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广东省汕头市2018_2019学年度普通高中教学质量监测(期末)高一数学试题及解析word

广东省汕头市2018_2019学年度普通高中教学质量监测(期末)高一数学试题及解析word
广东省汕头市2018_2019学年度普通高中教学质量监测(期末)高一数学试题及解析word

试卷类型A

汕头市2018~2019学年度普通高中教学质量监测

高—数学

本试卷共4页,22小题,满分150分.考试用时120分钟.

考生注意:

1.答卷前,考生务必将自己的准考证号、姓名填写在答题卡上.考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名、考试科目”与考生本人准考证号、姓名是否一致.

2.回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑.如需改动,用橡皮擦干净后,再选涂其它答案标号.回答非选择题时,将答案写在答题卡上.写在本试卷上无效. 3.考试结束后,监考员将试题卷和答题卡一并交回.

第I 卷 选择题

一、选择题:本大题共12小题,每小题5分,满分60分.在每小题给出的四个选项中,只有项是符合题

目要求的.

1.已知集合{0,1,2,3,4}M =,{|(2)(5)0}N x x x =--<,则M N =I ( )

A .{3,4}

B .{2,3,4,5}

C .{2,3,4}

D .{3,4,5}

2.已知平行四边形ABCD 对角线AC 与BD 交于点O ,设AB a =u u u r r ,BC b =u u u r r ,则1()2

a b -=r

r ( )

A .OA u u u r

B .OB uuu r

C .OC u u u r

D .OD u u u r

3.同时掷两个骰子,向上的点数之和是6的概率是( )

A .

118

B .

19

C .

536

D .

12

4.下列函数中,在区间(0,)+∞上为增函数的是( )

A .y =

B .2

(1)y x =-

C .2x

y -=

D .0.5log (1)y x =+

5.已知等差数列{}n a 的前n 项和为n S ,4867a a a +=+,则11S =( )

A .77

B .8

C .154

D .176

6.已知角α的顶点为坐标原点,始边与x 轴的非负半轴重合,终边上有一点(P -,则

cos sin 2παα??

+-= ???

( )

A .3

-

B .

3

C .

3

D .3

-

7.棉花的纤维长度是棉花质量的重要指标,在一批棉花中抽测了60根棉花的纤维长度(单位:mm ),将样本数据作成如下的频率分布直方图:

下列关于这批棉花质量状况的分析,不合理...的是( ) A .这批棉花的纤维长度不是特别均匀 B .有一部分棉花的纤维长度比较短

C .有超过一半的棉花纤维长度能达到300mm 以上

D .这批棉花有可能混进了一些次品

8.若22log log 1x y +=,则2x y +的最小值为( )

A .1

B .2

C .

D .4

9.设0,

2πθ?

?

∈ ??

?,且tan 24πθ??+=- ???,则cos 12πθ?

?-= ??

?( )

A

B

C

D .

10.已知向量(1,1)a =-r

,(1,)b m =r .若向量a -r 与b a -r r 的夹角为,则实数m =( )

A

B .1

C .-1

D .

11.将函数2

()2cos cos 1f x x x x =+-的图象向右平移

4

π

个单位长度后得到函数()g x 的图象,若当0,4x x π??

∈??

??

时,()g x 的图象与直线(12)y a a =≤<恰有两个公共点,则0x 的取值范围为( ) A .75,124ππ??

??

??

B .7,412ππ??

?

???

C .75,124ππ??

??

?

D .5,34ππ??

??

?

12.设()f x 是定义在R 上的偶函数,对任意的x R ∈,都有(2)(2)f x f x -=+,且当[2,0]x ∈-时,

1()12x

f x ??

=- ???

若在区间(2,6]-内关于x 的方程()log (2)0(1)a f x x a -+=>恰有3个不同的实数根,则a 的取值范围是( ) A .(1,2)

B .(2,)+∞

C

D

.2)

第Ⅱ卷 非选择题

二、填空题:本大题共4小题,每小题5分,满分20分.

