The domain of `y=(x)/(sqrt(x^(2)-5x+6))` is :

Options

A. `R-{2, 3}`
B. `R-(2, 3)`
C. `R-[2, 3]`
D. none of these

Answer

To find the domain of the function \( y = \frac{x}{\sqrt{x^2 - 5x + 6}} \), we need to ensure that the denominator is defined and not equal to zero. The steps to find the domain are as follows:

### Step 1: Identify the denominator
The denominator of the function is \( \sqrt{x^2 - 5x + 6} \). For the function to be defined, this expression must be greater than zero.

### Step 2: Set the condition for the square root
We need to find when the expression inside the square root is greater than zero:
\[ x^2 - 5x + 6 > 0 \]

### Step 3: Factor the quadratic expression
To solve the inequality, we first factor the quadratic:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
Thus, we need to solve:
\[ (x - 2)(x - 3) > 0 \]

### Step 4: Determine the critical points
The critical points from the factors are \( x = 2 \) and \( x = 3 \). These points divide the number line into intervals. We will test each interval to see where the product is positive.

### Step 5: Test intervals
The intervals to test are:
1. \( (-\infty, 2) \)
2. \( (2, 3) \)
3. \( (3, \infty) \)

- **Interval 1: \( (-\infty, 2) \)**
Choose \( x = 0 \):
\((0 - 2)(0 - 3) = (−2)(−3) = 6 > 0\) (Positive)

- **Interval 2: \( (2, 3) \)**
Choose \( x = 2.5 \):
\((2.5 - 2)(2.5 - 3) = (0.5)(−0.5) = -0.25 < 0\) (Negative)

- **Interval 3: \( (3, \infty) \)**
Choose \( x = 4 \):
\((4 - 2)(4 - 3) = (2)(1) = 2 > 0\) (Positive)

### Step 6: Combine results
From our tests, we find that the expression \( (x - 2)(x - 3) > 0 \) is true in the intervals \( (-\infty, 2) \) and \( (3, \infty) \).

### Step 7: Exclude critical points
Since the original function has a square root in the denominator, we must exclude the points where the denominator equals zero:
- \( x = 2 \) and \( x = 3 \) must be excluded.

### Final Domain
Thus, the domain of the function is:
\[ (-\infty, 2) \cup (3, \infty) \]