5、在空间四边形ABCD的边AB、BC、CD、DA上分别取E、F、G、H四点，如果EF∩HG＝M，则点M(　 ).

(A)一定在AC上　　　　　　　 　　　　　　　(B)一定在BD上

(C)可能在AC上，也可能在BD上 　　　　　　(D)既不在AC上，也不在BD上

4、已知M表示平面，a、b表示直线，给出下面四个命题：

①若a//b，a⊥M，则b⊥M；　　　　　　　　　 ②若a⊥M，b⊥M，则a//b；

③若a⊥M，a⊥b，则b//M；　　　　　　　　　 ④若a//M ，a⊥b，则b⊥M.

其中正确的命题的选项是(　 ).

(A) ①② 　　　　(B) ①②③ 　　　　(C) ②③④ 　　　　(D) ①②④

3、已知直线a平面α，则直线b//a是b//α的(　 ).

(A)充分不必要条件　　　 (B)必要不充分条件

(C)充要条件　　　　　　 (D)既不充分又不必要条件

2、若a、b、c为三条直线，且a⊥c，b⊥c，则a、b位置关系是(　 ).

(A) 相交　 (B)平行　 　(C)异面　 　(D)相交、平行或异面

1、一条直线和梯形两腰垂直，则这条直线和底边的位置关系是(　 ).

(A)相交不垂直　 (B)平行　 (C)垂直　 (D)不确定

21．解：

(Ⅰ)．·································· 2分

当()时，，即；

当()时，，即．

因此在每一个区间()是增函数，

在每一个区间()是减函数．·································· 6分

(Ⅱ)令，则

．

故当时，．

又，所以当时，，即．···························· 9分

当时，令，则．

故当时，．

因此在上单调增加．

故当时，，

即．

于是，当时，．

当时，有．

因此，的取值范围是．　　 12分

21．(Ⅰ)解：依题设得椭圆的方程为，

直线的方程分别为，．············································ 2分

如图，设，其中，

且满足方程，

故．①

由知，得；

由在上知，得．

所以，化简得，

解得或．···································································································· 6分

(Ⅱ)解法一：根据点到直线的距离公式和①式知，点到的距离分别为，

．································································ 9分

又，所以四边形的面积为

，

当，即当时，上式取等号．所以的最大值为．····························· 12分

解法二：由题设，，．

设，，由①得，，

故四边形的面积为

···················································································································· 9分

，

当时，上式取等号．所以的最大值为．·············································· 12分

20．解：

(Ⅰ)依题意，，即，

由此得．··················································································· 4分

因此，所求通项公式为

，．①········································································ 6分

(Ⅱ)由①知，，

于是，当时，

，

，

当时，

．又．

综上，所求的的取值范围是．··································································· 12分

18．解法一：

依题设知，．

(Ⅰ)连结交于点，则．

由三垂线定理知，．····················································································· 3分

在平面内，连结交于点，

由于，

故，，

与互余．

于是．

与平面内两条相交直线都垂直，

所以平面．······························································································· 6分

(Ⅱ)作，垂足为，连结．由三垂线定理知，

故是二面角的平面角．································································· 8分

，

，．

，．

又，．

．

所以二面角的大小为．·························································· 12分

解法二：

以为坐标原点，射线为轴的正半轴，

建立如图所示直角坐标系．

依题设，．

，

．················································································ 3分

(Ⅰ)因为，，

故，．

又，

所以平面．································································································ 6分

(Ⅱ)设向量是平面的法向量，则

，．

故，．

令，则，，．······························································ 9分

等于二面角的平面角，

．

所以二面角的大小为．························································· 12分

17．解：

各投保人是否出险互相独立，且出险的概率都是，记投保的10 000人中出险的人数为，

则．

(Ⅰ)记表示事件：保险公司为该险种至少支付10 000元赔偿金，则发生当且仅当，　　　　 2分

，又，

故．·············································································································· 5分

(Ⅱ)该险种总收入为元，支出是赔偿金总额与成本的和．

支出　　　　　 ，

盈利　　　　　 ，

盈利的期望为　 ，················································· 9分

由知，，

．

(元)．

故每位投保人应交纳的最低保费为15元．··································································· 12分