Question

Let \( a \in \mathbb{R} \). If \( f(x) = \begin{cases} (x + a)^2, & x \leq 0 \\ (x + a)^3, & x > 0 \end{cases} \), then
 

• A. \( \dfrac{d^2 f}{dx^2} \) does not exist at \( x = 0 \) for any value of \( a \)
• B. \( \dfrac{d^2 f}{dx^2} \) exists at \( x = 0 \) for exactly one value of \( a \)
• C. \( \dfrac{d^2 f}{dx^2} \) exists at \( x = 0 \) for exactly two values of \( a \)
• D. \( \dfrac{d^2 f}{dx^2} \) exists at \( x = 0 \) for infinitely many values of \( a \)

Hint:

For piecewise functions, ensure continuity, differentiability, and equality of higher derivatives at the junction to find valid parameter values.

Answer 1

The Correct Option is A

Step 1: Compute first derivatives. 
For \( x < 0, \, f(x) = (x + a)^2 \Rightarrow f'(x) = 2(x + a). \) 
For \( x > 0, \, f(x) = (x + a)^3 \Rightarrow f'(x) = 3(x + a)^2. \) 
At \( x = 0^-: f'(0^-) = 2a. \) 
At \( x = 0^+: f'(0^+) = 3a^2. \) 
 

Step 2: Condition for differentiability at \( x = 0 \). 
For \( f'(x) \) to exist at \( x = 0 \), \( f'(0^-) = f'(0^+) \): \[ 2a = 3a^2 \Rightarrow a(3a - 2) = 0 \Rightarrow a = 0 \text{ or } a = \dfrac{2}{3}. \]

Step 3: Compute second derivatives. 
For \( x < 0, \, f''(x) = 2. \) For \( x > 0, \, f''(x) = 6(x + a). \) 
At \( x = 0^-: f''(0^-) = 2. \) At \( x = 0^+: f''(0^+) = 6a. \) 
 

Step 4: Condition for second derivative to exist. 
For \( f''(x) \) to exist at \( x = 0 \), \[ f''(0^-) = f''(0^+) \Rightarrow 2 = 6a \Rightarrow a = \dfrac{1}{3}. \]

Step 5: Check compatibility. 
For \( f''(0) \) to exist, \( f'(x) \) must be continuous. At \( a = \dfrac{1}{3} \), the derivative continuity condition \( 2a = 3a^2 \) is not satisfied, so we must find \( a \) satisfying both: \[ 2a = 3a^2 \text{ and } 2 = 6a. \] Solving gives \( a = \dfrac{1}{3} \) (only one consistent value). 
 

Final Answer: \( \dfrac{d^2 f}{dx^2} \) exists at \( x = 0 \) for exactly one value of \( a = \dfrac{1}{3}. \)