海淀区高三年级第二学期期末练习
数学(理科)2018.5
第一部分(选择题共40分)
一、选择题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要
求的一项.
(1)已知全集集合,则= (A)(B)(C)(D)
(2)已知复数在复平面上对应的点为,则
(A)是实数(B)是纯虚数
(C)是实数(D)是纯虚数
(3)已知,则
(A)(B)
(C)(D)
(4)若直线是圆的一条对称轴,则的值为(A)(B)(C)(D)
(5)设曲线是双曲线,则“的方程为”是“的渐近线方程为”
的
(A)充分而不必要条件(B)必要而不充分条件
(C)充分必要条件(D)既不充分也不必要条件
(6)关于函数,下列说法错误的是
(A)是奇函数
(B)不是的极值点
(C)在上有且仅有个零点
(D)的值域是
(7) 已知某算法的程序框图如图所示,则该算法的功能是
(A )求首项为1,公比为2的等比数列的前2017项的和 (B )求首项为1,公比为2的等比数列的前2018项的和 (C )求首项为1,公比为4的等比数列的前1009项的和 (D )求首项为1,公比为4的等比数列的前1010项的和
(8)已知集合
,集合满足
① 每个集合都恰有个元素 ② .
集合中元素的最大值与最小值之和称为集合
的特征数,记为(),
则
的值不可能为( ). (A )
(B )
(C )
(D )
第二部分 (非选择题 共110分)
二、填空题共6小题,每小题5分,共30分。 (9)极坐标系中,点到直线
的距离为________. (10)在
的二项展开式中,
的系数为 . (11)已知平面向量,的夹角为
,且满足
,
,则
,
.
(12)在
中,
,则 . (13)能够使得命题“曲线上存在四个点
,
,
,
满足四边
形
是正方形”为真命题的一个实数的值为 .
(14)如图,棱长为2的正方体中,
是
棱的中点,点
在侧面
内,若
垂直于
,则的面积的最小值为_________.
开始S = 0,n = 1
S = S + 2n - 1n = n + 2
n > 2018输出 S 结束
是
否
B
C
D
A 1
B 1
C 1
D 1
M
P
三、解答题共6小题,共80分.解答应写出文字说明,演算步骤或证明过程. (15)(本小题13分)
如图,已知函数
在一个周期内的图象经过
,
,
三点. (Ⅰ)写出,
,
的值; (Ⅱ)若
,且
,求
的值.
16. (本小题共13分)
某中学为了解高二年级中华传统文化经典阅读的整体情况,从高二年级随机抽取10名学生进行了两轮测试,并把两轮测试成绩的平均分作为该名学生的考核成绩.记录的数据如下:
1号 2号 3号 4号 5号 6号 7号 8号 9号 10号 第一轮测试成绩 96 89 88 88 92 90 87 90 92 90 第二轮测试成绩
9
96
92
89
92
(Ⅰ)从该校高二年级随机选取一名学生,试估计这名学生考核成绩大于等于90分的概率; (Ⅱ)从考核成绩大于等于90分的学生中再随机抽取两名同学,求这两名同学两轮测试成绩均大于等于90分的概率;
(Ⅲ)记抽取的10名学生第一轮测试成绩的平均数和方差分别为,
,考核成绩的平均
数和方差分别为
,
,试比较
与
,
与
的大小. (只需写出结论)
17. (本小题共14分)
如图,在三棱柱中,,
⊥
平面,
,
,
分别是
,
的
中点. (Ⅰ)证明:
(Ⅱ)证明:平面;
(Ⅲ)求
与平面
所成角的正弦值.
x
y
D
C
B O A
C 1A 1
C
B 1
B
D
E
18. (本小题共14分)
已知椭圆:,为右焦点,圆:,为椭圆上一点
,且位于第一象限,过点作与圆相切于点,使得点,在两侧. (Ⅰ)求椭圆的焦距及离心率;
(Ⅱ)求四边形面积的最大值.
19. (本小题共13分)
已知函数()
(Ⅰ)求的极值;
(Ⅱ)当时,设.求证:曲线存在两条斜率为
且不重合的切线.
20. (本小题共13分)
如果数列满足“对任意正整数,,都存在正整数,使得”,则称数列具有“性质P”.已知数列是无穷项的等差数列,公差为.
(Ⅰ)若,公差,判断数列是否具有“性质P”,并说明理由;
(Ⅱ)若数列具有“性质P”,求证:且;
(Ⅲ)若数列具有“性质P”,且存在正整数,使得,这样的数列共有
多少个?并说明理由
海淀区高三年级第二学期期末练习参考答案及评分标准
数学(理科)2018.5
第一部分(选择题共40分)
一、选择题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要
求的一项.
1 2 3 4 5 6 7 8
B C D B A C C A
第二部分(非选择题共110分)
二、填空题共6小题,每小题5分,共30分.
(9)1 (10)10
(11)1;(12)
(13)答案不唯一,或的任意实数(14)
三、解答题共6小题,共80分.解答应写出文字说明,演算步骤或证明过程.
(15)(本小题13分)
解:(Ⅰ),,.·····································································7分(Ⅱ)由(Ⅰ)得,.
因为,所以.·········································· 8分
因为,所以. ································· 9分
所以, ··································································· 11分
所以, ········································································· 12分
所以.····················································· 13分16. (本小题共13分)
解:(Ⅰ)这10名学生的考核成绩(单位:分)分别为:
93,89.5,89,88,90,88.5,91.5,91,90.5,91.
