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17.已知條件p:A={x|x2+ax+1≤0},條件q:B={x|x2-3x+2≤0},若p是q的充分不必要條件,求實數(shù)a的取值范圍.
集合與簡易邏輯參考答案
1.D (驗證)若a=-2,則A={1,7,9} I A={3,5}不合條件,若a=2,則A={1,3,9}, I A={5,7},滿足條件;若a=8則A={1,3,9},仍符合條件,故選D.
2.B (直接計算)由x2-x>0且x≥1得x>1,故選B.
3.A (驗證) I A={-2,-1,-}, I B={-1,-,1,3},故選A.
4.DM=(-∞,m),N =[-1,+∞),由m<-1選D.
5.D(檢驗)若m=-1則B={1}不合條件,若m=0則B= 符合條件,故選D.
6.A(逐一檢驗)選A.
7.C 構(gòu)造函數(shù)f (x)=(1-m2)x2+2mx-1, f (0)=-1,開口向上,由f (1)<0得1-m2+2m-1<0m>2或m<0.
8.C 當(dāng)A2B2C2≠0時,l1∥l2.
9.A 因丁丙乙甲,故丁甲(傳遞性)
10.C 若Δ=0則4-4a=0,a=1滿足條件,當(dāng)Δ>0時,4-4a>0a<1.綜合即得.
11.(例舉)M={1,5}, M={2,4}, M={3}, M={1,3,5}, M={2,3,4}, M={1,2,4,5}, M={1,2,3,4,5}7個.
12.a21+a22+a23+…+a22004≠0(偶數(shù)次冪之和不等于0).
13.a=-2(畫圖即知)
14.必要
15.證明:①設(shè)t∈A,則存在a、b∈Z,使得t=6a+8b=2(3a+4b)
∵3a+4b∈Z,∴t∈B即aB.
②設(shè)t∈B,則存在m∈Z使得x=2m=6(-5m)+8(4m).
∵-5m∈Z,4m∈Z,∴x∈A即BA,由①②知A=B.
16.解:∵ S A={0},∴0∈S但0A,∴x3+3x2+2x=0故x=0,-1,-2
當(dāng)x=0時,|2x-1|=1, A中已有元素1,
當(dāng)x=-1時,|2x-1|=3,3∈S;
當(dāng)x=-2時,|2x-1|=5,但5S
故實數(shù)x的值存在,它只能是-1.
17.由條件知B=[1,2],∵AB且A≠B,或者A= , 故方程x2+ax+1=0無實根或者兩根滿足:1≤x1,x2≤2,當(dāng)Δ<0時,a 2-4<0-2<a<2,當(dāng)時,a=-2,故a的取值范圍是[-2,2].
18.證明:(1)充分性:∵m=x2-y2=(x+y)(x-y)且x∈Z,y∈Z,而(x+y)與(x-y)具有相同的奇偶性.
故當(dāng)x+y與x-y都為偶數(shù)時,m是4的倍數(shù),即存在k∈Z,使m=4k;
當(dāng)(x+y)與(x-y)都為奇數(shù)時,則其乘積仍為奇數(shù),即存在k∈Z,使m=2k+1,∴pq.
(2)必要性:當(dāng)m=4k時m=(k+1)2-(k-1)2,故存在整數(shù)x=k+1, y=k-1使m=x2-y2;
當(dāng)m=2k+1時,則m=(k+1)2-k2=x2-y2,∴qp.