Q228 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let $f(x) = \begin{cases} x^3, & x^2 < 1 \\ x^2, & x^2 \geq 1 \end{cases}$.
What is $\lim_{x \to 0} f'(x)$ equal to?
Let $f(x) = \begin{cases} x^3, & x^2 < 1 \\ x^2, & x^2 \geq 1 \end{cases}$.
What is $\lim_{x \to 0} f'(x)$ equal to?
- A$2$
- B$1$
- Correct answerC$0$
- DLimit does not exist
Solution
Given $f(x) = \begin{cases} x^3, & x^2 < 1 \\ x^2, & x^2 \geq 1 \end{cases}$
For $x \to 0$, the relevant domain is $x^2 < 1$, which means $-1 < x < 1$.
In this interval, the function is $f(x) = x^3$.
Differentiating with respect to $x$, we get:
$f'(x) = 3x^2$
Taking the limit as $x \to 0$:
$\lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0$
Answer: $0$
For $x \to 0$, the relevant domain is $x^2 < 1$, which means $-1 < x < 1$.
In this interval, the function is $f(x) = x^3$.
Differentiating with respect to $x$, we get:
$f'(x) = 3x^2$
Taking the limit as $x \to 0$:
$\lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0$
Answer: $0$
Hence, the correct answer is option: 'C'.