文档文本内容
绵阳市高中2021级第三次诊断性考试
理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．
CBACB DDCBC DA
二、填空题：本大题共4小题，每小题5分，共20分．
13．10 14． 15． 16．2
三、解答题：本大题共6小题，共70分．
17．解：（1）由列联表可计算 ，·······4分
∴有95%的把握认为参数调试能够改变产品合格率．····························5分
（2）根据题意，设备更新后的合格概率为0.8，淘汰品概率为0.2．········6分
可以认为从生产线中抽出的6件产品是否合格是相互独立的，················8分
设X表示这6件产品中淘汰品的件数，则 ，·······················9分
可得： ·································10分
．······················································12分
18．解：（1）设 的公差为d，则1，1+d，2+2d成等比数列，················1分
∴ ，解得：d=1或d=−1，·····································3分
而d=−1，不满足 ， ， 成等比数列，
∴d=1，·······················································································4分
∴数列 的通项公式 ．·······················································5分
（2）令 ，·····························6分
∴ ，······················7分
两式相减有： ， ···················8分
∴数列 的前n+1项和为 ，即 ，································9分
又 ，所以 ，·························································10分
∴ ，·····················································11分
数学（理科）评分标准 第 1 页 共 9 页∴ ．············································································12分
19．解：（1）过C作CH⊥ 交 于H，············································1分
∵ 在平面 内的射影落在棱 上，
∴ 平面 ，又 平面 ，···································2分
∴ ，··············································································3分
又 ，且 ，·····················································4分
∴ 平面 ；···································································5分
2）∵ ，则 ，·························6分
（
过C作 交 于 ，连结 ，
AA 与CC 的距离为 则 ，
∵ 1 1
平面 ，则 ，···········································7分
又∵
Rt△CHQ中： ，则 ，
在
且 ，
又
∴ 平面 ∴
（1）知： 平面 ，∴ ，
又由
，则四边形ABHQ为矩形，
∴
∴ ，
ABBA 的面积为3，则BB=3，············································8分
又四边形 1 1 1
分别以 为 轴、 轴、 轴建立如图所示
空间直角坐标系，设 ，
， ， ，
∴
∴ ，
∵
解得 ，············································9分
B(2，0，0)， ， ，
∴
， ，
∴
的法向量为n ，
设平面 1
数学（理科）评分标准 第 2 页 共 9 页∴
，令
，
则n
1
，·························10分
的法向量n ，········································11分
易知平面 2
，
∴
∴平面 与平面 所成锐二面角的余弦值为 ．················12分
20．解：（1）离心率 ，则 ，①·······························1分
当x=1， ，则 ，②····························3分
联立①②得: ，···························································4分
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
数学（理科）评分标准 第 3 页 共 9 页故 ，解得： ，
∴ ，···············································································11分
∴直线AB的方程为： ．············································12分
数学（理科）评分标准 第 4 页 共 9 页1 1
f(x)=( x2 +x)lnx− x2 −x
21．解：（1）当a=1时， 2 4 ，
∴ f' (x)=(x+1)lnx ，则切线斜率k=e+1，······································2分
1
y− e2 =(e+1)(x−e)
f(x) 4
∴曲线 在(e， )处的切线方程： ，···········4分
3
(e+1)x−y− e2 −e=0
4
即： ，························································5分
（2）证明方法一：因为
f' (x)=(x+a)(lnx−lna)，·····························6分
由 得到x>a；由 得到 0<x<a ．
(0，a) (a，+∞)
∴ 在 单调递减，在 单调递增．
5
f(x) =f(a)=− a2
min 4
∴ ，·····························································7分
要证 ，即证： ，
只需证： （*）·········································8分
2x2
g(x)= −lnx−2(1<x<2)
设
ex−1
，
则 ，···········································9分
设 ，
则 ，
易知： 在(1，2)上单调递减，
而 ， ，
x ∈(1，2)
故必存在唯一 0 ，使得 ，·······································10分
x∈(1，x )
∴当 0 时， ，即 ；
x∈(x ，2)
当 0 时， ，即 ，
(x ，2)
∴ 在 上单调递增，在 0 上单调递减．························11分
8
g(2)= −ln2−2>0
g(1)=0 e
而 ， ，
∴ 在(1，2)上恒成立，即（*）式成立，原命题得证．··············12分
数学（理科）评分标准 第 5 页 共 9 页方法二：因为
f' (x)=(x+a)(lnx−lna)，···········································6分
由 得到x>a；由 得到 0<x<a ．
(0，a) (a，+∞)
∴ 在 单调递减，在 单调递增．
5
f(x) =f(a)=− a2
min 4
∴ ，·····························································7分
要证 ，即证： ，
只需证： ，················································8分
设 ，即证 在 (1，2)恒成立.
则 ，
令 ，则 ，·························9分
又∵ ，
∴ ，
∴ 在(1，2)上恒成立.······························10分
∴ 在(1，2)单调递减，又 ，
∴存在 ，使得 在 单调递增，在 单调递减.
又 ，·············································11分
∴ 在 (1，2)恒成立，得证．···········································12分
数学（理科）评分标准 第 6 页 共 9 页方法三：因为
f' (x)=(x+a)(lnx−lna)，···········································6分
由 得到x>a；由 得到0<x<a．
(0，a) (a，+∞)
∴ 在 单调递减，在 单调递增．
5
f(x) =f(a)=− a2
min 4
∴ ，·····························································7分
要证 ，即证： ，
只需证： ，··············································8分
令 ，
则 ，
设 ，··········································9分
∴ ，易知 在(1，2)单调递增．
∴ ，·································································10分
∴ 在(1，2)单调递增，
又 ，
∴存在唯一 ，使得当 ， 单调递减，
当 ， 单调递增，···························11分
又 ，
∴ 在 恒成立，原不等式得证．·································12分
数学（理科）评分标准 第 7 页 共 9 页22．（1）方法一：
√3
tanα=−
令x=0，即 cosα+√3sinα=0 ，解得 3 ，····························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
∴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
sinθ(sinθ+√3cosθ)=1⇒ + sin2θ=1
2 2
从而 ，·····················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
数学（理科）评分标准 第 8 页 共 9 页π π π
= − =
∴∠AOB 2 6 3 ．································································10分
3 3 3(a+b)
a+b= + a+b= ⇒ab=3
23.（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
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≤
2
√1−t=cosθ
令 ，则 ，········································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（理科）评分标准 第 9 页 共 9 页
