Fie functia \( f:[0,1]\rightarrow R \) derivabila cu derivata continua astfel incat \( \int_0^1(f^{\prime}(x))^{2}dx \le 2 \int_0^1f(x)dx \). Sa se determine \( f \) stiind ca \( f(1)=-\frac{1}{6} \).
Radu Gologan
ONM 2009, problema 1
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
bogdanl_yex
- Pitagora
- Posts: 91
- Joined: Thu Jan 31, 2008 9:58 pm
- Location: Bucuresti
ONM 2009, problema 1
"Don't worry about your difficulties in mathematics; I can assure you that mine are still greater"(Albert Einstein)
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Indicatie:
Sa cauta \( a \) real astfel incat \( \int_0^1(f^{\prime}(x)+ax)^2dx\le 0 \)
Alte probleme de acest fel aici.
Sa cauta \( a \) real astfel incat \( \int_0^1(f^{\prime}(x)+ax)^2dx\le 0 \)
Alte probleme de acest fel aici.
-
Laurentiu Tucaa
- Thales
- Posts: 145
- Joined: Sun Mar 22, 2009 6:22 pm
- Location: Pitesti
Exista si o solutie diferita de ceea ce e in barem :
Avem:\( \int_0^1 x^2dx\int_0^1(f^{\prime}(x))^2dx\ge \(\int_0^1 xf^{\prime}(x)dx\)^2 \).(din CBS)
Tinand cont de ipoteza si integrand prin parti avem \( \int_0^1(f^{\prime}(x))^2dx\ge 3F^2(1)+F(1)+\frac{1}{12}\ge 3F^2(1)+\frac{1}{2}\int_0^1(f^{\prime}(x))^2dx+\frac{1}{12} \).
De aici rezulta \( \int_0^1(f^{\prime}(x))^2dx\ge 6F^2(1)+\frac{1}{6}\ge 2F(1) \),ultima inegalitate fiind AM-GM,tinand cont si de ipoteza.
Deci avem egalitate in CBS .In concluzie exista \( a\in\mathbb{R} \) a.i. \( f^{\prime}(x)=ax \).De aici totul vine de la sine pt ca obtinem \( f(x)=a\frac{x^2}{2}+c =>\frac{a}{2}+c=-\frac{1}{6} \)Introducem totul in inegalitatea din ipoteza care am demonstrat ca e egalitate si rezulta \( f(x)=-\frac{x^2}{2}+\frac{1}{3} \).
Avem:\( \int_0^1 x^2dx\int_0^1(f^{\prime}(x))^2dx\ge \(\int_0^1 xf^{\prime}(x)dx\)^2 \).(din CBS)
Tinand cont de ipoteza si integrand prin parti avem \( \int_0^1(f^{\prime}(x))^2dx\ge 3F^2(1)+F(1)+\frac{1}{12}\ge 3F^2(1)+\frac{1}{2}\int_0^1(f^{\prime}(x))^2dx+\frac{1}{12} \).
De aici rezulta \( \int_0^1(f^{\prime}(x))^2dx\ge 6F^2(1)+\frac{1}{6}\ge 2F(1) \),ultima inegalitate fiind AM-GM,tinand cont si de ipoteza.
Deci avem egalitate in CBS .In concluzie exista \( a\in\mathbb{R} \) a.i. \( f^{\prime}(x)=ax \).De aici totul vine de la sine pt ca obtinem \( f(x)=a\frac{x^2}{2}+c =>\frac{a}{2}+c=-\frac{1}{6} \)Introducem totul in inegalitatea din ipoteza care am demonstrat ca e egalitate si rezulta \( f(x)=-\frac{x^2}{2}+\frac{1}{3} \).