Question

The function $f(x)=\begin{cases}\dfrac{3x^2-12}{x-2}, & x\neq 2 \\ \lambda, & x=2 \end{cases}$ is continuous for $x\in\mathbb{R}$, then the value of $\lambda$ is:

• A. \( 0 \)
• B. \( 4 \)
• C. \( 6 \)
• D. \( 8 \)
• E. \( 12 \)

Hint:

For continuity questions involving a removable discontinuity, first factor the numerator, cancel the common factor, and then evaluate the limit at the problematic point.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
For a function to be continuous at a point \( x = c \), the following three conditions must be met:
1. \( f(c) \) is defined.
2. \( \lim_{x\to c} f(x) \) exists.
3. \( \lim_{x\to c} f(x) = f(c) \).
In this problem, the function is defined for all real numbers. For it to be continuous at \( x=2 \), the limit of \( f(x) \) as \( x \) approaches 2 must be equal to the value of the function at \( x=2 \), which is \( A \).
Step 2: Key Formula or Approach:
We need to calculate \( \lim_{x\to 2} f(x) \) and set it equal to \( f(2) \).
\[ A = f(2) = \lim_{x\to 2} \frac{3x^2-12}{x-2} \] Step 3: Detailed Explanation:
First, let's try to substitute \( x=2 \) into the expression to check for an indeterminate form.
Numerator: \( 3(2)^2 - 12 = 3(4) - 12 = 12 - 12 = 0 \).
Denominator: \( 2 - 2 = 0 \).
Since we have the indeterminate form \( \frac{0}{0} \), we can simplify the expression by factoring the numerator.
\[ \frac{3x^2-12}{x-2} = \frac{3(x^2-4)}{x-2} \] Using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \):
\[ = \frac{3(x-2)(x+2)}{x-2} \] For \( x \neq 2 \), we can cancel the \( (x-2) \) terms:
\[ = 3(x+2) \] Now, we can find the limit:
\[ \lim_{x\to 2} 3(x+2) = 3(2+2) = 3(4) = 12 \] For the function to be continuous at \( x=2 \), we must have \( A = f(2) = \lim_{x\to 2} f(x) \).
Therefore, \( A = 12 \).
Step 4: Final Answer:
The value of \( A \) that makes the function continuous is 12. This corresponds to option (E).