2019-2020年高三上学期开学摸底测试理数试题解析(解析版)含解斩

一、选择题（本大题共12个小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.）

1.已知集合2

{|230}A x x x =--≤，{|ln(2)}B x y x ==-，则A

B =（ ）

A ．(1,3)

B ．(1,3]

C ． [1,2)-

D ．(1,2)- 【答案】C

考点：集合交集.

【易错点晴】集合的三要素是：确定性、互异性和无序性.研究一个集合，我们首先要看清楚它的研究对象，是实数还是点的坐标还是其它的一些元素，这是很关键的一步.第二步常常是解一元二次不等式，我们首先用十字相乘法分解因式，求得不等式的解集.在解分式不等式的过程中，要注意分母不能为零.元素与集合之间是属于和不属于的关系，集合与集合间有包含关系.

2.已知2016

z =（是虚数单位），则等于（ ） A ．-1 B ．1 C ．0 D ． 【答案】B 【解析】

试题分析：()2

2,1

i i =--=，即504

201641??==????

.

考点：复数概念及运算.

3.设变量,x y 满足约束条件220

22010

x y x y x y --≤??

-+≥??+-≥?

，则1y x s x -=+的取值范围是（ ）

A ．3[1,]4

B ．1[,1]2

C ．1[,2]2

D ．1[,1]2

-

【答案】D 【解析】

试题分析：画出可行域如下图所示，分别将,,A B C 三点坐标代入1

y x

s x -=

+，可得最小值为1

2

-，最大值为

.

考点：线性规划.

4.等比数列{}n a 中，182,4a a ==，函数128()()()()f x x x a x a x a =---，则'(0)f =

（ ）

A ．6

2 B ．92 C ．12

2 D ．15

2 【答案】

C

考点：等比数列的基本概念.

5.已知函数()sin()(0,||)2

f x x π

ω?ω?=+><

的最小正周期为π，且其图像向左平移

3

π个单位后得到函数()cos g x x ω=的图象，则函数()f x 的图象（ ） A ．关于直线12

x π

=对称 B ．关于直线512

x π

=

对称 C ．关于点(,0)12

π

对称 D ．关于点5(

,0)12

π

对称 【答案】C 【解析】

考点：三角函数图象与性质.

6. 已知边长为ABCD 中，60BAD ∠=，沿对角线BD 折成二面角A BD C --为120的四面体ABCD ，则四面体的外接球的表面积为（ ）

A ．25π

B ．26π

C ．27π

D ．28π 【答案】D 【解析】

试题分析：如图所示，设两三角形外心分别为23,O O ,球心为O ，1120AO C ∠=,故

132,OO OO ==OC ==28π.

考点：几何体外接球.

7.执行如图所示的程序框图，若输出的值为8，则判断框内可填入的条件是（ ）

A ．11?12S ≤

B ．

3?4S ≤ C ．25?24S ≤ D ．137

?120

S ≤

【答案】A

考点：算法.

8.某几何体的三视图如图所示，则该几何体的体积为（ ）

A C ．3

D

【答案】B 【解析】

试题分析：由三视图可知,该几何体是由正三棱柱截取一部分所得,故体积为

21122224

V =???=考点：三视图.

9.已知2

2

1

)a ex dx π-=?

，若2016220160122016(1)()ax b b x b x b x x R -=++++∈，

则

2016

12

22016

22

2

b b b +++

的值为（ ） A ．0 B ．-1 C ．1 D ． 【答案】B 【解析】

试题分析：()2

2

2

-21

1

1

11

)422

a ex dx ex dx ππ

π

π

π-=

=

-

=

??=?

?

?.即2016(12)x -.令0x =，得01b =，令1

2x =

，得2016

122

2016

011222

b b b +++

=-=-. 考点：定积分.

10.一个不透明的袋子装有4个完全相同的小球，球上分别标有数字为0,1,2,2，现甲从中摸出一个球后便放回，乙再从中摸出一个球，若摸出的球上数字大即获胜（若数字相同则为平局），则在甲获胜的条件下，乙摸1号球的概率为（ ） A ．516 B ．916 C ．1

5

D ．

25

【答案】D

考点：1.古典概型；2.条件概型.

