Question

Given \( f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \le x \le a \\[6pt] \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a<x \le b \\[6pt] \frac{1}{3} \cdot \frac{b^3 - a^3}{x}, & x>b \end{cases} \). Then:

• A. $f''(a) = 2b$
• B. $f''(a) = 1$
• C. $f''(a) = b^2 - a^2$
• D. $f'(x)$ is not differentiable at $x = a$

Hint:

Second Derivative at Transition Point:

• Continuity $\ne$ Differentiability.
• $f'(x)$ may be continuous, but $f''(x)$ may not be.
• Always check left/right limits of $f''(x)$.

Answer 1

The Correct Option is D

• $f(x)$ is continuous at $x=a$.
• $f'(x)$ from left = 0; from right = $-\frac{a}{3} + \frac{a}{3} = 0$
• So $f'(x)$ is continuous at $x=a$
• But $f''(x)$: from left = 0, from right = $-1$

$\Rightarrow f''(a)$ does not exist. So $f'(x)$ is not differentiable at $x=a$.