Question

If $f\left(x\right)=\sin x-\cos x$, the function decreasing in $0\le x\le 2\mathrm{\pi }$ is

• A. $\left[\frac{5\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{4}\right]$
• B. $\left[\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right]$
• C. $\left[\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{2}\right]$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option.

Find the range:

Given,

$f\left(x\right)=\sin x-\cos x$.

Now, differentiate the $f\left(x\right)$ w.r.t. $x$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(\sin x-\cos x\right)\\ & =& \cos x+\sin x\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{sin}\mathbf{x}\mathbf{=}\mathbf{cos}\mathbf{x}\mathbf{}\mathbf{\&}\mathbf{\&}\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{cos}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{sin}\mathbf{x}\mathbf{]}\\ & =& \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x\right)\\ & =& \sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}\cos x+\sin \frac{\mathrm{\pi }}{4}\sin x\right)\\ & =& \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)\mathbf{\left[}\mathbf{}\mathbf{\because }\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{cos}\mathbf{a}\mathbf{·}\mathbf{cos}\mathbf{b}\mathbf{+}\mathbf{sin}\mathbf{a}\mathbf{·}\mathbf{sin}\mathbf{b}\mathbf{\right]}\\ & & \end{array}$

The function is decreasing.

$\begin{array}{rcl}& \Rightarrow & f\text{'}\left(x\right)<0\\ \sqrt{2}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)& <& 0\\ \cos \left(x-\frac{\mathrm{\pi }}{4}\right)& <& 0\end{array}$

$$\begin{gathered} \Rightarrow \frac{\mathrm{\pi }}{2}<\left(x-\frac{\mathrm{\pi }}{4}\right)<\frac{3\mathrm{\pi }}{2} \\ \frac{3\mathrm{\pi }}{4}<x<\frac{7\mathrm{\pi }}{4} \end{gathered}$$

Therefore, $x\in \begin{array}{l}\left[\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right]\end{array}$

Hence, the correct option is D.