河南省许昌市普通高中2017届高三下学期高中毕业班高考适应性测试(二模)数学(理)试题含答案

2017年河南省普通高中毕业班高考适应性测试

理科数学

第Ⅰ卷（选择题 共60分）

一、选择题：本大题共12小题，每小题5分，共60分.在每个小题给出的四个选项中，有且只有一项符合题目要求.

1.已知集合{}(){}

2|230|lg 20A x x x B x x =-->=-≤，则()R C A B = A. ()1,12- B. ()2,3 C. (]2,3 D.[]1,12-

2.欧拉（Leonhard Euler,国籍瑞士）是科学史上最多产的一位杰出的数学家，他发明的公式cos sin ix e x i x =+（i 为虚数单位），将指数函数的定义域扩大到复数，建立了三角函数和指数函数的关系，这个公式在复变函数理论中占有非常重要的地位，被誉为“数学中的天桥”.根据此公式可知，表示的复数i e π-在复平面内位于

A. 第一象限

B. 第二象限

C. 第三象限

D.第四象限

3.下列命题中，正确的是 A. 0003,sin cos 2x R x x ?∈+=

B. 0x ?≥且x R ∈，22x x >

C. 已知,a b 为实数，则2,2a b >>是4ab >的充分条件

D. 已知,a b 为实数，则0a b +=的充要条件是1a b

=- 4.已知圆22

:4O x y +=（O 为坐标原点）经过椭圆()22

22:10x y C a b a b +=>>的短轴端点和两个焦点，则椭圆C 的标准方程为 A. 22142x y += B. 22184x y += C.221164x y += D. 22

13216

x y +=

5.已知等差数列{}n a 满足121,6n n a a a +=-=，则11a 等于

A. 31

B. 32

C. 61

D.62

6.某几何体的三视图如图所示，则该几何体的体积为

A.

7.已知函数()132221

x x x f x +++=+的最大值为M ，最小值为m ，则M m +等于 A. 0 B. 2 C. 4 D. 8

8.如图所示的程序框图的算法思路来源于我国古代数学名著《九

章算术》中的“更相减损术”，执行该程序框图，若输入,a b 的

值分别为21,28，则输出a 的值为

A. 14

B. 7

C. 1

D. 0

9.已知函数1ln y x x =++在点()1,2A 处的切线为l ，若l 与二

次函数()2

21y ax a x =+++的图象也相切，则实数a 的取值范围为

A. 12

B. 8

C. 0

D.4

10.已知ABC ?的三个顶点坐标为

()()()0,1,1,0,0,2,A B C O -为坐标原点，动点M 满足

1CM = ，则OA OB OM ++ 的最大值是

1111

11.已知双曲线()22

22:10,0x y C a b a b

-=>>的左、右焦点分别为12,,F F O 为坐标原点，点P 是双曲线在第一象限内的点，直线2,PO PF 分别交双曲线C 的左、右支于另一点M,N ，若122PF PF =，且2120MF N ∠= ，则双曲线的离心率为

A. 312.定义在R 上的函数()f x ，当[]0,2x ∈时，()()411f x x =--，且对任意实数

()122,22,2n n x n N n +*??∈--∈≥??，都有()1

122x f x f ??=- ???

.若()()log a g x f x x =-有且仅有三个零点，则a 的取值范围是

A. []2,10

B.

C. ()2,10

D.[)2,10

第Ⅱ卷（非选择题 共90分）

二、填空题：本大题共4小题，每小题5分，共20分.

13.已知实数,x y 满足条件2420x x y x y m ≥??+≤??-++≥?

，若目标函数2z x y =+的最小值为3，则其最

大值为 .

14.

设二项式6

x ? ?展开式中的常数项为a ，则20cos 5ax dx π?的值为 . 15.已知A,B,C 是球O

的球面上三点，且3,AB AC BC D ===为该球面上的动点，球心O 到平面ABC 的距离为球半径的一半，则三棱锥D ABC -体积的最大值为 .

