Question

If $f\left(x\right)=\sin x-\frac{x}{2}$​ is increasing function, then

• A. $0<x<\frac{\mathrm{\pi }}{3}$
• B. $-\frac{\mathrm{\pi }}{3}<x<0$
• C. $-\frac{\mathrm{\pi }}{3}<x<\frac{\mathrm{\pi }}{3}$
• D. $x=\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is C

$-\frac{\mathrm{\pi }}{3}<x<\frac{\mathrm{\pi }}{3}$

Explanation for the correct option.

Find the range of the function:

Given,

$f\left(x\right)=\sin x-\frac{x}{2}$

For increasing function,

$\begin{array}{rcl}f\text{'}\left(x\right)& >& 0\\ \frac{d}{dx}\left(\sin x-\frac{x}{2}\right)& >& 0\\ \cos x-\frac{1}{2}& >& 0\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{s}\mathbf{i}\mathbf{n}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathit{x}\mathbf{]}\\ \cos x& >& \frac{1}{2}\end{array}$

Therefore, $x\in \left[-\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right]$

Hence, the correct option is C.