If x satisfies the equation `t^2-t-2 gt 0`, then there exists a value for

Options

A. `sin^(-1)x`
B. `cos^(-1)x`
C. `sec^(-1)x`
D. none

Answer

To solve the inequality \( t^2 - t - 2 > 0 \), we will follow these steps:

### Step 1: Factor the quadratic expression
We start with the quadratic inequality:
\[ t^2 - t - 2 > 0 \]
To factor this, we look for two numbers that multiply to \(-2\) (the constant term) and add to \(-1\) (the coefficient of \(t\)). The numbers \(-2\) and \(1\) satisfy these conditions. Thus, we can factor the expression as:
\[ (t - 2)(t + 1) > 0 \]

### Step 2: Find the critical points
Next, we find the critical points by setting each factor equal to zero:
\[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \]
The critical points are \(t = -1\) and \(t = 2\).

### Step 3: Test intervals
We will test the intervals determined by the critical points to see where the product \((t - 2)(t + 1)\) is positive. The intervals to test are:
1. \( (-\infty, -1) \)
2. \( (-1, 2) \)
3. \( (2, \infty) \)

- **Interval 1: \( (-\infty, -1) \)**
Choose \(t = -2\):
\[ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0 \]

- **Interval 2: \( (-1, 2) \)**
Choose \(t = 0\):
\[ (0 - 2)(0 + 1) = (-2)(1) = -2 < 0 \]

- **Interval 3: \( (2, \infty) \)**
Choose \(t = 3\):
\[ (3 - 2)(3 + 1) = (1)(4) = 4 > 0 \]

### Step 4: Conclusion
The inequality \( (t - 2)(t + 1) > 0 \) holds true in the intervals:
\[ (-\infty, -1) \quad \text{and} \quad (2, \infty) \]
Thus, the solution set for the inequality \( t^2 - t - 2 > 0 \) is:
\[ t \in (-\infty, -1) \cup (2, \infty) \]