Question

The range of$f\left(x\right)=\cos x-\sin x$ is

• A. $[-1,1]$
• B. $(-1,2)$
• C. $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
• D. $\left(-\sqrt{2},\sqrt{2}\right)$

Answer 1

The correct option is D

$\left(-\sqrt{2},\sqrt{2}\right)$

The explanation for the correct answer.

Solve for the range of$f\left(x\right)=\cos x-\sin x$

$f\left(x\right)=\cos x–\sin x$

$=\sqrt{2}\left[\cos x\left(\frac{1}{\sqrt{2}}\right)–\sin x\left(\frac{1}{\sqrt{2}}\right)\right]$

$=\sqrt{2}[\cos x\times \cos \left(\frac{\mathrm{\pi }}{4}\right)–\sin x\times \sin \left(\frac{\mathrm{\pi }}{4}\right)]$

$=\sqrt{2}\left[\cos (x+\left[\frac{\mathrm{\pi }}{4}\right]\right]$

$$\begin{gathered} \Rightarrow -1\le \left[\cos \left(x+\left[\frac{\mathrm{\pi }}{4}\right]\right)\right]\le 1 \\ \Rightarrow -\sqrt{2}\le \sqrt{2}\left[\cos \left(x+\left[\frac{\mathrm{\pi }}{4}\right]\right)\right]\le \sqrt{2} \\ \Rightarrow -\sqrt{2}\le f\left(x\right)\le \sqrt{2} \end{gathered}$$

The required range is $[-\sqrt{2},\sqrt{2}]$

Hence, option (D) is the correct answer.