# The application of critical state in Mathematics

2021-07-21 15:04:43

## Preface

The critical state Equality and inequality ; Parallel and oblique 、 vertical ;

In a two-dimensional plane , Line demarcation , Special point localization ; In 3D space , Plane delimitation , Special point localization ;

## Specific application

• 【 Closed curves --- round 】

spot $$p(x_0,y_0)$$ In the circle $$x^2+y^2=1$$ On , be $$x_0^2+y_0^2=1$$;

spot $$p(x_0,y_0)$$ In the circle $$x^2+y^2=1$$ Inside , be $$x_0^2+y_0^2<1$$;

spot $$p(x_0,y_0)$$ In the circle $$x^2+y^2=1$$ external , be $$x_0^2+y_0^2>1$$;

• 【 Closed curves --- The ellipse 】

spot $$p(x_0,y_0)$$ In the ellipse $$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$$ On , be $$\cfrac{x_0^2}{a^2}+\cfrac{y_0^2}{b^2}=1$$;

spot $$p(x_0,y_0)$$ In the ellipse $$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$$ Inside , be $$\cfrac{x_0^2}{a^2}+\cfrac{y_0^2}{b^2}<1$$;

spot $$p(x_0,y_0)$$ In the ellipse $$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$$ external , be $$\cfrac{x_0^2}{a^2}+\cfrac{y_0^2}{b^2}>1$$;

• 【 Unclosed curves --- A straight line 】

spot $$p(x_0,y_0)$$ In a straight line $$ax+by+c=0$$ On , be $$ax_0+by_0+c=0$$;

spot $$p(x_0,y_0)$$ In a straight line $$ax+by+c=0$$ Outside ( Or both sides ), be $$ax_0+by_0+c\neq 0(>0 or <0)$$;

• 【 Unclosed curves --- curve 】

spot $$p(x_0,y_0)$$ On the curve $$y^2=2px$$ On , be $$y_0^2=2px_0$$;

spot $$p(x_0,y_0)$$ On the curve $$y^2=2px$$ Outside ( Or both sides ), be $$y_0^2\neq 2px_0(>2px_0 or <2px_0)$$;

## Analysis of typical examples

【2019 In the third year of senior high school, there are many problems in math homework 】 Known as a pyramid $$S-ABC$$ The length of the bottom side of is 4, High for 3, Take any point in a pyramid $$P$$, bring $$V_{P-ABC}<\cfrac{1}{2}V_{S-ABC}$$ Is the probability that 【】

$A.\cfrac{3}{4}$ $B.\cfrac{7}{8}$ $C.\cfrac{1}{2}$ $D.\cfrac{1}{4}$

analysis ： Make a pyramid $$S-ABC$$ As shown in the figure , Let the high line be $$SO=h$$, Let's have a triangular pyramid $$P-ABC$$ The height of $$h_1$$,

First, rewrite the unequal relation to the equal relation , namely $$V_{P-ABC}=\cfrac{1}{2}V_{S-ABC}$$, That is to find the point in the critical state $$P$$ The location of .

By $$\cfrac{1}{3}\cdot S_{\triangle ABC}\cdot h_1=\cfrac{1}{2}\cdot \cfrac{1}{3}\cdot S_{\triangle ABC}\cdot h$$, obtain $$h_1=\cfrac{1}{2}h$$,

That is, in a critical state , spot $$P$$ It should be in a pyramid $$S-ABC$$ The middle cross section of $$MND$$ Inside [ In 3D space , Plane delimitation , Special point localization ],

And then we can easily analyze what we want to satisfy $$V_{P-ABC}<\cfrac{1}{2}V_{S-ABC}$$, The point of $$P$$ It should be on a triangular pyramid $$NDM-ABC$$ Inside ,

So the probability is $$P=1-\cfrac{V_{S-MND}}{V_{S-ABC}}=1-\cfrac{\frac{1}{3}\cdot \cfrac{\sqrt{3}}{4}\cdot 2^2\cdot h_1}{\frac{1}{3}\cdot \cfrac{\sqrt{3}}{4}\cdot 4^2\cdot h}=1-\cfrac{1}{8}=\cfrac{7}{8}$$, So choose $$B$$.

The edge length is 2 The cube of $$ABCD-A_1B_1C_1D_1$$ in , spot $$O$$ For the bottom $$ABCD$$ Center of , In Cube $$ABCD$$ $$-A_1B_1C_1D_1$$ Take a little bit at random $$P$$, The point of $$P$$ point-to-point $$O$$ The distance is greater than 1 The probability of is _____________.

analysis ： spot $$P$$ All the results are measured by the volume of a cube , Therefore, it should be a volumetric geometric type .

