--- lang: en title: Class Review 1 subtitle: date: 2026-06-05T20:28:07+08:00 author: name: link: email: avatar: description: keywords: comment: false weight: 0 tags: - draft categories: - draft hiddenFromHomePage: false hiddenFromSearch: false hiddenFromRelated: false hiddenFromFeed: false summary: featuredImagePreview: featuredImage: password: message: repost: enable: false url: # See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter --- ### Example 1: Analytical Algebra It is known that the ellipse $C:\frac{x^2}{4}+\frac{y^2}{3}=1$ and the function $y=m^{x+1}+n(m\ge 2)$ intersect the ellipse at two points $A,B$, and find the maximum value of $n$. Assume $A(x_0,y_0),B(-x_0,-y_0),x_0\in(0,2),y_0\in(0,\sqrt{3})$, there are: $\begin{cases} y_0=m^{x_0+1}+n,(1)\\ -y_0=m^{-x_0+1}+n,(2)\\ \frac{x_0^2}{4}+\frac{y_0^2}{3}=1,(3) \end{cases} $ Considering that there are a total of four variables $x_0,y_0,m,n$ and three constraints, there are $4-3=1$ degrees of freedom. Obviously if the scope of one of the variables can be determined, the scope of the other variables will be easily determined: In order to **use symmetry** while obtaining **identical deformation**, consider (1)+(2) and (1)-(2), then the **sufficient and necessary conditions** are: $$\begin{cases} -2n=m(m^{x_0}+m^{-x_0}),(4)\\ 2y_0=m(m^{x_0}-m^{-x_0}),(5)\\ \frac{x_0^2}{4}+\frac{y_0^2}{3}=1,(3) \end{cases}$$ Make a few simple observations: - It is easy to see from (4) that $n\lt 0$. - In (4), $m^{x_0}\gt 1$, then $m$ is larger, $-2n$ is smaller, and $n$ is larger - In (5), the larger m is, the smaller 2y_0 is, and $m\ge 2$ can be eliminated by scaling $m$ So, we further perform **lossless scaling** (m=2): $$\begin{cases} -2n\ge 2(2^{x_0}+2^{-x_0}),(6)\\ 2y_0\ge 2(2^{x_0}-2^{-x_0}),(7)\\ \frac{x_0^2}{4}+\frac{y_0^2}{3}=1\ge \frac{x_0^2}{4}+\frac{2^{2x_0}+2^{-2x_0}-2}{3},(\alpha) \end{cases}$$ ($\alpha$) is the key to solving the problem! It is not difficult to see that $\frac{x_0^2}{4}+\frac{2^{2x_0}+2^{-2x_0}-2}{3}$ is monotonically increasing with respect to $x_0$, and the solution is $x_0\le 1$. So according to (6) we have: $$\begin{gathered} n\le -(2^{x_0}+2^{-x_0})\le -\frac{5}{2} \end{gathered}$$ ## Mathematical background: hyperbolic trigonometric functions ![alt text](image.png) ![alt text](image-1.png) Hyperbolic trigonometric functions are derived from hyperbola $x^2-y^2=1$, so they are called **hyperbolic functions** $$\begin{cases} \sinh x=\frac{e^x-e^{-x}}{2},\\ \cosh x=\frac{e^x+e^{-x}}{2}\ge 1,\\ \tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}\lt 1 \end{cases}$$ Similar to trigonometric functions, hyperbolic functions have corresponding identity transformation formulas: $$\begin{gathered} \cosh^2 x-\sinh^2 x=1,\\ \cosh^2 x+\sinh^2 x=\cosh 2x,\\ \sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y\\ \sinh(x-y)=\sinh x\cosh y-\cosh x\sinh y\\ \cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y\\ \cosh(x-y)=\cosh x\cosh y-\sinh x\sinh y\\ \tanh(x+y)=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}\\ \tanh(x-y)=\frac{\tanh x-\tanh y}{1-\tanh x\tanh y}\\ \end{gathered}$$ For details, see [Wikipedia](https://en.wikipedia.org/wiki/Hyperbolic_functions). When dealing with Example 1 (4) (5), you can also use the analogy of $\cosh^2 x-\sinh^2 x=1$ and consider the squares and then add them. ### Example 2 If $f(x)=a^{x-1}$ and $g(x)=\log_a x+1$ have three intersection points, find the value range of $a$: When $a\gt 1$, $f(x)=a^{x-1}$ increases monotonically, and $f(x),g(x)$ are inverse functions of each other, then if the intersection point is not on $y=x$: Suppose the intersection point is $(x,y)$, then there must be a pair of intersection points $(y,x)$, and as long as $x\neq y$, it is inconsistent with monotonicity. This means that $y=a^{x-1}$ as a downward convex function has two intersection points with $y=x$, which obviously leads to a contradiction: Therefore, $0\lt a\lt 1$, at this time, there is obviously an intersection point $(1,1)$, and the other two intersection points cannot be on $y=x$, then the two paired intersection points are not on $y=x$, and there is a pair of $(x,y),(y,x)$ that is an intersection point that meets the meaning of the question: $$\begin{cases} a^{x-1}=y,(1)\\ a^{y-1}=x,(2) \end{cases}$$ $$\begin{gathered} a^{x-1}+a^{y-1}=x+y,\\ a^{x-1}-a^{y-1}=y-x \end{gathered}$$ Let $k=\ln a\lt 0,x-1=u,y-1=v$, then: $$\begin{gathered} e^{ku}+e^{kv}=u+v+2,(3)\\ e^{ku}-e^{kv}=v-u,(4) \end{gathered}$$ Order $u+v=S,u-v=D$ again $$\begin{cases} e^{\frac{kS}{2}}(e^{\frac{kD}{2}}+e^{-\frac{kD}{2}})=S+2,(5)\\ e^{\frac{kS}{2}}(e^{\frac{kD}{2}}-e^{-\frac{kD}{2}})=-D,(6) \end{cases}$$ (5)/(6): $$\boxed{\tanh(\frac{kD}{2})=-\frac{D}{S+2}}$$ The above are the necessary conditions for the existence of $x,y$. ![alt text](image-2.png) When $S$ is fixed, $y=-\frac{1}{S+2}D$ is a straight line, considering that it has three intersection points with $y=\tanh(\frac{kD}{2})$. Obviously, it can be seen from the image that the slope of $y=\tanh \frac{kD}{2}$ at $D=0$ is $\frac{k}{2}$. Then if $-\frac{1}{S+2}\gt \frac{k}{2}$, then the straight line and the hyperbolic tangent function will have three intersection points. If $-\frac{1}{S+2}\le \frac{k}{2}$, then there is only one intersection point between the straight line and the hyperbolic tangent function. From (3): $$\begin{gathered} S+2=e^{ku}+e^{kv}\ge 2e^{k\frac{S}{2}} \end{gathered}$$ If $S\lt 0$, then $2\gt S+2\ge 2e^{k\frac{S}{2}}\gt 2$, resulting in a contradiction. So $S\ge 0$. Critical condition $S\to 0,k\to -1,a\to \frac{1}{e}$ Replace "ratio slope at $S\to0$" with a **single variable, global** argument to completely avoid the trouble of entanglement of two unknown quantities. The cleanest thing is to go back to a fixed point language. Assume $f(x)=a^{x-1}$ ($01$ (that is, $a<\frac1e$, because $\ln a<0$ is $\ln a<-1$): then $\psi'(1)>0$, $\psi$ crosses the zero point at $x=1$. However, $\psi(x)\to+\infty\ (x\to-\infty)$ and $\psi(x)\to-\infty\ (x\to+\infty)$ (use $01\ \text{and}\ 0