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

Options

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

Answer

To determine the value of \( a \) such that the function

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

is continuous at \( x = 2 \), we need to ensure that

\[ f(2) = \lim_{x \to 2} f(x). \]

### Step 1: Find \( f(2) \)

From the definition of the function, we have:

\[ f(2) = 2. \]

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

Next, we need to evaluate the limit:

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

### Step 3: Substitute \( x = 2 \)

Substituting \( x = 2 \) directly into the expression gives us:

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

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

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

Thus, we have:

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

### Step 5: Evaluate the limit

Now substituting \( x = 2 \):

\[ \lim_{x \to 2} (2x - (a + 2)) = 2(2) - (a + 2) = 4 - (a + 2) = 4 - a - 2 = 2 - a. \]

### Step 6: Set the limit equal to \( f(2) \)

For continuity at \( x = 2 \), we set the limit equal to \( f(2) \):

\[ 2 - a = 2. \]

### Step 7: Solve for \( a \)

Solving the equation:

\[ 2 - a = 2 \implies -a = 0 \implies a = 0. \]

### Conclusion

Thus, the value of \( a \) for which \( f(x) \) is continuous at \( x = 2 \) is

\[ \boxed{0}. \]

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