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绝密★启用前 卷类型:A
2020年茂名市高三级第二次综合测试
文科数学 2020.5
本试卷分选择题和非选择题两部分,共6页,23小题,满分150分,考试时间120分钟.
注意事项:
1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上.
2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑.写在试题卷、草稿纸和答题卡上的非答题区域均无效.
3.填空题和解答题的作答:用签字笔直接答在答题卡上对应的答题区域内. 写在试题卷草稿纸和答题卡上的非答题区域均无效.
4.选考题的作答:先把所选题目的题号在答题卡上指定的位置用2B铅笔涂黑. 答案写在答题卡上对应的答题区域内,写在试题卷、草稿纸和答题卡上的非答题区域均无效.
5.考试结束后,请将答题卡上交.
一、选择题:(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,有且只有一项是符合题目要求的)
1.已知集合U={1, 2, 3, 4, 5},A={2, 3, 5},B={2, 5},则( )
A. AÌB B. ∁UB={1, 3, 4} C. A∪B={2, 5} D. A∩B ={3}
2.若,则复数的虚部为( )
A.2 B.1 C. D.−1
3.已知函数f(x)在点(1, f(1))处的切线方程为x+2y−2=0,则f(1)+f ′(1) =( )
x
O
y
2
–2
第4题图
A. B.1 C. D.0
4.函数的图象如图
所示,则的值为( )
A. B.1 C. D.
5.下列命题错误的是( )
A.“x=2”是“x2−4x+4=0”的充要条件
B.命题“若,则方程x2+x−m=0有实根”的逆命题为真命题
C.在△ABC中,若“A>B”,则“sinA>sinB”
第6题图
D.若等比数列{an}公比为q,则“q>1”是“{an}为递增数列”的充要条件
6.《易·系辞上》有“河出图,洛出书”之说,河图、洛书是中国
古代流传下来的两幅神秘图案,蕴含了深奥的宇宙星象之理,被誉
为“宇宙魔方”,是中华文化阴阳术数之源。河图的排列结构如图
所示,一与六共宗居下,二与七为朋居上,三与八同道居左,四与
九为友居右,五与十相守居中,其中白圈为阳数,黑点为阴数,若
从阳数和阴数中各取一数,则其差的绝对值为5的概率为:( )
A. B. C. D.
否
结束
输出m
是
r>0?
r=1
开始
输入m, n
求m除以n的余数r
m=n
n=r
第7题图
7.“辗转相除法”是欧几里得《原本》中记录的一个算法,是由欧几里得在公元前300年左右首先提出的,因而又叫欧几里得算法.如图所示是一个当型循环结构的“辗转相除法”程序框图.当输入m=2020,n=303时,则输出的m是( )
A. 2 B. 6 C. 101 D. 202
8.已知双曲线(a>0, b>0)的离心率为2,
其一条渐近线被圆(x−m)2+y2=4(m>0)截得的线段长
为2,则实数m的值为( )
A. B. C.2 D.1
9.已知函数是定义在R上的偶函数,当时,
.则使不等式成立的
x取值范围是( )
A. B. C. D.
10.A
O
y
5
−5
C
O
y
5
−5
x
x
B
O
y
5
−5
x
D
O
y
5
−5
x
函数在[−5, 5]的图形大致是( )
11.已知三棱锥 中, 且
平面PAB⊥平面ABC,则该三棱锥的外接球的表面积为( )
A. B. C. D.
12.已知函数,对于函数有下述四个结论:①函数在其定义域上为增函数;②对于任意的,都有成立;③有且仅有两个零点;④若y=ex在点处的切线也是y=lnx的切线,则x0必是零点.
其中所有正确的结论序号是
A.①②③ B.①② C.②③④ D.②③
二、填空题:(本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置)
13.已知向量,,若,则 .
B
C
A
B1
C1
A1
D
D1
E
F
第15题图
14. 为了贯彻落实十九大提出的“精准扶贫”政策,某地政府投入16万元帮助当地贫困户通过购买机器办厂的形式脱贫,假设该厂第一年需投入运营成本3万元,从第二年起每年投入运营成本比上一年增加2万元,该厂每年可以收入20万元,若该厂n(n∈N*)年后,年平均盈利额达到最大值,则n等于 .
