四川省绵阳市高中2025级第一学年末教学质量测试数学+答案

高中2025级第一学年末教学质量测试

数学参考答案及评分标准

一、选择题：本题共8小题，每小题5分，共40分．

1．D2．B3．C4．A5．B6．D7．C8．D

二、选择题：本大题共3小题，每小题6分，共18分．

9．AC10．BCD11．BCD

三、填空题：本题共3个小题，每小题5分，共15分．

12．1；13．3；14．6

四、解答题：共77分。

15．解：（1）由已知得a==(3，2)，b==(_3，3)，c==(0，_5)，··········3分

由c=ma+nb，得(0，−5)=m(3，2)+n(−3，3)=(3m−3n，2m+3n)，·····················4分

于是···············································································5分

解得m=_1，n=_1；················································································6分

（2）因为线段BC的三等分点为M（点M靠近点C），

···········································································8分

设M(x，y)，即(x_4，y_3)=(_2，2)，····················································10分

得到，解得x=2，y=5，························································12分

∴M点的坐标为(2，5)．·········································································13分

16．解：（1）由正弦定理可得，

由asinB=·3bcosA，得sinAsinB=·3sinBcosA，············································4分

∵sinB≠0，····························································································5分

∴tanA=·3，则A．·········································································7分

（2）由2sinB=3sinC及A=，································································9分

∴2sinB=3sinsinBcosB．·············································13分

又sin2B+cos2B=1，············································································14分

∴sinB．··················································································15分

高一数学参考答案第1页(共4页)

17．（1）证明：设AB=a，则AAa，圆柱底面半径为a，

由已知可得：ax，可得a=6，·····································1分

∴AB=6，AAa=9；·······································································2分

∵点G为C1D1与平面A1BE上的交点，

∴点G在平面A1BE上，且平面A1BE∩平面CDD1C1=EG，······························3分

又∵平面ABB1A1//平面CDD1C1，平面A1BE∩平面ABB1A1=BA1，

∴GE//A1B，··························································································4分

易知GE//CD1，则GE//CD1//A1B，

···········································································5分

由EG//CD1//A1B，

∴A1GC平面A1BE，··············································································6分

∵CF/C1G且CF=C1G，

∴FG/DD1/AA1，且FG=DD1=AA1，

∴四边形AFGA1是平行四边形，AF/A1G，···············································7分

又∵AF丈平面A1BE，

∴AF/平面A1BE；···············································································8分

（2）解：连接BD，设BDAF=K，

则BDBK，·······················································9分

设点K到平面A1BE的距离为d，则点D到平面A1BE的距离为d，··············10分

由（1）∵AF/平面A1BE，

∴点A到平面A1BE的距离为d，·····························································11分

∴V=Sxd=xSxd=V=V，分

棱锥D-A1BEΔA1BEΔA1BE棱锥A-A1BE棱锥E-AA1B··············13

∵V=S.BC=xx6x9x6=54，分

棱锥E-AA1BΔA1AB·········································14

∴=．分

VD-A1BEVE-AA1B90·······································································15

方法2：连接BD，设BDAC=O，连接EO，A1O，

∵BD⊥AC且BD⊥AA1，则BD⊥平面A1OE，·············································9分

∴+=+=分

VD-A1BE=VB-EOA1VD-EOA1SEEOABOSEEOAODSEEOABD·················10

∴=---E分

VD-A1BE=SEOA1BD(SACC1A1SECOSA1AOSA1C1)BD··························12

∴分

VD-A1BE···················14

∴VD-A1BE=90．·····················································································15分

高一数学参考答案第2页(共4页)

18．解：（1）∵LABD+LC=90°,∴LCBD+LA=90°,

在△ABD中，························································1分

在△CBD中·······················································2分

又AD=CD，则····················································3分

即

∴sin2LBAD=sin2LBCD，·····································································4分

∵2LBAD，2LBCD∈(0，2π)，

∴2LBAD=2LBCD，或2LBAD+2LBCD=π,············································5分

∴A=C或A+C··············································································6分

（2）∵△ABC为直角三角形，由（1）知B=，AC=8为斜边，·····················7分

由勾股定理可知a2+c2=64，···································································8分

面积Sacx64=16，······················································9分

当且仅当a=c=4（即A=C时取等号，

∴△ABC面积的最大值为16；·································································10分

（3）法一：△ABC中AB≠BC，

∴LABC······················································································11分

令LACB=θ,则CB=8cosθ.

∴CE=8cosθ.···················································································12分

在△DCE中，LDCE

由余弦定理，DE2=DC2+CE2_2DC.CEcos

∴DE2=42+2_2.4.8cosθcos········································13分

∴DE2=64cos2θ+32sin2θ+16，则DE2=32(sin2θ+cos2θ)+48，··············14分

∴DE2=32(sin2θ+cos2θ)+48，

∴DEsin······························································15分

当sin时，DE2的最大值为32+48，········································16分

∴DE的最大值为42+4．····································································17分

高一数学参考答案第3页(共4页)

19．证明：∵PA丄AB，且平面PAB丄平面ABCD，平面PAB平面ABCD=AB，

∴PA丄平面ABCD，

∴PA丄CD，························································································1分

在△ACD中，AC2=AD2+CD2_2AD.CDcos60O=4+1_2=3，

可得AC2+CD2=3+1=4=AD2，

∴AC丄CD，·······················································································2分

又∵PAAC=A，则CD丄平面PAC，

∴PC丄CD；·······················································································3分

（2）①AB=PA.tan30O=3，由（1）可得LBAC=90O_LCAD=60O，

∵AB丄PA，AB丄AD，

∴LPAD=120O

∵AB丄PA，AB丄AD，

∴AB丄平面PAD，

∴平面PAD丄平面ABCD，

过点P作PE垂直于DA延长线于点E，

33

∴PE丄平面ABCD，且PE=PA.sin60O=，·······································5分

2

可得

V棱锥P_ABCSΔABC.PE

设点C到平面PAB的距离为d，

由V棱锥C_PAB=V棱锥P_ABC可得：SΔPAB.d解得d·······················7分

连接CE，在△CDE中，DE=AD+AE

在Rt△PCE中，PCEC2P+E

设PC与平面PAB所成角为θ,则sin

即PC与平面PAB所成角的正弦值为··············································9分

②设Q为BA与DC的交点，连接PQ，平面PAB平面PCD=PQ．

由翻折前，PA丄AB，AD丄AB，翻折后不变，则BA丄平面PAD，