11.已知直线980x y --=与曲线3

2

:3C y x px x =-+相交于,A B ，且曲线C 在,A B 处的切线平行，则实数p 的值为（ ）

A ．4

B ．4或-3

C ．-3或-1

D ．-3 【答案】B 【解析】

试题分析：'

2

323y x px =-+，设()()1122,,,A x y B x y ，切线平行，即斜率相等，即可令

221122323323x px x px m -+=-+=，12,x x 是方程23230x px m -+-=的两个根，则

1223

x x p +=

，下证线段

AB 的中点在曲线C

上，因为

323

23

111222332227

x px x x px x p p -++-+=-，而

32

31212122322227x x x x x x p p p +++????

-+?=- ? ?

????

，所以线段AB 的中点在曲线C 上，由1223x x p +=

知，线段的中点为111,8393

p p ??

??- ? ?????，所以381292727p p p -+=-，解得1,3,4p =--，经验证，1p =-时，不符合题意，故选B.

考点：导数与切线.

【思路点晴】本题考察利用导数研究曲线上某点的切线方程，求解该题的关键是利用AB 中点的坐标相等，关键是证明AB 中点在曲线C 上.求函数切线的步骤如下：第一先求函数的导数，然后求出在该点的导数，接着求出切点，然后利用点斜式()()()'

000y f x f

x x x -=-，即可

得到切线方程.读题时要注意是“在某点的切线”，还是“过某点的切线”. 12.数列{}n a 满足143

a =，*

11(1)()n n n a a a n N +-=-∈且12111

n n

S a a a =

+++

，则n S 的整数部

分的所有可能值构成的集合是（ ）

A ．{0,1,2}

B ．{0,1,2,3} C

．{1,2} D ．{0,2} 【答案】A

考点：数列.

【思路点晴】这个是递推数列求通项的问题，首先由11(1)n n n a a a +-=-两边取倒数，得

111111n n n a a a +-=--，累加得111111

3111

n n n S a a a ++=-=----，然后通过列举123413133,,3981

a a a ===

，和2

11(1)0,n n n n n a a a a a ++-=-≥≥，n a 为单调递增数列，判断出可以取0,1，由于11331

n n S a +=-

<-，故不能取，根据选项可有A 正确.

2019-2020年高三上学期开学摸底测试理数试题解析（解析版）含解斩 二、填空题（本大题共4小题，每题5分，满分20分．）

13.已知0

3

sin m xdx π

=?

，则二项式(23)m a b c +-的展开式中23m ab c -的系数为 .

【答案】6480-

考点：二项式定理.

14.已知等差数列{}n a 的前项和n S 满足350,5S S ==，数列2121

1

{}n n a a -+的前2016项的和

为 . 【答案】2016

4031

- 【解析】 试题分析：

111330,5105,1,1,2n a d a d a d a n +=+==-==-,

()()

212111

2321n n a a n n -+=?-?-

()()11122321n n ??

=

-??--??,故201611140322016124031240314031S ??=--=-?=- ?

??

. 考点：裂项求和法.

15.已知AD 是ABC ?的中线，(,)AD AB AC R λμλμ=+∈，0

120,2A AB AC ∠=?=-，

则||AD 的最小值是 .

【答案】 【解析】 试

题分析：

c o s A B A C b c ?==-，4

bc =，

()

()()2

2

2211142242AD AB AC c b bc ??

=+=+-≥-????

1=．

考点：向量运算.

【思路点晴】AD 是ABC ?的中线，则11

22

AD AB AC =

+，这个公式可以作为一个常用的结论记忆下来.利用两个向量数量积的概念，可将cos1202,4AB AC bc bc ?==-=，要求的是||AD 的最小值，要能够运算，必须先对其进行平方，化为

()

()2

2

2211424AD AB AC c b ??

=+=+-????

，然后考虑基本不等式，有

()()2211

42142

c b bc +-≥-=，最后两边开方. 16.已知函数1

()3(3)ln f x mx m x x

=--+，若对任意的(4,5)m ∈，12,[1,3]x x ∈，恒有

12(ln 3)3ln 3|()()|a m f x f x -->-成立，则实数的取值范围是 .

【答案】37,6??