(盈利额=总收入−总成本)
15.在棱长为2的正方体ABCD–A1B1C1D1中,E是棱DD1
的中点,则平面A1EC截该正方体所得截面面积为:
.
16.过点作圆的切线,已知A,B分别为切点,直线AB恰好经过椭圆的右焦点和下顶点,则直线AB方程为 ;椭圆的标准方程是 . (第一空2分,第二空3分)
三、解答题:(共70分. 解答应写出文字说明、证明过程或演算步骤. 第17~21题为必考题,每个试题考生都必须作答,第22、23题为选考题,考生根据要求作答)
(一)必考题:共60分
17.(分)
在中,角,,的对边分别为,,,已知,.
(1)求;
(2)若,求的面积.
18.(分)
某种治疗新型冠状病毒感染肺炎的复方中药产品的质量以其质量指标值衡量,质量指标越大表明质量越好,为了提高产品质量,我国医疗科研专家攻坚克难,新研发出A、B两种新配方,在两种新配方生产的产品中随机抽取数量相同的样本,测量这些产品的质量指标值,规定指标值小于85时为废品,指标值在[85,115)为一等品,大于115为特等品. 现把测量数据整理如下, 其中B配方废品有6件.
A配方的频数分布表
质量指标值分组
[75,85)
[85,95)
[95,105)
[105,115)
[115,125)
频数
8
a
36
24
8
B配方的频频率分布直方图
75
85
95
105
115
125
质量指标值
O
0.008
0.006
b
0.022
0.038
第18题图
(1)求a, b的值;
(2)试确定A配方和B配方哪一种好?
(说明:在统计方法中,同一组数据常用
该组区间的中点值作为代表)
19.(分)
如图1,在□ABCD中,AD=4,AB=2,∠DAB=45°,E为边AD的中点,以BE为折痕将△ABE折起,使点A到达P的位置, 得到图2几何体P−EBCD.
(1)证明: ;
E
B
C
A
D
第19题图1
P
E
B
C
D
第19题图2
(2)当BC⊥平面PEB时,求三棱锥C−PBD的体积.
20.(分)
已知抛物线C: y2=2px (p>0)与直线l: x+y+1=0相切于点A,点B与A关于x轴对称.
(1)求抛物线C的方程,及点B的坐标;
(2)设M、N是x轴上两个不同的动点,且满足∠BMN=∠BNM,直线BM、BN与抛物线C的另一个交点分别为P、Q,试判断直线PQ与直线l的位置关系,并说明理由.
如果相交,求出的交点的坐标.
21.(分)
设函数.
(1)讨论的单调性;
(2)若,当m=1,且时,,求的取值范围.
(二)选考题:共10分
请考生在第22、23两题中任选一题作答,如果多做,则按所做的第一题计分,作答时,
请用2B铅笔在答题卡上把所选题目对应的题号涂黑.
22.[选修4−4:坐标系与参数方程] (10分)
在平面直角坐标系xOy中,已知曲线C: (q为参数),以原点O为极点,x轴正半轴为极轴建立极坐标系,直线l的极坐标方程,点M()
在直线l上,直线l与曲线C交于A,B两点.
(1)求曲线C的普通方程及直线l的参数方程;
(2)求△OAB的面积.
23.[选修4-5:不等式选讲](10分)
已知函数f(x)=|x+1|−|x−2|.
(1)若f(x)≤1,求x的取值范围;
(2)若f(x)最大值为M,且a+b+c=M,求证:a2+b2+c2≥3.
绝密★启用前 卷类型:A
2020年茂名市高三级第二次综合测试
文科数学参考答案及评分标准
一、选择题(本大题共12小题,每小题5分,共60分)
题号
1
2
3
4
5
6
7
8
9
10
11
12
答案
B
B
D
B
D
A
C
C
A
A
B
C
提示:
1. B【解析】∵U={1, 2, 3, 4, 5},B={2, 5},∴∁UB={1, 3, 4}. 故选B.
2. B【解析】,所以的虚部,故选B .
3. D【解析】切点(1, f(1))在切线x+2y−2=0上,∴1+2f(1)−2=0,得f(1)=,又切线斜率
x
O
y
2
–2
第4题图
故选D.
