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

Given function:

$$f(x) = \begin{cases} (x+a)^2, & x \leq 0 \ (x+a)^3, & x > 0 \end{cases}$$

Finding conditions for $\frac{d^2f}{dx^2}$ to exist at $x = 0$:

For the second derivative to exist at $x = 0$, the function must be twice differentiable at $x = 0$, which requires:

1. $f$ is continuous at $x = 0$
2. $f'$ is continuous at $x = 0$
3. $f''$ exists from both sides at $x = 0$

Step 1: Check continuity of $f$ at $x = 0$

$$\lim_{x \to 0^-} f(x) = a^2, \quad \lim_{x \to 0^+} f(x) = a^3, \quad f(0) = a^2$$

For continuity: $a^2 = a^3 \Rightarrow a^2(1-a) = 0 \Rightarrow a = 0$ or $a = 1$

Step 2: Find $f'(x)$

For $x < 0$: $f'(x) = 2(x+a)$

For $x > 0$: $f'(x) = 3(x+a)^2$

Step 3: Check continuity of $f'$ at $x = 0$

$$\lim_{x \to 0^-} f'(x) = 2a, \quad \lim_{x \to 0^+} f'(x) = 3a^2$$

For $f'$ to be continuous: $2a = 3a^2 \Rightarrow 2a - 3a^2 = 0 \Rightarrow a(2-3a) = 0$

So $a = 0$ or $a = \frac{2}{3}$

Step 4: Combining conditions

From Step 1: $a \in {0, 1}$

From Step 3: $a \in {0, \frac{2}{3}}$

Common value: $a = 0$

Step 5: Verify $f''$ exists at $x = 0$ when $a = 0$

For $a = 0$:

• $f(x) = x^2$ for $x \leq 0$, so $f''(x) = 2$
• $f(x) = x^3$ for $x > 0$, so $f''(x) = 6x$

At $x = 0$: $$\lim_{x \to 0^-} f''(x) = 2, \quad \lim_{x \to 0^+} f''(x) = 0$$

Since $2 \neq 0$, the second derivative does not exist at $x = 0$ even when $a = 0$.

Answer: (A) $\frac{d^2f}{dx^2}$ does not exist at $x = 0$ for any value of $a$