高 一 数 学 试 卷 答 案
一. 单选题：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.
题号 1 2 3 4 5 6 7 8
答案 D A B D C D A B
二. 多选题：本大题共 4 小题，每小题 5 分，共 20 分.在每小题给出的选项中，有多项符合题目要求，全部选
对的得 5 分，有选错的得 0 分，部分选对得 2 分.
题号 9 10 11 12
答案 AC ACD BD ABD
三. 填空题：本大题共 4 小题，每小题 5 分，共 20 分. 把正确答案写在答题卡相应题的横线上.
2 3
13 14 3 1、 、 15、[1,2] 16、② ④
3 2
四. 解答题：共 70 分，解答应写出文字说明、解答过程或演算步骤.
17.（10 分）
sin( 3 )cos( )sin( )
(1). f ( ) 2 ( sin )( cos )cos 解： ············································3 分
cos( ) tan( )cos(2 ) ( cos )( tan )cos
sin sin
sin cos f ( ) cos ······················································································································5 分tan
cos
(2). cos( 3 ) sin sin 1 ······················································································································7 分
2 3
2 2为第三象限角， cos 1 sin2 ·····························································································9 分
3
f ( ) 2 2 的值为 ··························································································································································10 分
3
18.（12 分）
2
解： f (x) x 2m m 3 ,其中 2 m 2,m Z
当m 1时 f (x) x2 ···························································································································································1 分
当m 0时 f (x) x3 ·······························································································································································2 分
当m 1时 f (x) x0 1,(x 0) ·········································································································································3 分
高一数学试卷答案 第 1 页（共 4页）
f (x)在区间 (0， )上单调递增 m 1,或m 0 ······························································································4 分
选①时 (1). f (x)的解析式为 f (x) x2 ···························································································································6 分
(2). f (x) x2 , x [ 3,3]
易知 f (x) x2在[ 3,0]上单调递减，在[0,3]上单调递增························································································9 分
当 x 0时， f (x)min 0 ······················································································································································10 分
当 x 3时， f (x)max 9 ··················································································································································11 分
f (x)的值域为[0,9] ···························································································································································12 分
选②时 (1). f (x)的解析式为 f (x) x3 ·····························································································································6 分
(2). f (x) x3, x [ 3,3]
易知 f (x) x3 在[ 3,3]上是增函数··································································································································9 分
当 x 3时， f (x)min 27 ·············································································································································10 分
当 x 3时， f (x)max 27 ··················································································································································11 分
f (x)的值域为[ 27,27] ················································································································································12 分
19.（12 分）
a
解：b克糖水中有 a克糖，则糖的质量与糖水的质量的比为 ···············································································2 分
b
若再添加m a m克糖，则糖的质量与糖水的质量的比为 ·······················································································4 分
b m
a m a
根据生活常识可知，加糖后的糖水更甜，所以糖在糖水中占的“比例”就越大，因此 ················5 分
b m b
a m a
所以这一事实表示为一个不等式是： (b a 0,m 0) ·······································································6 分
b m b
证明过程如下：
a m a b(a m) a(b m) m(b a)
················································································································8 分
b m b b(b m) b(b m)
b a 0 b a 0 m 0 b m 0 m(b a) 0 ·················································································10 分
b(b m)
a m a 0 a m a ···········································································································································12 分
b m b b m b
高一数学试卷答案 第 2 页（共 4页）
20.（12 分）
解： (1). f (0) 3 3 sin ···································································································································1 分
2
(0, ) ······························································································································································2 分
2 3
f (x) 2sin(1 x ) ·························································································································································3 分
2 3
1 x 3 令 [2k ,2k ],k Z ·······················································································································4 分
2 3 2 2
