If A denotes the area bounded by `f(x)=|("sin"x + "cos"x)/(x)|`, X-axis, `x=pi` and `x=3 pi`,then

Options

A. `1 lt A lt 2`
B. `0 lt A lt 2`
C. `2 lt A lt 3`
D. None of these

Answer

To find the area \( A \) bounded by the function \( f(x) = \left| \frac{\sin x + \cos x}{x} \right| \), the x-axis, and the vertical lines \( x = \pi \) and \( x = 3\pi \), we can follow these steps:

### Step 1: Determine the function and the interval
The function we are dealing with is:
\[ f(x) = \left| \frac{\sin x + \cos x}{x} \right| \]
We need to find the area from \( x = \pi \) to \( x = 3\pi \).

### Step 2: Analyze the function on the interval
We need to analyze \( \sin x + \cos x \) on the interval \( [\pi, 3\pi] \). The maximum value of \( \sin x + \cos x \) can be found using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The maximum value of \( \sin x + \cos x \) is \( \sqrt{2} \) and occurs at certain points in the interval.

### Step 3: Calculate the area from \( \pi \) to \( 2\pi \)
For \( x \) in \( [\pi, 2\pi] \):
\[ \text{Area}_1 = \int_{\pi}^{2\pi} \left| \frac{\sin x + \cos x}{x} \right| \, dx \]
Since \( x \) is positive in this interval, we can drop the absolute value:
\[ \text{Area}_1 = \int_{\pi}^{2\pi} \frac{\sin x + \cos x}{x} \, dx \]

### Step 4: Calculate the area from \( 2\pi \) to \( 3\pi \)
For \( x \) in \( [2\pi, 3\pi] \):
\[ \text{Area}_2 = \int_{2\pi}^{3\pi} \left| \frac{\sin x + \cos x}{x} \right| \, dx \]
Again, since \( x \) is positive, we can drop the absolute value:
\[ \text{Area}_2 = \int_{2\pi}^{3\pi} \frac{\sin x + \cos x}{x} \, dx \]

### Step 5: Combine the areas
The total area \( A \) is the sum of the two areas:
\[ A = \text{Area}_1 + \text{Area}_2 = \int_{\pi}^{2\pi} \frac{\sin x + \cos x}{x} \, dx + \int_{2\pi}^{3\pi} \frac{\sin x + \cos x}{x} \, dx \]

### Step 6: Estimate the bounds of the area
We can estimate the maximum value of \( \frac{\sin x + \cos x}{x} \) in the intervals:
- For \( x \in [\pi, 2\pi] \), the maximum value of \( \sin x + \cos x \) is \( \sqrt{2} \), hence:
\[ \frac{\sqrt{2}}{x} \text{ where } x \text{ is between } \pi \text{ and } 2\pi \]
- For \( x \in [2\pi, 3\pi] \), the maximum value is again \( \sqrt{2} \) but divided by \( 2\pi \) and \( 3\pi \) respectively.

### Step 7: Final bounds
Thus, we can bound the area \( A \):
\[ \frac{2\sqrt{2}}{2\pi} < A < \frac{2\sqrt{2}}{3\pi} \]
This gives us an approximate range for \( A \).

### Conclusion
The area \( A \) is bounded and can be estimated using the integrals calculated above.