Out of the two roots of `x^(2)-2 lambda x + lambda^(2)-1= 0`, one is greater than 4 and the other root is less than 4, then the limits of `lambda` are

Options

A. `3 lt lambda lt 5`
B. `lambda lt 3`
C. `lambda lt 5`
D. `5 lt lambda lt 6`

Answer

To solve the problem, we need to analyze the quadratic equation given by:

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

We are told that one root is greater than 4 and the other root is less than 4. We will use this information to find the limits of \(\lambda\).

### Step 1: Identify the roots using the quadratic formula
The roots of the quadratic equation \(ax^2 + bx + c = 0\) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, \(a = 1\), \(b = -2\lambda\), and \(c = \lambda^2 - 1\). Therefore, the roots are:

\[ x = \frac{2\lambda \pm \sqrt{(-2\lambda)^2 - 4 \cdot 1 \cdot (\lambda^2 - 1)}}{2 \cdot 1} \]

### Step 2: Simplify the expression for the roots
Calculating the discriminant:

\[ (-2\lambda)^2 - 4 \cdot 1 \cdot (\lambda^2 - 1) = 4\lambda^2 - 4(\lambda^2 - 1) = 4\lambda^2 - 4\lambda^2 + 4 = 4 \]

Now substituting back into the roots formula:

\[ x = \frac{2\lambda \pm \sqrt{4}}{2} = \frac{2\lambda \pm 2}{2} = \lambda \pm 1 \]

Thus, the roots are:

\[ x_1 = \lambda + 1 \quad \text{and} \quad x_2 = \lambda - 1 \]

### Step 3: Set up inequalities based on the roots
According to the problem, one root is greater than 4 and the other root is less than 4. This gives us two inequalities:

1. \(x_1 > 4\) (i.e., \(\lambda + 1 > 4\))
2. \(x_2 < 4\) (i.e., \(\lambda - 1 < 4\))

### Step 4: Solve the inequalities
**For the first inequality:**

\[ \lambda + 1 > 4 \implies \lambda > 3 \]

**For the second inequality:**

\[ \lambda - 1 < 4 \implies \lambda < 5 \]

### Step 5: Combine the results
From the two inequalities, we have:

\[ 3 < \lambda < 5 \]

### Conclusion
The limits of \(\lambda\) are:

\[ \lambda \in (3, 5) \]