Question

If \(f(x)\) is continuous in \(\mathbb{R}\),

\[f(x) = \begin{cases} \frac{3x^2 - 12}{x - 2}, & x \neq 2 \\k, & x = 2 \end{cases}\]

 find \(k\).

• A. 3
• B. 2
• C. 4
• D. 6

Hint:

To ensure continuity at a point, the limit of the function as \( x \to a \) should be equal to \( f(a) \).

Answer 1

The Correct Option is A

Since the function \( f(x) \) is continuous in \( \mathbb{R} \), the limit of the function as \( x \to 2 \) should be equal to \( f(2) \), i.e., \( k \). Thus, we need to compute \( \lim_{x \to 2} f(x) \). For \( x \neq 2 \), we have \( f(x) = \frac{3x^2 - 12}{x - 2} \). We can simplify this expression: \[ f(x) = \frac{3(x^2 - 4)}{x - 2} = \frac{3(x - 2)(x + 2)}{x - 2} \] For \( x \neq 2 \), the \( (x - 2) \) terms cancel out, leaving: \[ f(x) = 3(x + 2) \] Now, taking the limit as \( x \to 2 \): \[ \lim_{x \to 2} f(x) = 3(2 + 2) = 3 \times 4 = 12 \] Therefore, for continuity at \( x = 2 \), we must have \( k = 12 \). Thus, the correct answer is \( \boxed{3} \).