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高中 2022 级第二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.A 2.D 3.B 4.C 5.D 6.D 7.B 8.C
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.ABD 10.CD 11.ABC
三、填空题:本题共3个小题,每小题5分,共15分.
12. ; 13. ; 14.4.
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)高二年级至少2名同学入选校队包括以下情况:
高二年级仅2名同学入选校队有 种;········································2分
高二年级仅3名同学入选校队有 种;·········································3分
高二年级4名同学入选校队有 种;·············································4分
高二年级至少2名同学入选校队共有18+12+1=31种选法.··························6分
(2)由题意可知,随机变量X的取值为0,1,2,3,································7分
校队由0个女生4个男生组成时, ,···························8分
校队由1个女生3个男生组成时, , ··························9分
校队由2个女生2个男生组成时, ,·························10分
校队由3个女生1个男生组成时, , ························11分
所以,随机变量X的分布列为
X 0 1 2 3
数学试题卷 第1页(共6页)
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···································································································12分
随机变量X的均值为: .··························13分
16.解:(1) ,令 ,则 , ,·············2分
①当a=0时, ,所以 为增函数,故 无极值点;·······3分
②当a>0时,当x变化时, 及 变化如下表:
x −a
+ 0 − 0 +
递增 极大值 递减 极小值 递增
由此表可知 的极值小点为 ,其极大值点−a;···································5分
③当a<0时,当x变化时, 及 变化如下表:
x −a
+ 0 − 0 +
极小
递增 极大值 递减 递增
值
由此表可知 的极值小点为−a,其极大值点 .···································7分
综上所述,当a=0时, 无极值点;当a>0时, 的极值小点为 ,极大值点
−a;当a<0时, 的极值小点为−a,其极大值点 .·····························8分
(2)方法一:假设存在实数a,使得在区间[0,1]的最小值为0,且最大值为1,
则 [0,1], ;································································9分
由已知可得, ,则 ,·············································10分
数学试题卷 第2页(共6页)
学科网(北京)股份有限公司由(1)②可知, 在区间[0, ]上单调递减,在[ ,1]上单调递增,
∴ ,···········································11分
······················································································13分
∴ ,
, ,则 成立,解得: ,
∵
∵ ,
∴当 时, ,即 的最大值为 ,···14分
所述,满足题意的 .···························································15分
综上
方法二:假设存在实数a,使得在区间[0,1]的最小值为0,且最大值为1,
则 [0,1], ;·································································9分
由已知可得, ,则 ,···········································10分
由(1)②可知, 在区间[0, ]上单调递减,在[ ,1]上单调递增,
∴ ,···········································11分
∴ ······················································································13分
,
∵ , ,
令 ,则 的零点为 ,且 在 上单调
递增,
数学试题卷 第3页(共6页)
学科网(北京)股份有限公司∵ ,则 ,
∴当 时,则 成立,则 ,即 的最大值为
,符合题意,·········································································14分
综上所述, .··········································································15分
17.解:(1)由 ,可知当 时, ;··················1分
当 时, ,即 ,其中 也满足;
综上, .·····························································3分
又数列 满足 ,且 ,
当 时,可得:
,
········································································································5分
当 时, 适合上式,
所以数列 的通项公式为 .·······························7分
(2)由于 ,·····················································8分
则 ,·············10分
即 , ·································12分
两式相减得: ,
,·······················································14分
所以 .············································15分
数学试题卷 第4页(共6页)
学科网(北京)股份有限公司18.解析:(1) ,则 ,····························1分
∴切线斜率为: ,又 ,··················································2分
∴所求切线方程为 ;·····························································4分
(2)方法一:函数 的定义域是 ,
∴ ,······································································5分
①若a≤0,则 , 在 上单调递增,
, ,
∵ , , ,则 ,
则 仅有一个零点,且零点位于(1, ];···········································7分
②当 , 在 单调递减, 在 单调递增;
因为 的最小值为: ,
若a>1时, ,此时 无零点;··········································8分
若a=1时, ,此时 仅有一个零点;·································9分
若01.···············································································10分
方法二:令 ,则 ,····················································5分
设 ,则 ,·······················································7分
∴ 在(0,1)上单调递增,在(1,+∞)上单调递减,
∴ 的最大值为 ,且x>e, ,·········································9分
∴要使 在定义域上无零点,则a>1.···············································10分
(3)令 (x≥1),
则 ························································11分
数学试题卷 第5页(共6页)
学科网(北京)股份有限公司①当a<1时,x−a>0,∴ 时, , 在 上单调递减,
此时, ,不符合题意;·············································13分
②当a=1时,
∴ 时, , 在 上单调递减,
∴ ,即x=1时, ,符合题意;····································15分
③当a>1时,
∴ 时, , 在(1,a)上单调递增;
时, , 在 上单调递减,
∴ , ,符合题意;
综上所述,a≥1.··············································································17分
19.解:设事件 表示:第 天中午去A餐厅用餐,
事件 :第i天中午去B餐厅用餐,其中 , ,…….···························1分
(1)小王第2天中午去A餐厅用餐的概率为:
∴ ;····················4分
(2)设 ,依题可知, , ,
∵如果小王第1天中午去A餐厅,那么第2天中午去A餐厅的概率为0.8,
即 ,而 ,
∴ ,·······························································5分
∵如果第1天中午去B餐厅,那么第2天中午去A餐厅的概率为0.4,
∴ .·································································6分
由全概率公式可知 ,
即 ,··········································································7分
∴ ,而 ,························································8分
∴数列 是以 为首项,以 为公比的等比数列,······························9分
∴ ,即 ;·····················································10分
(3)设王某第i天去B餐厅的次数为Xi ,则Xi 的所有可能取值为0,1,·····11分
数学试题卷 第6页(共6页)
学科网(北京)股份有限公司当Xi=0时表示王某第i天没去B餐厅,当Xi=1时表示王某第i天去B餐厅,
∵ , ,
∴ ,··················································13分
∵ , , ,2,……,······································15分
∴当 n∈ N*时,
,·······················16分
故 .··································································17分
数学试题卷 第7页(共6页)
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