The value of `lamda` for. Which `2x^(2)-2(2 lamda+1) x+lamda (lamda +1)=0` may have one root less than `lamda` and other root greater than `lamda` are given by

Options

A. `1 gt lamda gt 0`
B. `-1 lt lamda lt 0`
C. `lamda ge 0`
D. `lamda gt 0" or "lamda lt -1`

Answer

To solve the problem, we need to find the values of \( \lambda \) for which the quadratic equation

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

has one root less than \( \lambda \) and the other root greater than \( \lambda \). This means that \( \lambda \) must be between the two roots of the quadratic equation.

### Step 1: Identify the condition for \( \lambda \)

For \( \lambda \) to be between the roots, we must have:

\[ f(\lambda) < 0 \]

where \( f(x) = 2x^2 - 2(2\lambda + 1)x + \lambda(\lambda + 1) \).

### Step 2: Substitute \( \lambda \) into the quadratic

Substituting \( x = \lambda \) into the quadratic equation gives:

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

### Step 3: Simplify the expression

Now, simplify \( f(\lambda) \):

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

Combine like terms:

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

### Step 4: Set the inequality

We need to solve:

\[ -\lambda^2 - \lambda < 0 \]

This can be rewritten as:

\[ \lambda^2 + \lambda > 0 \]

### Step 5: Factor the inequality

Factoring gives:

\[ \lambda(\lambda + 1) > 0 \]

### Step 6: Determine the intervals

To find the intervals where this inequality holds, we find the roots:

1. \( \lambda = 0 \)
2. \( \lambda + 1 = 0 \) which gives \( \lambda = -1 \)

The critical points are \( -1 \) and \( 0 \). We can test intervals around these points:

- For \( \lambda < -1 \) (e.g., \( \lambda = -2 \)):
\(-2(-2 + 1) = -2(-1) = 2 > 0\) (satisfied)

- For \( -1 < \lambda < 0 \) (e.g., \( \lambda = -0.5 \)):
\(-0.5(-0.5 + 1) = -0.5(0.5) = -0.25 < 0\) (not satisfied)

- For \( \lambda > 0 \) (e.g., \( \lambda = 1 \)):
\(1(1 + 1) = 1 \cdot 2 = 2 > 0\) (satisfied)

### Step 7: Conclusion

The solution to the inequality \( \lambda(\lambda + 1) > 0 \) is:

\[ \lambda \in (-\infty, -1) \cup (0, \infty) \]

Thus, the values of \( \lambda \) for which the quadratic equation has one root less than \( \lambda \) and the other root greater than \( \lambda \) are:

\[ \lambda < -1 \quad \text{or} \quad \lambda > 0 \]