(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 1 页( 12 页)
保密★启前
市普通高中 2019 级高学期教学质量踪监测 数学
(新教材必修第 1 册部分)
202001
单项选择题:题 10 题题 5 分 50 分题出四选项中项符
合题目求
1已知集合 { | 2 2}M x x { | 1}N x x M N
A.{ | 2 1}x x B { |1 2}x x C { | 2}x x D { | 2}x x
答案:A
2函数 2( ) ( 0xf x a a 1)a 图象恒定点该定点
A. (20) B (21) C (30) D (31)
答案:B
3命题 0
π(0 )4x 0 0sin cosx x 否定
A π(0 )4x sin cosx x B π(0 )4x sin cosx x
C 0
π(0 )4x 0 0sin cosx x D π(0 )4x sin cosx x
答案:D
4 四边形角线互相垂直四边形菱形
A.充分必条件 B.必充分条件
C.充条件 D.充分必条件
答案:B
解析: 四边形角线互相垂直法推出四边形菱形 四边形角线互相垂直
充分条件反四边形菱形推出四边形角线互相垂直四边形角线互相垂直必
条件
5命题正确
A bcac ba B 0)( 2 cba ba
C 1 1
a b
a b D ba ba
答案:B准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 2 页( 12 页)
解析:选项 A 中 0c ba A 错误
选项 C 中取 1a 2b 满足 1 1
a b
ba C 错误
选项 D 中取 2a 1b 满足 ba ba D 错误
选项 B 中易 0)( 2 cba 02 c ba B 正确
6 四变量 1y 2y 3y 4y 变量 x 变化数表:
x 1 2 4 6 8 10 12
1y 16 29 55 81 107 133 159
2y 9 15 87 735 6567 59055 531447
3y 1 8 64 216 512 1000 1728
4y 2000 3710 5419 6419 7129 7679 8129
中关 x 呈指数增长变量
A. 1y B. 2y C. 3y D. 4y
答案:B
解析:题中表格出四变量 1y 2y 3y 4y 越越
增长速度中变量 2y 增长速度快画出散点图:
知变量 2y 呈指数增长选择 B 答案
7 函数 ( ) tanf x x x π π( )2 2x 图象致准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 3 页( 12 页)
A B C D
答案:A
解析:函数定义域关原点称 ( ) ( ) tan( ) tan ( )f x x x x x f x
函数 ( )f x 奇函数排 BC 选项
π π π π( ) tan 1 04 4 4 4f 排 D 选项
(法: π0 2x 时 tanx x ( ) tan 0f x x x 排 BD 选项 )
8 代数式 24sin 1 意义锐角 取值范围
A π(0 ]6
B π(0 ]3
C π π[ )6 2
D π π[ )3 2
答案:C
解析:题意 24sin 1 0 结合 π0 2
0 sin 1
解 1 sin 12
取值范围 π π[ )6 2
选 C
9 已知函数 2( ) ln( 1)f x x ax [23]单调递减 a 取值范围
A ( 4] B [6 ) C 10( 4]3 D 10[ 4]3
答案:C
解析: 2( ) ln( 1)f x x ax [23]单调递减
应满足 2 1y x ax [23]恒零单调递减
2
22
3 3 1 0
a
a
解:10 43 a
10 设 2log 6a 3log 15b 02515c
A b a c B a c b C c b a D c a b
答案:C
解析:注意 025 02515 16 2c
