Let `f(x)={{:(3x-4",",0lexle2),(2x+l",",2ltxle9):}`
If is continuous at x=2, then what is the value of l ?

Options

A. 0
B. 2
C. -2
D. -1

Explanation

To solve the problem, we need to find the value of \( l \) such that the function \( f(x) \) is continuous at \( x = 2 \). The function is defined as follows:

\[ f(x) = \begin{cases} 3x - 4 & \text{if } 0 \leq x \leq 2 \\ 2x + l & \text{if } 2 < x \leq 9 \end{cases} \]

### Step-by-Step Solution:

1. **Determine \( f(2) \)**:
Since \( x = 2 \) falls within the first case of the piecewise function, we calculate:
\[ f(2) = 3(2) - 4 = 6 - 4 = 2 \]

2. **Calculate the Right-Hand Limit as \( x \) approaches 2**:
We need to find the limit of \( f(x) \) as \( x \) approaches 2 from the right (i.e., \( x \to 2^+ \)):
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x + l) \]
Substituting \( x = 2 \):
\[ \lim_{x \to 2^+} f(x) = 2(2) + l = 4 + l \]

3. **Set the Left-Hand Limit Equal to the Right-Hand Limit**:
For the function to be continuous at \( x = 2 \), the left-hand limit must equal the right-hand limit, which must also equal \( f(2) \):
\[ f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \]
Therefore, we have:
\[ 2 = 4 + l \]

4. **Solve for \( l \)**:
Rearranging the equation gives:
\[ l = 2 - 4 = -2 \]

### Conclusion:
The value of \( l \) that makes the function continuous at \( x = 2 \) is:
\[ \boxed{-2} \]