文档内容
绵阳市高中2021级第三次诊断性考试
文科数学参考答案及评分意见
一、选择题：本大题共12小题，每小题5分，共60分．
CDACC ACCBD CA
二、填空题：本大题共4小题，每小题5分，共20分．
13．2 14． 15． 16．
三、解答题：本大题共6小题，共70分．
17．解：（1）①当 时， ，则 ．································1分
②当 时，由 ，得 ，······························2分
两式相减，得 ，·······················································3分
，即 ，··················································4分
∴
∴数列 是以 为首项， 为公比的等比数列，························5分
∴ ．·········································································6分
（2）由（1）得 ，······························8分
可知数列 是以 为首项，4为公比的等比数列，····························9分
∴ ······································································10分
········································································11分
．（也可不计算到此步）·····································12分
18．解：（1）调试前，电池的平均放电时间为：
2.5×0.02×5+7.5×0.06×5+12.5×0.08×5+17.5×0.04×5=11小时，···················4分
调试后的合格率为：0.1×5+0.06×5=0.8，则 ，·······················5分
∴a=48；·····················································································6分
数学（文科）评分标准 第 1 页 共 7 页（2）由列联表可计算 ，·················10分
∴有95%的把握认为参数调试能够改变产品合格率．··························12分
19．解：（1）∵E是BP的中点，AB=AP，
∴AE⊥PB，·················································································1分
又平面PAB∩平面PBC=PB，且平面PAB⊥平面PBC，
∴AE⊥平面PBC，·········································································2分
过D作DF⊥PC交PC于F，
∵平面PCD⊥平面PBC，且平面PCD∩平面PBC=PC，
∴DF⊥平面PBC，········································································4分
∴AE∥DF，·················································································5分
又DF 平面PCD，AE 平面PCD，
∴AE∥平面PCD；········································································6分
（2）∵AD∥BC
∴V =V =V =V ，
C-PBD D-PBC A-PBC C-PAB
∴V =V = S ·d，·······························································8分
C-PBD C-PAB △ABP
又∵平面PBC⊥平面PAB，过C作CH⊥PB交PB于H，
∴CH⊥平面PAB，·········································································9分
在直角△CHB中： ，·····················10分
∴ ，·································11分
∴当sin∠BAP=1时，体积的最大值为 ．········································12分
数学（文科）评分标准 第 2 页 共 7 页20．解：（1）解：当 时， ，·······················1分
a=1
，··········································································2分
此时切线斜率为： ；····························································3分
所以曲线 在(e， )处的切线方程： ···············4分
即： ；····························································5分
（2）证明方法一：因为 ，·································6分
由 得到x>a；由 得到0a；由 得到0b>0，所以a−b=2， ②·······················································4分
数学（文科）评分标准 第 6 页 共 7 页由①②得：
a=3,b=1
；··································································5分
√3−at+√bt=√3−3t+√t=√3√1−t+√t
（2） ，·································6分
π
√t=sinθ,0≤θ≤
令 2 ，则
√1−t=cosθ
，··········································7分
π
√3√1−t+√t=√3cosθ+sinθ=2sin(θ+ )
∴ 3 ，··································9分
π 1
θ= t=
∴当 6 时，即 4 时，
√3−at+√bt
的最大值为2．·····················10分
数学（文科）评分标准 第 7 页 共 7 页