252 4 2 4 15 6 25
2 2log 6 log 2 25a
253 9 3 9 17 153 15 23 15 2 25
3 32 log 3 log 3 25b
综 c b a .准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 4 页( 12 页)
二项选择题:题 2 题题 5 分 10 分题出四选项中少 2 选
项符合题目求作出选择中选含错误选项 0 分选出部分正确选项 2 分正
确选项全部选出 5 分
11 已知 1( )A x m 2( )B x m 函数 ( ) 2sin 3
xf x 图象两点 2 1| | π 12345x x k k
m 值
A 0 B 1 C 2 D 3
答案:ABD
解析:已知 ( )f x 周期 6π
3k 时图示时 0m
2k 4k 时图示结合称性时 1m
1k 5k 时图示结合称性时 3m
综题答案 ABD
12 已知函数 )(xf 定义 00 偶函数
0x 时
222
1
2012
)(
|1|
xxf
x
xf
x
.说法正确
A. 2 4x 时 3 1 1( ) 2 2
xf x B. 1(2 1) 2
n
f n n N
C.存 0 0 0x 0( ) 2f x D.函数 ( ) 4 ( ) 1g x f x 零点数10
答案:AD
解析: 2 4x 时 0 2 2x 3( 2) 2 1xf x 准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 5 页( 12 页)
3 11 1( ) 2 22 2
xf x f x A 选项正确
0n 时
01(1) 12f
|1 1|(1) 2 1 0f 矛盾 B 选项错误
f x 偶函数作出正半轴图:
观察图 f x 值域 01 C 选项错误
g x 零点数 1
4f x 根数 f x 1
4y 交点数
观察图 0x 时 5 交点
根称性 0x 时 5 交点.计 10 交点 D 选项正确.
三填空题:题 4 题题 5 分中第问 2 分第 2 问 3 分 20 分答案填答题
卡相应位置
13 面直角坐标系中角 顶点原点始边 x 轴非负半轴重合角 终边点 3 4( )5 5P
sin tan
答案: 4sin 5
4tan 3
14已知 2 3a 3log 2b a ab .
答案: 2log 3 1
解析: 2log 3a 2 3log 3 log 2 1ab .
15 已知 0 0 2 4a b a b> > + 1a a
+ 值 1 1
a b
+ 值________.
答案: 2 3 2 2
4
解析:第(1)问: 1 12 2a aa a
+ 仅 1a a
1a 时=成立
第(2)问: 1 1 1 1 1 1 2 1 2 (3 2 2)( )( 2 ) (1 2 ) (3 2 )4 4 4 4
b a b aa ba b a b a b a b
++ + + + + + ³ + × 准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 6 页( 12 页)
仅 2b a
a b
4 2 4
4 2 2
a
b
时=成立
16 已知函数 2
2
4 4 sin 1( ) 2 2
x x xf x x x
( 2) ( ) 0f x f x a a 取值范围_______
( )f x 值值________.
答案: 2 2
解析:第(1)问:
2
2
1 1 sin 1( )
1 1
x xf x
x
( 2) ( )f x f x
2 2
2 2
2 1 1 sin 2 1 1 1 sin 1
2 1 1 1 1
x x x x
x x
2
2
2 4 4 sin 1 sin 1 22 2
x x x x
x x
2 0a a 取值范围 2 .
第(2)问:法: ( 2) ( ) 2f x f x 知 ( )f x 关点 11 成中心称图形
max min( ) ( ) 2f x f x .
