名校
解题方法
1 . 我国古代数学家僧一行应用“九服晷影算法”在《大衍历》中建立了晷影长l与太阳天顶距θ(
)的对应数表,这是世界数学史上较早的正切函数表.根据三角学知识可知,晷影长l等于表高h与太阳天顶距θ正切值的乘积,即
.对同一“表高”测量两次,第一次和第二次太阳天顶距分别为
,
,第二次的“晷影长”是“表高”的2倍,且
,则
的值为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/33b13a855bec1e9c264742bbed4685ed.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/bd7ba1eba7fba0b56e4e9b4d032e24da.png)
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A.![]() | B.![]() | C.![]() | D.![]() |
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2 . 如果方程
能确定
是
的函数,那么称这种方式表示的函数为隐函数.隐函数的求导方法如下:在方程
中,把
看成
的函数
,则方程可看成关于
的恒等式
,在等式两边同时对
求导,然后解出
即可.例如,求由方程
所确定的隐函数的导数
,将方程
的两边同时对
求导,则有
(
是
的函数,需要用复合函数的求导法则求导),得
.利用隐函数求导方法可求得曲线
在点
处的切线方程为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5bbbf52d1f9d61b41bdd4acfc9fac268.png)
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C.![]() | D.![]() |
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解题方法
3 . 我国南宋时期著名的数学家秦九韶在其著作《数书九章》中,提出了已知三角形三边长求其面积的公式,求法是:“以小斜幂并大斜幂减中斜幂,余半之,自乘于上以小斜幂乘大斜幂减上,余四约之,为实.一为从隅,开平方得积”翻译成公式,即
,其中
,
,
分别为
中角
,
,
的对边,
为
的面积.现有面积为
的
满足
,则其内切圆的半径是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8f2b188be1fde51a349c10f5ec492734.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
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4 . 任取一个正整数,若是奇数,就将该数乘3再加上1若是偶数,就将该数除以2.反复进行上述两种运算,经过有限次步骤后,必进入循环圈
,这就是数学史上著名的“冰雹猜想”(又称“角谷猜想”),参照“冰雹猜想”,提出了如下问题:设各项均为正整数的数列
满足
,若
,则
的取值可以为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1a16f78ce0dab1ac8fa6abbd70f2b008.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/83cf38189d5cbf627d2b82ac0eb76006.png)
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A.1 | B.3 | C.6 | D.7 |
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5 . “四平方和定理”最早由欧拉提出,后被拉格朗日等数学家证明.“四平方和定理”的内容是:任意正整数都可以表示为不超过四个自然数的平方和,例如正整数
.设
,其中
均为自然数,则满足条件的有序数组
的个数是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/19e9edd49b95d101473211fa54acfcdc.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cd6f5f4751622b599216b655a679cdd8.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8e5e4be004a34cfce346c12feea0a696.png)
A.26 | B.28 | C.29 | D.30 |
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6 . 中国数学家华罗庚倡导的“0.618优选法”在各领域都应用广泛,0.618就是黄金分割比
的近似值,古希腊的数学家毕达哥拉斯通过研究正五边形和正十边形的作图,发现了黄金分割率,黄金分割率的值也可以用
表示,即
,则
的值为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5c8382dcdb655ab1d049f8dba22fa467.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0d3caf448beca2df4d2427360e93b599.png)
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A.![]() | B.1 | C.![]() | D.![]() |
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名校
解题方法
7 . 中国南北朝时期的著作《孙子算经》中,对同余除法有较深的研究.设
为整数,若
和
被
除得的余数相同,则称
和
对模
同余,记为
.若
,
,则
的值可以是( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a5924004836cc5973c0a701a67c50d4e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
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A.2022 | B.2023 | C.2024 | D.2025 |
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8 . 17世纪法国数学家费马在给朋友的一封信中曾提出一个关于三角形的有趣问题:在三角形所在平面内,求一点,使它到三角形每个顶点的距离之和最小.现已证明:在
中,若三个内角均小于120°,则当点
满足
时,点
到
三个顶点的距离之和最小,点
被人们称为费马点.根据以上知识,已知在
中,
,
,
,
为
内一点,则
的最小值为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15c0dbe3c080c4c4636c64803e5c1f76.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
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名校
解题方法
9 . 《几何补编》是清代梅文鼎撰算书,其中卷一就给出了正四面体,正六面体(立方体)、正八面体、正十二面体、正二十面体这五种正多面体的体积求法.若正四面体
的棱长为
,
为棱
上的动点,则当三棱锥
的外接球的体积最小时,三棱锥
的体积为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/63397cda22cb1fad59cf966dfb588643.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/38387ba1cadfd3dfc4dea4ca9f613cea.png)
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5卷引用:河北省沧州市部分示范性高中2024届高三下学期三模数学试题
河北省沧州市部分示范性高中2024届高三下学期三模数学试题河北省沧州市盐山中学2024届高三三模数学试题(已下线)核心考点8 立体几何中综合问题 A基础卷 (高一期末考试必考的10大核心考点) 海南省2020-2021学年高二下学期期末考试数学试题(已下线)第1套 全真模拟卷 (中等)【高一期末复习全真模拟】
10 . 国家二级文化保护遗址玉皇阁的台基可近似看作上、下底面边长分别为
,
,侧棱长为
的正四棱台,则该台基的体积约为( )
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f71a41641aa0d0e45a3c03d3d2c1196b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1e7854968bbf6576a1fd9926ee0d4d63.png)
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