福建省三明市普通高中2022-2023学年高一上学期期末质量检测数学试题含答案

（在此卷上答题无效）

三明市普通高中2022-2023学年第一学期期末质量检测

高一数学试题本试卷共5页．满分150分．注意事项：1．答题前，考生务必在试题卷、答题卡规定的地方填写自己的姓名、准考证号．考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致．2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其它答案标号．非选择题用0.5毫米黑色签字笔在答题卡上书写作答．在试题卷上作答，答案无效．一、选择题：本题共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.已知集合AxZxx20，Bx0x2，则AB

A．1,0,1,2B．0,1,2C．0,2D．1,222．设a30.7，b30.4，clog30.7，则a,b,c的大小关系是A．bac3．函数fxe

A．0,1x1

B．acbC．cabD．abc



1

2的零点所在区间为x1B．1,2C．2,3D．3,44.在平面直角坐标系中，角的顶点为坐标原点，始边与x轴的非负半轴重合，若角的终边经过点P(m,2m)(m0)，则A．4

5x

3sin2cos的值为2sincosC．5

D．

B．5

451x

5.函数2图象的大致形状是y

xA．B．试卷第5页，共6页C．D．6．大气压强P=

压力

，它的单位是“帕斯卡”（Pa,1Pa1Nm2），大气压强PPa受力面积

kh

随海拔高度hm的变化规律是PP0e(k0.000126m1),P0是海平面大气压P11强.已知在某高山A1,A2测得的大气压强分别为P，那么A1,A2两处1,P2，且P22的海拔高度的差约为（参考数据：ln20.693）A.550mB.1818mC.5500m

D.8732m

log（2x),x0,7．若函数fx为奇函数，则fg2

gx,x0.A．2

B．1

C．0

D．1

8.已知函数f(x)sin(x)(0,0)，若f(x)为奇函数，f(x)为偶函数，且f(x)A．10

π

2π8π8π2在(0,)至多有2个实根，则的最大值为62B．14C．15

D．18

二、选择题：本题共4小题，每小题5分，共20分。在每小题给出的四个选项中，有多个选项符合题目要求。全部选对的得5分，有选错的得0分，部分选对的得2分。9.已知ab1，c0，则下列四个不等式中，一定成立的是A．ccab

B．acbcD．abc

C．abcbac10.下列说法正确的是A．命题“a1，a210”的否定是“a1，a210”B．“lnalnb”是“ab”的充分不必要条件C．fx

x1x1与gx

2x1x1表示同一函数D．函数fx2xmx1在区间1,单调递增，则实数m的取值范围是4,11.函数f(x)2sin(x)(0,||π)的部分图象如图所示，下列结论正确是试卷第5页，共6页A．f(x)2sin(x)B．不等式fx1的解集为x6kππx6kπ3π,kZC．若把函数f(x)的图象向左平移13π3π

个单位长度，得到函数h(x)的图象，则函数2h(x)是奇函数D．f(x)图象上所有点的横坐标缩短到原来的图象，则函数g(x)在

1

，纵坐标不变，得到函数g(x)的32π5π

,上是减函数33

2

cosx的结论正确的有x1e

B．在0,

12.下列关于函数fx1





A．图象关于原点对称

π

上单调递增2

C．在

π

,π上单调递减2

D．值域为1,1三、填空题：本题共4小题，每小题5分，共20分。13.9log24

124

．14.函数fxloga(x1)1(a0且a1)的图象恒过定点

π33ππ<<,则cos=352

．．15．已知cos

23x,x0,1

,若faf(a)，则a=16.已知函数f(x)21x,0x1

．四、解答题：本题共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．（10分）已知集合Ax|





1

2x116,Bx(xm)(xm1)0.2

（1）求集合A；（2）若ABB，求实数m的取值范围.试卷第5页，共6页18.（12分）已知sin(π)2cos(2π).（1）若为锐角，求cos()的值；（2）求tan(2

π6π

)的值.419.（12分）某革命老区县因地制宜的将该县打造成“生态水果特色小县”．该县某水果树的单株产量（单位：千克）与施用肥料x（单位：千克）满足如下关系：x230,0x3,

(x)，且单株施用肥料及其它成本总投入为10x元．己知这4

,3x643

x2

种水果的市场售价为10元/千克.在国务院关于新时代支持革命老区振兴发展的意见，支持发展特色农业产业的保障下，该县水果销路畅通．记该水果树的单株利润为fx（单位：元）．（1）求函数fx的解析式；（2）当施用肥料为多少千克时，该水果树的单株利润最大？最大利润是多少？20.（12分）已知函数fxsinx(cosx

33sinx)cos2x，xR.22（1）求函数fx的单调递增区间；（2）若fxm2在

ππ

,上恒成立，求实数m的取值范围．43

试卷第5页，共6页21.（12分）已知函数f(x)log281

x3x.2（1）判断fx的奇偶性，并加以证明；2

（2）判断函数fx的单调性（无需证明）；若xR，都有f(1ax)f4x，求实数a的取值范围.22．（12分）“函数(x)的图象关于点m,n对称”的充要条件是“对于函数(x)定义域内的任=2n”.已知函数f(x)的图象关于点2,2对称，且意x，都有(x)(2mx）

当x0,2时，f(x)x22ax4a2.（1）求f(0)f(4)的值；（2）设函数g(x)

115x

,x2（i）证明函数g(x)的图象关于点2,5对称；（ii）若对任意x10,4，总存在x2

数a的取值范围.27

,3,使得f(x1)g(x2)成立，求实13

试卷第5页，共6页数学试题第6页6页）（共2022-2023学年第一学期三明市期末质量检测

高一数学参考答案及评分细则

评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、选择题：本题共8小题，每小题5分，共60分。1．B2．D3．B4．A5．D6．C7．C8．A二、选择题：本题共4小题，每小题5分，共20分。全部选对的得5分，有选错的得0分，部分选对的得2分。9．BC13．215．10．AB11．BCD14．2,116．12．ACD三、填空题：本题共4小题，每小题5分，满分20分．3431014四、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.（10分）解:（1）因为1

2x116，2所以212x124，······················································································2分所以1x14，························································································3分即2x3,·································································································4分所以集合Ax|2x3.············································································5分（2）依题意Bxm1xm,····································································6分因为ABB，所以BA.···········································································7分m12,所以····························································································9分m3.

