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绵阳市高 2022 级第一次诊断考试
数学参考答案和评分标准
一、选择题：本题共8小题，每小题5分，共40分．
1．B 2．A 3．D 4．D 5．C 6．B 7．C 8．A
二、选择题：本大题共3小题，每小题6分，共18分。在每小题给出的四个选项中，
有多项符合题目要求。全部选对的得6分，选对但不全的得部分分，有选错的得0
分．
9．AB 10．ABD 11．BCD
三、填空题：本题共3个小题，每小题5分，共15分．
12． ； 13．−8； 14．
四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17
分，共77分．解答应写出文字说明、证明过程或演算步骤．
15．解：（1）
有报考意
无报考意向 合计
向
男学生 100 400 500
女学生 100 300 400
合计 200 700 900
··································································································· 2分
男学生有报考意向的概率 ，··············································· 4分
女学生有报考意向的概率 ；·············································· 6分
（2）零假设为H：有报考军事类院校意向与性别独立，
0
∴
，······························ 9分
根据小概率值α＝0.1的独立性检验，我们推断H 不成立，即认为报考军事类院校
0
数学试题 第1页，共5页与
性别有关联，此推断犯错误的概率不大于0.1，故能认为报考军事类院校意愿与性
别有关．······················································································ 13分
16．解：（1）∵ ，由余弦定理可得：
，·······················································2分
∴ ，则b=1，······································································· 3分
又 ，
∴ = = ；·································································· 6分
b=1，B= ，
（2）∵
在
中，由正弦定理： ，························· 8分
∴ ，又∵ ，
∴ ，则 ，·································· 10分
∴ ，则 ，································ 12分
∴ ，又∵A是三角形内角，
∴2A= 或 ，··········································································· 14分
∴A= 或 ．············································································· 15分
17．解：（1）由已知可得， ， ， ，则 ，····· 2分
∴ ， ，
数学试题 第2页，共5页∴ ，············································································· 3分
又∵ + =4，
∴ ；····················································································· 4分
（2）（i）由题意得 ，得 ，
····························································································6分
∴ ，············································································· 7分
∴ (n≥2)，·············· 8分
∴ ，则 (n∈N*)，
∴ ，·········································· 10分
（ii）由（i）可得： ，························ 11分
因此， 是以2为首项，1为公差的等差数列，·························· 12分
∴
．····································································· 15分
18．解：（1） ，·················································· 1分
设切点为 ，
则切线斜率 ，············································ 3分
切线方程为： ，
数学试题 第3页，共5页∵切线过(0，2)．将(0，2)代入上式整理得： ，············· 5分
即： 该方程有三个实数解，故切线有三条，
其中一解 ．故一切线方程为： ；······························· 7分
（2） ，
（i）当a=0时， 成立，f(x)在R上单调递增；··················· 8分
（ii）当a>0时，由 ，得 或 ，································9分
由 ，得 ，························································· 10分
∴ 在 上单调递减，在 和 上单调递增；······11分
（iii）当a<0时， 在 上单调递减；在 和 上单调递
增．·····························································································13分
（3）由（2）可知，
当a=0时， 在R上单调递增，有唯一零点；··································14分
当a≠0时，由唯一零点得 ，即 且a≠0，··········16分
综上所述： 有唯一零点，即 ．····································17分
19．解：（1） ，·····················1分
， ， 单调递增；
， ， 单调递减，··············································3分
∴ ，
∴a=2；··························································································4分
数学试题 第4页，共5页（2）（i） ，
，
可得
∴
，·································································6分
，则 ，
设
令
， ，···············································8分
为增函数，且 ，∴ 为增函数，
∴
又 ，则 ，即当 时， ，
∵ ，则 ，
∴ ， ，……， ，
∴ ；·····················································································11分
（ii）可证明得出 （当x=1时，“=”成立），························12分
，由 得：
∴
，即 ，··································13分
，且 ，
即
∴ ，
∴ （n≥2），···························································14分
∴
，······15分
即 ．··········································································17分
数学试题 第5页，共5页
