Let `f(x)={{:(3x-4,0lexle2),(2x+1,2ltxle9):}`
If f is continuous at x = 2, then what is the value of l ?

Options

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

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 2 \). This means that the left-hand limit as \( x \) approaches 2 must equal the right-hand limit as \( x \) approaches 2, and both must equal \( f(2) \).

Given the piecewise function:

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

### Step 1: Calculate the left-hand limit as \( x \) approaches 2

For \( x \) approaching 2 from the left (i.e., \( x \to 2^- \)), we use the first piece of the function:

\[ f(2) = 3(2) - 4 = 6 - 4 = 2 \]

### Step 2: Calculate the right-hand limit as \( x \) approaches 2

For \( x \) approaching 2 from the right (i.e., \( x \to 2^+ \)), we use the second piece of the function:

\[ f(2) = 2(2) + l = 4 + l \]

### Step 3: Set the left-hand limit equal to the right-hand limit

For continuity at \( x = 2 \), we set the left-hand limit equal to the right-hand limit:

\[ 2 = 4 + l \]

### Step 4: Solve for \( l \)

To find \( l \), we rearrange the equation:

\[ l = 2 - 4 = -2 \]

### Conclusion

Thus, the value of \( l \) that makes the function continuous at \( x = 2 \) is:

\[ \boxed{-2} \]