法二: 2
2 2
4 4 sin 1 2 2 sin 1( ) 12 2 2 2
x x x x xf x x x x x
记
2
2 2 sin 1( ) 2 2
x xg x x x
2
sin( 1) 1
x xg x x
R 奇函数
max min max min( ) ( ) ( 1) ( 1) 0g x g x g x g x
max min max min( ) ( ) 1 ( ) 1 ( ) 2f x f x g x g x .准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 7 页( 12 页)
四解答题:题 6 题 70 分.解答应出文字说明证明程演算步
17 (题满分 10 分)
(1)化简求值: 2 0lg5 lg2 (e 3) ln(π 2)
(2) 已知 3tan 4
求 sin( 3π ) cos 2π+
πsin( ) sin( )2
值
解析(1) 2 0lg5 lg2 (e 3) ln(π 2)
原式 lg10 | e 3| 0 ···········································································3 分
1 3 e
4 e ························································································· 4 分
(2)解法:原式 sin cos
sin cos
········································································· 8 分
tan 1
tan 1
············································································· 9 分
3tan 4
原式
3 14
3 14
1
7
·········································· 10 分
解法二:原式 sin cos
sin cos
···········································································8 分
3tan 4
3sin cos4
··················································9 分
原式
3 cos cos4
3 cos cos4
1
7
······················································· 10 分准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 8 页( 12 页)
18 (题满分 12 分)
已知函数 )(xf 二次函数 0)1( f ( 3) (1) 4f f
(1)求 )(xf 解析式
(2)函数 ( ) ( ) ln 1h x f x x R 连续断试探究否存 n nZ 函数 ( )h x 区
间 1n n 存零点存求出符合题意 n 存请说明
解析:(1) ( 3) (1)f f 知二次函数图象称轴 1x ······························ 1 分
( 1) 0f ( 10) ( )f x 顶点········································· 2 分
设 2( ) ( 1)f x a x ··········································································· 3 分
(1) 4f 2(1 1) 4a ··································································· 4 分
设 1a ·························································································· 5 分
2( ) ( 1)f x x ················································································6 分
(2)(1)知 2( ) ( 1) ln 1h x x x + ···························································· 7 分
2( 1) ( 1 1) ln 1 1 ln(2) 0h + ·············································8 分
2(0) (0 1) ln 0 1 1 0h + ······························································· 9 分
(0) (1) 0h h (跳步)···································································· 10 分
函数 ( ) ( ) ln 1h x f x x R 连续断····································11 分
零点存性定理函数 ( )h x (10)存零点····························· 12 分
( 2( 3) ( 3 1) ln 3 1 4 ln(4) 0h + ·········································8 分
2( 2) ( 2 1) ln 2 1 1 ln3 0h + ··············································· 9 分
( 3) ( 2) 0h h (跳步)································································ 10 分
函数 ( ) ( ) ln 1h x f x x R 连续断·································11 分
零点存性定理函数 ( )h x (32)存零点························ 12 分准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 9 页( 12 页)
19 (题满分 12 分)
已知函数 ( ) sin 3 sinf x x x
(1)分段函数形式出 ( )f x [02π]x 解析式画出图象
(2)直接出 ( )f x xR 正周期单调递增区间
解析:(1) [0 π]x 时 sin 0| sin | sinx x x ( ) 4sinf x x ·······································
(π2π]x 时sin 0| sin | sinx x x ( ) 2sinf x x ································ 2 分
4sin [0 π]( ) 2sin (π2π]
x xf x x x
····································································3 分
图象图示 ···························································································6 分
(2) ( 2π) sin( 2π)+3|sin( 2π) | sin 3| sin | ( )f x x x x x f x
知 2π 函数 ( )f x 周期··············································································· 7 分
结合图象 2π 函数 ( )f x 正周期································································8 分
(直接出答案满分)
图 [02π]x 时函数 ( )f x 递增区间 π 3π[0 ][π ]2 2 ···································9 分
( )f x 正周期 2π 函数 ( )f x 递增区间 π[ π π]( )2k k k Z ···················12 分准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 10 页( 12 页)
