Question
If $$f\left( x \right) = \left| {1 - x} \right|,$$ then the points where $${\sin ^{ - 1}}\left( {f\left| x \right|} \right)$$ is non-differentiable are :
A.
$$\left\{ {0,\,1} \right\}$$
B.
$$\left\{ {0,\, - 1} \right\}$$
C.
$$\left\{ {0,\,1,\, - 1} \right\}$$
D.
none of these
Answer :
$$\left\{ {0,\,1,\, - 1} \right\}$$
Solution :
Given that $$f\left( x \right) = \left| {1 - x} \right|$$
\[\therefore \,f\left( {\left| x \right|} \right) = \left\{ \begin{array}{l}
\,\,\,\,x - 1,\,\,\,\,\,\,\,\,\,\,\,\,x > 1\\
\,\,\,\,1 - x,\,\,\,\,\,\,\,\,0 < x \le 1\\
\,\,\,\,1 + x,\,\,\,\,\, - 1 \le x \le 0\\
- x - 1,\,\,\,\,\,\,\,\,\,\,x < - 1
\end{array} \right.\]
Clearly, the domain of $${\sin ^{ - 1}}\left( {f\left| x \right|} \right)$$ is $$\left[ { - 2,\,2} \right].$$
Therefore, it is non-differentiable at the points $$\left\{ { - 1,\,0,\,1} \right\}.$$