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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as:
\[ f(x) = \sin x - \sqrt{3} \cos x + 1 \]

2. **Rewrite the trigonometric expression**:
We can express \( \sin x - \sqrt{3} \cos x \) in a single sine function using the formula:
\[ a \sin x + b \cos x = R \sin(x + \phi) \]
where \( R = \sqrt{a^2 + b^2} \) and \( \tan \phi = \frac{b}{a} \).

Here, \( a = 1 \) and \( b = -\sqrt{3} \).

3. **Calculate \( R \)**:
\[ R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

4. **Determine the range of the sine function**:
The expression \( \sin x - \sqrt{3} \cos x \) can take values from \( -R \) to \( R \):
\[ -2 \leq \sin x - \sqrt{3} \cos x \leq 2 \]

5. **Adjust for the constant term**:
Now, we add 1 to the entire inequality to find the range of \( f(x) \):
\[ -2 + 1 \leq \sin x - \sqrt{3} \cos x + 1 \leq 2 + 1 \]
Simplifying this gives:
\[ -1 \leq f(x) \leq 3 \]

6. **Conclusion**:
Therefore, the range of \( f(x) \) is:
\[ f(x) \in [-1, 3] \]
Since the function is onto, the interval of \( S \) is:
\[ S = [-1, 3] \]

### Final Answer:
The interval of \( S \) is \( [-1, 3] \).