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

Options

A. 4
B. 3
C. 2
D. 1

Answer

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

The function \( f(x) \) is defined as follows:
- \( f(x) = x + \lambda \) for \( x < 3 \)
- \( f(3) = 4 \)
- \( f(x) = 3x - 5 \) for \( x > 3 \)

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

### Step 2: Determine the value of \( f(3) \)
From the definition of the function, we have:
\[ f(3) = 4 \]

### Step 3: Set the left-hand limit equal to the function value at \( x = 3 \)
For continuity at \( x = 3 \), we set the left-hand limit equal to the function value:
\[ 3 + \lambda = 4 \]

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

### Step 5: Verify with the right-hand limit
Now, let's check the right-hand limit as \( x \) approaches 3:
\[ f(3^+) = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 5) = 3(3) - 5 = 9 - 5 = 4 \]

### Conclusion
Both the left-hand limit and right-hand limit at \( x = 3 \) equal the function value at \( x = 3 \), confirming that \( f(x) \) is continuous at \( x = 3 \) when \( \lambda = 1 \).

Thus, the value of \( \lambda \) is:
\[ \lambda = 1 \]