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Question Number 164027 by HongKing last updated on 13/Jan/22

Commented by HongKing last updated on 13/Jan/22

$$\mathrm{Yes}\mathrm{my}\mathrm{dear}\mathrm{Sir}$$

Commented by mr W last updated on 13/Jan/22

$$nowigotit.$$$$Aisthenumberofrootsof$$$$f\left(f\left(f(....)\right)\right)=x.$$$$A=2^{2020-1}.$$

Answered by mr W last updated on 13/Jan/22

$$duetosymmetrywejustconsiderx⩾0.$$$$f\left(x\right)=2x\sqrt{1-x^2}⩾0$$$$letx=\sin \theta with0⩽\theta ⩽\frac{\pi }{2}$$$$f\left(x\right)=2\sin \theta \sqrt{1-{\sin }^2\theta }=2\sin \theta \cos \theta =\sin 2\theta$$$$f\left(f\left(x\right)\right)=\sin \left(2\times 2\theta \right)=\sin \left(2^2\theta \right)$$$$\text{Extra left or missing right}$$$$\text{Extra left or missing right}$$$$\Rightarrow \sin \left(2^n\theta \right)=\sin \theta$$$$\Rightarrow 2^n\theta =2k\pi +\theta \Rightarrow \theta =\frac{2k\pi }{2^n-1}or$$$$\Rightarrow 2^n\theta =(2m+1)\pi -\theta \Rightarrow \theta =\frac{(2m+1)\pi }{2^n+1}$$$$\theta =\frac{2k\pi }{2^n-1}⩽\frac{\pi }{2}\Rightarrow 0⩽k⩽\frac{2^n-1}{4}$$$$\theta =\frac{(2m+1)\pi }{2^n+1}⩽\frac{\pi }{2}\Rightarrow 0⩽m⩽\frac{2^n+1}{4}-\frac{1}{2}$$$$x=\sin \frac{2k\pi }{2^n-1}or$$$$x=\sin \frac{(2k+1)\pi }{2^n+1}$$$$with0⩽k⩽2^{n-2}-1andn=2020$$$$$$$$totallywehaveA=2^{n-1}=2^{2019}roots.$$$$$$$$A\left(mod1000\right)=288$$

Commented by HongKing last updated on 13/Jan/22

$$\mathrm{perfect}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by Rasheed.Sindhi last updated on 13/Jan/22

$${\mathbb{G}}_{\mathbb{S}\underset{!}{\mathbb{I}}\mathbb{R}}^{{\mathbb{R}}^{\mathbb{E}}\mathbb{A}}\mathbb{T}$$

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