即1m3.所以m的取值范围为1,3.············································································10分1/618.（12分）解：（1）由已知得sin2cos，···································································1分又sin2cos21，且为锐角，·································································2分解得cos

525，······································································4分，sin

55∴cos



πππ

coscossinsin···························································5分666

532511525；··········································6分

525210（2）由（1）得tan2，··············································································7分所以tan2

2tan，·············································································8分1tan2224············································································9分2123所以tan(2

tan21)41tan2······································································10分4

1

1

3

4713··································································12分19.（12分）10x23010x,0x3,



解：（1）f(x)4

104310x,3x6.

x2

10x210x300,0x3,



即f(x)·····················································4分40

10x,3x6.430

x2

（2）当0x3时，f(x)10x10x300，211

f(x)在0,单调递减，在,3单调递增，·····················································5分22

则当x3时，f(x)取到最大值为360.······························································7分2/6当3x6时，f(x)430404

x2.·····················8分10x41010

x2x2

4x2370，····························10分x2因为x20，所以fx41020当且仅当4

x2，即x4时,f(x)取到最大值为370,······························11分x2因为370360，所以当施用肥料为4千克时，该水果树的单株利润最大，最大利润是370元.···········12分20.（12分）（1）fxsinxcosx

32313sinxcos2xsin2xcos2xsin2222π

2x.

3

····················································································································3分πππ

2x2kπ,kZ,································································4分2325ππ

xkπ,kZ,·····································································5分解得kπ1212由2kπ

5ππ

所以函数fx的单调增区间为kπ,kπ,kZ.································6分1212ππππ

（2）由x,得2x,π，······················································7分3643π1

所以sin2x,1，

32

1

即fx,1，·······················································································9分2

因为fxm2在

ππ

,上恒成立，所以mfx2min.·························10分43

5

fx2又因为，········································································11分min

255

则m，所以m的取值范围为,.··················································12分22

21.（12分）（1）f(x)是偶函数.······················································································1分3/6证明：f(x)log281因为f(x)log28

x3

x，定义域为R关于原点对称，································2分2x31x

2

8x13

log2xx···········································································3分82log28x1log28xlog28x1

3

xfx,233

xlog28x13xx··························5分22所以fx是偶函数.·······················································································6分（2）fx在是减函数；在是增函数，········································7分（，0）（0，+）又因为fx是偶函数，所以xR，都有f(1ax)f4x

2

，等价于|1ax||4x

2

|在R上恒成立，····················································································································8分即|1ax|4x2在R上恒成立，即4x

2

1ax4x

2

在R上恒成立.························································9分2

xax30所以2在R上恒成立，································································10分xax502a120所以，解得23a23.2

a200

所以a的取值范围是23,23.···································································12分22.（12分）（1）解：因为函数f(x)的图象关于点2,2对称，所以f(x)f(4x)4，············································································1分所以f(0)f(4)=4.····················································································2分（2）（i）证明：因为g(x)所以g(4x)

115x

，x,22,，x2115(4x)5x9

=,·····························································3分(4x)22x4/6所以g(x)g(4x)

115x5x92010x

10.x22xx2即对任意x,22,，都有g(x)g(4x)10成立.所以函数g(x)的图象关于点2,5对称.·······················································4分（ii）由g(x)=

115x15,x2x2易知g(x)在

27

,3上单调递减，13

27

,3上的值域为4,8.·····················································5分13

所以g(x)在x

设函数yf(x),x0,4的值域为A.27xx0,4若对任意1，总存在2,3，使得f(x1)g(x2)成立，则A4,8.13

因为x0,2时，f(x)x22ax4a2，所以f(2)2，即函数f(x)的图象过对称中心2,2.①当a0时，函数f(x)在0,2上单调递增.因为函数f(x)的图象关于点2,2对称，所以f(x)在2,4上单调递增，所以函数f(x)在0,4上单调递增.易知f(0)4a2，又f(0)f(4)4，所以f(4)64a，则A=4a2,64a.4a24,

又因为A4,8，所以64a8,

4a264a.

解得

1

a0.······················································································7分2②当0a2时，函数f(x)在0,a上单调递减，在a,2上单调递增.由函数f(x)的图象关于点2,2对称,知f(x)在2,4a上单调递增，在4a,4上单调递减.所以函数f(x)在0,a上单调递减，在a,4a上单调递增，在4a,4上单调递5/6减.因为，f(0)4a2(2,6)，f(a)a4a22,2，由函数f(x)的图象关于点2,2对称得2

f(0)f(40)4，f(a)f(4a)4，所以f(4a)2,6，f(4)(2,6)，a2时A4,8恒成立.····················································9分所以，当0

③当a2时，函数f(x)在0,2上单调递减.由函数f(x)的图象关于点2,2对称,知f(x)在2,4上单调递减.所以函数f(x)在0,4上单调递减.易知f(0)=4a2，又f(0)f(4)4，所以f(4)64a，则A=64a,4a2.64a4,由A4,8，得4a28,64a4a2.解得2a

5

.·······················································································11分215

,.···················································12分22

综上所述，实数a的取值范围为

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