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高一阶段性教学质量检测数学试题2020.05一、单选题：本题共8个小题，每小题5分，共40分.1.若复数：满足()12i43iz−=−，则z在复平面内对应的点在（）A.第一象限B.第二象限C.第三象限D.第四象限2.已知角α的终边在直线2yx=上，则tan4−=（）

A.13B.13−C.3D.3−3.在ABC△中，角A，B，C的对边分别为a，b，c，4a=，22b=，45A=，则B=（）A.30°B.60°C.60°或120°D.30°或150°4.已知a，b是夹角为60°的单位向量，则23ab−=（）A.7B.13C.7D.1

35.某种浮标是一个半球，其直径为0.2米，如果在浮标的表面涂一层防水漆，每平方米需要0.5kg涂料，那么给1000个这样的浮标涂防水漆需要涂料（）（取3.14）A.47.1kgB.94.2kgC.

125.6kgD.157kg6.已知函数()()sinfxAx=+（0A，0，2）的部分图象如图所示，则12f=（）A.12B.1C.2D.37.已知一个圆柱的侧面积等于表面积的23，且其轴截面的周长是16，则该圆柱的体积是（）A.54B.36

C.27D.168.已知3sin63+=，则2cos23−的值为（）A.19−B.19C.13−D.13二、多选题：本题共4个小题，每小题5分，共20分.全部选对得5分，部分选对得3分，有选错的得0分.9.用一个平面去截一个几

何体，截面的形状是三角形，那么这个几何体可能是A.圆锥B.圆柱C.三棱锥D.正方体10.已知，是两个不同的平面，m，n是两条不同的直线，则下列说法中正确的是A.若m⊥，//mn，n，则⊥B.若m，n，//m，//n，则//C.若//，m，n，则//mnD

.若⊥，m，n=，mn⊥，则m⊥11.如图，四边形ABCD为直角梯形，90D=，//ABCD，2ABCD=，M，N分别为AB，CD的中点，则下列结论正确的是A.12ACADAB=+B.1122MCACBC=+C.12BCADAB=−D.14MNADAB=+

12.将函数()2sin223cos3fxxx=+−图象向左平移12个单位，再把各点的横坐标伸长到原来的2倍（纵坐标不变），得到函数()gx的图象，则下列说法中正确的是A.()fx的最大值为3B.()gx是奇函数C.()fx的图

象关于点,06−对称D.()gx在2,63上单调递减三、填空题：本题共4个小题，每小题5分，共20分.13.sin72cos27sin18sin27−的值为_____________

_.14.若正方体的外接球的体积为43，则此正方体的棱长为____________.15.在ABC△中，角A，B，C所对的边分别为a，b，c，若27a=，2b=，60A=，则c=__________.16.已知向量()2sin,1a

x=，()1,cosbx=，则ab的最大值为________；若//ab且(),0x−，则x的值为__________.（第一个空2分，第二个空3分）四、解答题：本题共6个小题，共70分.17.（10分）已知复数()2i42iiz=

+−+.（1）求复数z的模z；（2）若223izmzn−−=+（m，nR），求m和n的值.18.（12分）已知向量()2,1a=，()3,1b=−.（1）求向量a与b的夹角；（2）若()3,cm=（mR），且()2abc−⊥，求m的值19.（12分）已知向量cos2,2sin

34axx=−−，1,sin4bx=+，函数()fxab=.（1）求()fx的最小正周期和()fx的图象的对称轴方程；（2）求()fx在区间,122−上的值域.20.（12分）如图，在四棱锥PABCD−中

，PDABCD⊥平面，//ADBC，2BCAD=，ADCD⊥，E，F分别是BC和PB的中点，（1）证明：ADPC⊥；（2）证明：//AEFPCD平面平面.21.（12分）在ABC△中，角A，B，C的对边分别为a，b，c，coscos2cos0aBbAcA++=.（

1）求A；（2）若2abc=+，ABC△的外接圆半径为1，求ABC△的面积.22.（12分）如图，在三棱柱111ABCABC−中，1AAABC⊥平面，16ABAA==，8AC=，D，E分别是AC，1CC的中点.（1）求证：11//ABBDC平面；（2）若异面直线AB与11AC所成的角为

30°，求三棱锥1CBDE−的体积.高一阶段性教学质量检测数学试题参考答案及评分标准2020.05一、单选题：本题共8个小题，每小题5分，共40分.1.D2.B3.A4.C5.A6.D7.D8.C二、多选题：本题共4个小题，每小题5分，共20分.全部

选对得5分，部分选对得3分，有选错的得0分.9.ACD10.AD11.ABC12.CD三、填空题：本题共4个小题，每小题5分，共20分.13.2214.215.616.5；34−四、解答题：本题共6个小题，共70分.17.（10分）解：（1）

