The value of a for which both the roots of the equation `(1-a^(2))x^(2)+2ax-1=0` lie between 0 and 1, will always be greater than

Answer

To solve the problem, we need to determine the values of \( a \) for which both roots of the equation

\[ (1 - a^2)x^2 + 2ax - 1 = 0 \]

lie between 0 and 1.

### Step 1: Identify the coefficients
The coefficients of the quadratic equation can be identified as follows:
- \( A = 1 - a^2 \)
- \( B = 2a \)
- \( C = -1 \)

### Step 2: Condition for real roots
For the quadratic equation to have real roots, the discriminant must be non-negative:

\[ D = B^2 - 4AC \geq 0 \]

Substituting the values of \( A \), \( B \), and \( C \):

\[ D = (2a)^2 - 4(1 - a^2)(-1) \geq 0 \]

This simplifies to:

\[ 4a^2 + 4(1 - a^2) \geq 0 \]

\[ 4a^2 + 4 - 4a^2 \geq 0 \]

\[ 4 \geq 0 \]

This condition is always satisfied, so there are no restrictions from the discriminant.

### Step 3: Roots must lie between 0 and 1
Next, we need both roots to lie between 0 and 1. For this, we will use the conditions:

1. \( f(0) > 0 \)
2. \( f(1) < 0 \)

#### Condition 1: \( f(0) > 0 \)

Substituting \( x = 0 \) into the equation:

\[ f(0) = (1 - a^2)(0)^2 + 2a(0) - 1 = -1 \]

This condition is not satisfied since \( -1 \) is not greater than 0. Therefore, we need to analyze the behavior of \( A \).

#### Condition 2: \( f(1) < 0 \)

Substituting \( x = 1 \):

\[ f(1) = (1 - a^2)(1)^2 + 2a(1) - 1 = 1 - a^2 + 2a - 1 = -a^2 + 2a \]

We want:

\[ -a^2 + 2a < 0 \]

This can be rearranged to:

\[ a^2 - 2a > 0 \]

Factoring gives:

\[ a(a - 2) > 0 \]

The critical points are \( a = 0 \) and \( a = 2 \). Analyzing the sign of the expression, we find:

- \( a < 0 \) (positive)
- \( 0 < a < 2 \) (negative)
- \( a > 2 \) (positive)

Thus, the roots of the quadratic must be either \( a < 0 \) or \( a > 2 \).

### Conclusion
The value of \( a \) for which both roots of the equation lie between 0 and 1 will always be greater than 2.