If both the roots of the equation `4x^(2)-2x+m=0` lie in the interval `(-1, 1)`, then

Options

A. `-3 lt m lt -2`
B. `0 lt m lt 2`
C. `2lt m lt (5)/(2)`
D. `-2 lt m le (1)/(4)`

Answer

To solve the problem, we need to find the values of \( m \) such that both roots of the quadratic equation \( 4x^2 - 2x + m = 0 \) lie in the interval \( (-1, 1) \).

### Step 1: Identify the conditions for the roots
For the roots of the quadratic equation to lie within the interval \( (-1, 1) \), the function must be positive at the endpoints of the interval. Therefore, we need to check:
1. \( f(-1) > 0 \)
2. \( f(1) > 0 \)

### Step 2: Calculate \( f(-1) \)
Substituting \( x = -1 \) into the function:
\[ f(-1) = 4(-1)^2 - 2(-1) + m = 4(1) + 2 + m = 4 + 2 + m = m + 6 \]
For \( f(-1) > 0 \):
\[ m + 6 > 0 \implies m > -6 \]

### Step 3: Calculate \( f(1) \)
Substituting \( x = 1 \) into the function:
\[ f(1) = 4(1)^2 - 2(1) + m = 4(1) - 2 + m = 4 - 2 + m = m + 2 \]
For \( f(1) > 0 \):
\[ m + 2 > 0 \implies m > -2 \]

### Step 4: Calculate the discriminant
For the roots to be real, the discriminant of the quadratic must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4(4)(m) = 4 - 16m \]
Setting the discriminant greater than or equal to zero:
\[ 4 - 16m \geq 0 \implies 4 \geq 16m \implies \frac{1}{4} \geq m \implies m \leq \frac{1}{4} \]

### Step 5: Combine the conditions
We have three conditions:
1. \( m > -6 \)
2. \( m > -2 \)
3. \( m \leq \frac{1}{4} \)

The most restrictive conditions are \( m > -2 \) and \( m \leq \frac{1}{4} \).

### Step 6: Final result
Thus, the values of \( m \) must satisfy:
\[ -2 < m \leq \frac{1}{4} \]

### Conclusion
The final answer is:
\[ m \in (-2, \frac{1}{4}] \]