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

Answer 1

Answer: (D)

$[-1,3]$

We know that,

$-\sqrt{a^2+b^2}\le a\cos \theta +b\sin \theta \le \sqrt{(a^2+b^2)}$

Since, $-2\le \sin x-\sqrt{3}\cos x\le 2$

$\Rightarrow -1\le \sin x-\sqrt{3}\cos x+1\le 3$

Therefore, $S=$ Range of $f\left(x\right)=[-1,3]$