2023一2024学年度第一学期教学质量检查
高一数学
一、单项选择题：本大题共8小题，每小题5分，共40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.请把正确选项在答题卡中的相应位置涂黑.
1.已知集合A={-2,-1,0l,2},B={xy=gx},则A∩B=
A.{0l,2}
B.{1,2}
c.{2)
D.{-1,0,l}
2.使>1”成立的一个充分不必要条件是
A.x>1
B.x<1
C.-1D.x>-1
3.己知函数f(x)=
2*,x≤0
1og2x,x>0'
则ff(兮》的值是
A.4
B.
1
C.2
D.
2
4.函数了)=具图象的大致形状是
xe
B一
5.扇形的弧长为15，圆心角为3江，则该扇形的面积为
A
300
A.
B.J50
C.150π
D.300元
6.已知a,b,c为正实数，满足a+a2=b+4=c+log2c=4,则实数a,b,c之间的大小关系为
A.a>b>c
B.a>c>b
C.c>a>b
D.c>b>a
7.在平面直角坐标系xOy中，半径为2的圆O与y轴非负半轴的交点为。，动点P从P出发，
以1rad/s的角速度按顺时针方向在圆O上做匀速圆周运动，则2s时点P的坐标为
A.(2cos(-2),2sin(-2)
B.(2cos 2,2sin 2)
c.(2sin(-2),2cos(-2)
D.(2sin 2,2 cos2)
8.某企业从2010年开始实施新政策后，年产值逐年增加，下表给出了该企业2010年至2020年
高一数学第1页（共4页）
0000000
的年产值（万元），为了描述该企业年产值y(万元)与新政策实施年数x(年)的关系，现有以
下三种函数模型：y=+b,y=ka(a>0,且a≠1)，y=klog。x+b(a>0,且a≠1)，选出你
认为最符合实际的函数模型，预测该企业2023年的年产值约为
年份
2010
2011
2012
2013
201420152016
2017
2018
2019
2020
年产值278
309
344
383
427
475528
588
655
729
811
A.924万元
B.976万元
C.1109万元
D.1231万元
二、多项选择题：本大题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项
符合题目要求，全部选对的得5分，有选错的得0分，部分选对的得2分.请把正确选项在答题
卡中的相应位置涂黑.
9.下列运算正确的是
A.3lo815=5
B.1
3+In-=1
c.b6=6
27
10.己知a,b,c∈R,则下列说法正确的是
A.若a>b,则a2>b2
B.若a>b,则a>
C.若a>b>0,则ac2>bc
D.若a>b>0,则aa+l
1.已知函数f()=4in(@x+p)4>0,@>0<习的
w
调递增，则实数m的取值可以是
5π
5π
3π
A.
B.元
c.
D.
6
4
2
12.用[x]表示不超过实数x的最大整数，如：【-1.2]=-2，[1.5]=1.己知函数f(x)=cosx+lsin,
函数g(x)=[(x],则下列结论正确的是
A.函数(x)的图象关于y轴对称
B,函数∫(x)是周期函数
C.函数g(x)的值域是{-2，-1,0,1}
D.方程g(x)=2x只有一个实数根
三、填空题：本大题共4小题，每小题5分，共20分，请把答案填在答题卡的相应位置上，
13.
己知sina=。,则cos2a=
高一数学第2页（共4页）
00000002023－2024学年度第一学期教学质量检查
高一数学 参考答案
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 B A D A B C D C
二、多项选择题（全部选对的得 5分，选对但不全的得 2分，有选错的得 0分）
题号 9 10 11 12
答案 AD BD AB ABD
三、填空题（16题第一空 2分，第二空 3分）
7 1 12
13. 14. 15. f x ln x（ f x log a x ( a 1) 均可） 16. 3； 9 4 5
四、解答题
17. 解：（1）当m 2时， A (x, y) | y x 2 ························································ 1分
y x 2 A B (x, y) 则 2 ············································································ 2分
y x
y x 2
方程组 消去 y后得 x22 x 2 0，得 x 1或 x 2 ····································· 3分
y x
当 x 1时， y 1；当 x 2时， y 4 .·································································· 4分
故 A B ( 1,1)，(2,4) ；················································································· 5分
y x m
（2） A B ,即方程组 2 无解，····························································6分
y x
即方程 x2 x m 0无解，·················································································7分
1
则 1 4m 0，得m ，···········································································9分
4
1
所以实数 m的取值范围为 m |m .·······························································10分
4
tan cos tan cos
18.解：（1）已知 f tan
sin cos
2
高一数学参考答案 第 1 页 （共 5 页）
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······················································································································· 3分
所以 f ( ) tan 3；·················································································· 5分
3 3
（2）由（1） f tan ，即 tan sin 2，·············································· 6分
cos
π
又 π，所以 sin 0,cos 0，······························································ 8分
2
且 sin 2 cos2 1··························································································9分
解得 sin 2 5 5 , cos ··········································································11分
5 5
sin cos 3 5 .······················································································12分
5
19. 解：（1）因为函数 f (x)是定义在 R 上的奇函数，所以 f 0 0 .···························· 1分
设 x 0，则 x 0，
1 1
因为 x 0时， f (x) x 1，所以 f ( x) x 1 ·····································3分
x x
又因为 f (x) 1是定义R 上的奇函数，所以 f (x) f x x 1 ··························· 4分
x
x
1
1, x 0
x
故函数 f (x)的解析式为 f x 0, x 0 ;······················································ 5分
x 1 1, x 0
x
（2）函数 f (x)在区间 ， 1 上单调递增；·························································6分
1
