Let `f(x)={((a|x^(2)-x-2|)/(2+x-x^(2)), , xlt2),(b, , x=2),((x-[x])/(x-2), , xgt2):}`
([.]denotes thet greatest integer functional). If `f(x)` is continuous at x=2 then:

Options

A. `a=1, b=2`
B. `a=1, b=1`
C. `a=0, b=1`
D. `a=2, b=1`

Answer

To determine the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous at \( x = 2 \), we will analyze the function piece by piece.

### Step 1: Define the function at \( x = 2 \)
The function \( f(x) \) is defined as follows:
- For \( x < 2 \):
\[ f(x) = \frac{a |x^2 - x - 2|}{2 + x - x^2} \]
- For \( x = 2 \):
\[ f(x) = b \]
- For \( x > 2 \):
\[ f(x) = \frac{x - [x]}{x - 2} \]

### Step 2: Find the limit as \( x \) approaches 2 from the left
First, we need to evaluate the limit of \( f(x) \) as \( x \) approaches 2 from the left (\( x \to 2^- \)):
\[ f(2^-) = \lim_{x \to 2^-} \frac{a |x^2 - x - 2|}{2 + x - x^2} \]
We simplify \( |x^2 - x - 2| \):
\[ x^2 - x - 2 = (x - 2)(x + 1) \]
For \( x < 2 \), \( |x^2 - x - 2| = -(x^2 - x - 2) = -((x - 2)(x + 1)) \).

Now, substituting this into the limit:
\[ f(2^-) = \lim_{x \to 2^-} \frac{a(-(x - 2)(x + 1))}{2 + x - x^2} \]
Next, simplify the denominator:
\[ 2 + x - x^2 = 2 + 2 - 4 = 0 \quad \text{(as \( x \to 2 \))} \]
Thus, we need to factor the denominator:
\[ 2 + x - x^2 = -(x^2 - x - 2) = -(x - 2)(x + 1) \]
Now we have:
\[ f(2^-) = \lim_{x \to 2^-} \frac{a(-(x - 2)(x + 1))}{-(x - 2)(x + 1)} = a \]

### Step 3: Find the limit as \( x \) approaches 2 from the right
Next, we evaluate the limit of \( f(x) \) as \( x \) approaches 2 from the right (\( x \to 2^+ \)):
\[ f(2^+) = \lim_{x \to 2^+} \frac{x - [x]}{x - 2} \]
For \( 2 < x < 3 \), \( [x] = 2 \), so:
\[ f(2^+) = \lim_{x \to 2^+} \frac{x - 2}{x - 2} = 1 \]

### Step 4: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ f(2^-) = f(2) = f(2^+) \]
This gives us:
\[ a = b = 1 \]

### Conclusion
Thus, the values of \( a \) and \( b \) are:
\[ \boxed{1} \]