Let `f(x)=5-[x-2]`
`g(x)=[x+1]+3`
If maximum value of `f(x)` is `alpha`
`&` minimum value of `f(x)` is `beta`
then `underset(xrarr(alpha-beta))lim ((x-3)(x^(2)-5x+6))/((x-1)(x^(2)-6x+8))` is (A) -1/2 (B)1/2 (C)3/2 (D)-3/2

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. `(3)/(2)`
D. `-(3)/(2)`

Answer

To solve the problem step by step, let's break it down as follows:

### Step 1: Determine the maximum and minimum values of \( f(x) \)
The function \( f(x) = 5 - [x - 2] \) involves the greatest integer function. The greatest integer function, denoted by \( [x] \), gives the largest integer less than or equal to \( x \).

- To find the maximum value of \( f(x) \), we need to minimize \( [x - 2] \). The minimum value of \( [x - 2] \) occurs when \( x \) is exactly 2, giving:
\[ [2 - 2] = [0] = 0 \]
Thus, the maximum value of \( f(x) \) is:
\[ f(2) = 5 - 0 = 5 \]
Therefore, \( \alpha = 5 \).

### Step 2: Determine the minimum value of \( g(x) \)
The function \( g(x) = [x + 1] + 3 \) also involves the greatest integer function.

- To find the minimum value of \( g(x) \), we need to minimize \( [x + 1] \). The minimum value of \( [x + 1] \) occurs when \( x = -1 \), giving:
\[ [-1 + 1] = [0] = 0 \]
Thus, the minimum value of \( g(x) \) is:
\[ g(-1) = 0 + 3 = 3 \]
Therefore, \( \beta = 3 \).

### Step 3: Calculate \( \alpha - \beta \)
Now we compute:
\[ \alpha - \beta = 5 - 3 = 2 \]

### Step 4: Evaluate the limit
We need to evaluate the limit:
\[ \lim_{x \to 2} \frac{(x - 3)(x^2 - 5x + 6)}{(x - 1)(x^2 - 6x + 8)} \]

#### Step 4.1: Factor the expressions
First, we factor the quadratic expressions:
- \( x^2 - 5x + 6 = (x - 2)(x - 3) \)
- \( x^2 - 6x + 8 = (x - 2)(x - 4) \)

Substituting these factorizations into the limit gives:
\[ \lim_{x \to 2} \frac{(x - 3)(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 4)} \]

#### Step 4.2: Cancel common factors
We can cancel \( (x - 2) \) from the numerator and denominator:
\[ \lim_{x \to 2} \frac{(x - 3)^2}{(x - 1)(x - 4)} \]

#### Step 4.3: Substitute \( x = 2 \)
Now we substitute \( x = 2 \):
\[ = \frac{(2 - 3)^2}{(2 - 1)(2 - 4)} = \frac{(-1)^2}{(1)(-2)} = \frac{1}{-2} = -\frac{1}{2} \]

### Final Answer
Thus, the final answer is:
\[ \boxed{-\frac{1}{2}} \]