PUCPR 2019/2 - conjuntos

por brunohenrq Dom 15 Set 2024, 20:33

Considerando os conjuntos A={a1,a2,...,a9988} e B={b1,b1,...,b10013}, onde PUCPR 2019/2 - conjuntos AdbI+NMZYTdUAAAAAElFTkSuQmCC

são números reais, admita que:

I. AxB é o conjunto de elementos distintos que contém todos os pares ordenados do tipo (a,b), onde "a" pertence a A e "b" pertence a B.
II. O conjunto AxB contém apenas pares ordenados do tipo (a,b), onde "a" pertence a A e "b" pertence a B.
III. A^2=AxA

IV. PUCPR 2019/2 - conjuntos W7oCCiC2M1D+R1AEHiJRQA5R4tkdBOCjOPHK1eu2E9B34MCLUbp2bOn8tCgZ47SixgffkppQtFwBZ4Hb4W+hAv8cNf+18dRhWjorJnHHmCKbB1MAolrv6BAyyA4GOWTSP6TQjahZMGhIlmbcgyPhiMnpiz0J235t9Yi01Ip3RUhkr0EDdpIFkksY+FUNCz8ExQtAyYW3RLrPqCUVEk7IovGPv8p0GIwuCo80KMI0YBPwnNQGfnXLtoJ4p8DbcJm9p7GggU80MaCl7w9+ljAA62PObwvsWABD7Sx4CVvjz4W+B9NBPv2Ksv07QAAAABJRU5ErkJggg==

PUCPR 2019/2 - conjuntos W7oCCiC2M1D+R1AEHiJRQA5R4tkdBOCjOPHK1eu2E9B34MCLUbp2bOn8tCgZ47SixgffkppQtFwBZ4Hb4W+hAv8cNf+18dRhWjorJnHHmCKbB1MAolrv6BAyyA4GOWTSP6TQjahZMGhIlmbcgyPhiMnpiz0J235t9Yi01Ip3RUhkr0EDdpIFkksY+FUNCz8ExQtAyYW3RLrPqCUVEk7IovGPv8p0GIwuCo80KMI0YBPwnNQGfnXLtoJ4p8DbcJm9p7GggU80MaCl7w9+ljAA62PObwvsWABD7Sx4CVvjz4W+B9NBPv2Ksv07QAAAABJRU5ErkJggg==

V.

PUCPR 2019/2 - conjuntos N8AZrwYzXRErDwAAAAASUVORK5CYII=

VI.

PUCPR 2019/2 - conjuntos 7WH8Gg8CakQhMDhZgoGAB7hdITEx6wSlOyo9EoWB9mQPkEz1Gn8HaB3vHOOzCSFdxidwClDSkPKv7Ao0YEEiBOkT7cyKQ+ojdYHKm5mi3Szj6UQtt3rxZ60PbZAQaJ+aAZAuF5ZKDTW2+WLyHYsOYBVIsOjaa1xQHUjR7JwXpFgdSCnJWNKsaB1I0eycF6RYHUgpyVjSrGgdSNHsnBen2L0p96OdL0HMkAAAAAElFTkSuQmCC

Se a51=113, b1=17 e a2-a1=a3-a2=...=a9988-a9987=b2/b1=b3/b2=...=b10013/b10012=2, entao, eh correto afirmar que

PUCPR 2019/2 - conjuntos Sm0KAgWBgkBBoCBQEOgSgUJmugSsHF4QKAgUBAoCBYGCwPhCoJCZ8dUfpTYFgYJAQaAgUBAoCHSJwP8H2z4QZRbBdEEAAAAASUVORK5CYII=

PUCPR 2019/2 - conjuntos Sm0KAgWBgkBBoCBQEOgSgUJmugSsHF4QKAgUBAoCBYGCwPhCoJCZ8dUfpTYFgYJAQaAgUBAoCHSJwP8H2z4QZRbBdEEAAAAASUVORK5CYII=

gabarito: letra C

por **Elcioschin** Seg 16 Set 2024, 09:32

Note que:

1) Os elementos de A formam uma PA de razão r = 2, a51 = 113 e n = 9988

a51 = a1 + 50.r --> 113 = a1 + 50.2 ---> a1 = 13

a9998 = a1 + 9997.2 ---> a9988 = 13+ 19976 ---> a19988 = 19989

PA ---> 13, 15, 17 ............... 113, .................., 19989

2) Os elementos de B formam uma PG com b1 = 17, razão q = 2 e n' = 10013

b2013 = b1.q²⁰¹²= 17.2²⁰¹²

PG ---> 17, 34, 68, 136, .............., b10013

Encontre os termos comuns entre Ae B e complete

PUCPR 2019/2 - conjuntos

PUCPR 2019/2 - conjuntos

Re: PUCPR 2019/2 - conjuntos

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