Question

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

• A. 2
• B. 1
• C. 0
• D. 3

Hint:

For a piecewise function to be differentiable at a join point, it must first be continuous at that point.
Then, the left-hand derivative must equal the right-hand derivative at that point.
Calculate derivatives of each piece and compare values at the join points.

Answer 1

The Correct Option is C

The function is defined piecewise: \( f(x) = 4 \) for \( x<-\sqrt{5} \) \( f(x) = x^2-1 \) for \( -\sqrt{5} \le x \le \sqrt{5} \) \( f(x) = 4 \) for \( x>\sqrt{5} \) We need to check for differentiability at the points where the definition changes: \(x = -\sqrt{5}\) and \(x = \sqrt{5}\). For differentiability at a point, the function must first be continuous at that point, and the left-hand derivative must equal the right-hand derivative. Continuity at \(x = -\sqrt{5}\): Left-hand limit: \( \lim_{x \to (-\sqrt{5})^-} f(x) = \lim_{x \to (-\sqrt{5})^-} 4 = 4 \). Right-hand limit: \( \lim_{x \to (-\sqrt{5})^+} f(x) = \lim_{x \to (-\sqrt{5})^+} (x^2-1) = (-\sqrt{5})^2-1 = 5-1 = 4 \). Function value: \( f(-\sqrt{5}) = (-\sqrt{5})^2-1 = 5-1 = 4 \). Since LHL = RHL = f(-\(\sqrt{5}\)), the function is continuous at \(x = -\sqrt{5}\). Differentiability at \(x = -\sqrt{5}\): Left-hand derivative (LHD): For \(x<-\sqrt{5}\), \(f(x)=4\), so \(f'(x)=0\). LHD = 0. Right-hand derivative (RHD): For \(-\sqrt{5}<x<\sqrt{5}\), \(f(x)=x^2-1\), so \(f'(x)=2x\). RHD at \(x=-\sqrt{5}\) is \(2(-\sqrt{5}) = -2\sqrt{5}\). Since LHD (0) \(\neq\) RHD (\(-2\sqrt{5}\)), \(f(x)\) is not differentiable at \(x = -\sqrt{5}\). Continuity at \(x = \sqrt{5}\): Left-hand limit: \( \lim_{x \to (\sqrt{5})^-} f(x) = \lim_{x \to (\sqrt{5})^-} (x^2-1) = (\sqrt{5})^2-1 = 5-1 = 4 \). Right-hand limit: \( \lim_{x \to (\sqrt{5})^+} f(x) = \lim_{x \to (\sqrt{5})^+} 4 = 4 \). Function value: \( f(\sqrt{5}) = (\sqrt{5})^2-1 = 5-1 = 4 \). Since LHL = RHL = f(\(\sqrt{5}\)), the function is continuous at \(x = \sqrt{5}\). Differentiability at \(x = \sqrt{5}\): Left-hand derivative (LHD): For \(-\sqrt{5}<x<\sqrt{5}\), \(f(x)=x^2-1\), so \(f'(x)=2x\). LHD at \(x=\sqrt{5}\) is \(2(\sqrt{5}) = 2\sqrt{5}\). Right-hand derivative (RHD): For \(x>\sqrt{5}\), \(f(x)=4\), so \(f'(x)=0\). RHD = 0. Since LHD (\(2\sqrt{5}\)) \(\neq\) RHD (0), \(f(x)\) is not differentiable at \(x = \sqrt{5}\). The function is not differentiable at two points: \(x = -\sqrt{5}\) and \(x = \sqrt{5}\). So, the number of points where \(f(x)\) is not differentiable is \(k=2\). We need to find \(k-2\). \(k-2 = 2-2 = 0\). This matches option (c). \[ \boxed{0} \]