Statement-1: `int_(0)^(npi+v)|sin x|dx=2n+1-cos v` where `n in N` and `0 le v lt pi`.
Stetement-2: If f(x) is a periodic function with period T, then
(i) `int_(0)^(nT) f(x)dx=n int_(0)^(T) f(x)dx`, where `n in N`
and (ii) `int_(nT)^(nt+a) f(x)dx=int_(0)^(a) f(x) dx`, where `n in N`

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to evaluate the integral \( \int_{0}^{n\pi + v} |\sin x| \, dx \) and show that it equals \( 2n + 1 - \cos v \) for \( n \in \mathbb{N} \) and \( 0 \leq v < \pi \).

### Step-by-Step Solution

1. **Understanding the Integral**:
The function \( |\sin x| \) is periodic with a period of \( 2\pi \). We can break the integral into two parts:
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = \int_{0}^{n\pi} |\sin x| \, dx + \int_{n\pi}^{n\pi + v} |\sin x| \, dx \]

2. **Evaluating the First Integral**:
For \( n\pi \), we can observe that the integral from \( 0 \) to \( n\pi \) consists of \( n \) full periods of \( |\sin x| \):
\[ \int_{0}^{n\pi} |\sin x| \, dx = n \int_{0}^{\pi} |\sin x| \, dx \]
The integral \( \int_{0}^{\pi} |\sin x| \, dx \) can be calculated as follows:
\[ \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]
Thus,
\[ \int_{0}^{n\pi} |\sin x| \, dx = n \cdot 2 = 2n \]

3. **Evaluating the Second Integral**:
Now, we evaluate the second part:
\[ \int_{n\pi}^{n\pi + v} |\sin x| \, dx \]
Since \( n\pi \) is an integer multiple of \( \pi \), we have:
\[ |\sin(n\pi + x)| = |\sin x| \quad \text{for } 0 \leq x < v \]
Therefore,
\[ \int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_{0}^{v} |\sin x| \, dx = \int_{0}^{v} \sin x \, dx \]
Calculating this integral:
\[ \int_{0}^{v} \sin x \, dx = [-\cos x]_{0}^{v} = -\cos(v) - (-\cos(0)) = 1 - \cos(v) \]

4. **Combining the Results**:
Now we combine both parts:
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = 2n + (1 - \cos v) = 2n + 1 - \cos v \]

5. **Conclusion**:
We have shown that:
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v \]
Hence, Statement-1 is true.