If `f:A rarr B` defined as `f(x)=2sinx-2 cos x+3sqrt2` is an invertible function, then the correct statement can be

Options

A. `A=[(pi)/(4), (5pi)/(4)], B=[2sqrt2, 5sqrt5]`
B. `A=[(-pi)/(4),(5pi)/(4)], B=[sqrt2, 5sqrt2]`
C. `A=[(-pi)/(4), (3pi)/(4)],B=[sqrt2, 4,sqrt2]`
D. `A=[(-pi)/(4),(3pi)/(4)], B=[sqrt2, 5sqrt2]`

Answer

To solve the problem of determining the correct statement regarding the function \( f(x) = 2\sin x - 2\cos x + 3\sqrt{2} \) being an invertible function, we need to analyze the function's properties of being one-to-one (injective) and onto (surjective).

### Step-by-Step Solution:

1. **Understanding the Function**:
The given function is \( f(x) = 2\sin x - 2\cos x + 3\sqrt{2} \). We need to check the behavior of this function over different intervals to determine if it is one-to-one and onto.

**Hint**: Start by rewriting the function in a more manageable form.

2. **Rewriting the Function**:
We can rewrite the trigonometric part of the function using the sine addition formula. Notice that:
\[ 2\sin x - 2\cos x = 2(\sin x - \cos x) \]
We can express \( \sin x - \cos x \) in terms of sine:
\[ \sin x - \cos x = \sqrt{2}\left(\sin x \cdot \frac{1}{\sqrt{2}} - \cos x \cdot \frac{1}{\sqrt{2}}\right) = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \]
Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = 2\sqrt{2}\sin\left(x - \frac{\pi}{4}\right) + 3\sqrt{2} \]

**Hint**: Identify the range of the sine function to understand the behavior of \( f(x) \).

3. **Analyzing the Range**:
The sine function oscillates between -1 and 1. Thus,
\[ 2\sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \text{ oscillates between } -2\sqrt{2} \text{ and } 2\sqrt{2}. \]
Adding \( 3\sqrt{2} \) shifts this range:
\[ f(x) \text{ oscillates between } 3\sqrt{2} - 2\sqrt{2} \text{ and } 3\sqrt{2} + 2\sqrt{2} = \sqrt{2} \text{ and } 5\sqrt{2}. \]

**Hint**: Determine the intervals where the function is one-to-one.

4. **Finding Intervals for Injectivity**:
To ensure that \( f(x) \) is one-to-one, we need to restrict \( x \) to intervals where the sine function does not repeat values. A suitable interval is \( \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \), where the sine function is decreasing.

**Hint**: Check if the function is onto in the chosen interval.

5. **Checking Surjectivity**:
Since \( f(x) \) maps from the interval \( \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \) to the range \( [\sqrt{2}, 5\sqrt{2}] \), we need to ensure that every value in this range can be achieved by some \( x \) in the interval. Given the continuity of sine and the fact that it covers all values in this range, \( f(x) \) is onto.

**Hint**: Conclude the properties of the function.

6. **Conclusion**:
Since \( f(x) \) is both one-to-one and onto in the interval \( \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \), it is invertible in this interval.

### Final Answer:
The correct statement regarding the function \( f(x) \) being invertible is that it is bijective (both one-to-one and onto) in the interval \( \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \).