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冲刺强基计划数学物理专题一
2022-06-13 02:32:00 【傲天居士】
1. 强基计划数学物理模拟卷
本人出了一套强基计划数学物理模拟试题,适合有一定基础的高考生。由于本人水平有限,若存在疏漏还请诸位大佬批评指正! 



2. 高斯函数专题
对应到模拟卷第一题第一问。解方程:
移项得:
由高斯函数的性质:
因此综合可得:
解不等式得:或
下面分情况讨论:
当 时, ,此时原方程即为 ,解得 .结合 的范围,取 . 当 时, ,此时原方程即为 ,解得 .结合 的范围,取 . 当 时, ,此时代入原方程得 成立。 综上, 或 或 .
3. 复数与单位根专题
对应到模拟卷第一题第二问。
已知,且.
则容易知道是方程在复数域上的全部根。那么方程又可以写成:
对比系数不难得出:
因此,,则有
4. 椭圆面积与坐标系变换
4.1 伸缩变换
对应到模拟卷第二题第一问。
记椭圆(或圆)的面积为
方法一:
由对称性可得:
方法二:
利用坐标系变换:
则椭圆(或圆)可化为:
设变化后的面积为,则有对应关系:
因此可得椭圆(或圆)的面积为
4.2 旋转变换
对应到模拟卷第二题第二问。
坐标系上任一点绕坐标原点逆时针旋转后的坐标为:
利用三角函数公式可展开为:
在矩阵的视角下,可以写为:
在直角坐标系下,假设变换前的坐标为(x,y),绕坐标原点旋转后的坐标记为.
则有:
首先将圆锥曲线化为:
记矩阵
则矩阵可以对角化:
则有:
令:
则等式左边可化简为:
等式右边即为:
而由椭圆方程:
因此有:
记为椭圆
由此可知,椭圆逆时针旋转(即顺时针旋转)后可得到椭圆
由模拟卷第二题第一问所得椭圆(或圆)面积公式可知,椭圆的面积为
由于面积的旋转不变性,可得圆锥曲线的面积也为.
未完待续
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