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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.135.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.47.1

kgB.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⊥，//m

n，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.sin72cos27sin18

sin27−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,1ax=，()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7

0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cos2,2sin34axx=−−，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13cos

cos2sinsin22sincoscos2sin2sin23344222xxxxxxx=++−−=++−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23kx=+（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，ADABCD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）∵coscos2cos0aBbAc

A++=，∴由正弦定理得sincossincos2sincos0ABBACA++=，················································2分∴()sin2sincos0ABC

A++=，···················································································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2sin3aR

A==，23bc+=，·····························································8分由余弦定理得()222222ccos3abcbbcb

c=+−=+−，····················································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.∴1111661233BCDCBCDVSCC−===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