If `f:A rarr B` defined by `f(x)=sinx-cosx+3sqrt2` is an invertible function, then the correct statement can be

Options

A. `A=[(pi)/(4),(5pi)/(4)],B=[3sqrt2, 4sqrt2]`
B. `A=[(-pi)/(4),(5pi)/(4)],B=[2sqrt2, 4sqrt2]`
C. `A=[(-pi)/(4),(3pi)/(4)], B=[sqrt2, 4sqrt2]`
D. `A=[(-pi)/(4),(3pi)/(4)], B=[2sqrt2, 4sqrt2]`

Answer

To determine if the function \( f(x) = \sin x - \cos x + 3\sqrt{2} \) is invertible, we need to check if it is both one-to-one (1-1) and onto (onto).

### Step-by-Step Solution:

1. **Rewrite the Function**:
We can rewrite the function in a more manageable form:
\[ f(x) = \sin x - \cos x + 3\sqrt{2} \]
We can factor out \( \sqrt{2} \) to help with the analysis:
\[ f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right) + 3\sqrt{2} \]

2. **Use Trigonometric Identities**:
Recognizing that \( \frac{1}{\sqrt{2}} \) corresponds to \( \sin\left(\frac{\pi}{4}\right) \) and \( \cos\left(\frac{\pi}{4}\right) \), we can use the sine subtraction formula:
\[ f(x) = \sqrt{2} \sin\left(x - \frac{\pi}{4}\right) + 3\sqrt{2} \]

3. **Determine the Interval for 1-1**:
The sine function is one-to-one in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, we need to restrict \( x \) to this interval:
\[ x - \frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
This gives us:
\[ -\frac{\pi}{2} + \frac{\pi}{4} < x < \frac{\pi}{2} + \frac{\pi}{4} \]
Simplifying this:
\[ -\frac{\pi}{4} < x < \frac{3\pi}{4} \]

4. **Find the Range of the Function**:
Now we need to find the minimum and maximum values of \( f(x) \) in the interval \( \left[-\frac{\pi}{4}, \frac{3\pi}{4}\right] \).

- **At \( x = -\frac{\pi}{4} \)**:
\[ f\left(-\frac{\pi}{4}\right) = \sqrt{2} \sin\left(-\frac{\pi}{4} - \frac{\pi}{4}\right) + 3\sqrt{2} = \sqrt{2} \sin\left(-\frac{\pi}{2}\right) + 3\sqrt{2} = 0 + 3\sqrt{2} = 3\sqrt{2} \]

- **At \( x = \frac{3\pi}{4} \)**:
\[ f\left(\frac{3\pi}{4}\right) = \sqrt{2} \sin\left(\frac{3\pi}{4} - \frac{\pi}{4}\right) + 3\sqrt{2} = \sqrt{2} \sin\left(\frac{\pi}{2}\right) + 3\sqrt{2} = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2} \]

5. **Conclusion**:
Since \( f(x) \) is one-to-one in the interval \( \left[-\frac{\pi}{4}, \frac{3\pi}{4}\right] \) and the range of \( f(x) \) is \( [3\sqrt{2}, 4\sqrt{2}] \), we conclude that the function is invertible on this interval.