()2i42ii2i4i243iz=+−+=+−+=−，··························································3分则22435z=+=.·······················

···············································································5分（2）由（1）知43iz=−，43iz=+，························

····················································6分∴()243i43i223izmznmn−−=−−+−=+，即()44233i23imnm−−+−−=+，·············

··································································8分∴4422333mnm−−=−−=，········································

····························································9分解得25mn=−=.··················································

···························································10分18.（12分）解：（1）∵()2,1a=，()3,1b=−，∴.()23115abx=+−=，·····························

·····························································1分2215a=+=，····································································

···································2分()223110b=+−=，·····························································································

··3分设向量a与b的夹角为，则52cos2510abab===，···················································································5分∵0,，4=，即向量a与

b的夹角为4.·······························································································6分（

2）∵()2,1a=，()3,1b=−，∴()24,3ab−=−，······································································································7分∵

()2abc−⊥，∴()20abc−=，··················································································

·····················9分∵()3,cm=，∴()4330m−+=，·································································································

·11分解得4m=.············································································································

····12分19.（12分）解：（1）()cos22sinsin344fxabxxx==−+−+···········································1分13coscos2sinsin

22sincoscos2sin2sin23344222xxxxxxx=++−−=++−31sin2cos2sin2226xxx=−=−，即()sin26fxx=−，·············

··················································································4分∴()fx的最小正周期22T==，··············

···································································5分令262xk−=+（kZ），得23kx=+（kZ），∴()fx的对称轴方程为23

kx=+（kZ）.·································································7分（2）∵122x−剟，52366x−−剟，·····················

··················································8分∴当262x−=，即3x=时，()sin26fxx=−取得最大值1；···························

··········9分当263x−=−，即12x=−时，()sin26fxx=−取得最小值32−，··························11分∴()fx在区间,122−

上的值域为3,12−.·······························································12分20.（12分）证明：（1）∵PDABCD⊥平面，ADA

BCD平面，∴ADPD⊥，·············································································································2分又ADCD⊥，

PDCDD=，∴ADPCD⊥平面，····································································································4分∵PCPCD平面，∵ADPC

⊥.···············································································································

·6分（2）2BCAD=，E为BC的中点，∴ADCE=，································································································

··············7分又∵//ADCE，∴四边形AECD为平行四边形，·································································8分∴

//AECD.··················································································································9分∵

在BCP△中，E，F分别是BC和PB的中点，∴//EFPC，·············································································································10分∵EFAEE

=，PCCDC=，∴//AEFPCD平面平面.···········································································

···················12分21.（12分）解：（1）∵coscos2cos0aBbAcA++=，∴由正弦定理得sincossincos2sincos0ABBACA++=，············

····································2分∴()sin2sincos0ABCA++=，······································································

·············3分∵ABC+=−，∴()()sinsinsinABCC+=−=，∴sin2sincos0CCA+=，·······································································

····················4分又C为三角形的内角，sin0C，∴12cos0A+=，∴1cos2A=−，·········································································

·········5分又A为三角形内角，∴23A=；·························································································

·······················6分（2）设ABC△的外接圆半径为R，则1R=，∴由正弦定理得2sin3aRA==，23bc+=，···················································

··········8分由余弦定理得()222222ccos3abcbbcbc=+−=+−，····················································10分∴312bc=−，∴9bc=.·····

·························································································11分∴ABC△的面积为：193sin24SbcA==.····················

················································12分22.（12分）解：（1）证明：如图，连接B，C，交BC，于点F，连接DF，··················

···························1分在1ACB△中，由于D为AC的中点，F为1BC的中点，∴DF为1ACB△的中位线，∴1//DFAB，···············································

································································3分∵1DFBDC平面，11ABBDC平面，∴11//ABBDC平面；·················

···················································································5分（2）∵11//ACAC，∴BAC即为异面直线AB与11AC所成的角，·······

·····························································6分∵异面直线AB与11AC所成的角为30°，∴30BAC=，·······························

··········································································7分∴111sin6812222ABCSABACBAC===△，·

·······················································8分∵D是AC的中点.∴162DBCABCSS==△△，·······························

································································9分又∵1CCABC⊥平面，16CC=，E是1CC的中点.∴1111661233BCDCBCD

VSCC−===三棱锥△.··································································10分1163633BCDEBCDVSCE−===三棱锥△，································

·····································11分∴111266CBDECBCDEBCDVVV−−−=−=−=三棱锥三棱锥三棱锥即三棱锥1CBDE−的体积为6.····················

···································································12分