Limits Continuity And Differentiability Question 44
Question: If $ f(x)=| 1-x |, $ then the points where $ {{\sin }^{-1}}(f| x |) $ is non-differentiable are
Options:
A) $ { 0,1 } $
B) $ { 0,-1 } $
C) $ { 0,1,-1 } $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ f(x)=| 1-x | $
$ \therefore f(| x |)= \begin{vmatrix} x-1, & x>1 \\ 1-x, & 0<x\le 1 \\ 1+x, & -1\le x\le 0 \\ -x-1, & x<-1 \\ \end{vmatrix} . $
Clearly, the domain of $ {{\sin }^{-1}}(f| x |) $ is $ [-2,2] $ .
Therefore, it is non-differentiable at the points $ {-1,0,1} $ .
BETA
