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: Recall the condition for continuity of a piecewise function.
For the function to be continuous at \( x=2 \), we must have \[ \lim_{x\to 2} f(x) = f(2) \] Here, \[ f(2)=\lambda \] So we need \[ \lambda=\lim_{x\to 2}\frac{3x^2-12}{x-2} \]

Step 2: Simplify the numerator.
Take out the common factor \( 3 \): \[ 3x^2-12 = 3(x^2-4) \] Now factor the difference of squares: \[ x^2-4=(x-2)(x+2) \] Hence, \[ 3x^2-12 = 3(x-2)(x+2) \]

Step 3: Substitute this factorization into the function.
For \( x\neq 2 \), \[ \frac{3x^2-12}{x-2} = \frac{3(x-2)(x+2)}{x-2} \] Since \( x\neq 2 \), we can cancel \( x-2 \): \[ \frac{3x^2-12}{x-2}=3(x+2) \]

Step 4: Find the limit as \( x\to 2 \).
Now, \[ \lim_{x\to 2}\frac{3x^2-12}{x-2} = \lim_{x\to 2} 3(x+2) \] \[ =3(2+2)=3\cdot 4=12 \]

Step 5: Use the continuity condition.
For continuity at \( x=2 \), we must set \[ \lambda=\lim_{x\to 2}\frac{3x^2-12}{x-2} \] Therefore, \[ \lambda=12 \]

Step 6: Verify the result.
If we choose \[ \lambda=12 \] then \[ f(2)=12 \] and the left-hand and right-hand behaviour of the rational part near \( x=2 \) also approaches \( 12 \).
Hence the function becomes continuous at \( x=2 \).

Step 7: Final conclusion.
Thus, the required value of \( \lambda \) is \[ \boxed{12} \] Therefore, the correct option is \[ \boxed{(5)\ 12} \]