If the roots of `x^(2) - 2kx + k^(2)-1 = 0` lie between -2 and 4 then the interval in which k lies, is

Options

A. (-3,1)
B. (-3, 3)
C. (-1, 5)
D. (-1, 3)

Answer

To solve the problem, we need to determine the interval in which \( k \) lies, given that the roots of the quadratic equation \( x^2 - 2kx + (k^2 - 1) = 0 \) are between -2 and 4.

### Step-by-Step Solution:

1. **Identify the Quadratic Equation**:
The given quadratic equation is:
\[ x^2 - 2kx + (k^2 - 1) = 0 \]

2. **Use the Quadratic Formula**:
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2k \), and \( c = k^2 - 1 \).
Plugging in these values, we get:
\[ x = \frac{2k \pm \sqrt{(-2k)^2 - 4 \cdot 1 \cdot (k^2 - 1)}}{2 \cdot 1} \]

3. **Simplify the Expression**:
Calculate the discriminant:
\[ (-2k)^2 - 4 \cdot 1 \cdot (k^2 - 1) = 4k^2 - 4(k^2 - 1) = 4k^2 - 4k^2 + 4 = 4 \]
Therefore, the roots simplify to:
\[ x = \frac{2k \pm 2}{2} = k \pm 1 \]
Thus, the roots are \( k + 1 \) and \( k - 1 \).

4. **Set Up Inequalities for the Roots**:
We know that both roots must lie between -2 and 4:
\[ -2 < k - 1 < 4 \quad \text{and} \quad -2 < k + 1 < 4 \]

5. **Solve the First Inequality**:
For \( k - 1 \):
\[ -2 < k - 1 \implies k > -1 \]
\[ k - 1 < 4 \implies k < 5 \]
Thus, from the first root, we have:
\[ -1 < k < 5 \]

6. **Solve the Second Inequality**:
For \( k + 1 \):
\[ -2 < k + 1 \implies k > -3 \]
\[ k + 1 < 4 \implies k < 3 \]
Thus, from the second root, we have:
\[ -3 < k < 3 \]

7. **Find the Common Interval**:
Now we need to find the intersection of the two intervals:
\[ (-1, 5) \quad \text{and} \quad (-3, 3) \]
The common interval is:
\[ -1 < k < 3 \]

### Final Answer:
Thus, the interval in which \( k \) lies is:
\[ \boxed{(-1, 3)} \]