Question

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

• A. $\pi$
• B. $0$
• C. $\frac{\sqrt{3}}{2}+\frac{\pi }{3}$
• D. $-\frac{\sqrt{3}}{2}+\frac{2\pi }{3}$

Answer 1

The correct option is A $\pi$

$\begin{array}{c}f\left(x\right)=\sin 2x-x\\ f^{\prime }\left(x\right)=2\cos 2x-1=0\\ \Rightarrow \cos 2x=\frac{1}{2}\\ \Rightarrow \cos 2x=\cos \frac{\pi }{3}\\ x=\frac{\pi }{6}\\ f^{\prime \prime }\left(x\right)=-2\sin 2x<0\\ putx=\frac{\pi }{6}\\ f^{\prime \prime }\left(x\right)=-2\sin \frac{\pi }{2}=2\times \frac{\sqrt{3}}{2}=-\sqrt{3}<0\\ \Rightarrow x=\frac{\pi }{6}pointpofmaxima\\ Thanmaximumvalueis\\ \Rightarrow \sin 2\times \frac{\pi }{6}-\frac{\pi }{6}=\frac{\sqrt{3}}{2}-\frac{\pi }{6}\\ Theminimumvalueatx=-\frac{\pi }{2}\\ \Rightarrow putx=-\frac{\pi }{2}in{f}^n\\ \sin 2\frac{\pi }{2}+\frac{\pi }{2}=-\sin \pi +\frac{\pi }{2}=\frac{\pi }{2}\\ Minimumvalueis=\frac{\pi }{2}\\ anddifferncebetweenthenis\\ \Rightarrow \frac{\pi }{2}\cdot \left(\frac{\sqrt{3}}{2}-\frac{\pi }{6}\right)\\ \Rightarrow \frac{2\pi }{3}-\frac{\sqrt{3}}{2}\end{array}$