The value of 'k' for which roots of the equation `4x^2-2x+k=0 ` are complctely in (-1,1) may be equal to

Options

A. `-1`
B. 0
C. 2
D. `-3`

Answer

To find the value of 'k' for which the roots of the equation \(4x^2 - 2x + k = 0\) are completely within the interval (-1, 1), we need to satisfy two conditions:

1. The discriminant of the quadratic must be non-negative (for real roots).
2. The function values at the endpoints of the interval (-1 and 1) must be positive.

### Step 1: Ensure the Discriminant is Non-negative

The discriminant \(D\) of the quadratic equation \(ax^2 + bx + c = 0\) is given by:

\[ D = b^2 - 4ac \]

For our equation \(4x^2 - 2x + k\), we have \(a = 4\), \(b = -2\), and \(c = k\). Thus, the discriminant is:

\[ D = (-2)^2 - 4 \cdot 4 \cdot k = 4 - 16k \]

For the roots to be real, we need:

\[ D \geq 0 \implies 4 - 16k \geq 0 \]

Solving this inequality:

\[ 4 \geq 16k \implies k \leq \frac{1}{4} \]

### Step 2: Evaluate the Function at the Endpoints

Next, we need to ensure that \(f(-1) > 0\) and \(f(1) > 0\).

**Evaluate at \(x = 1\):**

\[ f(1) = 4(1)^2 - 2(1) + k = 4 - 2 + k = 2 + k \]

For \(f(1) > 0\):

\[ 2 + k > 0 \implies k > -2 \]

**Evaluate at \(x = -1\):**

\[ f(-1) = 4(-1)^2 - 2(-1) + k = 4 + 2 + k = 6 + k \]

For \(f(-1) > 0\):

\[ 6 + k > 0 \implies k > -6 \]

### Step 3: Combine the Conditions

From the inequalities we derived:

1. \(k \leq \frac{1}{4}\)
2. \(k > -2\)
3. \(k > -6\)

The most restrictive conditions are \(k > -2\) and \(k \leq \frac{1}{4}\). Thus, we combine these to find the range of \(k\):

\[ -2 < k \leq \frac{1}{4} \]

### Conclusion

The values of \(k\) for which the roots of the equation \(4x^2 - 2x + k = 0\) are completely in the interval (-1, 1) are:

\[ k \in (-2, \frac{1}{4}] \]