题目内容
设f(n)=2n+1(n∈N),P={1,2,3,4,5},Q={3,4,5,6,7},记(A) {0,3} (B){1,2} (C) (3,4,5) (D){1,2,6,7}
A
解析:P={n∈N|f(n)∈P}={0,1,2},
={n∈N|f(n)∈Q}={1,2,3},则
∩
N
={0},
∩
N
={3},因而(
∩
N
)∪(
∩
N
)={0,3}.
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题目内容
设f(n)=2n+1(n∈N),P={1,2,3,4,5},Q={3,4,5,6,7},记(A) {0,3} (B){1,2} (C) (3,4,5) (D){1,2,6,7}
A
解析:P={n∈N|f(n)∈P}={0,1,2},
={n∈N|f(n)∈Q}={1,2,3},则
∩
N
={0},
∩
N
={3},因而(
∩
N
)∪(
∩
N
)={0,3}.