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

Options

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

Answer

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

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

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

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

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

Since \( f(2) \) is defined as \( 2 \), we have:

\[ f(2) = 2. \]

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

We need to find:

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

### Step 3: Substitute \( x = 2 \) into the limit

Substituting \( x = 2 \) directly into the function gives us an indeterminate form \( \frac{0}{0} \):

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

### Step 4: Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and denominator:

\[ \lim_{x \to 2} \frac{d}{dx}(x^2 - (A + 2)x + A) \Big/ \frac{d}{dx}(x - 2). \]

The derivative of the numerator is:

\[ \frac{d}{dx}(x^2 - (A + 2)x + A) = 2x - (A + 2), \]

and the derivative of the denominator is:

\[ \frac{d}{dx}(x - 2) = 1. \]

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 \( x = 2 \):

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

### 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 the function is continuous at \( x = 2 \) is:

\[ \boxed{0}. \]