4. B【解析】根据图象可得,,即,
根据,得, ∴,
又的图象过点,∴,
即,
∴,,又因,∴ ,
∴,,故选B.
5.D.【解析】由 x2−4x+4=0Û(x−2)2=0Û x−2=0Û x=2,∴A正确;
命题“若, 则方程x2+x−m=0有实根”的逆命题为命题“若方程x2+x−m=0有
实根,则”,∵方程x2+x−m=0有实根Þ△=1+4m≥0Þ,∴B正确;
在△ABC中,若A>BÞa>bÞsinA>sinB(根据正弦定理) ∴C正确;故选D.
(事实上等比数列{an}公比为q,则“q>1”是“{an}为递增数列”的既不充分也不必要条件)
6. A.【解析】∵阳数为:1, 3, 5, 7, 9;阴数为:2, 4, 6, 8, 10,∴从阳数和阴数中各取一数
的所有组合共有:个,满足差的绝对值为5的有:(1, 6), (3, 8), (5, 10), (7, 2),
(9, 4)共5个, 则, 故选A.
7. C【解析】输入m=2020,n=303,又r=1.
①r=1>0,2020÷303=6··············202,
r=202,m=303,n=202;
②r=202>0,303÷202=1············101
r=101,m=202,n=101;
③r=101>0,202÷101=2··············0.
r=0,m=101,n=0;
④r=0,则r>0否,输出m=101,故选C.
8.C.【解析】依题意: ∴双曲线渐近线方程为
不妨取渐近线l1:,则圆心(m,0) (m>0)到l1的距离.
由勾股定理得,解得,∵m>0,∴m=2,故选C.
9. A.【解析】∵,由得,.又∵为偶函数,
, 易知在上为单调递减,
∴或,即或,故选A.
10. A【解析】易知f(−x)= −f(x),即函数f(x)是奇函数,图象关于原点对称,排除D,
f(x)在y轴右侧第一个零点为.
当时,,∴f(x)<0排除B,
当x (>0)®0时,则, 且∴y®−¥.故选A.
(当时,.
,排除C)
11. B【解析】在中,由余弦定理得 ,又,
∴为直角三角形, ,又平面PAB⊥平面ABC且交于AB,
∴CB⊥平面PAB,∴几何体的外接球的球心到平面PAB的距离为,
B
C
A
P
设的外接圆半径为,则∴
设几何体的外接球半径为R,则,
所求外接球的表面积 故选B.
12.解析:依题意定义域为(−¥, 1)∪(1, +¥),且,
∴在区间(−¥, 1)和(1, +¥)上是增函数,①错;
∵当时,则,因此成立,②对;
∵在区间(−¥, 1)上单调递增,且
∴,即在区间(−¥, 1)上有且仅有1个零点.
∵在区间(1, +¥)上单调递增,且,
∴,(也可以利用当时,,)
得在区间(1, +¥)上有且仅有1个零点. 因此,有且仅有两个零点;③对
∵y=ex在点处的切线方程l为.
又l也是y=lnx的切线,设其切点为,则l的斜率,
从而直线l的斜率,∴,即切点为,又点在l上.
∴,即x0必是零点.④对.
二、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置.
13.3 14. 4 15. 16. 2x−y−2=0(2分); (3分).
提示:
13.【答案】3【解析】∵ ,∴,即,
由已知得,∴
14.【答案】4【解析】设每年的营运成本为数列,依题意该数列为等差数列,且 所以n年后总营运成本,因此,年平均盈利额为: 当且仅当时等号成立.
B
C
A
B1
C1
A1
D
D1
E
F
15.【答案】【解析】如图,在正方体ABCD–A1B1C1D1中,
∵平面A1D1DA∥平面B1C1CB,
∴平面A1EC与平面B1C1CB的交线必过C且平行于A1E,
故平面A1EC经过B1B的中点F,连接A1F,得截面A1ECF,
易知截面A1ECF是边长为的菱形,其对角线EF=BD=2,
A1C=,截面面积S=A1C´EF=´2´=.
16.【答案】2x−y−2=0,.
