
$f(x) = \left| {\left| x \right| - 1} \right|$ is not differentiable at
  • A
    $0$
  • $ \pm 1,\,0$
  • C
    $1$
  • D
    $ \pm \,1$

Answer

Correct option: B.
$ \pm 1,\,0$
b
(b )$ = \left\{ \begin{array}{l}|x| - 1,\,\,\,\,\,\,\,|x| - 1 \ge 0\\ - |x| + 1,\,\,\,|x| - 1 < 0\end{array} \right.$

$ = \left\{ \begin{array}{l}|x| - 1,\,\,\,x \le - 1\,\,{\rm{or}}\,x \ge 1\\ - |x| + 1,\,\,\,\,\,\,\, - 1 < x < 1\end{array} \right.$

$ = \left\{ \begin{array}{l} - x - 1,\,\,\,\,\,x \le - 1\\x + 1,\,\,\,\,\,\, - 1 < x < 0\\ - x + 1,\,\,\,\,\,0 \le x < 1\\\,\,x - 1,\,\,\,\,\,\,\,x \ge 1\end{array} \right.$

From the graph. It is clear that $f(x)$ is not differentiable at $x = - 1,\,0$ and $1$.

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