Consider the critical state first , When $$P$$ point-to-point $$O$$ The distance is equal to 1 when , spot $$P$$ In the center of the ball is $$O$$ On the hemispherical surface of ,

Then when the point $$P$$ point-to-point $$O$$ The distance is greater than 1 when , spot $$P$$ In the center of the ball is $$O$$ Outside the hemisphere and inside the cube ,

So what we want $$P=\cfrac{2^3-\cfrac{1}{2}\times \cfrac{4}{3}\times \pi\times 1^3}{2^3}=1-\cfrac{\pi}{12}$$.

Reflection summary ： Critical state delimitation , The inequality relation is localized , It's the same as the line demarcation in linear programming , The localization of special points is interlinked .

【2019 The third year of science mathematics quality inspection in Baoji senior high school Ⅱ The first 12 topic 】 It's defined in $$R$$ The function on $$y=f(x)$$, Satisfy $$f(3-x)=f(x)$$,$$f'(x)$$ For the function $$f(x)$$ The derivative of , And $$(x-\cfrac{3}{2})\cdot f'(x)<0$$, if $$x_1<x_2$$, And $$x_1+x_2>3$$, Then there are 【】

$A.f(x_1) >f(x_2)$ $B.f(x_1)=f(x_2)$ $C.f(x_1) analysis ： from $$(x-\cfrac{3}{2})\cdot f'(x)<0$$, Get when $$x>\cfrac{3}{2}$$ when , There must be $$f'(x)<0$$, When $$x<\cfrac{3}{2}$$ when , There must be $$f'(x)>0$$, namely $$x\in (\cfrac{3}{2},+\infty)$$ when ,$$f'(x)<0$$, function $$f(x)$$ Monotonic decline , $$x\in (-\infty,\cfrac{3}{2})$$ when ,$$f'(x)>0$$, function $$f(x)$$ Monotone increasing , Again by $$f(3-x)=f(x)$$ The axis of symmetry of the function is $$x=\cfrac{3}{2}$$, Using $$x_1<x_2$$,$$x_1+x_2>3$$ On this condition , We can consider the critical state first to reduce the difficulty , Make $$x_1+x_2=3$$, be $$x_1$$ and $$x_2$$ Equidistant from the axis of symmetry , Then there will be $$f(x_1)=f(x_2)$$, So when $$x_1<x_2$$,$$x_1+x_2>3$$ when , There must be $$x_1$$,$$x_2$$ On both sides of the axis of symmetry , And $$x_2$$ Farther from the axis of symmetry , There are $$f(x_1)>f(x_2)$$, So choose $$A$$. How to understand expressions ：$$(|x|- 1)^2+(|y|-1)^2\leq 4$$ analysis ： If you don't think clearly , So let's change the original title to $$(|x|- 1)^2+(|y|-1)^2=4$$, So the critical state is on the circle ; And the original question is $$(|x|- 1)^2+(|y|-1)^2\leq 4$$, So it should be related to the inside of the circle . And because there are absolute values in the title , Therefore, it is necessary to classify and discuss to remove the absolute value sign , In this way, the problem is transformed into a feasible region problem , It's a nonlinear feasible region, of course . After thinking about this , We can make the following transformation smoothly ： The original inequality $$\Leftrightarrow$$ $$\left\{\begin{array}{l}{x\ge 0}\\{y\ge 0}\\{(x-1)^2+(y-1)^2\leq 4}\end{array}\right.$$ or $$\left\{\begin{array}{l}{x\ge 0}\\{y< 0}\\{(x-1)^2+(-y-1)^2\leq 4}\end{array}\right.$$ or $$\left\{\begin{array}{l}{x< 0}\\{y\ge 0}\\{(-x-1)^2+(y-1)^2\leq 4}\end{array}\right.$$ or $$\left\{\begin{array}{l}{x< 0}\\{y< 0}\\{(-x-1)^2+(-y-1)^2\leq 4}\end{array}\right.$$ In this way, the hand working diagram is used , We can also make the following figure . 【2018 Fujian Quanzhou is the first mock exam 】 As shown in the figure , Regular hexagon $$ABCDEF$$ in ,$$N$$ For the line segment $$AE$$ The midpoint of , In the line segment $$DE$$ I'm going to take a random number of points $$G$$, Incident light $$NG$$ the $$DE$$ Reflection , Then the light is reflected on the line segment $$BC$$ The probability of intersection is 【】$A.\cfrac{1}{4}B.\cfrac{1}{3}C.\cfrac{5}{12}D.