+∞??

??

考点：函数导数.

【思路点晴】恒成立问题主要解题思路是划归与转化的思想.本题中，任意的(4,5)m ∈，

12,[1,3]x x ∈，恒有12(ln 3)3ln 3|()()|a m f x f x -->-成立，等价于

()()max min (ln3)3ln3a m f x f x -->-.经过划归之后，

问题就转化为求函数()f x 的最大值和最小值问题，可以通过导数来解决.在问题的最后，还需要用分离常数的方法来计算.

三、解答题（本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.）

17.（本小题满分12分）

已知ABC ?中，(01)BD BC λλ=<<，3

cos 5

C =，cos ADC ∠=.

（1）若5,7AC BC ==，求AB 的大小； （2）若7,10AC BD ==，求ABC ?的面积.

【答案】（1）（2）42.

考点：解三角形.

18.（本小题满分12分）

长郡中学为研究学生的身体素质与课外体育锻炼时间的关系，对该校200名高三学生的课外体育锻炼平

均每天运动的时间进行调查，如下表：（平均每天锻炼的时间单位：分钟）

将学生日均课外体育运动时间在[40,60)上的学生评价为“课外体育达标”.

（1）请根据上述表格中的统计数据填写下面22

?列联表，并通过计算判断是否能在犯错误的概率不超过

0.01的前提下认为“课外体育达标”与性别有关？

（2）将上述调查所得到的频率视为概率，现在从该校高三学生中，抽取3名学生，记被抽取的3名学生

中的“课外体育达标”学生人数为X，若每次抽取的结果是相互独立的，求X的数学期望和方差.

参考公式：

2

2

()

()()()()

n ad bc

k

a b c d a c b d

-

=

++++

，其中n a b c d

=+++.

参考数据：

【答案】（1）列联表见解析，不能；在犯错误的概率不超过0.01的前提下认为“课外体育达

标”与性别有关；（2）

13

()3

44

E X=?=，

139

()3

4416

D X=??=.

【解析】

试题分析：（1）计算22

200(60203090)200

6.060 6.635150509011033

K ?-?=

=≈

X ～1(3,)4B ，故13()344E X =?=，139

()34416

D X =??=.

试题解析：

（1）22

200(60203090)200

6.060 6.635150509011033

K ?-?=

=≈

B ， ∴13()344E X =?

=，139()34416

D X =??=. 考点：1.独立性检验；2.二项分布. 19.（本小题满分12分）

在四棱锥P ABCD -中，设底面ABCD 是边长为1的正方形，PA ⊥面ABCD . （1）求证：PC BD ⊥；

（2）过BD 且与直线PC 垂直的平面与PC 交于点E ，当三棱锥E BCD -的体积最大时，求二面角

E BD C --的大小.

【答案】（1）证明见解析；（2）

4

π

.

（2）设PA x =，三棱锥E BCD -的底面积为定值，求得它的高2

2

x

h x =+，

考点：空间向量与立体几何. 20.（本小题满分12分）

已知点C 为圆22

(1)8x y ++=的圆心，P 是圆上的动点，点Q 在圆的半径CP 上，且有点

(1,0)A 和AP

上的点M ，满足0MQ AP ?=， 2AP AM =. （1）当点P 在圆上运动时，求点Q 的轨迹方程；

（2）若斜率为的直线与圆2

2

1x y +=相切，直线与（1）中所求点Q 的轨迹交于不同的两点

F ，H ，

O 是坐标原点，且34

45OF OH ≤?≤时，求的取值范围.

【答案】（1）2

212

x y +=；（2||2k ≤≤. 【解析】

试题分析：（1）由题意知：MQ 中线段AP 的垂直平分线，所以

||||||||||||2CP QC QP QC QA CA =+=+=>=，所以点Q 的轨迹是以点,C A 为焦点，

焦距为，长轴为Q 的轨迹方程是2

212x y +=；（2）设出直线方程，联立直线方程和椭圆方程，写出根与系数关系，代入求得22

1

12k OF OH k +?=+，

22

231411412532

k k k +≤≤?≤≤+||k ≤≤.

?≤≤?≤≤k

||k k

考点：直线与圆锥曲线位置关系.