证明如下：由（1） x ， 1 时， f (x) x 1············································· 7分
x
x1, x2 , 1 ，且 x1 x2 时···········································································8分
1 1 f x f x 1 1 1 2 x1 1x
x2 1 x1 x2
1 x 2 x 1 x2
x x x2 x1 x 1 x1x2 1 1 2 x x 1 x2
1 x1 x2 ·····································10分
1 2 x1x2 x1x2
由 x1, x2 , 1 ，得 x1x2 1, x1x2 1 0，又由 x1 x2 ，得 x1 x2 0 ··················11分
高一数学参考答案 第 2 页 （共 5 页）
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于是 f x1 f x2 ，所以函数 f (x)在区间 ， 1 上单调递增.······························ 12分
20. 解：（1）因为 f x 3 cos2 x sin x cos x 3 3 1 cos 2 x sin 2 x 3 ，
2 2 2 2
所以 f x cos 2 x ，··············································································· 2分
6
π y cos 向左平移 个单位后，得到函数 2 x
cos
2 x
··········3分
6 6 6 3 6
因为新函数图象关于原点对称，所以 n n Z ,·································· 4分
3 6 2
即 3n 1，又 0，所以 所有可能取值组成的集合为 | 3n 1，n N ；·· 5分
（2）由（1）知： f x cos 2 x ,因为 0， 2 x 随着 x的增大而增大，···6分
6 6
π
所以当 x (0 ， )时， 2 x ， ······················································· 7分
6 6 6 3 6
f (x) π 5因为函数 在 (0， )单调递减,所以 ，解得0 ·····················8分
6 6 3 6 2
又由（1）得 3n 1， n N ,所以 n 0时， 1 ···············································9分
f x cos 2x 此时 ，令 f (x) 1 2x ，则 2k , 2k ·············10分
6 2 6 3 3
解得 x k , k k Z ，································································· 11分
4 12
1
所以 f (x) 的解集为 x | k x k ,k Z .···································· 12分
2 4 12
21. 解：（1）作图如下，（说明：对于作图方法和过程不要求说明，作对即得满分）
···················································································· 2分
高一数学参考答案 第 3 页 （共 5 页）
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由作图过程知： AP A1P1 ·················································································3分
（3）由三角函数定义得 P1 cos , sin ， A1 cos , sin ，P cos , sin ，
······················································································································· 4分
由（1）知 AP A1P1 ，根据两点间的距离公式，
得 [cos 1]2 [sin 0]2 cos cos 2 sin sin 2 ,············ 5分
化简得： cos cos cos sin sin ，····················································7分
当角 ， 终边重合，即 2k k Z ，容易证明上式仍然成立,························8分
所以对于任意角 ， ，有 cos cos cos sin sin .······························9分
sin cos （3）
cos
········································10分 2 2
cos cos sin
sin ;·······························································11分
2 2
sin cos cos sin .················································································ 12分
22. 1 4
x
解：（1）易知函数 f (x)的定义域为 R ,且 f ( x) log (4 x2 1) bx log2 ( x ) bx4
log2 (1 4
x ) 2x bx log2 (1 4
x ) (b 2)x，·························································2分
f (x)为偶函数， f ( x) f (x)，则 (b 2)x bx对 x R成立，·························· 3分
得 b 2 b，则 b 1 .························································································· 4分
x
（2）由（1）知，f (x) log (4x2 1) x，设h(x) f (x) x sin log2 4x 1 2x sin x 2 2
log (4
x 1
2 ) sin
x log (1 1 2 ) sin
x
，·····················································6分
4x 2 4x 2
当 x 0,1 时， h x 单调递减，（无需证明）····························································7分
且 h 0 1 0， h 1 log 52 1 0 ·····································································8分4
x 0,1 log 1 1 x0根据函数零点存在定理知： h x 有唯一的零点 0 ，且 2 x sin 0， 4 0 2
······················································································································· 9分
高一数学参考答案 第 4 页 （共 5 页）
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1
sin x0 log2 1 x
所以 2 2 2 4 0 1 1 ，······································································ 10分
4x0
1 1
而函数 y 1 x 在 0,1 上单调递减，所以当 x 1时， y 1 x 取得最小值，即 x (0,1)时4 4
1 1 5 x ，··································································································· 11分4 4
sin x0
所以 2 2 5 .································································································12分
4
高一数学参考答案 第 5 页 （共 5 页）
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