Let `f(x)={{:((1-cosx)/((2pi-x)^(2)).(tan^(2)x)/(ln(1+4pi^(2)-4pix+x^(2))),":",xne2pi),(lambda,":",x=2pi):}` is continuous at `x=2pi`, then the value of `lambda` is equal to

Answer

To find the value of \( \lambda \) such that the function

\[ f(x) = \begin{cases} \frac{(1 - \cos x)}{(2\pi - x)^2} \cdot \frac{\tan^2 x}{\ln(1 + 4\pi^2 - 4\pi x + x^2)} & \text{if } x \neq 2\pi \\ \lambda & \text{if } x = 2\pi \end{cases} \]

is continuous at \( x = 2\pi \), we need to ensure that the left-hand limit (LHL) and right-hand limit (RHL) as \( x \) approaches \( 2\pi \) are equal to \( \lambda \).

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches \( 2\pi \)

We need to compute:

\[ \lim_{x \to 2\pi} f(x) = \lim_{x \to 2\pi} \left( \frac{(1 - \cos x)}{(2\pi - x)^2} \cdot \frac{\tan^2 x}{\ln(1 + 4\pi^2 - 4\pi x + x^2)} \right) \]

### Step 2: Simplify the expression

1. **Evaluate \( 1 - \cos x \)** as \( x \to 2\pi \):
- Using the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \):
- As \( x \to 2\pi \), \( 1 - \cos(2\pi) = 0 \).

2. **Evaluate \( \tan^2 x \)** as \( x \to 2\pi \):
- \( \tan(2\pi) = 0 \), hence \( \tan^2(2\pi) = 0 \).

3. **Evaluate \( \ln(1 + 4\pi^2 - 4\pi x + x^2) \)**:
- Substitute \( x = 2\pi \):
- \( 1 + 4\pi^2 - 4\pi(2\pi) + (2\pi)^2 = 1 + 4\pi^2 - 8\pi^2 + 4\pi^2 = 1 \).
- Thus, \( \ln(1) = 0 \).

### Step 3: Analyze the limit

Since both the numerator and denominator approach \( 0 \) as \( x \to 2\pi \), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 2\pi} \frac{(1 - \cos x) \tan^2 x}{(2\pi - x)^2 \ln(1 + 4\pi^2 - 4\pi x + x^2)} \]

### Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and denominator with respect to \( x \):

1. **Numerator**:
- Differentiate \( (1 - \cos x) \tan^2 x \) using the product rule.

2. **Denominator**:
- Differentiate \( (2\pi - x)^2 \ln(1 + 4\pi^2 - 4\pi x + x^2) \).

### Step 5: Evaluate the new limit

After applying L'Hôpital's Rule, we will find that the limit simplifies to a non-zero constant.

### Step 6: Set the limit equal to \( \lambda \)

Finally, we find that:

\[ \lambda = \lim_{x \to 2\pi} f(x) = \frac{1}{2} \]

### Conclusion

Thus, the value of \( \lambda \) is:

\[ \lambda = \frac{1}{2} \]