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益阳市2020年上学期高一数学期中考试时量：120分钟满分150分第Ⅰ卷（选择题共60分）一、选择题（本大题共12个小题，每小题5分，共60分，在每小题给出的四个选项中，只有一项是最符合题目要求的.）1.下列

命题正确的是（）A.终边相同的角都相等B.钝角比第三象限角小C.第一象限角都是锐角D.锐角都是第一象限角2.若角的终边经过点34,55P−，则sintan的值是（）A.1615B.1615−C.35

D.35−3.已知2sinsin1+=，则24coscos+=（）A.1B.2C.2D.34.化简()()()()sincossin2cossin−−++−等于A.s

inB.cosC.tan−D.cos−5.若向量a，b满足1a=，2b=，且()aab⊥+，则a与b的夹角为（）A.2B.23C.34D.566.在函数①cos2yx=，②cosyx=

，③cos26yx=+，④tan24yx=−中，最小正周期为的所有函数为（）A.①②③B.①③④C.②④D.①③7.函数()()sinfxx=+（0，2）的图像如图所示，先将图像上所有点的横坐标

伸长到原来的6倍，纵坐标不变，再将所得的图像向左平移72个单位长度，得到函数()gx的图像，下列结论正确的是（）A.()gx是奇函数B.()gx在2,0−上单调递增C.()gx的图像关于()3,0对称D.()gx的图像关于3

x=−对称8.函数()tan23fxx=−的单调递增区间是（）A.5,212212kk−+（kZ）B.5,212212kk−+（kZ）C.2,63kk++

（kZ）D.5,1212kk−+（kZ）9.()()12sin3cos3+−+化简的结果是（）A.sin3cos3−B.cos3sin3−C.()sin3cos3−D.以上都不对10.在ABC△中，A

D、BE、CF分别是BC、CA、AB上的中线，它们交于点G，则下列各等式中不．正确．．的是（）A.23BGBE=B.2CGGF=C.12DGAG=D.0GAGBGC++=11.已知函数()()1cos31xxefxxae−=++，其中0,a，则()f

x的大致图像不可能是（）A.B.C.D.12.已知函数()2sin6fxx=+（*N）有一条对称轴为23x=，当取最小值时，关于x的方程()fxa=在区间,63−上

有且只有一个根，则实数a的取值范围是（）A.1,1−B.)1,1−C.1,0−D.以上都不对第Ⅱ卷（非选择题共90分）二、填空题（本大题共4小题，每小题5分，共20分.把答案填在题中的横线上.）13.7cos6−=_______________.14.已知扇形的圆心角

为6，面积为3，则扇形的弧长等于_______________.15.若1tan2=，则sincos2sin3cos+=−_______________.16.已知点()1,0A，()3,4B，O为坐标原点，点C在AOB的平分线上，且2OC=，则点C的坐

标为_______________.三、解答题：第17小题满分10分，第18至第22小题满分各12分，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10分）已知向量()3,1a=−

，()1,2b=−，()1,1c=.（1）求向量a与b的夹角的大小；（2）若()//cakb+，求实数k的值.18.（本小题满分12分）已知sin，cos是关于x的方程21370xxt−+=的两根，()0,，（1）求t的值

；（2）求tan的值.19.（本小题满分12分）如图，在OAB△中，P为线段AB上一点，且OPxOAyOB=+.（1）若APPB=，求x，y的值；（2）若3APPB=，||4OA=，||2OB=，且OA与

OB的夹角为60°，求OPAB的值.20.（本小题满分12分）已知函数()()sinfxx=+（0，22−剟）的图象上相邻的最高点和最低点的距离为22，且()fx的图像过点12,2−，（1）求函数()fx的解析式；（2）求函数()fx的单调递

减区间（3）求()fx在区间1,2−上的值域.21.（本小题满分12分）在平面直角坐标系xOy中，已知四边形OABC是等腰梯形，()6,0A，()1,3C，点M满足12OMOA=，点P在线段BC上运动（包括端点）

，如图所示.（1）求OCM的余弦值；（2）是否存在实数，使()OAOPCM−⊥？若存在，求出实数的取值范围；若不存在，请说明理由.22.（本小题满分12分）函数()2122cos2sinfxaaxx=−−−的最小值为()ga，aR.函数()21hxx=−（1）求()ga（结论写成分段

