Question

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

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

Answer 1

The correct option is D $[-1,3]$

$\begin{array}{c}f\left(x\right)=\sin x-\sqrt{3}\cos x+1\\ =2\left(\frac{1}{2}\sin x-\frac{1}{2}\sqrt{3}\cos x+1\right)\\ =2\left(\sin \left(x-\frac{\pi }{3}\right)\right)+1\\ Weknowthat-1\le \sin \left(x-\frac{\pi }{3}\right)\le 1\\ \Rightarrow -2\le 2\sin \left(x-\frac{\pi }{3}\right)\le 2\\ \Rightarrow -1\le 2\sin \left(x-\frac{\pi }{3}\right)+1\le 3\\ rangeoff\left(x\right)=S=\left[-1,3\right]\\ Hence,TheoptinDisthecorrectanswer.\end{array}$