If `f:[-pi//2,pi//2] cupR` given by `f(x)=cosx+sin[(x+1)/(lambda)]` is an even function. Then the set of values of ` lambda(lambda gt0)` is Here, `[*]` denotes the greatest integer function.

Options

A. `((-pi)/(2),(pi)/(2))`-{0}
B. `((pi+2)/(2),oo)`
C. `(0,(pi+2)/(2)) cup ((pi+2)/(2),oo)`
D. `((-pi)/(2),(pi+2)/(2))-{0}`

Explanation

To determine the set of values of \( \lambda \) such that the function \( f(x) = \cos x + \sin\left(\frac{x+1}{\lambda}\right) \) is an even function, we need to analyze the properties of even functions.

### Step-by-Step Solution:

1. **Understanding Even Functions**:
A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \) in its domain.

2. **Setting up the Equation**:
We need to find \( \lambda \) such that:
\[ f(-x) = f(x) \]
This gives us:
\[ \cos(-x) + \sin\left(\frac{-x + 1}{\lambda}\right) = \cos x + \sin\left(\frac{x + 1}{\lambda}\right) \]

3. **Using Properties of Cosine and Sine**:
Since \( \cos(-x) = \cos x \), we can simplify our equation to:
\[ \cos x + \sin\left(\frac{-x + 1}{\lambda}\right) = \cos x + \sin\left(\frac{x + 1}{\lambda}\right) \]
This implies:
\[ \sin\left(\frac{-x + 1}{\lambda}\right) = \sin\left(\frac{x + 1}{\lambda}\right) \]

4. **Using the Sine Function Property**:
The sine function has the property that \( \sin A = \sin B \) implies:
\[ A = B + 2k\pi \quad \text{or} \quad A = \pi - B + 2k\pi \quad (k \in \mathbb{Z}) \]
Applying this to our equation:
\[ \frac{-x + 1}{\lambda} = \frac{x + 1}{\lambda} + 2k\pi \quad \text{or} \quad \frac{-x + 1}{\lambda} = \pi - \frac{x + 1}{\lambda} + 2k\pi \]

5. **Solving the First Case**:
From the first case:
\[ -x + 1 = x + 1 + 2k\lambda\pi \implies -2x = 2k\lambda\pi \implies x = -k\lambda\pi \]
This must hold for all \( x \), which is not possible unless \( k = 0 \).

6. **Solving the Second Case**:
From the second case:
\[ -x + 1 = \pi - (x + 1) + 2k\lambda\pi \implies -x + 1 = \pi - x - 1 + 2k\lambda\pi \]
Simplifying gives:
\[ 2 = \pi + 2k\lambda\pi \implies 2k\lambda\pi = 2 - \pi \implies k\lambda = \frac{2 - \pi}{2\pi} \]

7. **Finding the Range for \( \lambda \)**:
Since \( \lambda > 0 \), we need \( k \) to be non-negative. Thus, we find that:
\[ \lambda > \frac{2 - \pi}{2k\pi} \]
The maximum value of \( x + 1 \) occurs when \( x = \frac{\pi}{2} \), giving us:
\[ x + 1 = \frac{\pi}{2} + 1 \]
Therefore, we conclude:
\[ \lambda > \frac{\pi}{2} + 1 \]

### Conclusion:
The set of values of \( \lambda \) such that \( \lambda > 0 \) and \( f(x) \) is an even function is:
\[ \lambda \in \left( \frac{\pi}{2} + 1, \infty \right) \]