13.已知2

()()g x f x x =+是奇函数,且(1)1f =,则(1)f -=________. 14.抽样调查某地区120名教师的年龄和学历状况,情况如下饼图

本科学历的构成人数

35岁以下人员学历构成比例

则估计该地区35岁以下具有研究生学历的教师百分比为________.

15.已知n S 为数列{}n a 的前n 项和,()

*1112,n n n S S a n n N --+=+≥∈且844S S =,则11a =________. 16.在ABC V 中,角A ,B ,C 的对边分别为a ,b ,c

,b =且ABC V

面积为)2

2212

S b a c =

--,则面积S 的最大值为________.

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11分)

已知函数()2sin()f x x ωφ=+(其中0,ω>||2

π

φ<

)的最小正周期为π,且图象经过点,26π??

???

. (1)求函数()f x 的解析式; (2)求函数()f x 的单调递增区间. 18.(本题满分11分)

已知数列12n n a -??

?

???

是以2为首项,2为公比的等比数列. (1)求数列{}n a 的通项公式; (2)若()

*2log n n b a n N =∈,求数列11n n b b +??

????

,下的前n 项和n T . 19.(本题满分为12分)

某厂家生产一种产品的固定成本为4万元,并且每生产1百台产品需增加投入0.8万元.已知销售收

入()R x (万元)满足20.610.4(010)

()44(10)

x x x R x x ?-+≤≤=?>?,其中x 是该产品的月(产量(单位:百

台).假定生产的产品都能卖掉,请完成下列问题: (1)将利润表示为月产量x 的函数()y f x =;

(2)当月产量x 为何值时,公司所获利润最大?最大利润为多少万元? 20.(本题满分12分)

在凸四边形ABCD 中,22CD AD ==. (1)若3

2

AB =

,75C ∠=?,60D ∠=?,求sin B 的大小; (2)若2AB =,3BC =且2

A C π

∠-∠=,求四边形ABCD 的面积.

21.(本题满分为12分)

为了解人们对某种食材营养价值的认识程度,某档健康养生电视节目组织8名营养专家和8名现场观众各组成一个评分小组,给食材的营养价值打分(十分制),下面是两个小组的打分数据: 第一小组 8.2 7.5 6.4 9.5 8.3 8.0 1.5 6.6 第二小组 8.8 8.5 9.5 8.6 9.2 8.2 8.9 8.7

(1)求第一小组数据的中位数与平均数,用这两个数字特征中的哪一种来描述第一小组打分的情况更

合适?说明你的理由

(2)你能否判断第一小组与第二小组哪一个更像是由营养专家组成的吗?请比较数字特征并说明理由. (3)节目组收集了烹饪该食材的加热时间t (单位:min )与其营养成分保留百分比y 的有关数据:

在答题卡上画出散点图,求y 关于t 的线性回归方程(系数精确到0.01),并说明回归方程中斜率

?b

的含义.

附注:参考数据:

6

1

1817i i

i t y

==∑,6

21

1235i i t ==∑.

参考公式:回归方程y a b t =+?中斜率和截距的最小二乘估计公式分别为:1

22

1

?,n

i i

i n

i

i t y nt y

b

t

nt =-=-=-∑∑

?a y b

t =-?. 22.(本题满分为12分)

设a R ∈,已知函数()||f x x x a a =--,()

()2()x F x e f x =-?. (1)若0x =是()F x 的零点,求不等式()0F x >的解集; (2)当[2,3]x ∈时,()0F x ≥,求a 的取值范围.

汕头市2018~2019学年度普通高中教学质量监测

高一数学参考答案

一、选择题

二、填空题

16.【解析】)222

1(2cos )sin 2

S b a c ac B ac B =

--=-=Q ,

tan B ∴=5,6B π=cos B =1sin 2

B =,

又b =Q 2

2

8(2a c ac =++≥+,

8(2

ac ∴≤

=,

ABC V 面积111

sin 8(24222

S ac B =≤??=-

ABC ∴V 的面积S 的最大值为4-.

