Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 191623 by Shrinava last updated on 27/Apr/23

	Answered by mehdee42 last updated on 27/Apr/23

	$l e t : f (x) = a x + b$ $f (f (x)) + f (x) = - x \Rightarrow (a^{2} + a) x + a b + 2 b = - x \Rightarrow a = \frac{- 1 + \sqrt{3} i}{2} = e^{\frac{2 π}{3} i} & b = 0 \Rightarrow f (x) = e^{\frac{2 π}{3} i} x$ $f (f (f (\frac{1}{2 + c o s x}))) = \frac{1}{2 + c o s x}$ $\Rightarrow Ω = \int_{0}^{π} \frac{d x}{2 + c o s x} = \int_{0}^{π} \frac{1 + t a n \frac{x}{2}}{{t a n}^{2} \frac{x}{2} + 3} d x$ $l e t : t a n \frac{x}{2} = u \Rightarrow Ω = 2 \int_{0}^{\infty} \frac{d u}{u^{2} + 3} = {[\frac{2}{\sqrt{3}} {t a n}^{- 1} (\frac{u}{\sqrt{3}})]}_{0}^{\infty}$ $= \frac{π}{\sqrt{3}}$
	Answered by witcher3 last updated on 28/Apr/23
	$f(f(0))+f(0)=0;f(0)=a f(f(x))+f(x)+x=0 (fof(x)+f(x)+x)o.f(x)−{fof(x)+f(x)+x}=0of(x)−0=0 fofof(x)+fof(x)+f(x)−fof(x)−f(x)−x=0 ⇒fofof(x)=x Ω=∫_0 ^π (1/(2+cos(x)))=∫_0 ^π (dx/(2cos^2 ((x/2))+1)) =2∫_0 ^(π/2) (dy/(cos^2 (y)(3+tg^2 (y)))) =(2/( (√3))).[tan^(−1) (((tg(y))/( (√3))))]_0 ^(π/2) =(π/( (√3)))$
	$f (f (0)) + f (0) = 0; f (0) = a$ $f (f (x)) + f (x) + x = 0$ $(fof (x) + f (x) + x) o . f (x) - {fof (x) + f (x) + x} = 0 of (x) - 0 = 0$ $fofof (x) + fof (x) + f (x) - fof (x) - f (x) - x = 0$ $\Rightarrow fofof (x) = x$ $Ω = \int_{0}^{π} \frac{1}{2 + \cos (x)} = \int_{0}^{π} \frac{dx}{{2 \cos}^{2} (\frac{x}{2}) + 1}$ $= 2 \int_{0}^{\frac{π}{2}} \frac{dy}{\cos^{2} (y) (3 + {tg}^{2} (y))}$ $= \frac{2}{\sqrt{3}} . {[\tan^{- 1} (\frac{tg (y)}{\sqrt{3}})]}_{0}^{\frac{π}{2}} = \frac{π}{\sqrt{3}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum