$f(x)=\left\{\begin{array}{lc}4 & ,-\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} < x \leq \sqrt{5} \\ 4, &…
$f(x)=\left\{\begin{array}{lc}4 & ,-\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} < x \leq \sqrt{5} \\ 4, & \sqrt{5} \leq x < \infty\end{array}\right.$
If $k$ is the number of points where $f(x)$ is not differentiable, then $k-2=$
$2$
$1$
$0$
$3$
Solution
Points to check, $x=-\sqrt{5}, x=\sqrt{5}$
$
f(x)=\left\{\begin{array}{l}
4,-\infty < x < -\sqrt{5} \\
x^2-1,-\sqrt{5} \leq x \leq \sqrt{5} \\
4, \sqrt{5} \leq x < \infty
\end{array}\right.
$
At $x=-\sqrt{5}$,
$\mathrm{LHL}=\mathrm{RHL}=f(-\sqrt{5})=4$
$\mathrm{LHD}=0, \mathrm{RHD}=2(-\sqrt{5})=-2 \sqrt{5}$
LHD $\neq$ RHD
$\Rightarrow$ Not differentiable at $x=-: \sqrt{5}$
At $x=\sqrt{5}$
$\mathrm{LHD}=2 \sqrt{5}, \mathrm{RHD}=0$
$\Rightarrow$ Not differentiable at $x=\sqrt{5}$
$\therefore k=2, k-2=0$