Question

Which of the following statements is true for the function \[ f(x) = \begin{cases} x^2 + 3, & x \neq 0, \\ 1, & x = 0? \end{cases} \]

• A. \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \)
• B. \( f(x) \) is continuous \( \forall x \in \mathbb{R} \)
• C. \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \setminus \{0\} \)
• D. \( f(x) \) is discontinuous at infinitely many points

Hint:

To check continuity, equate \( \lim_{x \to a} f(x) \) with \( f(a) \). To check differentiability, verify the existence of \( f'(x) \) in the neighborhood of the point.

Answer 1

The Correct Option is C

The given function is: \[ f(x) = \begin{cases} x^2 + 3, & x \neq 0, \\ 1, & x = 0? \end{cases} \] 

Step 1: Check for continuity at \( x = 0 \) 
For \( f(x) \) to be continuous at \( x = 0 \), the following condition must hold: \[ \lim_{x \to 0} f(x) = f(0). \] Here, \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 + 3) = 3, \quad f(0) = 1. \] Since \( \lim_{x \to 0} f(x) \neq f(0) \), the function is discontinuous at \( x = 0 \). 

Step 2: Check for differentiability at \( x \neq 0 \) 
For \( x \neq 0 \), \( f(x) = x^2 + 3 \), which is a polynomial function. Polynomial functions are differentiable everywhere on \( \mathbb{R} \), so \( f(x) \) is differentiable for all \( x \neq 0 \). 

Step 3: Verify other points of continuity and differentiability 
Since \( f(x) = x^2 + 3 \) for \( x \neq 0 \), it is both continuous and differentiable for \( x \in \mathbb{R} \setminus \{0\} \). Hence, the function \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \setminus \{0\} \).