三、解答题

17.【解析】(1)因为2||

T π

πω=

=,所以2ω=, 因为函数()f x 的图象经过点,26π??

???

, 则2sin 23πφ??+=

???,即sin 13πφ??

+= ???

, 因为2

2

π

π

φ-

<<

,即56

3

6

π

π

π

φ-

<

+<

, 则

3

2

π

π

φ+=

,所以6

π

φ=

所以函数()f x 的解析式为:()2sin 26f x x π??

=+

??

?

. (2)令2222

6

2

k x k π

π

π

ππ-

≤+

≤+

,k Z ∈,

得3

6

k x k π

π

ππ-

≤≤

+,k Z ∈.

所以函数()f x 的单调递增区间为,

36k k ππ

ππ??

-

+???

?

,k Z ∈. 18.【解析】(1)由已知数列12n n a -??

?

???

是以2为首项、2为公比的等比数列, 则

11

2222

n n n

n a --=?=,即212n n a -=. (2)因为21

2log 2

21n n b n -==-,

111111(21)(21)22121n n b b n n n n +??

==- ??-+-+??

所以12231

111

n n n T b b b b b b +=

++???+

??? 11111

1123352121n n ????????=

-+-+???+- ? ? ???-+????????

11122121

n

n n ??=-= ?

++??. 19.【解析】(1)由已知,该产品的生产成本()40.8G x x =+,则利润函数

20.610.40.84,(010)

()()()4440.8,(10)x x x x y f x R x G x x x ?-+--≤≤==-=?-->?,

即20.69.64,(010)

()0.840,(10)

x x x f x x x ?-+-≤≤=?-+>?.

(2)当010x ≤≤时,2

2

()0.69.440.6(8)34.4f x x x x =-+-=--+,

为开口向下的抛物线,对称轴为8x =.

所以当8x =时,()y f x =的最大值为34.4万元; 当10x >时,()400.8y f x x ==-为减函数, 则()(10)40832f x f <=-=万元,

综上所述,当月产量为8百台时,公司所获利润最大,最大利润为34.4万元.

20.【解析】(1)连接AC ,在ACD V 中,1AD =,2CD =,60D ∠=?.

由余弦定理得,2222cos 3AC AD DC AD DC D =+-??=

,即AC =

则222AC AD DC =+,90DAC ∠=?,则30DCA ∠=?. 所以,在ACB V 中,3

2

AB =

,753045ACB ∠=?-?=?.

由正弦定理

sin 45sin AB AC B

?=得,sin 3B =.

(2)连接BD ,在ABD V 中,由余弦定理得,

254cos 54cos 54sin 2BD A C C π?

?=-=-+=+ ??

?,

在BCD V 中,由余弦定理得,21312cos BD C =-. 则54sin 1312cos C C +=-,

即sin 3cos 2C C +=,联立221cos sin C C =+得,

210cos 12cos 30C C -+=,所以6cos 10

C =

, 由∠C 为锐角,有sin 23cos 0C C =->,

得2cos 3

C <

,所以cos C =

且sin 23cos C C =-=

则四边形ABCD 面积16(2sin 6sin )cos 3sin 25

S A C C C +=

+=+=. 21.【解析】(1)由已知,第一小组的打分从小到大可排序为:

1.5

6.4

6.6

7.5

8.0

8.2

8.3

9.5

则中位数为

7.58.0

7.752

+=. 平均数为111

(1.57.5 6.4 6.68.08.28.39.5)56788

x =

?+++++++=?=. 可发现第一小组中出现极端数据1.5,会造成平均数偏低,

则由以上算得的两个数字特征可知,选择中位数7.75更适合描述第一小组打分的情况. (2)第一小组:平均数为17x =.

方差:2

2222211(8.27)(7.57)(6.47)(9.57)(8.37)8

s ?=

?-+-+-+-+-? 2221(8.07)(1.57)(6.67)41.4 5.1758?+-+-+-=?=?.

