If f: `R to S ` defined by `f (x) = sin x - sqrt(3) cos x + 1` is onto, then the interval of S is:

Options

A. `[0,3]`
B. `[-1,1]`
C. `[0,1]`
D. `[-1,3]`

Answer

To find the interval of \( S \) for the function \( f(x) = \sin x - \sqrt{3} \cos x + 1 \), we need to determine the range of this function.

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sin x - \sqrt{3} \cos x + 1 \]
We can combine the sine and cosine terms. To do this, we can express it in the form \( R \sin(x + \phi) \).

### Step 2: Identify \( R \) and \( \phi \)
We know that:
\[ R = \sqrt{a^2 + b^2} \]
where \( a = 1 \) (coefficient of \( \sin x \)) and \( b = -\sqrt{3} \) (coefficient of \( \cos x \)). Thus,
\[ R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Next, we find \( \phi \) using:
\[ \cos \phi = \frac{a}{R} = \frac{1}{2}, \quad \sin \phi = \frac{-\sqrt{3}}{2} \]
This gives us \( \phi = -\frac{\pi}{3} \) (since \( \sin \) is negative and \( \cos \) is positive in the fourth quadrant).

### Step 3: Rewrite \( f(x) \)
Now we can rewrite \( f(x) \):
\[ f(x) = 2 \sin\left(x - \frac{\pi}{3}\right) + 1 \]

### Step 4: Determine the range of \( f(x) \)
The range of \( \sin \) function is from -1 to 1. Therefore,
\[ -1 \leq \sin\left(x - \frac{\pi}{3}\right) \leq 1 \]
Multiplying the entire inequality by 2:
\[ -2 \leq 2 \sin\left(x - \frac{\pi}{3}\right) \leq 2 \]
Adding 1 to all parts of the inequality:
\[ -2 + 1 \leq 2 \sin\left(x - \frac{\pi}{3}\right) + 1 \leq 2 + 1 \]
This simplifies to:
\[ -1 \leq f(x) \leq 3 \]

### Step 5: Conclusion
Thus, the range of \( f(x) \) is:
\[ [-1, 3] \]
Therefore, the interval of \( S \) is \( [-1, 3] \).