If the value of of the integral `I=int_(0)^(2pi)sgn(e^(x))dx` is equal to `kpi`, then the smallest prime number greatest than 2k is (where, sgn (x) represents the signum functiono f x)

Options

A. 3
B. 5
C. 7
D. 11

Answer

To solve the integral \( I = \int_{0}^{2\pi} \text{sgn}(e^x) \, dx \), we first need to understand the behavior of the function \( e^x \) and consequently the signum function \( \text{sgn}(x) \).

### Step 1: Evaluate the Signum Function
The signum function \( \text{sgn}(x) \) is defined as:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

Since \( e^x \) is always positive for all real numbers \( x \), we have:
\[ \text{sgn}(e^x) = 1 \quad \text{for all } x \in \mathbb{R} \]

### Step 2: Substitute into the Integral
Given that \( \text{sgn}(e^x) = 1 \) for all \( x \) in the interval \( [0, 2\pi] \), we can simplify the integral:
\[ I = \int_{0}^{2\pi} 1 \, dx \]

### Step 3: Compute the Integral
Now we compute the integral:
\[ I = \int_{0}^{2\pi} 1 \, dx = [x]_{0}^{2\pi} = 2\pi - 0 = 2\pi \]

### Step 4: Relate to \( k\pi \)
From the problem, we know that \( I = k\pi \). Since we found \( I = 2\pi \), we can equate:
\[ k\pi = 2\pi \implies k = 2 \]

### Step 5: Find the Smallest Prime Number Greater than \( 2k \)
Now we need to find the smallest prime number greater than \( 2k \):
\[ 2k = 2 \times 2 = 4 \]
The prime numbers greater than 4 are: 5, 7, 11, ...

Thus, the smallest prime number greater than 4 is:
\[ \text{Smallest prime} = 5 \]

### Final Answer
The smallest prime number greater than \( 2k \) is:
\[ \boxed{5} \]