consider f(X)= `cos2x+2xlambda^(2)+(2lambda+1)(lambda-1)x^(2),lambda in R`
If `alpha ne beta and f(alpha+beta)/(2)lt(f(alpha)+f(beta))/(2)` for `alpha` and `beta` then find the values of `lambda`

Explanation

To solve the problem, we need to analyze the function given and apply the condition provided. Let's break it down step by step.

### Step 1: Define the function
We are given the function:
\[ f(x) = \cos(2x) + 2x\lambda^2 + (2\lambda + 1)(\lambda - 1)x^2 \]

### Step 2: Find the second derivative
To determine the concavity of the function, we need to compute the second derivative \( f''(x) \).

1. **First derivative \( f'(x) \)**:
\[ f'(x) = -2\sin(2x) + 2\lambda^2 + 2(2\lambda + 1)(\lambda - 1)x \]

2. **Second derivative \( f''(x) \)**:
\[ f''(x) = -4\cos(2x) + 2(2\lambda + 1)(\lambda - 1) \]

### Step 3: Set the condition for concavity
According to the problem, we have the condition:
\[ f\left(\frac{\alpha + \beta}{2}\right) < \frac{f(\alpha) + f(\beta)}{2} \]
This implies that the function is concave down at the points \( \alpha \) and \( \beta \). For this to hold for all \( x \in \mathbb{R} \), we require:
\[ f''(x) \geq 0 \]
Thus, we need:
\[ -4\cos(2x) + 2(2\lambda + 1)(\lambda - 1) \geq 0 \]

### Step 4: Analyze the inequality
Since \( \cos(2x) \) oscillates between -1 and 1, we can find the maximum and minimum values of \( -4\cos(2x) \):
- Maximum: \( -4(-1) = 4 \)
- Minimum: \( -4(1) = -4 \)

Thus, we want:
\[ 2(2\lambda + 1)(\lambda - 1) \geq 4 \quad \text{and} \quad 2(2\lambda + 1)(\lambda - 1) \geq -4 \]

### Step 5: Solve the inequalities
1. **First inequality**:
\[ 2(2\lambda + 1)(\lambda - 1) \geq 4 \implies (2\lambda + 1)(\lambda - 1) \geq 2 \]
Expanding this:
\[ 2\lambda^2 - 2\lambda + \lambda - 1 \geq 2 \implies 2\lambda^2 - \lambda - 3 \geq 0 \]
Factoring gives:
\[ (2\lambda + 3)(\lambda - 1) \geq 0 \]
The critical points are \( \lambda = -\frac{3}{2} \) and \( \lambda = 1 \).

2. **Second inequality**:
\[ 2(2\lambda + 1)(\lambda - 1) \geq -4 \implies (2\lambda + 1)(\lambda - 1) \geq -2 \]
Expanding this:
\[ 2\lambda^2 - 2\lambda + \lambda - 1 \geq -2 \implies 2\lambda^2 - \lambda + 1 \geq 0 \]
This quadratic has no real roots (discriminant < 0), hence it is always positive.

### Step 6: Combine the results
From the first inequality, we analyze the intervals:
- \( \lambda \leq -\frac{3}{2} \) or \( \lambda \geq 1 \)

Thus, the values of \( \lambda \) that satisfy the conditions are:
\[ \lambda \in (-\infty, -\frac{3}{2}] \cup [1, \infty) \]

### Final Answer
The values of \( \lambda \) are:
\[ \lambda \in (-\infty, -\frac{3}{2}] \cup [1, \infty) \]