IIT JEE 2011 Solution1
21
Quest˜ ao do IIT-JEE-2011 Recorrˆ encia em Fun¸ c˜ ao Quadr´ atica Prof. Fabiano Ferreira Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
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IIT JEE 2011 Solution1
Transcript of IIT JEE 2011 Solution1
Questao do IIT-JEE-2011
Recorrencia em Funcao Quadratica
Prof. Fabiano Ferreira
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011
Sumario
1 Questao do IIT-JEE-2011
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011
Agradecimento
Rodrigo Carlos Silva de Lima
Ao Renji Rodrigo, sem o qual este vıdeo nesta forma deapresentacao nao seria possıvel.
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011
Agradecimento
Rodrigo Carlos Silva de Lima
Ao Renji Rodrigo, sem o qual este vıdeo nesta forma deapresentacao nao seria possıvel.
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011
Agradecimento
Rodrigo Carlos Silva de Lima
Ao Renji Rodrigo, sem o qual este vıdeo nesta forma deapresentacao nao seria possıvel.
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Introducao
Objetivo desta apresentacao e desenvolver a primeira solucaode uma questao da prova do IIT-JEE-2011, envolvendoRecorrencia em Funcao Quadratica.
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Enunciado
Sejam r1 e r2 raızes de x2 − 6x − 2 = 0, com r1 > r2. Sexn = rn1 − rn2 , calcule
m =x10 − 2x8
2x9.
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Vejamos uma solucao elementar direta da questao.
m =x10 − 2x8
2x9=
r101 − r102 − 2(r81 − r82 )
2(r91 − r92 )=
=r101 − 2r81 − (r102 − 2r82 )
2(r91 − r92 )=
=r81 (r21 − 2) − r82 (r22 − 2)
2(r91 − r92 )
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Vejamos uma solucao elementar direta da questao.
m =x10 − 2x8
2x9=
r101 − r102 − 2(r81 − r82 )
2(r91 − r92 )=
=r101 − 2r81 − (r102 − 2r82 )
2(r91 − r92 )=
=r81 (r21 − 2) − r82 (r22 − 2)
2(r91 − r92 )
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Vejamos uma solucao elementar direta da questao.
m =x10 − 2x8
2x9=
r101 − r102 − 2(r81 − r82 )
2(r91 − r92 )=
=r101 − 2r81 − (r102 − 2r82 )
2(r91 − r92 )=
=r81 (r21 − 2) − r82 (r22 − 2)
2(r91 − r92 )
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Vejamos uma solucao elementar direta da questao.
m =x10 − 2x8
2x9=
r101 − r102 − 2(r81 − r82 )
2(r91 − r92 )=
=r101 − 2r81 − (r102 − 2r82 )
2(r91 − r92 )=
=r81 (r21 − 2) − r82 (r22 − 2)
2(r91 − r92 )
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Vejamos uma solucao elementar direta da questao.
m =x10 − 2x8
2x9=
r101 − r102 − 2(r81 − r82 )
2(r91 − r92 )=
=r101 − 2r81 − (r102 − 2r82 )
2(r91 − r92 )=
=r81 (r21 − 2) − r82 (r22 − 2)
2(r91 − r92 )
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0.
Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2=
3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3
∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
Questao do IIT-JEE-2011 Solucao 1
Solucao 1
Agora usamos r21 − 2 = 6r1 e r22 − 2 = 6r2, poisx2 − 6x − 2 = 0. Substituindo na expressao temos:
m =r81
=6r1︷ ︸︸ ︷(r21 − 2)−r82
=6r2︷ ︸︸ ︷(r22 − 2)
2(r91 − r92 )=
=r81 (6r1) − r82 (6r2)
2(r91 − r92 )=
6r91 − 6r922(r91 − r92 )
=
=6(r91 − r92 )
2(r91 − r92 )=
6
2= 3 ∴ m = 3
Prof. Fabiano Ferreira IIT-JEE-Instituto Indiano de Tecnologia-2011
