The largest negative integer which satisfies `(x^2-1)/((x-2)(x-3))>0` is a. -4 b. -3 c. -2 d. -1

Options

A. `-4`
B. `-3`
C. `-2`
D. `-1`

Answer

To solve the inequality \(\frac{x^2 - 1}{(x - 2)(x - 3)} > 0\), we will follow these steps:

### Step 1: Factor the numerator
The numerator \(x^2 - 1\) can be factored using the difference of squares:
\[ x^2 - 1 = (x - 1)(x + 1) \]
Thus, we can rewrite the inequality as:
\[ \frac{(x - 1)(x + 1)}{(x - 2)(x - 3)} > 0 \]

### Step 2: Identify critical points
Next, we find the critical points where the expression is equal to zero or undefined. This occurs when the numerator or denominator is zero:
- From \(x - 1 = 0\), we get \(x = 1\)
- From \(x + 1 = 0\), we get \(x = -1\)
- From \(x - 2 = 0\), we get \(x = 2\)
- From \(x - 3 = 0\), we get \(x = 3\)

Thus, the critical points are \(x = -1, 1, 2, 3\).

### Step 3: Test intervals
We will test the sign of the expression in the intervals defined by these critical points:
1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, 2) \)
4. \( (2, 3) \)
5. \( (3, \infty) \)

**Interval 1: \( (-\infty, -1) \)**
Choose \(x = -5\):
\[ \frac{(-5 - 1)(-5 + 1)}{(-5 - 2)(-5 - 3)} = \frac{(-6)(-4)}{(-7)(-8)} = \frac{24}{56} > 0 \]

**Interval 2: \( (-1, 1) \)**
Choose \(x = 0\):
\[ \frac{(0 - 1)(0 + 1)}{(0 - 2)(0 - 3)} = \frac{(-1)(1)}{(-2)(-3)} = \frac{-1}{6} < 0 \]

**Interval 3: \( (1, 2) \)**
Choose \(x = 1.5\):
\[ \frac{(1.5 - 1)(1.5 + 1)}{(1.5 - 2)(1.5 - 3)} = \frac{(0.5)(2.5)}{(-0.5)(-1.5)} = \frac{1.25}{0.75} > 0 \]

**Interval 4: \( (2, 3) \)**
Choose \(x = 2.5\):
\[ \frac{(2.5 - 1)(2.5 + 1)}{(2.5 - 2)(2.5 - 3)} = \frac{(1.5)(3.5)}{(0.5)(-0.5)} = \frac{5.25}{-0.25} < 0 \]

**Interval 5: \( (3, \infty) \)**
Choose \(x = 4\):
\[ \frac{(4 - 1)(4 + 1)}{(4 - 2)(4 - 3)} = \frac{(3)(5)}{(2)(1)} = \frac{15}{2} > 0 \]

### Step 4: Combine results
The expression is positive in the intervals:
- \( (-\infty, -1) \)
- \( (1, 2) \)
- \( (3, \infty) \)

### Step 5: Identify the largest negative integer
We are asked for the largest negative integer satisfying the inequality. The only negative interval is \( (-\infty, -1) \). The largest negative integer in this interval is:
\[ -2 \]

### Final Answer
Thus, the largest negative integer which satisfies the inequality is:
\[ \boxed{-2} \]