Question-164027

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

Commented by HongKing last updated on 13/Jan/22

Yes my dear Sir

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

n o w i g o t i t .

A i s t h e n u m b e r o f r o o t s o f

f (f (f (\dots .))) = x .

A = 2^{2020 - 1} .

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

d u e t o s y m m e t r y w e j u s t c o n s i d e r x ⩾ 0 .

f (x) = 2 x \sqrt{1 - x^{2}} ⩾ 0

l e t x = \sin θ w i t h 0 ⩽ θ ⩽ \frac{π}{2}

f (x) = 2 \sin θ \sqrt{1 - \sin^{2} θ} = 2 \sin θ \cos θ = \sin 2 θ

f (f (x)) = \sin (2 \times 2 θ) = \sin (2^{2} θ)

Extra \left or missing \right

Extra \left or missing \right

\Rightarrow \sin (2^{n} θ) = \sin θ

\Rightarrow 2^{n} θ = 2 k π + θ \Rightarrow θ = \frac{2 k π}{2^{n} - 1} o r

\Rightarrow 2^{n} θ = (2 m + 1) π - θ \Rightarrow θ = \frac{(2 m + 1) π}{2^{n} + 1}

θ = \frac{2 k π}{2^{n} - 1} ⩽ \frac{π}{2} \Rightarrow 0 ⩽ k ⩽ \frac{2^{n} - 1}{4}

θ = \frac{(2 m + 1) π}{2^{n} + 1} ⩽ \frac{π}{2} \Rightarrow 0 ⩽ m ⩽ \frac{2^{n} + 1}{4} - \frac{1}{2}

x = \sin \frac{2 k π}{2^{n} - 1} o r

x = \sin \frac{(2 k + 1) π}{2^{n} + 1}

w i t h 0 ⩽ k ⩽ 2^{n - 2} - 1 a n d n = 2020

t o t a l l y w e h a v e A = 2^{n - 1} = 2^{2019} r o o t s .

A (m o d 1000) = 288

Commented by HongKing last updated on 13/Jan/22

perfect my dear Sir thank you so much

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

G_{S \underset{!}{I} R}^{R^{E} A} T