20 (题满分 12 分)
已知函数 2( ) ( )1
ax bf x a bx
R R 奇函数 (1) 1f
(1)定义证明 ( )f x 1 ( )单调性
(2)解等式 (2 3) (4 1)x xf f
解析:(1)函数 2( ) ( )1
ax bf x a bx
R R 奇函数 (0) 0f ···················· 1 分
0 00 1
11 1
b
a b
····························································································2 分
解 2
0
a
b
2
2( ) 1
xf x x
········································································3 分
1 2 (1 )x x 1 2x x ·········································································· 4 分
2
2 2
2 2
1 2 11 2
1 2 2 2 2 2
1 1
2 ( 1) 2 ( 1)2 2( ) ( ) 1 1 ( 1)( 1)
x x x xx xf x f x x x x x
·································· 5 分
2
1 2 2 1
2 2
1
2( 1)( )
( 1)( 1)
x x x x
x x
········································· 6 分
1 2 (1 )x x 1 2x x
2
2 2
1( 1)( 1) 0x x 1 2 1 0x x 2 1 0x x
1 2( ) ( ) 0f x f x 1 2( ) ( )f x f x ························································· 7 分
( )f x 1 ( )单调递减······································································ 8 分
(2) 2 3 1x 4 1 1x (1) 2 3 4 1x x ··········································· 9 分
等式化 2 2 2 2 0x x x (2 1)(2 2) 0x x ··································· 10 分
解 2 2x 1x ···················································································11 分
等式解集{ | 1}x x ······································································ 12 分准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 11 页( 12 页)
21 (题满分 12 分)
泉州全国休闲食品重生产基食品产业特色产业糖果产量占全国 20.现
拥中国驰名商标 17 件全国食品工业强县2 (晋江惠安)等荣誉称号涌现出达利盼盼
友臣金冠雅客安记回头客等批龙头企业.已知泉州某食品厂需定期购买食品配料该
厂天需食品配料 200 千克配料价格 1 元千克次购买配料需支付运费 90 元.设该厂
隔 x x N 天购买次配料.公司次购买配料均需支付保费标准:6 天(含 6
天)均 10 元天支付超出 6 天支付前 6 天保费外需支付剩余配料保费剩余配
料 3 5
200
x 元千克次性支付.
(1) 8x 时求该厂配料保费 P 元
(2)求该厂配料总费 y (元)关 x 函数关系式根均天支付费请出
合理建议隔少天购买次配料较.
附: 80f x x x
04 5 单调递减 4 5 单调递增.
解析:(1)第 6 天剩余配料 8 6 200 400 (千克)············································· 1 分
3 8 560 400 78200P
······························································ 3 分
(2) 6x 时 200 10 90 210 90y x x x ··············································4 分
6x 时 3 5200 90 60 200 6200
xy x x
23 167 240x x ····5 分
2
210 900 6
3 167 240 6
x x
y
x x x
中 Nx .··············································· 6 分
设均天支付费 f x 元
0 6x 时 210 90 90210xf x x x
·············································· 7 分
f x 06 单调递减 min 6 225f x f ···································· 8 分
6x 时
23 167 240 803 167x xf x xx x
······························· 9 分
知 f x 04 5 单调递减 4 5 单调递增······························· 10 分
8 4 5 9 8 221f 29 220 3f min
29 220 3f x f ·····11 分
综述该厂9天购买次配料均天支付费少 ······················ 12 分准考证号______________ 姓名______________
(卷答题效)
20192020学年度学期教学质量踪监测数学(新教材必修第1册)试题
第 12 页( 12 页)
22 (题满分 12 分)
题略
解析:(1)① )(xf 称轴
2
bx ········································································1 分
2|| b 2b 2b 12
b 12
b
12
b 2b 时 1 1 2 4M m f f b ································ 2 分
12
b 2b 时 1 1 2 4M m f f b ······························· 3 分
综述 mM 取值范围[4 ) (回答 4M m 扣分)·················· 4 分
②① 9
4M m 知| | 2b 1 12
b ········································· 5 分
时
2 91 1 2 2 4
02
b bM m f f
b
············································ 6 分
2 91 1 2 2 4
02
b bM m f f
b
············································ 7 分
1 1b ··················································································· 8 分
(2)
2 2 2
2 cos cos cos 3cos 1 1 12 4 4 4b c c c
·············· 9 分
( )f x 称轴 02
bx
( )f x 12 增函数···································································· 10 分
(1) 1f b c (2) 4 2f b c
11 4 2 2 84b c b c b c
··············································· 11 分
12 14b c b c 3
4b 3
4b
时 cos 1 1
2c
2 3 1( ) 4 2f x x x (2) 6f ·················································12 分
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