16.已知函数()212n n n f x a x a x a x =+++ ，且()()11,.n

n f n n N *-=-∈设函数(),,2n a n g n n g n ??=??? ????

?为奇数为偶数，若()24,n n b g n N *=+∈，则数列{}n b 的前()2n n ≥项和n S = .

三、解答题：本大题共6小题，共70分.解答应写出必要的文字说明或推理、验算过程.

17.（本题满分12分）已知向量(

)()

2cos ,sin ,cos a x x b x x == ，函数() 1.f x a b =?-

（1）求函数()f x 的单调递减区间；

（2）在锐角ABC ?中，内角A,B,C

的对边分别为，222

tan B a c b =

+-对任意满足条件的A,求()f A 的取值范围.

18.（本题满分12分）某品牌汽车的4S 店，对最近100份分期付款购车情况进行统计，统计情况如下表所示.已知分9期付款的频率为0.4,；该店经销一辆该品牌汽车，若顾客分3期付款，其利润为1万元；分6期或9期付款，其利润为2万元；分12期付款，其利润为3万元.

（1）若以上表计算出的频率近似替代概率，从该店采用分期付款购车的顾客（数量较大）中随机抽取3为顾客，求事件A:“至多有1位采用分6期付款”的概率();P A

（2）按分层抽样的方式从这100为顾客中抽取5人，再从抽取的5人中随机抽取3人，记该店在这3人身上赚取的总利润为随机变量η，求η的分布列和数学期望()E η.

19.（本题满分12分）如图所示，已知长方体ABCD 中，2AB AD M ==为DC 的中点.将ADM ?沿AM 折起，使得.AD BM ⊥

（1）求证：平面ADM ⊥平面ABCM ；

（2）是否存在满足()01BE tBD t =<< 的点E ，使得二面角E AM D --为大小为

4

π，？若存在，求出相应的实数t ；若不存在，请说明理由.

20.（本题满分12分）设抛物线的顶点在坐标原点，焦点F 在y 轴上，过点F 的直线交抛物线于A,B 两点，线段AB 的长度为8，AB 的中点到x 轴的距离为3.

（1）求抛物线的标准方程；

（2）设直线m 在y 轴上的截距为6，且与抛物线交于P,Q 两点，连结QF 并延长交抛物线的准线于点R,当直线PR 恰与抛物线相切时，求直线m 的方程.

21.（本题满分12分）已知函数()()()ln 1.1ax f x x a R x

=+-

∈- （1）当1a =时，求函数()f x 的单调区间；

（2）若11x -<<时，均有()0f x ≤成立，求实数a 的取值范围.

请考生在第22、23两题中任选一题作答，如果两题都做，则按照所做的第一题给分；作答时，请用2B 铅笔将答题卡上相应的题号涂黑。

22.（本题满分10分）选修4-4：参数方程与极坐标系

在平面直角坐标系xoy 中，曲线1C

的参数方程为2cos x y θθ=???=??

（θ为参数），以坐标原点O 为极点，x 轴的正半轴为极轴，与直角坐标系xoy 取相同的单位长度建立极坐标系，曲线2C 的极坐标方程为2cos 4sin .ρθθ=-

（1）化曲线1C ，2C 的方程为普通方程，并说明它们分别表示什么曲线；

（2）设曲线2C 与x 轴的一个交点的坐标为()(),00P m m >，经过点P 作斜率为1的直线，l 交曲线2C 于A,B 两点，求线段AB 的长.

23.（本题满分10分）选修4-5：不等式选讲

已知函数()1212f x x x =-++

的最小值为.m （1）求m 的值；

（2）若,,a b c 是正实数，且a b c m ++=，求证：()

22223.a b c ab bc ca abc ++≥++-

普通高考(天津卷)适应性测试数学试题

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2020届北京市高考适应性测试数学试题(解析版)

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北京市海淀区2017届高三英语二模试题答案(最新整理)

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