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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0<x<a．
∴ 在(0，a)单调递减，在(a，+∞)单调递增．
∴ ，······························································7分
要证 ，即证： ，
只需证： ，·······················································8分
设 ，即证： 在 恒成立.···············9分
则 ，
令 ，························································10分
∴ ，
∴ 在 上单调递增，则
∴ 在 上单调递增，则 ····························11分
∴ 在 恒成立，则 在 上单调递增，
∴ ，原不等式得证.·················································12分
数学（文科）评分标准 第 3 页 共 7 页方法2：因为f' (x)=(x+a)(lnx−lna)，············································6分
由 得到x>a；由 得到0<x<a．
∴ 在(0，a)单调递减，在(a，+∞)单调递增．
5
∴f(x) =f(a)=− a2 ，·····························································7分
min 4
要证 ，即证： ，
即证： ，即证： ，
只需证： ，
令 ，则 ，
即证： ，·····························································8分
又∵ 且 ，则 ，
∴ 在 单调递减，···················································9分
又 ， ，
∴即证 ，只需证： ，·································10分
令 ，
∴ ，则 在 单调递增，·····················11分
∴ ，即 ，所以原不等式得证．···············12分
21．解：（1）离心率 ，则 ，①·······························1分
当x=1， ，则 ，②····························3分
联立①②得: ，···························································4分
数学（文科）评分标准 第 4 页 共 7 页故椭圆C方程为： ；·······················································5分
（2）设过F，A，B三点的圆的圆心为Q (0，n)， ，
又 ，
则 ，即 ，·················6分
又 在椭圆 上，故 ，
带入上式化简得到： ，③··········································7分
同理，根据 可以得到： ，④ ···················8分
由③④可得： 是方程 的两个根，则 ，·····9分
设直线AB： ，联立方程： ，
整理得： ，⑤················································10分
故 ，解得： ，
∴ ，···············································································11分
∴直线l的斜率为： ．···························································12分
22．解：（1）方法一：令x=0，即
cosα+√3sinα=0
，
√3
tanα=−
3
解得 ，········································································1分
5π 11π
α= +2kπ α= +2kπ，k∈Z
∴ 6 或 6 ，·········································2分
5π 1 √3
α= +2kπ y=2+ −√3×(− )=4
当 6 时， 2 2 ；·································3分
11π 1 √3
α= +2kπ y=2− −√3× =0
当 6 时， 2 2 ，····································4分
数学（文科）评分标准 第 5 页 共 7 页∴曲线C 与y轴的交点坐标为（0，4），（0，0）．···························5分
1
方法二：消参：由C 的参数方程得：
1
x2 +(y−2) 2 =(cosα+√3sinα) 2 +(sinα−√3cosα) 2 =1+3=4
，·············1分
x2 +(y−2) 2 =4
即曲线C 的普通方程为： ，·······································2分
1
令x=0，得 y=0 或4，··································································4分
∴曲线C 与y轴的交点坐标为（0，4），（0，0）．···························5分
1
x2 +(y−2) 2 =4
（2）方法一：将曲线C ： 化为极坐标方程，
1
ρ=4sinθ
得： ，···········································································6分
{ ρsin(θ+ π )=2
3 π
¿sin(θ+ )=2
联立C ，C 的极坐标方程 ρ=4sinθ ，得4sinθ 3 ，
1 2
1−cos2θ √3
+ sin2θ=1
sinθ(sinθ+√3cosθ)=1
2 2
从而 ，则 ····················7分
π 1 π π 5π
sin(2θ− )= 2θ− = 或
整理得： 6 2，所以 6 6 6 ，······························8分
π π
θ= 或
即 6 2，·············································································9分
π π π
= − =
∴∠AOB 2 6 3 ．··································································10分
π
ρsin(θ+ )=2
方法二：将C 的极坐标方程 3 ，
2
√3x+y−4=0
化为直角坐标方程： ，················································6分
2π
∴C 是过点（0,4）且倾斜角为 3 的直线，·········································7分
2
π π
= =
不妨设B（0,4），则∠OBA 6 ，因为BO为直径，所以∠BAO 2 ， ···9分
π π π
= − =
∴∠AOB 2 6 3 ．··································································10分
3 3 3(a+b)
a+b= + a+b= ⇒ab=3
23．解：（1）由 a b，得 ab ， ①··················1分
f(x)=|x−a|+|x−b|≥|（x−a)−(x−b)|=|b−a|=2
又由 ，·························3分
且a>b>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
