`f(x)={{:(3k-2x", when "xlt1),(2k+1", when "xge1):}` at x = 1

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} 3k - 2x & \text{if } x < 1 \\ 2k + 1 & \text{if } x \geq 1 \end{cases} \]

### Step 1: Determine the value of \( f(1) \)

Since \( x = 1 \) falls under the case \( x \geq 1 \), we use the second part of the function:

\[ f(1) = 2k + 1 \]

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

We need to find the left-hand limit, which is calculated using the first part of the function:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3k - 2x) \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1^-} f(x) = 3k - 2(1) = 3k - 2 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1

Next, we find the right-hand limit, which is calculated using the second part of the function:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2k + 1) = 2k + 1 \]

### Step 4: Set the left-hand limit equal to the right-hand limit

For the function to be continuous at \( x = 1 \), the left-hand limit must equal the right-hand limit:

\[ 3k - 2 = 2k + 1 \]

### Step 5: Solve for \( k \)

Now, we solve the equation:

\[ 3k - 2 = 2k + 1 \]

Subtract \( 2k \) from both sides:

\[ 3k - 2k - 2 = 1 \]

This simplifies to:

\[ k - 2 = 1 \]

Adding 2 to both sides gives:

\[ k = 3 \]

### Conclusion

The value of \( k \) that makes the function continuous at \( x = 1 \) is:

\[ \boxed{3} \]