if `f(x) = {{:(x + lambda, -1 lt x lt 3),(4, x =3),(3x-5, x gt 3):}`, is continous at 3 then `lambda` =

Options

A. 1
B. `-1`
C. 0
D. none

Answer

To determine the value of \(\lambda\) such that the function \(f(x)\) is continuous at \(x = 3\), we need to ensure that the left-hand limit as \(x\) approaches 3 is equal to the right-hand limit and the function value at \(x = 3\).

Given the function:
\[ f(x) = \begin{cases} x + \lambda & \text{for } -1 < x < 3 \\ 4 & \text{for } x = 3 \\ 3x - 5 & \text{for } x > 3 \end{cases} \]

### Step 1: Calculate the left-hand limit as \(x\) approaches 3
The left-hand limit is given by:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x + \lambda) \]
Substituting \(x = 3\):
\[ \lim_{x \to 3^-} f(x) = 3 + \lambda \]

### Step 2: Calculate the right-hand limit as \(x\) approaches 3
The right-hand limit is given by:
\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 5) \]
Substituting \(x = 3\):
\[ \lim_{x \to 3^+} f(x) = 3(3) - 5 = 9 - 5 = 4 \]

### Step 3: Set the left-hand limit equal to the function value at \(x = 3\)
For continuity at \(x = 3\), we need:
\[ \lim_{x \to 3^-} f(x) = f(3) \]
Thus, we have:
\[ 3 + \lambda = 4 \]

### Step 4: Solve for \(\lambda\)
Rearranging the equation gives:
\[ \lambda = 4 - 3 = 1 \]

### Conclusion
The value of \(\lambda\) that makes the function continuous at \(x = 3\) is:
\[ \lambda = 1 \]