【解析】①当过点的直线斜率不存在时, 直线方程为:x=1, 切点的坐标;
②当直线斜率存在时, 设方程为, 根据直线与圆相切, 圆心(0,0)到切线的距离等于半径1, 可以得到切线斜率, 即:.直线方程与圆方程的联立可以得切点的坐标;根据A、B两点坐标可以得到直线AB方程为2x−y−2=0,(或利用过圆外一点作圆的两条切线,则过两切点的直线方程为)
依题意,AB与x轴的交点即为椭圆右焦点,得,与y轴的交点即为椭圆下顶点坐标,所以,根据公式得,因此,椭圆方程为:.
三、解答题:共70分. 解答应写出文字说明、证明过程或演算步骤. 第17~21题为必考题,每个试题考生都必须作答,第22、23题为选考题,考生根据要求作答.
(一)必考题:共60分
17. 解:(1)依题意,由正弦定理得:.·····································1分
∵,∴, ·······················································2分
∴, ······························································3分
∴,,··················4分 ∴. ···············5分
(2)解法一:由题意得:. ··················································6分
∵,∴, ··········································7分
∴, ···············································8分
, ···············································9分
∴. ···········10分
················································11分
∴. ·····································12分
解法二:由题意及(1)得:. ··································6分
∵,∴, ···········································7分
由余弦定理得:, ························8分
即, 解得. ···············································9分
若,又 则A=C,又B=2C,得△ABC为直角三角形,而三边为
的三角形不构成直角三角形,矛盾. ∴. ·················11分
∴. ·······································12分
18.解:(1)依题意,A、B配方样本容量相同,设为n,又B配方废品有6件.
由B配方的频频率分布直方图,得废品的频率为, ·················1分
解得n=100. ···················2分 ∴a=100−(8+36+24+8)=24. ···············3分
由(0.006+b+0.038+0.022+0.008)´10=1 ······························4分
解得b=0.026.因此a, b的值分别为24, 0.026; ································5分
(2)由(1)及A配方的频数分布表得,A配方质量指标值的样本平均数为
····7分
质量指标值的样本方差为[(−20)2´8+(−10)2´24+0´36+102´24+202´8]=112.···8分
由B配方的频频率分布直方图得,B配方质量指标值的样本平均数为
=80´0.06+90´0.26+100´0.38+110´0.22+120´0.08=100. ··············9分
质量指标值的样本方差为
=(−20)2´0.06+(−10)2´0.26+0´0.38+102´0.22+202´0.08=104. ········10分
综上,>, ···································11分
即两种配方质量指标值的样本平均数相等,但A配方质量指标值不够稳定,
所以选择B配方比较好. ···········································································12分
(2)当BC⊥平面PEB时,求三棱锥C−PBD的体积.
E
B
C
A
D
第19题图1
P
E
B
C
D
第19题图2
19. 证明:(1)依题意,在△ABE中(图1),AE=2,
AB=2,∠EAB=45°,由余弦定理得
EB2=AB2+AE2−2AB·AEcos45°
=8+4−2´2´2´=4,·······························································2分
∴AB2= AE2+EB2, ···········································································3分
即在□ABCD中,EB⊥AD. ····································································4分
以BE为折痕将△ABE折起,由翻折不变性得,在几何体P−EBCD中,
EB⊥PE,EB⊥ED. 又ED∩PE=E,∴BE⊥平面PED, ···························5分
又BEÌ平面PEB,∴; ·······················································6分
(2)∵BC⊥平面PEB,PEÌ平面PEB,∴ BC⊥PE. ····································7分
由(1)得 EB⊥PE,同理可得PE⊥平面BCE,·············································8分
即PE⊥平面BCD,PE就是三棱锥P−CBD的高. ········································9分
又∠DCB=∠DAB=45°,BC=AD=4,CD=AB=2,PE=AE=2,
∴S△CBD=´BC´CD´sin45°=´4´2´=4. ·································10分
VC−PBD=VP−CBD=S△BCD´PE=´4´2=.
