Find all values of k for which the inequality, `2x^(2)-4k^(2)x-k^(2)+1gt0` is valid for all real x which do not exceed unity in the absolute value.

Answer

To solve the inequality \( 2x^2 - 4k^2x - k^2 + 1 > 0 \) for all real \( x \) such that \( |x| \leq 1 \), we can follow these steps:

### Step 1: Rewrite the inequality
We want the quadratic expression \( 2x^2 - 4k^2x - k^2 + 1 \) to be greater than zero for all \( x \) in the interval \([-1, 1]\).

### Step 2: Identify the quadratic function
Let \( f(x) = 2x^2 - 4k^2x - k^2 + 1 \). This is a quadratic function in the form \( ax^2 + bx + c \), where:
- \( a = 2 \)
- \( b = -4k^2 \)
- \( c = -k^2 + 1 \)

### Step 3: Check the discriminant
For the quadratic \( f(x) \) to be positive for all \( x \) in the interval \([-1, 1]\), it must not have any real roots. This means that the discriminant \( D \) must be less than zero.

The discriminant \( D \) of a quadratic \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
Substituting our values:
\[ D = (-4k^2)^2 - 4(2)(-k^2 + 1) \]
\[ D = 16k^4 - 8(-k^2 + 1) \]
\[ D = 16k^4 + 8k^2 - 8 \]

### Step 4: Set the discriminant less than zero
We need to solve the inequality:
\[ 16k^4 + 8k^2 - 8 < 0 \]

### Step 5: Factor the inequality
Let \( u = k^2 \). Then the inequality becomes:
\[ 16u^2 + 8u - 8 < 0 \]
Dividing the entire inequality by 8 gives:
\[ 2u^2 + u - 1 < 0 \]

### Step 6: Factor the quadratic
We can factor \( 2u^2 + u - 1 \):
\[ (2u - 1)(u + 1) < 0 \]

### Step 7: Find the critical points
Setting each factor to zero gives the critical points:
1. \( 2u - 1 = 0 \) → \( u = \frac{1}{2} \)
2. \( u + 1 = 0 \) → \( u = -1 \) (not relevant since \( u = k^2 \geq 0 \))

### Step 8: Test intervals
We test the intervals determined by the critical point \( u = \frac{1}{2} \):
- For \( u < \frac{1}{2} \): Choose \( u = 0 \) → \( (2(0) - 1)(0 + 1) = -1 < 0 \) (valid)
- For \( u > \frac{1}{2} \): Choose \( u = 1 \) → \( (2(1) - 1)(1 + 1) = 2 > 0 \) (not valid)

### Step 9: Conclusion
Thus, the solution for \( u \) is:
\[ 0 \leq u < \frac{1}{2} \]
Since \( u = k^2 \), we have:
\[ 0 \leq k^2 < \frac{1}{2} \]
Taking the square root gives:
\[ -\frac{1}{\sqrt{2}} < k < \frac{1}{\sqrt{2}} \]

### Final Answer
The values of \( k \) for which the inequality is valid for all \( x \) such that \( |x| \leq 1 \) are:
\[ k \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]