名校
解题方法
1 . (1)证明:当
时,
;
(2)已知正项数列
满足
.
(i)证明:数列
为递增数列;
(ii)证明:若
,则对任意正整数
,都有
.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/08115d6d9f876dea921a4d32260ff1fb.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ca947e8ea00b7a485097ecafd2dfcae9.png)
(2)已知正项数列
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/83cf38189d5cbf627d2b82ac0eb76006.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6f91f8b7476e67db488d85c3a14ffa6d.png)
(i)证明:数列
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0496f142d8ae5acb06e83526eaa3ef87.png)
(ii)证明:若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/90f2db682457da2d4abd0e7cca1bdf40.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/38d8a310323506a3c2f3626dec8d781f.png)
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解题方法
2 . 在一种新能源产品的客户调查活动中发现,某小区10位客户有4人是该产品的潜在用户,小刘负责这10人的联系工作,他先随机选择其中5人安排在上午联系,剩余5人下午联系.
(1)设上午联系的这5人中有
个潜在用户,求的
分布列与期望;
(2)小刘逐一依次联系,直至确定所有潜在用户为止,求小刘6次内即可确定所有潜在用户的概率.
(1)设上午联系的这5人中有
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5b734e8f1546481e3eb4976008a045de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5b734e8f1546481e3eb4976008a045de.png)
(2)小刘逐一依次联系,直至确定所有潜在用户为止,求小刘6次内即可确定所有潜在用户的概率.
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解题方法
3 . 为丰富第二课堂,拓展素质教育,某校鼓励学生参加书法兴趣小组和绘画兴趣小组,开展相关实践活动.该校共有3000名学生,为了解学生的参加情况,从全校学生中随机抽取150名学生进行调查,发现有5人没有参加兴趣小组,且样本中仅参加书法兴趣小组和仅参加绘画兴趣小组的学生每周投入时间情况如下表:
(1)用频率估计概率,试估计全校学生中书法兴趣小组和绘画兴趣小组都参加的人数;
(2)从仅参加书法兴趣小组和仅参加绘画兴趣小组的学生中各抽1人,以X表示2人中每周投入时间大于5小时的人数,求X的分布列和数学期望;
(3)根据公式
计算仅参加书法兴趣小组和仅参加绘画兴趣小组的学生在各投入时间段人数的样本相关系数,并推断它们的相关程度,其中
分别为仅参加书法兴趣小组的学生在各投入时间段人数的均值和标准差,
分别为仅参加绘画兴趣小组的学生在各投入时间段人数的均值和标准差.
附:
兴趣小组活动类别 | 投入时间(小时/周) | |||
大于10 | ||||
仅参加书法兴趣小组人数z | 25 | 30 | 15 | 10 |
仅参加绘画兴趣小组人数y | 10 | 20 | 25 | 5 |
(2)从仅参加书法兴趣小组和仅参加绘画兴趣小组的学生中各抽1人,以X表示2人中每周投入时间大于5小时的人数,求X的分布列和数学期望;
(3)根据公式
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a06fcaa42bb392779de51abfc941b222.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dd399feedaac8848cee39095509b2625.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b977dff2630a81f0509102190dad26a6.png)
附:
相关系数r | |||
相关程度 | 低度线性相关 | 显著性相关 | 高度线性相关 |
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4 . 对于数列
,定义
,满足
,记
,称
为由数列
生成的“
函数”.
(1)试写出“
函数”
,并求
的值;
(2)若“
函数”
,求n的最大值;
(3)记函数
,其导函数为
,证明:“
函数”
.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/83cf38189d5cbf627d2b82ac0eb76006.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c214af101c995038e5229277692350e0.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5a9dfaa821691b4a0f548d3a57cafb6d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6f3862a345e98a4a34b3101bffe283be.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0ad01b28c2aa9d86daa50de5a9fe698a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/83cf38189d5cbf627d2b82ac0eb76006.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bb654dbe976f077495105b21b7963d0f.png)
(1)试写出“
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2837786afdd7b9b8bc37823040d7dd64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e167bc147c8a470e8060815303b6313a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/abdb56b20415e38b25c6cc9c4319dc53.png)
(2)若“
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e4e0c84de10f0f2186313169c3dc997b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/502cb9fcc607d8622546867ccf60407f.png)
(3)记函数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3ab95fa0eb7b774e15babe6854f4d8e4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c5840bbbca91e9057cc99a5b3cd85fe7.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bb654dbe976f077495105b21b7963d0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/506f4aa695eb5785abc58cf6c254c492.png)
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5 . 某校开设了“五子棋”社团课,甲乙两位同学进行五子棋比赛,每局有一人先手(每局中先走第一颗棋),规则如下:每局输者下一局先手.已知甲先手时,甲赢的概率为
;乙先手时,乙赢的概率为
.假设每局无平局,且每局比赛的输赢相互独立,第一局甲先手.