因此,三棱锥C−PBD的体积为.··························································12分
(写出VC−PBD=VP−CBD得1分,结果正确并作答得1分)
x
O
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N
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Q
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20.解: (1)联立·········································1分
消去x得y2+2py+2p=0,···········································2分
∵直线与抛物线相切,∴△=4p2−8p=0,
又p>0,解得p=2,∴抛物线C的方程为y2=4x.·········3分
由y2+4y+4=0,得y=−2,∴切点为A(1, −2),
∵点B与A关于x轴对称,点B的坐标B(1, 2). ···········4分
(2)直线PQ∥l. ····························5分 理由如下:
依题意直线BM的斜率不为0, 设M(t, 0)(t≠1), 直线BM的方程为x=my+t, ·····6分
由(1)B(1, 2),1=2m+t,∴直线BM的方程为x=y+t, ·························7分
代入y2=4x.解得y=2(舍)或y=−2t,∴P(t 2,−2t). ·······························8分
∵∠BMN=∠BNM,∴ M、N关于AB对称,得N(2−t, 0) . ·····················9分
同理得BN的方程为x=y+2−t,代入y2=4x.得Q((t −2)2, 2t−4). ···········10分
, ·······················································11分
直线l的斜率为−1,因此PQ∥l. ·······················································12分
21. 解: (1)依题得,定义域为R,,,··········1分
令,.
①若,即,则恒成立,
从而恒成立,当且仅当,时,.
所以在R上单调递增. ································································2分
②若,即,令,得或.
当时,; ····································3分
当时,. ·····················4分
综合上述:当时,在R上单调递增;
当时,在区间上单调递减,
在区间上单调递增. ···················5分
(2)依题意可知: ···················6分
令,可得, ···························································7分
.
设,则.·····························8分
当时, ,单调递减, ······································9分
故. ······················································10分
要使在时恒成立,需要在上单调递减,
所以需要. ······················································11分
即,此时,故.
综上所述, 的取值范围是. ······································12分
(二)选考题:共10分
请考生在第22、23两题中任选一题作答,如果多做,则按所做的第一题计分,作答时,
请用2B铅笔在答题卡上把所选题目对应的题号涂黑.
22. 解:(1)将曲线C:消去参数q得, 曲线C的普通方程为:.·····1分
∵点M()在直线上,∴.············2分
∴,展开得, 又x=rcosq,y=rsinq,
∴直线l的直角坐标方程为x+y−2=0, ························································4分
显然l过点(1, 1), 倾斜角为.
∴直线l的参数方程为 (t为参数). ······································5分
(2)解法一:由(1),将直线l的参数方程代入曲线C的普通方程得:
, ····························································6分·
整理得,显然△>0.
设A, B对应的参数为t1, t2, 则由韦达定理得,.········7分
由参数t的几何意义得
|AB|=| t1−t2|==, ························8分
又原点O(0,0)到直线l的距离为. ····································9分
因此,△OAB的面积为. ···················10分
(2)解法二: 由(1),联立消去y得:, 显然△>0. ····6分
设,则由韦达定理得,.············· ·········7分
由弦长公式得
|AB|==, ··········· ·········8分
又原点O(0,0)到直线l的距离为. ··································9分
因此,△OAB的面积为. ··················10分
(2)解法三:由(1),联立消去y得:, 显然△>0. ····6分
设,则由韦达定理得,. ·····················7分
∵直线l过椭圆右顶点(2,0),∴,∴ ······················8分
把代入直线l的方程得, ······················9分
因此,△OAB的面积为. ··························10分
23.解:(1)由已知 ·················································1分
当x≥2时,f(x)=3,不符合; ···························································2分
当−1≤x<2时, f(x)=2x−1, 由f(x)≤1, 即2x−1≤1, 解得x≤1, ∴−1≤x≤1. ······3分
当x<−1时,f(x)= −3,f(x)≤1恒成立. · ··················································4分
综上,x的取值范围是x≤1. ·····························································5分
(2)由(1)知f(x)≤3,当且仅当x≥2时,f(x)=3, ········································6分
∴M= f(x)Max=3.即a+b+c=3, ·······················································7分·
∵a2+b2≥2ab,a2+c2≥2ac,c2+b2≥2cb, ·············································8分
∴2(a2+b2+c2)≥2(ab+ac+cb)
∴3(a2+b2+c2)≥a2+b2+c2+2ab+2ac+2cb=(a+b+c)2=9, ·································9分
因此(a2+b2+c2)≥3. ··············································································10分