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 need to analyze the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches 2.

### Step 1: Define the function segments
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \frac{a |x^2 - x - 2|}{2 + x - x^2} & \text{if } x < 2 \\ b & \text{if } x = 2 \\ \frac{x - [x]}{x - 2} & \text{if } x > 2 \end{cases} \]

### Step 2: Calculate the left-hand limit as \( x \) approaches 2
For \( x < 2 \), we need to find:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a |x^2 - x - 2|}{2 + x - x^2} \]
First, we simplify \( |x^2 - x - 2| \):
\[ x^2 - x - 2 = (x - 2)(x + 1) \]
Since \( x < 2 \), \( x - 2 < 0 \) and thus:
\[ |x^2 - x - 2| = -(x^2 - x - 2) = -(x - 2)(x + 1) = (2 - x)(x + 1) \]
Now substituting this into the limit:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a(2 - x)(x + 1)}{2 + x - x^2} \]
Next, simplify the denominator:
\[ 2 + x - x^2 = 2 + 2 - 2^2 = 2 + 2 - 4 = 0 \text{ (as } x \to 2\text{)} \]
To analyze the limit, we can factor the denominator:
\[ 2 + x - x^2 = -(x - 2)(x + 2) \]
Thus, we can rewrite the limit:
\[ \lim_{x \to 2^-} \frac{a(2 - x)(x + 1)}{-(x - 2)(x + 2)} = \lim_{x \to 2^-} \frac{-a(2 - x)(x + 1)}{(x - 2)(x + 2)} \]
As \( x \to 2 \), this limit approaches:
\[ \lim_{x \to 2^-} \frac{-a(2 - x)(3)}{-(0)(4)} \text{ (undefined, but we will evaluate the behavior)} \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 2
For \( x > 2 \):
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x - [x]}{x - 2} \]
Here, \( [x] = 2 \) for \( x \) just greater than 2, hence:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x - 2}{x - 2} = 1 \]

### Step 4: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( x = 2 \):
\[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) \]
Thus:
\[ b = 1 \]
And since the left-hand limit must also equal \( b \):
\[ \lim_{x \to 2^-} f(x) = 1 \]
This implies:
\[ a = 1 \]

### Conclusion
The values of \( a \) and \( b \) for which \( f(x) \) is continuous at \( x = 2 \) are:
\[ a = 1 \quad \text{and} \quad b = 1 \]