(1)甲乙两位同学比赛两局,求甲至少赢1局的概率;
(2)记
为第
局比赛中甲赢的概率,求
,并计算连续比赛20局中,甲赢的概率大于
的局数.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8b2a698891d42c70b597f0da4f215f09.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/eac97e6740365c85ad857aff85cefbe5.png)
(1)甲乙两位同学比赛两局,求甲至少赢1局的概率;
(2)记
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/59c709117ab1d3ef620883a732aed68b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c05b9832b09731a574d4a4adf7448de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/59c709117ab1d3ef620883a732aed68b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f89eef3148f2d4d09379767b4af69132.png)
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解题方法
6 . 已知函数
,将
的图像上所有点的横坐标变为原来的2倍,再向左平移
个单位长度,最后再把所有点的纵坐标伸长到原来的3倍,得到函数
的图象.
(1)求函数
的单调递增区间,并写出函数
的解析式;
(2)关于
的方程
在
内有两个不同的解
;
①求实数
的取值范围;
②用
的代数式表示
的值.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3b7ecff5c4200875c7655e505c57a103.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ac1a63ab608517bb10aa036783dfb51f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/028517e8bebe634441e0a5c79828e88a.png)
(1)求函数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d275fbb3ee5cd1177ca5a2ceecbbef0f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/028517e8bebe634441e0a5c79828e88a.png)
(2)关于
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7317f1f5bf19007cbf9bf46d1e4bcac4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5e13623a74057edba789cff4ba85c5fe.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9146b7c16d81ccb92aef1e1f1788f00f.png)
①求实数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
②用
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7241924141a25c43d7cd8288ddfd8501.png)
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7 . 如果存在实数对
使函数
,那么我们就称函数
为实数对
的“
型正余弦生成函数”,实数对
为函数
的“
型正余弦生成数对”.
(1)已知函数
的“4型正余弦生成数对”为
,求方程
在区间
上所有实根之和;
(2)若实数对
的“2型正余弦生成函数”
在
处取最大值,其中
,求
的取值范围.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a64d924836b4292239d9726c6473d7f5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8b660b878e6c0f3590c50bf52f896acc.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4fe7d5809da02c15a43a0e9a898b9086.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a64d924836b4292239d9726c6473d7f5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/595044a7750ab4f84519041979c3d780.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a64d924836b4292239d9726c6473d7f5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4fe7d5809da02c15a43a0e9a898b9086.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/595044a7750ab4f84519041979c3d780.png)
(1)已知函数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6a1cfb60420ff7e72c1b9d64f69ae063.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/425bdb703e21f599b68315be995b551b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/56c40363783d0e33e6a40e68957391ec.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/50e57682d2fe6f71ce86a38502e2f119.png)
(2)若实数对
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/46d552855e04ac50648a15e9b4aec868.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e915b67f8f747698b8b46d37bc453667.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11abb76da45ffa52b47c3a6b9a03ac7e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/88fc13c376b1423dc2cefb3018bdc26e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1ee0a78ef89b08120fcf8b7cb0d013b1.png)
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8 . 如图所示的几何体是一个半圆柱和一个三棱锥的组合体.
是半圆柱的母线,
分别是底面直径BC和
的中点,
是半圆
上一动点,
是半圆
上的动点,
是圆柱的母线,延长
至
点使得
为
的中点,连接
,
构成三棱锥
.
