Question

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

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

Answer 1

The correct option is D

$\left[-1,3\right]$

Explanation for the correct answer:

Finding the range of the function $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

Given function: $\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}+1$ is onto.

Therefore, the minimum of $f$ is, $-\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$ and the maximum of $f$ is, $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Where, $a$ is co-efficient of $\sin \left(x\right)\mathrm{and}b$ is co-efficient of $\cos \left(x\right).$

$$\begin{gathered} \mathrm{or},-\sqrt{1^2+{\left(\sqrt{-3}\right)}^2}\le \sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}\le \sqrt{1^2+{\left(\sqrt{-3}\right)}^2} \\ \mathrm{or},-\sqrt{4}\le \sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}\le \sqrt{4} \\ \mathrm{or},-2\le \sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}\le 2 \\ \mathrm{or},-2+1\le \sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}+1\le 2+1 \\ \mathrm{or},-1\le \sin \mathrm{x}-\sqrt{3}\cos \mathrm{x}\le 3 \end{gathered}$$

Therefore, the range of $f\left(x\right)$ is $\left[-1,3\right]$

Hence, the correct answer is an option (D).