Find the set of all possible real value of `a` such that the inequality `(x-(a-1))(x-(a^2+2))<0` holds for all `x in (-1,3)dot`

Options

A. `(0,1)`
B. `(-oo,-1]`
C. `(-oo,-1)`
D. `(1,oo)`

Answer

To solve the inequality \((x - (a - 1))(x - (a^2 + 2)) < 0\) for all \(x\) in the interval \((-1, 3)\), we will follow these steps:

### Step 1: Identify the roots of the inequality
The inequality can be rewritten as:
\[ (x - (a - 1))(x - (a^2 + 2)) < 0 \]
Let \(\alpha = a - 1\) and \(\beta = a^2 + 2\). We need to find conditions on \(a\) such that the product is negative in the interval \((-1, 3)\).

### Step 2: Determine the intervals for \(\alpha\) and \(\beta\)
For the product \((x - \alpha)(x - \beta) < 0\) to hold, one root must be less than -1 and the other must be greater than 3. This means:
1. \(\alpha < -1\) (i.e., \(a - 1 < -1\) or \(a < 0\))
2. \(\beta > 3\) (i.e., \(a^2 + 2 > 3\) or \(a^2 > 1\))

### Step 3: Solve the inequalities
1. From \(\alpha < -1\):
\[ a < 0 \]

2. From \(\beta > 3\):
\[ a^2 > 1 \]
This gives us two cases:
- \(a > 1\)
- \(a < -1\)

### Step 4: Combine the conditions
Now we need to combine the conditions:
- From \(\alpha < -1\), we have \(a < 0\).
- From \(\beta > 3\), we have \(a < -1\) or \(a > 1\).

Since \(a < 0\) and \(a < -1\) must both hold, we conclude that:
\[ a < -1 \]

### Step 5: Final solution
Thus, the set of all possible real values of \(a\) such that the inequality holds for all \(x\) in \((-1, 3)\) is:
\[ (-\infty, -1) \]