Let `f: [(2pi)/(3), (5pi)/(3)]to[0,4]` be a function difined as `f (x) =sqrt3 sin x - cos x +2,` then :

Options

A. `f ^(-1) (1) =(4pi)/(3)`
B. `f ^(-1) (1)=pi`
C. `f ^(-1) (2) =(5pi)/(6)`
D. `f ^(-1) (2) =(7pi)/(6)`

Answer

To solve the problem step by step, we will analyze the function \( f(x) = \sqrt{3} \sin x - \cos x + 2 \) defined on the interval \( \left[\frac{2\pi}{3}, \frac{5\pi}{3}\right] \) and find its inverse values for specific outputs.

### Step 1: Understanding the Function
The function is given as:
\[ f(x) = \sqrt{3} \sin x - \cos x + 2 \]
We need to find the range of this function over the specified interval.

### Step 2: Finding the Range of \( f(x) \)
To find the range, we will analyze the expression \( \sqrt{3} \sin x - \cos x \).

1. **Rewriting the Function**:
We can rewrite \( f(x) \) in a more manageable form. Notice that:
\[ f(x) = 2 + \sqrt{3} \sin x - \cos x \]
We can express \( \sqrt{3} \sin x - \cos x \) as a single sine function using the sine addition formula.

2. **Using the Sine Addition Formula**:
We know that:
\[ R \sin(x + \phi) = \sqrt{3} \sin x - \cos x \]
where \( R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2 \) and \( \phi = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \).

Therefore, we can rewrite:
\[ \sqrt{3} \sin x - \cos x = 2 \sin\left(x - \frac{\pi}{6}\right) \]

3. **Finding the Maximum and Minimum Values**:
The maximum value of \( \sin \) is 1 and the minimum value is -1. Thus:
\[ \text{Max: } 2 \cdot 1 = 2 \quad \text{and} \quad \text{Min: } 2 \cdot (-1) = -2 \]
Therefore,
\[ f(x) = 2 + 2 \sin\left(x - \frac{\pi}{6}\right) \]
will have a range of:
\[ [2 - 2, 2 + 2] = [0, 4] \]

### Step 3: Confirming the Range
Since the range of \( f(x) \) is \( [0, 4] \), it matches the codomain given in the problem.

### Step 4: Finding Inverse Values
Now, we will find specific inverse values:

1. **Finding \( f^{-1}(1) \)**:
Set \( f(x) = 1 \):
\[ 1 = \sqrt{3} \sin x - \cos x + 2 \]
Rearranging gives:
\[ \sqrt{3} \sin x - \cos x = -1 \]
This can be expressed as:
\[ 2 \sin\left(x - \frac{\pi}{6}\right) = -1 \implies \sin\left(x - \frac{\pi}{6}\right) = -\frac{1}{2} \]
The solutions in the interval \( \left[\frac{2\pi}{3}, \frac{5\pi}{3}\right] \) are:
\[ x - \frac{\pi}{6} = \frac{7\pi}{6} \implies x = \frac{7\pi}{6} + \frac{\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3} \]
Thus, \( f^{-1}(1) = \frac{4\pi}{3} \).

2. **Finding \( f^{-1}(2) \)**:
Set \( f(x) = 2 \):
\[ 2 = \sqrt{3} \sin x - \cos x + 2 \implies \sqrt{3} \sin x - \cos x = 0 \]
This implies:
\[ \tan x = \frac{1}{\sqrt{3}} \implies x = \frac{\pi}{6} + n\pi \]
In the interval \( \left[\frac{2\pi}{3}, \frac{5\pi}{3}\right] \), the solution is:
\[ x = \frac{7\pi}{6} \]
Thus, \( f^{-1}(2) = \frac{7\pi}{6} \).

### Conclusion
The inverse values are:
- \( f^{-1}(1) = \frac{4\pi}{3} \)
- \( f^{-1}(2) = \frac{7\pi}{6} \)