MCQ
The function given by $y = | |x| - 1|$ is differentiable for all real numbers except the points
- ✓$\{0, 1, -1\}$
- B$ \pm 1$
- C$1$
- D$-1$
Alternative method:
The graph of $y=|| x|-1|$ is as follows:
It has sharp turn at $x=-1,0$ and $1$ and hence, is not differentiable at $x=-1,0,1.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $f(x)$ is differentiable only in a finite interval containing zero
$(B)$ $f(x)$ is continuous $\forall x \in R$
$(C)$ $f^{\prime}(x)$ is constant $\forall x \in R$
$(D)$ $f(x)$ is differentiable except at finitely many points