Find the domain of `cos^-1((x^2-5x+6)/(x^2-9))/lnx^2`

Options

A. `x in (-oo,5/2]`
B. `x in (3/2,oo]-{0,1}`
C. `x in (-1/2,oo]-{0,1,3}`
D. `x in (-1/2,3] uu (5/2,oo)`

Answer

To find the domain of the expression \( \frac{\cos^{-1}\left(\frac{x^2 - 5x + 6}{x^2 - 9}\right)}{\ln(x^2)} \), we need to ensure that both the numerator and denominator are defined and valid.

### Step 1: Analyze the Denominator
The denominator is \( \ln(x^2) \). For the logarithm to be defined, \( x^2 \) must be greater than 0, which implies:
\[ x \neq 0 \]
Additionally, \( \ln(x^2) \) is equal to 0 when \( x^2 = 1 \), which means:
\[ x \neq \pm 1 \]

### Step 2: Analyze the Numerator
The numerator is \( \cos^{-1}\left(\frac{x^2 - 5x + 6}{x^2 - 9}\right) \). The argument of the \( \cos^{-1} \) function must be in the range \([-1, 1]\). Therefore, we need:
\[ -1 \leq \frac{x^2 - 5x + 6}{x^2 - 9} \leq 1 \]

#### Step 2.1: Solve the Inequality \( \frac{x^2 - 5x + 6}{x^2 - 9} \geq -1 \)
This can be rearranged as:
\[ \frac{x^2 - 5x + 6 + x^2 - 9}{x^2 - 9} \geq 0 \]
which simplifies to:
\[ \frac{2x^2 - 5x - 3}{x^2 - 9} \geq 0 \]

#### Step 2.2: Factor the Numerator
To factor \( 2x^2 - 5x - 3 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4} \]
This gives us the roots:
\[ x = 3 \quad \text{and} \quad x = -\frac{1}{2} \]
Thus, we can factor the numerator as:
\[ 2x^2 - 5x - 3 = 2(x - 3)(x + \frac{1}{2}) \]

#### Step 2.3: Factor the Denominator
The denominator \( x^2 - 9 \) can be factored as:
\[ (x - 3)(x + 3) \]

#### Step 2.4: Determine the Sign of the Expression
Now we analyze the sign of:
\[ \frac{2(x - 3)(x + \frac{1}{2})}{(x - 3)(x + 3)} \geq 0 \]
The critical points are \( x = 3, -\frac{1}{2}, -3 \).

### Step 3: Test Intervals
We test the intervals determined by the critical points:
1. \( (-\infty, -3) \)
2. \( (-3, -\frac{1}{2}) \)
3. \( (-\frac{1}{2}, 3) \)
4. \( (3, \infty) \)

- For \( x < -3 \): The expression is positive.
- For \( -3 < x < -\frac{1}{2} \): The expression is negative.
- For \( -\frac{1}{2} < x < 3 \): The expression is positive.
- For \( x > 3 \): The expression is positive.

### Step 4: Combine Conditions
From the analysis:
- \( x \neq 0 \)
- \( x \neq \pm 1 \)
- The valid intervals from the numerator are \( (-\infty, -3) \cup (-\frac{1}{2}, 3) \cup (3, \infty) \).

### Final Domain
Combining all conditions, the domain of the function is:
\[ (-\infty, -3) \cup (-\frac{1}{2}, -1) \cup (-1, 1) \cup (1, 3) \cup (3, \infty) \]