Consider the function `f(theta)=int_(0)^(1)(|sqrt(1-x^(2))-sintheta|)/(sqrt(1-x^(2)))dx`, where `0le theta le (pi)/2`, then

Options

A. `f_("min")=sqrt(2)-1`
B. `f_("min")=sqrt(2)+1`
C. `f_("max")=1`
D. `f_("max")=(pi)/2-1`

Explanation

To solve the problem, we need to evaluate the integral function defined as:

\[ f(\theta) = \int_{0}^{1} \frac{|\sqrt{1 - x^2} - \sin \theta|}{\sqrt{1 - x^2}} \, dx \]

where \(0 \leq \theta \leq \frac{\pi}{2}\).

### Step 1: Substitution
We can use the substitution \(x = \cos \phi\), which gives \(dx = -\sin \phi \, d\phi\). The limits change as follows:
- When \(x = 0\), \(\phi = \frac{\pi}{2}\)
- When \(x = 1\), \(\phi = 0\)

Thus, we rewrite the integral:

\[ f(\theta) = \int_{\frac{\pi}{2}}^{0} \frac{|\sqrt{1 - \cos^2 \phi} - \sin \theta|}{\sqrt{1 - \cos^2 \phi}} (-\sin \phi) \, d\phi \]

This simplifies to:

\[ f(\theta) = \int_{0}^{\frac{\pi}{2}} \frac{|\sin \phi - \sin \theta|}{\sin \phi} \sin \phi \, d\phi = \int_{0}^{\frac{\pi}{2}} |\sin \phi - \sin \theta| \, d\phi \]

### Step 2: Splitting the Integral
Next, we need to split the integral based on the value of \(\sin \theta\):
- For \(\phi\) in \([0, \theta]\), \(\sin \phi \leq \sin \theta\)
- For \(\phi\) in \([\theta, \frac{\pi}{2}]\), \(\sin \phi \geq \sin \theta\)

Thus, we can write:

\[ f(\theta) = \int_{0}^{\theta} (\sin \theta - \sin \phi) \, d\phi + \int_{\theta}^{\frac{\pi}{2}} (\sin \phi - \sin \theta) \, d\phi \]

### Step 3: Evaluating Each Integral
Now we evaluate each part separately.

1. **First Integral**:
\[ \int_{0}^{\theta} (\sin \theta - \sin \phi) \, d\phi = \sin \theta \cdot \theta - \int_{0}^{\theta} \sin \phi \, d\phi \]
The integral of \(\sin \phi\) is \(-\cos \phi\), so:
\[ = \sin \theta \cdot \theta + \cos(0) - \cos(\theta) = \sin \theta \cdot \theta + 1 - \cos \theta \]

2. **Second Integral**:
\[ \int_{\theta}^{\frac{\pi}{2}} (\sin \phi - \sin \theta) \, d\phi = \int_{\theta}^{\frac{\pi}{2}} \sin \phi \, d\phi - \sin \theta \cdot \left(\frac{\pi}{2} - \theta\right) \]
Again, the integral of \(\sin \phi\) gives:
\[ = [-\cos \phi]_{\theta}^{\frac{\pi}{2}} - \sin \theta \cdot \left(\frac{\pi}{2} - \theta\right) = (0 + \cos \theta) - \sin \theta \cdot \left(\frac{\pi}{2} - \theta\right) \]

### Step 4: Combine the Results
Now we combine both parts:

\[ f(\theta) = \left(\sin \theta \cdot \theta + 1 - \cos \theta\right) + \left(\cos \theta - \sin \theta \cdot \left(\frac{\pi}{2} - \theta\right)\right) \]

This simplifies to:

\[ f(\theta) = \sin \theta \cdot \theta + 1 - \sin \theta \cdot \left(\frac{\pi}{2} - \theta\right) \]

### Step 5: Finding Minimum and Maximum
To find the minimum and maximum values of \(f(\theta)\) over the interval \([0, \frac{\pi}{2}]\), we can evaluate \(f(0)\), \(f\left(\frac{\pi}{4}\right)\), and \(f\left(\frac{\pi}{2}\right)\):

1. **At \(\theta = 0\)**:
\[ f(0) = 1 \]

2. **At \(\theta = \frac{\pi}{4}\)**:
\[ f\left(\frac{\pi}{4}\right) = \sqrt{2} - 1 \]

3. **At \(\theta = \frac{\pi}{2}\)**:
\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1 \]

### Conclusion
From the evaluations, we find that:

- Minimum value occurs at \(\theta = \frac{\pi}{4}\) with \(f\left(\frac{\pi}{4}\right) = \sqrt{2} - 1\).
- Maximum value occurs at \(\theta = 0\) with \(f(0) = 1\).

Thus, the minimum value is \(\sqrt{2} - 1\) and the maximum value is \(1\).