If the function `f(x) ={((x^(2)-(k+2)x+2k)/(x-2),"for " x ne 2),(2," for x =2):}` is continuous at x =2, then k is equal to

Options

A. `-(1)/(2)`
B. `-1`
C. 0
D. `(1)/(2)`

Explanation

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{x^2 - (k + 2)x + 2k}{x - 2} & \text{for } x \neq 2 \\ 2 & \text{for } x = 2 \end{cases} \]

is continuous at \( x = 2 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 2 is equal to \( f(2) \).

### Step 1: Find \( f(2) \)
Since \( f(2) = 2 \), we need to find:

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

### Step 2: Evaluate the limit
Substituting \( x = 2 \) directly into the function gives:

\[ \frac{2^2 - (k + 2) \cdot 2 + 2k}{2 - 2} = \frac{4 - 2k - 4 + 2k}{0} = \frac{0}{0} \]

This indicates that we have an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator:

- The derivative of the numerator \( x^2 - (k + 2)x + 2k \) is \( 2x - (k + 2) \).
- The derivative of the denominator \( x - 2 \) is \( 1 \).

Thus, we have:

\[ \lim_{x \to 2} \frac{2x - (k + 2)}{1} \]

### Step 4: Substitute \( x = 2 \) into the limit
Now we substitute \( x = 2 \):

\[ \lim_{x \to 2} (2 \cdot 2 - (k + 2)) = 4 - (k + 2) = 4 - k - 2 = 2 - k \]

### Step 5: Set the limit equal to \( f(2) \)
For continuity at \( x = 2 \):

\[ 2 - k = 2 \]

### Step 6: Solve for \( k \)
Solving the equation:

\[ 2 - k = 2 \implies -k = 0 \implies k = 0 \]

Thus, the value of \( k \) is \( 0 \).

### Final Answer
The value of \( k \) is \( \boxed{0} \).

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