x [4k ,4k 7 ],k Z ·····································································································································5 分
3 3
f (x) 7 的单调递减区间为[4k ,4k ],k Z ·····························································································6 分
3 3
2 x 1 4 （） [0,2 ] z x [ , ] ·······················································································································7 分
2 3 3 3
y 2sin z 4 在[ , ]上单调递增，在[ , ]上单调递减····················································································8 分
3 2 2 3
1
当z x ,即x 时， f (x)的最大值为 2 ··························································································10 分
2 3 2 3
z 1 x 4 当 ,即x 2 时， f (x)的最小值为 3 ···············································································12 分
2 3 3
21.（12 分）
m m
解： (1). 每件产品的售价为64 150% 96 ··························································································1 分
2x 2x
该产品的月净利润 y x(96 m ) 64x m 3 y 32x m 3 ································································3 分
2x 2
x,0 x 1 63x 6
,0 x 1
2
又 m x 1 y 2 ·············································································6 分
,1 x 3 64x 197x 19 3 x ,1 x 3 6 2x
(2).①当 x (0,1]时， y单调递增， 当 x 1时， ymax 28.5 ············································································7 分
64x2x 197x 19②当 (1,3)时， y ，令t 3 x x 3 t(0 t 2)
2（3 x）
y 32t 2 187 ······························································································································································8 分
t 2
高一数学试卷答案 第 3 页（共 4页）
y 2 32t 2 187 77.5 ··········································································································································9 分
t 2
2 1
当且仅当32t ，即 t 时，等号成立
t 4
11
当x （万件）时， ymax 77.5（万元）···········································································································11 分4
综上所述：该产品月净利润 y的最大值为77.5万元.·································································································12 分
22.（12 分）
解： (1). p x 0 p x 3 x 4 p 4①·····································································································1 分
a 1 f (x) log 1 (x 1) log (x
3
1 ) log 1[(x 1)(x
3
)] log 1 ( p x) ·········································2 分2 2 2 2 2 2 2
(x 1)(x 3 ) ( p x) p 3 3 3 x2 x (x )2 15 ···············································································3 分
2 2 2 4 16
x (3,4) 6 23 p ②··················································································································································4 分
2
由①、②可知 p的取值范围为 (6, 23) ·······························································································································5 分
2
(2). f (x) loga (x 2a) loga (x 3a) loga[(x 2a)(x 3a)] 1 loga a ·················································6 分
x [a 3,a 4] a 3 3a 0 3 a ····················································································································7 分
2
0 a 5a①当 1时， (x 2a)(x 3a) a，函数 y (x 2a)(x 3a)开口向上，对称轴为 x a 3，
2
函数 y (x 2a)(x 3a)在[a 3,a 4]上单调递增，故 ymin (a 3 2a)(a 3 3a) a
a 5 7 a 5 7解得 或 0 a 1 ······················································································································9 分
2 2
②当1 a 3 时， (x 2a)(x 3a) 5a a，函数 y (x 2a)(x 3a)开口向上，对称轴为 x a 3，
2 2
函数 y (x 2a)(x 3a)在[a 3,a 4]上单调递增，故 ymax (a 4 2a)(a 4 3a) a
13 41 13 41 13 41 3
解得 a ，但 ， a无解·············································································11 分
4 4 4 2
综上所述： a的取值范围为0 a 1 ·······························································································································12 分
高一数学试卷答案 第 4 页（共 4页）高一数学试卷
本试卷分第|卷（选择题）和第川卷（非选择题）两部分
全卷共150分，考试时间120分钟，考生作答时，将答案答在答题卡上，在本试卷上答
题无效
注意事项：
1.答题前，考生先将自己的姓名、准考证号码填写清楚，将条形码准确粘贴在条形码
区域内，
2.选择题必须使用2B铅笔填涂：非选择题必须使用0.5毫米黑色字迹的签字笔书写，
字体工整、笔迹清楚，
3.请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草
稿纸、试题卷上答题无效
4.作图可先使用2B铅笔填涂；非选择题必须用黑色字迹的签字笔描黑
5保持卡面清洁，不要折叠、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀，
第I卷
一，单选题：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，
1.已知粜合A=1x-3xs0,B=x3S0,则集合4小、B之间的关系为
A.A∩B=
B.A=B
C.A军B
D.BSA
2.如图，扇形A0B的圆心角为120°，半径r=6,则图中阴影部分的面积为
A.12π-9√5
B.12π
C.
6m-9V3
D.12π-183
高一数学试卷
第1页共8页
3.已知函数f()=nx+x-2,则)的零点所在的区间为
A.（-1,1)
B.(1,2)
C.(2,e)
D.(e,3)
4.已知命题p:“3x∈R,x2-ar+l<0”为假命题，则实数a的取值范围是
A.（-∞，2]
B.(（-2,2)
C.(∞，-2)U(2,+∞)
D.[-2,2]
5.函数f(x)=a-a(a>0,且a≠1)的图象可能是
B
D
6.耳机的降噪效果成为衡量一个耳机好坏的标准之一，降噪的工作原理就是通
过麦克风采集周围环境的噪音，通过数字化分析，以反向声波进行处理，实现
声波间的抵消，使噪音降为0，完成降噪（如图所示），已知噪音的声波曲线
是y=3cos2x,通过主动降噪芯片生成的反向声波曲线是y=Asin(@r+p)(其
中A>0,0>0,0≤p<2π)，则p=
音声波
两种声波叠加后
火
用来降噪的反向声波
A.
-3
B.
C.
D.
3π
2
高一数学试卷第2页共8页
7、已知函数g()是定义域为R的偶函数，且在(-o,0)上单调递减，若a=g(2),
b=g(25),c=g(-1),则a,b,c的大小关系为
A.c
C.bD.b8.给定函数f(x)=2x2-1,g(x)=ar,xeR.用M(x)表示f(x),g(x)中的
较大者，记为M(=mxg》.若M凶的最小值为-之，则实数：的值
为
A.0
B.士1
C.±2
D.±2
二.多选题：本大题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符
合题目要求，全部选对的得5分，有选错的得0分，部分选对得2分，
9.将水注入不同形状的玻璃容器中，从空瓶到注满为止，如图所示（设单位时
间内进水量相同)，在每个容器下方给出的图像中，能正确反映该容器中水面的
高度随时间变化规律的是
度
时间
时间
时间
时间
A
B
D
10.下列命题不正确的是
A.第一象限角都是锐角
B.A=B是sinA=sinB的充分不必要条件
C.sin3<0
D.若cos0=cos01,则角0是第一或第四象限角
高一数学试卷第3页共8页