Q100 / 225 · JEE Main 2004 · Mathematics

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

  • A
    $[0,3]$
  • B
    $[-1,1]$
  • C
    $[0,1]$
  • Correct answer
    D
    $[-1,3]$

Solution

$-2 \leq \sin x-\sqrt{3} \cos x \leq 2 \Rightarrow-1 \leq \sin x-\sqrt{3} \cos x+1 \leq 3$ $\Rightarrow$ range of $f(x)$ is $[-1,3]$. Hence $S$ is $[-1,3]$

Hence, the correct answer is option: 'D'.

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