f(x) = begin{cases} 4, & -infty

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(C) To check for differentiability,we examine the points of transition: $x = -\sqrt{5}$ and $x = \sqrt{5}$.At $x = -\sqrt{5}$:Left-hand limit $(LHL)$

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(C) To check for differentiability,we examine the points of transition: $x = -\sqrt{5}$ and $x = \sqrt{5}$.At $x = -\sqrt{5}$:Left-hand limit $(LHL)$ = $4$.Right-hand limit $(RHL)$ = $(-\sqrt{5})^2 - 1 = 5 - 1 = 4$.Since $LHL = RHL = f(-\sqrt{5})$,the function is continuous at $x = -\sqrt{5}$.Left-hand derivative $(LHD)$ = $\frac{d}{dx}(4) = 0$.Right-hand derivative $(RHD)$ = $\frac{d}{dx}(x^2-1) = 2x = 2(-\sqrt{5}) = -2\sqrt{5}$.Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x = -\sqrt{5}$.At $x = \sqrt{5}$:$LHL$ = $(\sqrt{5})^2 - 1 = 4$.$RHL$ = $4$.Since $LHL = RHL = f(\sqrt{5})$,the function is continuous at $x = \sqrt{5}$.$LHD$ = $2x = 2(\sqrt{5}) = 2\sqrt{5}$.$RHD$ = $\frac{d}{dx}(4) = 0$.Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x = \sqrt{5}$.Thus,there are $k = 2$ points where the function is not differentiable.Therefore,$k - 2 = 2 - 2 = 0$.

$f(x) = \begin{cases} 4, & -\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} \leq x \leq \sqrt{5} \\ 4, & \sqrt{5} < x < \infty \end{cases}$
If $k$ is the number of points where $f(x)$ is not differentiable,then $k-2=$

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$f(x) = \begin{cases} 4, & -\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} \leq x \leq \sqrt{5} \\ 4, & \sqrt{5} < x < \infty \end{cases}$If $k$ is the number of points where $f(x)$ is not differentiable,then $k-2=$