If `x^2-ax+1-2a^2 > 0` for all `x in R,` then ....

Options

A. `a in (-2//3, 2//3)`
B. `a in [-2//3, 2//3]`
C. `a in (-2//3, 1)`
D. `a in (0, 2//3)`

Explanation

To solve the inequality \( x^2 - ax + 1 - 2a^2 > 0 \) for all \( x \in \mathbb{R} \), we need to ensure that the quadratic expression has no real roots. This can be achieved by ensuring that the discriminant of the quadratic is less than zero.

### Step-by-Step Solution:

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

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C \) is given by:
\[ D = B^2 - 4AC \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = (-a)^2 - 4 \cdot 1 \cdot (1 - 2a^2) \]
Simplifying this gives:
\[ D = a^2 - 4(1 - 2a^2) = a^2 - 4 + 8a^2 = 9a^2 - 4 \]

3. **Set the discriminant less than zero**:
For the quadratic to be positive for all \( x \), we require:
\[ 9a^2 - 4 < 0 \]

4. **Solve the inequality**:
Rearranging the inequality:
\[ 9a^2 < 4 \]
Dividing both sides by 9:
\[ a^2 < \frac{4}{9} \]

5. **Take the square root**:
Taking the square root of both sides gives:
\[ -\frac{2}{3} < a < \frac{2}{3} \]

6. **Conclusion**:
Therefore, the solution for \( a \) is:
\[ a \in \left(-\frac{2}{3}, \frac{2}{3}\right) \]