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

For the function to be onto, the set $S$ must be equal to the range of $f(x)$.
Consider the expression $A\sin x+B\cos x.$ Its range is $[-\sqrt{A^2+B^2},\sqrt{A^2+B^2}].$
For $\sin x-\sqrt{3}\cos x,$ $A=1$ and $B=-\sqrt{3}.$
Range $=[-\sqrt{1^2+(-\sqrt{3}{)}^2},\sqrt{1^2+(-\sqrt{3}{)}^2}]=[-\sqrt{4},\sqrt{4}]=[-2,2]$.
Therefore:
$-2\le \sin x-\sqrt{3}\cos x\le 2$.
Adding $1$ to the entire inequality:
$-1\le \sin x-\sqrt{3}\cos x+1\le 3$.
The range of $f(x)$ is $[-1,3]$.
Thus $S=[-1,3]$.