内容正文:
$雅安市高中2018级第三次诊断性考试
数学(文科)参考答案及评分标准
一.选择题
1.C 2.A 3.D 4.B 5.C 6.D 7.C 8.D 9.B 10.C 11.B 12.A
二.填空题
13.1 14. 15. 5 16. ②③④
3、 解答题
17解:∵是与的等差中项
∴=
∴ ∴........................................................3分
∵∴∴ ∴ ....................6分
(2) ∵
∴........................................9分
..................................11分
∴.............................................................................................................................12分
18解:(1)由已知可得,40岁及以下采用乘坐成雅高铁出行的有
人··························································1分
40岁及以下
40岁上
合计
乘成雅高铁
40
10
50
不乘成雅高铁
20
30
50
合计
60
40
100
列联表如表:
········································································4分
由列联表中的数据计算可得的观测值
····································6分
由于,故有的把握认为“采用乘坐成雅高铁出行与年龄有关”.7分
(2)采用分层抽样的方法,从“岁(含)以下”的人中抽取人,记为1.2.3,
从“岁以上”的人中抽取人,记为a.b. ··············8分
则基本事件为(1,2),(1,3),(1,a),(1,b),(2,3),(2,a),(2,b),(3,a),(3,b),(a,b),共10个.··································································10分
符合条件的共6种,故抽到2人中恰有一人为40岁以上的概率为6/10=0.6.·········12分
19.
解(1)由题意,可知在等腰梯形中,,
∵,分别为,的中点,
∴,.
∴折叠后,,.
∵,∴平面. ···································4分
又平面,∴. ······································6分
(2)易知,.
∵,∴.
又,∴四边形为平行四边形.
∴,故.
∵平面平面,平面平面,且,
∴平面. ·······························································9分
∴
.
即三棱锥的体积为. ···········································12分
20.解:(1)①,
且过点,②
③
由①②③解得:,
椭圆的标准方程,..........................................4分
(2)(i)若的斜率不存在,则
此时.........................................5分
(ii)若的斜率存在,设,设的方程为:,
,.........................6分
由韦达定理得:...........................7分
则:,......................................