Question

The difference between the greatest and least values of the function
f (x) = sin(2x) – x, on $\left(-\frac{\pi }{2},\frac{\pi }{2}\right]$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is C $\pi$

$f\left(x\right)=\sin \left(2x\right)-x\Rightarrow f\text{'}\left(x\right)=2\cos \left(2x\right)-1\therefore f\text{'}\left(x\right)=0\Rightarrow 2\cos \left(2x\right)-1=0\Rightarrow \cos 2x=\frac{1}{2}\Rightarrow 2x=-\frac{\pi }{3}\mathrm{or}\frac{\pi }{3}\Rightarrow x=-\frac{\pi }{6}\mathrm{or}\frac{\pi }{6}f\left(-\frac{\pi }{2}\right)=\sin \left(-\pi \right)+\frac{\pi }{2}=\frac{\pi }{2}f\left(\frac{\pi }{2}\right)=\sin \left(\pi \right)-\frac{\pi }{2}=-\frac{\pi }{2}f\left(-\frac{\pi }{6}\right)=\sin \left(-\frac{\pi }{3}\right)+\frac{\pi }{6}=-\frac{\sqrt{3}}{2}+\frac{\pi }{6}f\left(\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{3}\right)-\frac{\pi }{6}=\frac{\sqrt{3}}{2}-\frac{\pi }{6}\mathrm{Clearly},\frac{\pi }{2}\mathrm{is}\mathrm{the}\mathrm{greatest}\mathrm{value}\mathrm{and}–\frac{\pi }{2}\mathrm{is}\mathrm{the}\mathrm{least}.\mathrm{Therefore},\mathrm{difference}=\frac{\pi }{2}+\frac{\pi }{2}=\pi .$