the values of `a` for which `(a^2-1)x^2+2(a-1)x+2` is positive for all real `x` are.

Options

A. `a ge 1`
B. `a le 1`
C. `a gt - 3`
D. `a le - 3 or a ge 1`

Answer

To solve the problem, we need to determine the values of \( a \) for which the quadratic expression

\[ (a^2 - 1)x^2 + 2(a - 1)x + 2 \]

is positive for all real \( x \).

### Step-by-Step Solution:

1. **Identify the Quadratic Coefficients:**
The quadratic expression can be written in the standard form \( Ax^2 + Bx + C \), where:
- \( A = a^2 - 1 \)
- \( B = 2(a - 1) \)
- \( C = 2 \)

2. **Condition for Positivity:**
For the quadratic to be positive for all real \( x \), the following conditions must be satisfied:
- The coefficient \( A \) must be greater than 0: \( a^2 - 1 > 0 \)
- The discriminant \( D \) must be less than 0: \( D = B^2 - 4AC < 0 \)

3. **Solve for \( A > 0 \):**
\[ a^2 - 1 > 0 \]
This can be factored as:
\[ (a - 1)(a + 1) > 0 \]
The critical points are \( a = -1 \) and \( a = 1 \). Testing intervals:
- For \( a < -1 \): both factors are negative, product is positive.
- For \( -1 < a < 1 \): one factor is negative, one is positive, product is negative.
- For \( a > 1 \): both factors are positive, product is positive.

Thus, \( a < -1 \) or \( a > 1 \).

4. **Calculate the Discriminant \( D < 0 \):**
\[ D = [2(a - 1)]^2 - 4(a^2 - 1)(2) \]
Simplifying:
\[ D = 4(a - 1)^2 - 8(a^2 - 1) \]
Expanding:
\[ D = 4(a^2 - 2a + 1) - 8a^2 + 8 \]
\[ D = 4a^2 - 8a + 4 - 8a^2 + 8 \]
\[ D = -4a^2 - 8a + 12 \]
Setting \( D < 0 \):
\[ -4a^2 - 8a + 12 < 0 \]
Dividing by -4 (reversing the inequality):
\[ a^2 + 2a - 3 > 0 \]
Factoring:
\[ (a - 1)(a + 3) > 0 \]
The critical points are \( a = -3 \) and \( a = 1 \). Testing intervals:
- For \( a < -3 \): both factors are negative, product is positive.
- For \( -3 < a < 1 \): one factor is negative, one is positive, product is negative.
- For \( a > 1 \): both factors are positive, product is positive.

Thus, \( a < -3 \) or \( a > 1 \).

5. **Combine Conditions:**
From steps 3 and 4, we have:
- From \( A > 0 \): \( a < -1 \) or \( a > 1 \)
- From \( D < 0 \): \( a < -3 \) or \( a > 1 \)

The combined conditions are:
- \( a < -3 \) or \( a > 1 \)

### Final Answer:
The values of \( a \) for which the expression is positive for all real \( x \) are:
\[ a < -3 \quad \text{or} \quad a > 1 \]