Statement-1:For any `n in N`, we have
`int_(0)^(npi) |(sinx)/(x)|dx ge (2)/(pi)(1+(1)/(2)+(1)/(3)+....+(1)/(n))`
Statement-2:`(sin x)/(x)ge(2)/(pi)"on"(0,(pi)/(2))`

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 analyze both statements and prove their validity step by step.

### Step 1: Understanding Statement 1
We need to prove that:
\[ \int_0^{n\pi} \left| \frac{\sin x}{x} \right| dx \geq \frac{2}{\pi} \left( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \right) \]

### Step 2: Breaking Down the Integral
Let's denote the integral as:
\[ I = \int_0^{n\pi} \left| \frac{\sin x}{x} \right| dx \]
The function \(\frac{\sin x}{x}\) is positive for \(x > 0\), so we can drop the absolute value:
\[ I = \int_0^{n\pi} \frac{\sin x}{x} dx \]

### Step 3: Using the Properties of the Sine Function
The function \(\sin x\) has a period of \(2\pi\). Therefore, we can break the integral into \(n\) intervals of length \(\pi\):
\[ I = \sum_{r=0}^{n-1} \int_{r\pi}^{(r+1)\pi} \frac{\sin x}{x} dx \]

### Step 4: Change of Variables
For each interval, we can perform a change of variables:
Let \(x = r\pi + u\), where \(u\) varies from \(0\) to \(\pi\) as \(x\) varies from \(r\pi\) to \((r+1)\pi\). Then:
\[ I = \sum_{r=0}^{n-1} \int_0^{\pi} \frac{\sin(r\pi + u)}{r\pi + u} du \]
Using the property \(\sin(r\pi + u) = (-1)^r \sin u\), we can write:
\[ I = \sum_{r=0}^{n-1} (-1)^r \int_0^{\pi} \frac{\sin u}{r\pi + u} du \]

### Step 5: Estimating the Integral
We can estimate the integral \(\int_0^{\pi} \frac{\sin u}{r\pi + u} du\). Since \(\sin u\) is bounded between \(0\) and \(1\), we can say:
\[ \int_0^{\pi} \frac{\sin u}{r\pi + u} du \geq \int_0^{\pi} \frac{2}{\pi} \cdot \frac{\sin u}{u} du \]
Thus, we can conclude:
\[ I \geq \frac{2}{\pi} \sum_{r=1}^{n} \left( \int_0^{\pi} \frac{1}{r} du \right) = \frac{2}{\pi} \sum_{r=1}^{n} \frac{\pi}{r} = \frac{2}{\pi} \left( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \right) \]

### Step 6: Conclusion for Statement 1
Thus, we have shown that:
\[ I \geq \frac{2}{\pi} \left( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \right) \]
which confirms that Statement 1 is true.

### Step 7: Understanding Statement 2
Statement 2 claims that:
\[ \frac{\sin x}{x} \geq \frac{2}{\pi} \quad \text{for } x \in (0, \frac{\pi}{2}) \]
To prove this, we can analyze the function \(\frac{\sin x}{x}\) on the interval \((0, \frac{\pi}{2})\).

### Step 8: Analyzing the Function
The function \(\frac{\sin x}{x}\) is known to be decreasing on the interval \((0, \frac{\pi}{2})\). At \(x = \frac{\pi}{2}\):
\[ \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} = \frac{2}{\pi} \]
Since the function is decreasing, it follows that:
\[ \frac{\sin x}{x} \geq \frac{2}{\pi} \quad \text{for all } x \in (0, \frac{\pi}{2}) \]
Thus, Statement 2 is also true.

### Final Conclusion
Both statements are true, and Statement 2 serves as a correct explanation for Statement 1.

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