Question

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

• A. $\left[0.1\right]$
• B. $[-1,1]$
• C. $[0,3]$
• D. $[-1,3]$

Answer 1

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

We have

$-\sqrt{1+{\left(\sqrt{3}\right)}^2}\le \sin x-\sqrt{3}\cos x\le \sqrt{1+{\left(\sqrt{3}\right)}^2},\forall x\in R$

$\Rightarrow -2\le \sin x-\sqrt{3}\cos x\le 2,\forall x\in R$
$\Rightarrow -1\le \sin x-\sqrt{3}\cos x+1\le 3,\forall x\in R$
$\Rightarrow -1\le f\left(x\right)\le \sqrt{3},\forall x\in R$

$\therefore$ Range of $f=[-1,3]$

If $f:R\to S$ is onto, then $S=$ Range of $f=[-1,3]$