Mathematics - Application of Derivative Question with Solution | TestHub

$$\begin{array}{l}f(x)=2\sin x+\sin 2x\\ f^{\prime }(x)=2\cos x+2\cos 2x=2(\cos x+\cos 2x)\\ \therefore f^{\prime }(x)=0\Rightarrow 2{\cos }^{2}x+\cos x-1=0\\ \cos x=\frac{-1\pm 3}{4}=-1,\frac{1}{2}\therefore x=\pi ,\frac{\pi }{3}\end{array}$$ Now, $\mathrm{f}(0)=0,\mathrm{f}\left(\frac{3\pi }{2}\right)=-2$ $$\mathrm{f}(\pi )=0,\mathrm{f}\left(\frac{\pi }{3}\right)=2\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$$ $\therefore$ difference between greatest value and least value $$=\frac{3\sqrt{3}}{2}+2$$