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If $f: R \rightarrow S$, defined by $f(x)=\sin x-\sqrt{3} \cos x+1$, is onto, then the interval of $S$ is
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The correct answer is:
$[-1,3]$
$[-1,3]$
$-2 \leq \sin x-\sqrt{3} \cos x \leq 2 \Rightarrow-1 \leq \sin x-\sqrt{3} \cos x+1 \leq 3$
$\Rightarrow$ range of $f(x)$ is $[-1,3]$.
Hence $S$ is $[-1,3]$
$\Rightarrow$ range of $f(x)$ is $[-1,3]$.
Hence $S$ is $[-1,3]$
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