;
(2)当三棱锥
的体积最大时,求平面
与平面
的夹角.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8ebf67d880172b27fefacc3c5b808eae.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a4d6cde991efd85f03f8801e745127a6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/56f7ba05c54b3de1f4378f7c8eb58328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/683e17492ad37eeab880209057cb0feb.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1dde8112e8eb968fd042418dd632759e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a18722354086c42e62334983fc50eb6a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/23f919bd3dde10dbbc076f7ec5149699.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2777840758e70e7dbbc18cef8f3d6d2b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4a696a182fff038a86b2bbe8ca099442.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5963abe8f421bd99a2aaa94831a951e9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7d0ff310aabd2282b539537ebed3f788.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/30513ea48bc1ef3ae78adac83d894f14.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d4d260c4df7b0dc180af6980d21f3371.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63397cda22cb1fad59cf966dfb588643.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf5ae8d145c5ce43e4cfc95fe6f563ed.png)
(2)当三棱锥
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63397cda22cb1fad59cf966dfb588643.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e367b683581c7cbe018078168f69efc5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6b36a70bc52a720ba8750aee4924307.png)
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解题方法
9 . 我们知道,一个一元一次方程最多有一个根,一个一元二次方程最多有两个根,这些都是代数基本定理的简单表示,代数基本定理可以表述为:一元n次多项式方程最多有
个不同的根.由代数基本定理可以得到如下推论:若一个一元
次方程有不少于
个不同的根,则必有各项的系数均为0.已知函数
,函数
的图象上有四个不同的点A、B、C、D.利用代数基本定理及其推理回答下列问题:
(1)解关于x的方程
;
(2)是否存在实数
,使得关于
的方程
有三个以上不同的解,若存在,求出
的值,若不存在,请说明理由;
(3)若
按逆时针方向顺次构成菱形,设
,求代数式![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aa8a18548e00a131abe2eca8c4c815c2.png)
的值.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b6a24198bd04c29321ae5dc5a28fe421.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7a0876215b2fd463d151523cd3c6b447.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c7bc57a9ac3f82c3b8af4fe78e5c861b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4fe7d5809da02c15a43a0e9a898b9086.png)
(1)解关于x的方程
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b070bfc31cef4c001541af54d3c36cd3.png)
(2)是否存在实数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/280860dd039e1305a5ccc455f63e8223.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/81dea63b8ce3e51adf66cf7b9982a248.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/fb2158cfb945452be603a745510df299.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0877194ab8760f54c35527177b03ff93.png)
(3)若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e0b032796d46540441098204aa82c12a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/aa8a18548e00a131abe2eca8c4c815c2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c004d926a934cced9bc523a8ecde1df1.png)
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解题方法
10 . 炎炎夏日,上学路上若有一支冰淇淋该多么美妙啊!小明同学酷爱甜筒冰淇淋(图1),他想动手做一个甜筒模型(图2),若根据设计稿已知
为直角三角形,四边形
为直角梯形,
,
,曲线
是以
为圆心的四分之一圆弧,
,
,
,
,将平面图形
旋转一周得到小明设计的甜筒.
;
(2)小明准备将矩形
旋转所形成的几何体都用来盛装冰淇淋(如图2所示),该矩形内接于图形
,
在弧
上(不与端点
重合),点
在线段
上,
与
所在的直线重合,设
,求:
①盛装冰淇淋容器的体积
;(用
表示)
②炎热的天气下,若冰淇淋融化的时间与盛装的体积满足关系
,请计算这个冰淇淋完全融化需要的最长时间.
(3)小明想给甜筒一些新的装饰,如果修改后的甜筒俯视图如图3所示,且通过拼装后可以变成一个正四棱锥
(即俯视图可以看作一个正四棱锥
的展开图),我们记侧棱
的长为1,
,正四棱锥的表面积记作
,体积记作
.求
(将其表示为
的形式,其中
为常数).
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15c0dbe3c080c4c4636c64803e5c1f76.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e1333acc72211e3ddb9a0f8c726ce8a5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/45acdbac251ca6b76a166c1242e71df9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9f90126f831d6600522ecaa66c2a8b9c.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/49b50357a6545cae8348e3059312f520.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8455657dde27aabe6adb7b188e031c11.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/305a88d4e0249bd16d48eda01331d2d4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ef0402dd5ae3db10281f9f1e11738bcb.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/32f01c4faacedfe56f5127d6c0cc63cf.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f341b98caabf99bc683ce8407068735e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4465a96d65de0f859ea4ec9e548204d7.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c4764374bd2fb78e59cd0b283637baeb.png)
(2)小明准备将矩形
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cec780e86f57790cf88ec761e219bf5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8edaecff592cf677f45d8c54340a89f2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ac047e91852b91af639feec23a9598b2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/49b50357a6545cae8348e3059312f520.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3cf9b288c48c73463a2f214f02b6952a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/54a5d7d3b6b63fe5c24c3907b7a8eaa3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4eedae8d316c76e3d0b451256de03fb9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7a5f1641947153c80b987320885a2b57.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/6aa2b5e09f8ec785c59900a529390a02.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e1494be0926832f8cb95cbc7477ae685.png)
①盛装冰淇淋容器的体积
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c63055a5d6916f99d07fede49120753f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e170f206fdbbd834aad7580c727e2cc6.png)
②炎热的天气下,若冰淇淋融化的时间与盛装的体积满足关系
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/698295b8bd91909e0dbb581fe29a9bd0.png)
(3)小明想给甜筒一些新的装饰,如果修改后的甜筒俯视图如图3所示,且通过拼装后可以变成一个正四棱锥
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/faeb97acf19bd3b2c6c77c2814df4d2f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/faeb97acf19bd3b2c6c77c2814df4d2f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8a6e2867f32d3f1c3cd36cd3a11a8580.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/965466f8aae61f1a049e25f0325d4abf.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d887c94a751c97fc17d9ee6dba8590b8.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/46dc1a9783d9539ae3b0d2049bf1a123.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4bb84bef0f93107237d2860f477288d3.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/715f6103c80ca9f2c1c88b1308e33704.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/76f0649064a085fb74c997fb507a9b6d.png)
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