函数的形式）；（2）是否存在实数a满足：任给()10,2x，都存在2xR使得()()12hxfx=？若存在，求实数a的取值范围；若不存在，请说明理由.益阳市2020年上学期高一数学期中考试参考答案一、选择题123456789101112DA

ABCADBACBD二、填空题13.32−14.315.34−16.4525,55三、解答题17.【解析】（1）设向量a与b的夹角为，则322cos2105abab−−===−，····················

··························································3分又0,，所以34=，即向量a与b的夹角的大小为34.······························

··································5分（2）()3,12akbkk+=−+−，因为()//cakb+，所以1230kk−+−=，解得43k=，即实数k的值为43.····················

·································································10分18.【解析】（1）由根与系数的关系可得：7sincos13+=，sincos13

t=·······················3分∴()249sincos169+=，∴1202sincos169=−，··························································5分∴6016913t−=，∴6013t=

−·····························································································6分（2）由（

1）可知1202sincos0169=−，又()0,，则sin0，cos0，·········································································

·8分∴,2，故()217sincossincos4sincos13−=+−=，··················································10分由此得到方程组：7sin

cos1317sincos13+=−=·············································································11分解得12sin13

=，5cos13=−，12tan5=−.·································································12分19.【解析】（1）若APPB=，则1122OPOAOB=+，故12

xy==.······································4分（2）若3APPB=，则1344OPOAOB=+，············································································

······················8分13()44OPABOAOBOBOA=+−····································································

··········9分22113424OAOAOBOB=−−+·····················································································10分22113442cos602424

=−−+3=−.······································································································

··················12分20.【解析】（1）因为两个相邻最高点和最低点的距离为22，可得()2211222T++=，解得4T=，·····································

··································1分故22T==，即()sin2xfx=+.·····························································

············2分又函数图象过点12,2−，故()12sin2sin22f=+=−=−，又22−剟，解得6=，···································································

······················3分故()sin26xfx=+.···························································

·····································4分（2）令3222262xkk+++剟，得284433kxk++剟即：()fx的单调递减区间为284,433kk++（kZ）······

···············································8分（3）当1,2x−时，7,2636x+−，根据图像可知()()min312fxf=−=−，()ma

x213fxf==所以()fx在区间1,2−上的值域为3,12−.··································································12分21.【解析】（1）由题意，可得()6,0O

A=，()1,3OC=，()13,02OMOA==，()2,3CM=−，()1,3CO=−−.···················································4分∴7coscos,14||||COCMOCMCOCMCOCM=

==.···························································6分（2）设(),3Pt，其中15t剟，则(),3OPt=，()6,3OAOPt−=−−.若()OAOPCM−⊥，则()

0OAOPCM−=，即12230t−+=，∴()2312t−=，显然32t，∴1223t=−.······································································

························8分∵331,,522t.∴(12,12,7−−+.即满足条件的实数存在，实数的取值范围为(12,12,7−−+.····················

·············12分22.【解析】（1）()()2122cos21cosfxaaxx=−−−−22cos2cos12xaxa=−−−222cos2122aaxa=−−−−.·················································

········································2分若12a−，即2a−，则当cos1x=−时，()fx有最小值()221221122aagaa=−−−−−=；3分若112a−剟，即22a−剟，则当cos2ax=时，()fx

有最小值()2212agaa=−−−；··············4分若12a，即2a，则当cos1x=时，()fx有最小值()2212211422aagaaa=−−−−=−.·5分∴()21,221,22214,2aag

aaaaa−=−−−−−剟，················································································6分（2）函数(

)21hxx=−（()0,2x）的值域为()1,3−························································7分由于()22cos22122aafxxa

=−−−−，cos1,1x−∴当0a时，()max14fxa=−(cos1)x=当0a…时，()max1fx=(cos1)x=−······················································

··························8分∵任给()10,2x，都存在2xR使得()()12hxfx=，∴()hx的值域应为()fx值域的子集··········································

·······································9分①当2a−时，()1,31,14a−−，不成立；②当20a−„时，()21,321,142aaa−−−−−，不成立；③当02a剟时，()21,321,12aa

−−−−，不成立；④当2a时，()1,314,1a−−，不成立综上所述：题设的实数a不存在·············································································

·········12分