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 $x=a$.

Answer 2

Step 1: Understand the function definition
The function \( f(x) \) is defined piecewise as:
\[ f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \le x \le a \\[6pt] \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a < x \le b \\[6pt] \frac{1}{3} \cdot \frac{b^3 - a^3}{x}, & x > b \end{cases} \]

Step 2: Check continuity at \(x=a\)
From left side \(x \to a^-\), \( f(x) = \frac{1}{2}(b^2 - a^2) \) (constant).
From right side \(x \to a^+\), substitute \(x=a\) in second piece:
\[ f(a^+) = \frac{1}{2}b^2 - \frac{a^2}{6} - \frac{a^3}{3a} = \frac{1}{2}b^2 - \frac{a^2}{6} - \frac{a^2}{3} = \frac{1}{2}b^2 - \frac{a^2}{2} \]
Notice that:
\[ \frac{1}{2}(b^2 - a^2) = \frac{1}{2}b^2 - \frac{a^2}{2} \]
So, \( f \) is continuous at \(x=a\).

Step 3: Find derivatives from left and right at \(x=a\)
- For \(x <= a\), \(f(x) = \text{constant} = \frac{1}{2}(b^2 - a^2)\), so:
\[ f'(a^-) = 0 \]

- For \(a < x \le b\),
\[ f(x) = \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x} \]
Differentiate:
\[ f'(x) = -\frac{2x}{6} + \frac{a^3}{3x^2} = -\frac{x}{3} + \frac{a^3}{3x^2} \]
At \(x=a\),
\[ f'(a^+) = -\frac{a}{3} + \frac{a^3}{3a^2} = -\frac{a}{3} + \frac{a}{3} = 0 \]

Step 4: Check differentiability at \(x=a\)
Since \(f'(a^-)=0\) and \(f'(a^+)=0\), the first derivative exists and equals zero from both sides.

Step 5: Check if \(f'(x)\) is differentiable at \(x=a\) (i.e., if second derivative exists and is continuous)
- For \(x \le a\), \(f'(x) = 0\) constant, so second derivative:
\[ f''(a^-) = 0 \]

- For \(a < x \le b\),
\[ f'(x) = -\frac{x}{3} + \frac{a^3}{3x^2} \]
Differentiate:
\[ f''(x) = -\frac{1}{3} - \frac{2a^3}{3x^3} \]
At \(x=a\),
\[ f''(a^+) = -\frac{1}{3} - \frac{2a^3}{3a^3} = -\frac{1}{3} - \frac{2}{3} = -1 \]

Step 6: Conclusion
Since \(f''(a^-) = 0\) but \(f''(a^+) = -1\), the second derivative is not continuous at \(x=a\).
Therefore, \(f'(x)\) is not differentiable at \(x=a\).

Final answer:
\[ \boxed{f'(x) \text{ is not differentiable at } x = a} \]