其中大于等于90分的有1号、5号、7号、8号、9号、10号,共6人. ·········· 1分
所以样本中学生考核成绩大于等于90分的频率为:
, ································································3分从该校高二年级随机选取一名学生,估计这名学生考核成绩大于等于90分的概率为0 .6. ··································································································································4分
(Ⅱ)设事件:从上述考核成绩大于等于90分的学生中再随机抽取两名同学,这两名同学两轮测试成绩均大于等于90分. ························································ 5分
由(Ⅰ)知,上述考核成绩大于等于90分的学生共6人,其中两轮测试成绩均大于等于90分的学生有1号,8号,10号,共3人. ······················································ 6分
所以,. ························································· 9分(Ⅲ),. ··································································· 13分
17. (本小题共14分)
解:(Ⅰ)因为⊥平面,平面,
所以. ········································································· 1分
因为,,,平面,
所以平面. ······························································ 3分
因为平面,
所以.······································································· 4分(Ⅱ)法一:取的中点,连接、.
因为、分别是、的中点,
所以ME∥,且ME.·················································· 5分
在三棱柱中,,且,
所以ME∥AD,且ME=AD,
所以四边形ADEM是平行四边形,················6分
所以DE∥AM. ········································7分
又平面,平面,
所以平面. ··························9分
注:与此法类似,还可取AB的中点M,连接MD、MB1.
法二:取AB的中点,连接、.
因为D、分别是AC、AB的中点,
所以MD∥BC,且MD BC. ···················5分
在三棱柱中,,且,所以MD∥B1E,且MD=B1E,
A
C
1
A
1
B
1
B
D
E
M
A
C
1
A
1
B
1
D
E
M
A
C 1
A 1
C
B 1
B
D
E y
x
z
所以四边形B 1E DM 是平行四边形, ·············· 6分 所以DE ∥MB 1. ······································· 7分 又平面,
平面
,
所以平面. ·························· 9分 法三:取的中点
,连接
、
.
因为、分别是
、
的中点,
所以,. ···································································· 5分
在三棱柱中,,
,
因为
、
分别是
和
的中点, 所以,,
,
所以,四边形是平行四边形, ··········· 6分 所以,. ··································· 7分
又因为,
,
,
平面MDE ,BB 1,平面
,
所以,平面平面
. ·············· 8分
因为,平面,
所以,平面
. ······················· 9分 (Ⅲ)在三棱柱中,
,
因为,所以. 在平面内,过点作,
因为,平面,
所以,
平面
. ··························· 10分
建立空间直角坐标系C -xyz ,如图.则
,
,
,,
,
,
,
. ···························· 11分
设平面
的法向量为
,则 ,即
,
得
,令
,得
,故. ····························· 12分 设直线DE 与平面所成的角为θ,
则sin θ=
,
A
C 1
A 1
B 1
B D E
M
x
y
T
F
O
P
所以直线与平面所成角的正弦值为. ·························· 14分
18. (本小题共14分) 解:(Ⅰ)在椭圆
:
中,
,
,
所以, ····························································· 2分
故椭圆的焦距为
, ······················································ 3分
离心率. ··································································· 5分 (Ⅱ)法一:设(
,
),
则,故. ·················· 6分
所以,
所以
, ·································· 8分
. ··········· 9分
又,,故. ···················· 10分 因此
································ 11分
.
由,得,即,
所以, ·········································· 13分
当且仅当,即
,时等号成立. ················· 14分
(Ⅱ)法二:设(
), ······································ 6分
则,
所以
, ································································ 8分
. ········································· 9分
又,,故. ················ 10分 因此
························· 11分
,·········································· 13分当且仅当时,即,时等号成立.···················· 14分
19. (本小题共13分)
解:(Ⅰ)法一:,················· 1分令,得.······························································ 2分
①当时,与符号相同,
当变化时,,的变化情况如下表:
↘极小↗································································································ 4分
②当时,与符号相反,
当变化时,,的变化情况如下表:
↘极小↗································································································ 6分综上,在处取得极小值. ·································· 7分
法二:, ····························· 1分令,得.······························································ 2分令,则, ······································· 3分易知,故是上的增函数,
即是上的增函数. ················································· 4分所以,当变化时,,的变化情况如下表:
↘极小↗
因此,在处取得极小值. ·································· 7分(Ⅱ), ····································· 8分故. ······················································· 9分注意到,,,
所以,,,使得.
因此,曲线在点,处的切线斜率均为.
································································································ 11分
下面,只需证明曲线在点,处的切线不重合.
法一:曲线在点()处的切线方程为,即.假设曲线在点()处的切线重合,则.·································· 12分
法二:假设曲线在点(,)处的切线重合,则,整理得:. ····························· 12分法一:由,得,则
.
因为,故由可得.
而,,于是有,矛盾!
法二:令,则,且.
由(Ⅰ)知,当时,,故.
所以,在区间上单调递减,于是有,矛盾!
因此,曲线在点()处的切线不重合.········· 13分
20. (本小题13分)
解:(Ⅰ)若,公差,则数列不具有性质. ···················· 1分理由如下:
由题知,对于和,假设存在正整数k,使得,则有,解得,矛盾!所以对任意的,.··· 3分(Ⅱ)若数列具有“性质P”,则
①假设,,则对任意的,.
设,则,矛盾! ·············································· 4分
②假设,,则存在正整数,使得
设,,,…,,,,则,但数列中仅有项小于等于0,矛盾!···························································································· 6分
③假设,,则存在正整数,使得
设,,,…,,
,,则,但数列中仅有项大于等于0,矛盾! ·············································································· 8分综上,,.
(Ⅲ)设公差为的等差数列具有“性质P”,且存在正整数,使得.若,则为常数数列,此时恒成立,故对任意的正整数,