【方法点晴】解析几何解答题一般为试卷两个压轴题之一，“多考想，少考算”，但不是“不计算”.常用的解析几何题目中的简化运算的技巧有：利用圆锥曲线的概念简化运算，条件等价转化简化运算，用形助数简化运算，设而不求简化运算.圆锥曲线题目运算量较大时，要合理利用圆锥曲线的几何特征将所求的问题代数化.

21.（本小题满分12分）

已知函数ln ()(0)1

x x

f x a a x =

-<-. （1）当(0,1)x ∈时，求()f x 的单调性；

（2）若2

()()()h x x x f x =-，且方程()h x m =有两个不相等的实数根12,x x ，求证：

121x x +>.

【答案】（1）()f x 在(0,1)上单调递增；（2）证明见解析.

试题解析： （1）'

2

1ln ()(1)x x f x x --=

-，设()1ln g x x x =--，则'

1()1g x x

=-， ∴当(0,1)x ∈时，'

()0g x <，∴()(1)0g x g >=，∴'

()0f x >， ∴()f x 在(0,1)上单调递增.

（2）2

2

()ln (0)h x x x ax ax a =-+<，∴'

()2ln 2h x x x x ax a =+-+， ∴''

()2ln 23h x x a =-+，∴''

()h x 在(0,)+∞上单调递增，

考点：函数导数.

【方法点晴】极点偏移是16年全国乙卷压轴题考核的内容.本题就是基于极点偏移来命制的.

极点偏移题目在10年天津卷第一次出现，当时题目是已知()x

f x xe =，如果12x x ≠，且

()()12f x f x =，求证122x x +>．16年全国乙卷压轴题正是由这个题改编而成.在求解此类

问题中，主要利用极值点，和单调性来完成，对运算能力有较高的要求.

请考生在第22、23、24三题中任选一题作答，如果多做，则按所做的第一题记分.解答时请写清题号.

22.（本小题满分10分）选修4-1：几何证明选讲

如图，四边形ABCD 外接于圆，AC 是圆周角BAD ∠的角平分线，过点C 的切线与AD 延长线交于点E ，

AC 交BD 于点F .

（1）求证：//BD CE ；

（2）若AB 是圆的直径，4,1AB DE ==，求AD 的长.

【答案】（1）证明见解析；（2）.

（2）由（1）知，ECD BAC ∠=∠，CED ADB ∠=∠，

∵AB 是圆的直径，∴90ACB ADB ∠=∠=，∴90CED ACB ∠=∠=， ∴Rt CED ?～Rt ACB ?，∴

DE DC

BC BA

=

. ∵EAC DBC ∠=∠，由（1）知，EAC BDC ∠=∠，∴DBC BDC ∠=∠，∴DC BC =，

∴

DE DC BC BC BA AB

==

，则2

4BC AB DE =?=，∴2BC =. ∴在Rt ABC ?中，1

2

BC AB =，∴30BAC ∠=，∴60BAD ∠=，

∴在Rt ABD ?中，30ABD ∠=，所以1

22

AD AB ==.

考点：几何证明选讲.

23.（本小题满分10分）选修4-4：坐标系与参数方程

以直角坐标系的原点O 为极点，轴的正半轴为极轴建立极坐标系，已知点P 的直角坐标为

(1,2)，点M

的极坐标为(3,

)2

π

，若直线过点P ，且倾斜角为

6

π

，圆C 以M 为圆心，3为半径. （1）求直线的参数方程和圆C 的极坐标方程； （2）设直线与圆C 相交于,A B 两点，求||||PA PB .

【答案】（1）

直线的参数方程为12122

x y t ?=+????=+??（为参数），圆的极坐标方程为6sin ρθ=；（2）

. 考点：坐标系与参数方程.

24.（本小题满分10分）选修4-5：不等式选讲 （1）设函数5

()||||,2

f x x x a x R =-+-∈，若关于的不等式()f x a ≥在R 上恒成立，求实数的最大

值；

（2）已知正数,,x y z 满足231x y z ++=，求

321

x y z

++的最小值.

【答案】（1）

5

4

；（2）16+

考点：不等式选讲.

2019届高三上学期英语期末考试试卷第3套真题

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