第二小组: 平均数:211

(8.88.59.58.69.28.28.98.7)70.48.888

x =

?+++++++=?=. 方差:2

2222221(8.88.8)(8.58.8)(9.58.8)(8.68.8)(9.28.8)8

s ?=

?-+-+-+-+-? 2221(8.28.8)(8.98.8)(8.78.8) 1.160.1458?+-+-+-=?=?.

可知,22

12s s >,第一小组的方差远大于第二小组的方差,第二小组的打分相对集中, 故第二小组的打分人员更像是由营养专家组成的. (3)由已知数据,得散点图如下,

13,t =Q 27.5y =,且61

1817,i i i t y ==∑6

21

1235i i t ==∑,

则?27.5 2.9013.566.65a y b

t =-?=+?≈. 所以y 关于t 的线性回归方程为 2.966.65y x =-+.

回归方程中斜率?b

的含义:该食材烹饪时间每加热多1分钟, 则其营养成分大约会减少2.9%.

22.【解析】(1)因为0x =是()F x 的零点,则(0)(0)0F f =-=.

得0a =,则()||f x x x =.

不等式()()

()02||020x x F x e x x e x >?-?>?-?>, 解得0x <或ln 2x >,

即不等式()0F x >的解集为(,0)(ln 2,)-∞?+∞. (2)当[2,3]x ∈时,()0F x ≥即()

2()0x e f x -?≥,

因为23x ≤≤时,20x e e ≥>,

则()

2()0()0x e f x f x -?≥?≥恒成立,

当0a ≤时,由[2,3]x ∈,得()||0f x x x a a =--≥恒成立, 所以0a ≤符合题意;

当0a >时,由22,(),x ax a x a

f x x ax a x a

?-+-<=?--≥?,

易知()f x 在,2a ??-∞ ??

?上单调递增,在,2

a a ??????

上单调递减,在[,)a +∞上单调递增,

当02a <<时,min ()(2)2(2)430f x f a a a ==--=-≥, 解得403

a <≤

; 当23a ≤≤时,min ()()0f x f a a ==-<,不符题意,所以a 不存在; 当3a >时,min ()min{(2),(3)}f x f f =, 故令(2)2(2)40(3)3(3)290

f a a a f a a a =--=-≥??

=--=-≥?,解得9

2a ≥,

综上可得,49,,32

a ?

???∈-∞?+∞ ????

?

?

?

高一英语下学期期末考试试题(4)

湖北省宜昌市当阳一中2017-2018学年高一英语下学期期末考试试题 时间:120分钟分值:150分 第一部分听力 (满分30 分) 作题时,先将答案划在试卷上。录音结束后,你有两分钟的时间将答案转涂到答题卡上。 第一节 (共5小题, 每小题1.5分, 满分7.5分) 听下面5段对话。每段对话后有一个小题, 从题中所给的A、B、C三个选项中选出最佳答案, 并标在试卷的相应位置。听完每段对话后, 你都有10 秒钟的时间回答有关小题和阅读下一小题。每段对话仅读一遍。 1. What did the woman do in New York? A. She worked there. B. She studied there C. She visited there. 2. How old is Jerry? A. 25 years old. B. 30 years old. C. 40 years old. 3. What do we know about the woman? A. She is introducing somebody. B. She has seen the man’s sister before. C. She likes African art. 4. Who is the man? A. A taxi driver. B. A hotel worker. C. A restaurant waiter. 5. What are the speakers talking about? A. A park. B. A trip. C. A film. 第二节(共15小题;每小题1.5分, 满分22.5分) 听下面5段对话或独白。每段对话或独白后有几个小题, 从每题所给的A、B、C三个选项中选出最佳选项, 并标在试卷的相应位置。听每段对话或独白前, 你将有时间阅读各个小题, 每小题5秒钟;听完后, 各小题将给出5秒钟的作答时间。每段对话或独白读两遍。 听第6段材料,回答第6、7题。 6. Why did the man quit his job? A. He hated his boss. B. He disliked the work. C. He was overworked. 7. Who is Jim?

2017-2018学年高一下学期期末考试物理试卷 (1)

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