Question

2.813B ...am nevoie de ajutor. Multumesc!

Answer 1

$\displaystyle In~primul~rand~sa~observam~ca~f(x)=f(x+2\pi),~deci \\ \\ functia~data~este~periodica,~si~deci~isi~atinge~minimul~si~maximul. \\ \\ Atunci~punctele~in~care~se~atinge~maximul~se~afla~printre \\ \\ radacinile~derivatei. \\ \\ Sa~mai~observam~ca~M_n \ge f \left( \frac{\pi}{4} \right)\ \textgreater \ 0. \\ \\ f'(x)=n\sin^{n-1}x\cos^2x-\sin^{n+1}x= \\ \\ =\sin^{n-1}x(n\cos^2x-\sin^2x)= \\ \\ =\sin^{n-1}x(n-(n+1) \sin^2x).$
$\displaystyle Radacinile~derivatei~vor~fi~k \pi~(k \in \mathbb{Z})~precum~si~radacinile~ecuatiei \\ \\ n-(n+1)\sin^2x=0. ~(*)\\ \\ Pana~sa~studiem~ultima~ecuatie,~vom~analiza~primele~valori \\ \\ gasite,~adica~k \pi.~Avem~f(k \pi)=0,~deci~maximul~nu~se~realizeaza \\ \\ in~aceste~puncte. \\ \\ Revenind~la~(*):\sin^2x= \frac{n}{n+1},~deci~ \sin x \in \left \{ - \sqrt{\frac{n}{n+1}},\sqrt{\frac{n}{n+1}}\right \}. \\ \\ In~ambele~cazuri~|\cos x|= \sqrt{1- \sin^2x}= \frac{1}{\sqrt{n+1}}.$
$\displaystyle Deci~maximul~se~realizeaza~daca \left \{ {{\sin x=- \sqrt{\frac{n}{n+1}}} \atop {\cos x= \frac{1}{\sqrt{n+1}}}} \right. ~sau~ \\ \\ \left \{ {{\sin=\sqrt{ \frac{n}{n+1}}} \atop {\cos x= \frac{1}{\sqrt{n+1}}}} \right. . \\ \\ Pentru~aceste~valori~avem~f(x)=(-1)^n \frac{(\sqrt{n})^n}{(\sqrt{n+1})^{n+1}}~sau \\ \\ f(x)= \frac{(\sqrt{n})^n}{(\sqrt{n+1})^{n+1}}.~ \\ \\ Una~din~aceste~valori~este~M_n. ~Ele~au~acelasi~modul,~iar~M_n\ \textgreater \ 0,$
$\displaystyle deci~ \boxed{M_n=\frac{(\sqrt{n})^n}{(\sqrt{n+1})^{n+1}}}. \\ \\ Atunci~\lim_{n \to \infty} \sqrt{n}M_n= \lim_{n \to \infty} \left(\sqrt{\frac{n}{n+1}} \right)^{n+1}=\sqrt{\lim_{n \to \infty}\left(1- \frac{1}{n+1} \right)^{n+1}} \\ \\ =\frac{1}{\sqrt{e}}.$