Question

Let $f\left(x\right)=\{\begin{array}{c}\text{sgn}(x^2-1);|x|>1\\ x;|x|\le 1\end{array}$, then which of the following option is correct

• A. $f\left(x\right)$ is continuous and differentiable at $x=1$
• B. $f\left(x\right)$ is continuous but not differentiable at $x=1$
• C. $f\left(x\right)$ is continuous and differentiable at $x=-1$
• D. $f\left(x\right)$ is continuous but not differentiable at $x=-1$

Answer 1

The correct option is B $f\left(x\right)$ is continuous but not differentiable at $x=1$

Since, $x^2-1>0,\text{if}|x|>1$
$\Rightarrow f\left(x\right)=\{\begin{array}{c}1;x<-1\\ x;-1\le x\le 1\\ 1;x>1\end{array}$
Critical points for continuity are $x=\pm 1$
At $x=-1,L.H.L.=1,R.H.L.=-1$
$\Rightarrow$ discontinuous at $x=-1$, so not differentiable also.

At $x=1,L.H.L.=1=R.H.L.=f\left(1\right)$
$\Rightarrow$ continuous at $x=1$
also, $L.H.D.=f^{\prime }(1-h)=1$ and
$R.H.D.=f^{\prime }(1+h)=0$
So, not differentiable at $x=1$