The value of `a` for which the equation `(1-a^2)x^2+2ax-1=0` has roots belonging to `(0,1)` is

Options

A. (a) `agt0`
B. (b) `alt0`
C. (c) `agt2`
D. (d) none of these

Explanation

To find the value of \( a \) for which the equation \( (1-a^2)x^2 + 2ax - 1 = 0 \) has roots belonging to the interval \( (0, 1) \), we can follow these steps:

### Step 1: Identify the function
Let \( f(x) = (1-a^2)x^2 + 2ax - 1 \).

### Step 2: Evaluate the function at the endpoints of the interval
We need to evaluate \( f(0) \) and \( f(1) \).

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

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

### Step 3: Set up the condition for roots in the interval \( (0, 1) \)
For the roots to be in the interval \( (0, 1) \), the function must change signs between \( f(0) \) and \( f(1) \). This means:
\[ f(0) \cdot f(1) < 0 \]
Substituting the values we calculated:
\[ -1 \cdot (-a^2 + 2a) < 0 \]
This simplifies to:
\[ a^2 - 2a < 0 \]

### Step 4: Factor the inequality
Factoring gives us:
\[ a(a - 2) < 0 \]

### Step 5: Determine the intervals
The critical points are \( a = 0 \) and \( a = 2 \). We can test the intervals:
- For \( a < 0 \): Choose \( a = -1 \) → \( (-1)(-3) > 0 \) (not valid)
- For \( 0 < a < 2 \): Choose \( a = 1 \) → \( (1)(-1) < 0 \) (valid)
- For \( a > 2 \): Choose \( a = 3 \) → \( (3)(1) > 0 \) (not valid)

Thus, the solution to the inequality is:
\[ 0 < a < 2 \]

### Conclusion
The value of \( a \) for which the equation has roots in the interval \( (0, 1) \) is:
\[ a \in (0, 2) \]