Question

Let \[ f(x)= \begin{cases} 4, & -\infty\lt x\lt -\sqrt5 \\ x^2-1, & -\sqrt5\leq x\leq \sqrt5 \\ 4, & \sqrt5\lt x\lt \infty \end{cases} \] If \(k\) is the number of points where \(f(x)\) is not differentiable, then \(k-2=\)

• A. \(2\)
• B. \(1\)
• C. \(0\)
• D. \(3\)

Hint:

For a piecewise function, possible non-differentiability occurs at the joining points. Check the left hand derivative and right hand derivative at each joining point.

Answer 1

The Correct Option is C

Step 1: Check continuity at \(x = \pm\sqrt{5}\).
At \(x=\sqrt{5}\): left limit from \(x^2-1 = 5-1=4\), right value = 4. Continuous. At \(x=-\sqrt{5}\): \((-\sqrt{5})^2-1=4\), left value = 4. Continuous.

Step 2: Check differentiability.
At \(x=\sqrt{5}\): derivative from left = \(2\sqrt{5}\); derivative from right = 0. Not equal, so not differentiable. Similarly at \(x=-\sqrt{5}\): left derivative = 0, right derivative = \(-2\sqrt{5}\). Not differentiable.

Step 3: Conclude.
\(f\) is not differentiable at exactly \(k=2\) points. \(k-2=0\).
\[ \boxed{0} \]