\cfrac{2}{3}$analysis ： For convenience of calculation , Let's set up a regular hexagon $$ABCDEF$$ The length of $$2$$, be $$FN=1$$,$$NE=\sqrt{3}$$; Next, think about reflected light and line segments $$BC$$ The critical state of intersection ; Reflected light and line segments $$BC$$ One of the two critical states of intersection is crossing point $$B$$, The second one is over the point $$C$$; When the reflected light passes through the point $$B$$ when , The incident point is $$G$$, set up $$EG=x$$, By $$\angle NGM=\angle MGB$$, be $$tan\angle NGM=\cfrac{x}{\sqrt{3}}$$,$$tan\angle MGB=\cfrac{2-x}{2\sqrt{3}}$$, Then there are $$\cfrac{x}{\sqrt{3}}=\cfrac{2-x}{2\sqrt{3}}$$, namely $$\cfrac{2-x}{x}=\cfrac{2\sqrt{3}}{\sqrt{3}}$$, Using the sum ratio property, we can get ,$$\cfrac{2-x+x}{x}=\cfrac{2\sqrt{3}+\sqrt{3}}{\sqrt{3}}$$, namely $$\cfrac{2}{x}=\cfrac{3\sqrt{3}}{\sqrt{3}}$$, be $$x=\cfrac{2}{3}$$; When the reflected light passes through the point $$C$$ when , The incident point is $$G$$, set up $$EG=y$$, By $$\angle NGM=\angle MGC$$, be $$tan\angle NGM=\cfrac{y}{\sqrt{3}}$$,$$tan\angle MGC=\cfrac{3-y}{\sqrt{3}}$$, namely $$\cfrac{y}{\sqrt{3}}=\cfrac{3-y}{\sqrt{3}}$$, Solution $$y=\cfrac{3}{2}$$; So reflected light and lines $$BC$$ The intersection corresponds to the line segment $$DE$$ The length on is $$\cfrac{3}{2}-\cfrac{2}{3}=\cfrac{5}{6}$$; According to the geometric probability of length type , The probability is $$P=\cfrac{\frac{5}{6}}{2}=\cfrac{5}{12}$$. 【2019 Senior three science mathematics information problem 】 Known functions $$f(x)=x^2lnx+kx-1$$ There's zero , seek $$k$$ Value range of _________. analysis ： Known functions $$f(x)=x^2lnx+kx-1$$ There's zero , That's the equation $$f(x)=0$$ In the domain $$(0,+\infty)$$ There is a solution on the surface , The separation parameters are obtained $$k=\cfrac{x^2lnx+1}{x}=xlnx+\cfrac{1}{x}$$, Make $$h(x)=xlnx+\cfrac{1}{x}$$, Then the title becomes $$k=h(x)$$ stay $$(0,+\infty)$$ There is a solution on the surface , So we can either find the function from the angle of number $$h(x)$$ Range of values ; Or monotonicity , The image of the function , From the angle of shape, we can solve it by combining number with shape . Let's use the derivative to find the function $$h(x)$$ The monotonicity of .$$h'(x)=lnx+1-\cfrac{1}{x^2}$$, You need to pay attention to , There's something in the derivative $$lnx$$, So we artificially divide the above function into two parts ,$$y=lnx$$ and $$y=1-\cfrac{1}{x^2}$$, Shilling $$lnx=0$$ obtain $$x=1$$, Will be $$x=1$$ Plug in $$y=1-\cfrac{1}{x^2}$$ Verification is also its zero point , The zeros of these two functions coincide , So next we divide the domain into $$(0,1)$$ and $$(1,+\infty)$$ Two parts can be divided into two parts ： be $$0<x<1$$ when ,$$h'(x)<0$$,$$h(x)$$ Monotonic decline ,$$x>1$$ when ,$$h'(x)>0$$,$$f(x)$$ Monotone increasing , be $$h(x)_{min}=h(1)=1$$. namely $$h(x)$$ The value range of is $$[1,+\infty)$$, so $$k\ge 1$$, namely $$k\in [1,+\infty)$$. Or get a function by monotonicity $$h(x)$$ The image is as follows , Reuse functions $$y=k$$ And the function $$y=h(x)$$ There are intersections in the image of , obtain $$k$$ The value range of is $$k\in [1,+\infty)$$. 【2019 Senior three science mathematics starting paper ,2019 The second examination paper of Shaanxi Province No 12 topic 】 If two functions $$f(x)=x^2$$ And $$g(x)=a^x$$ $$(a>0,a\neq 1)$$ There is only one intersection point in the image of , Then the real number $$a$$ The range of phi is zero 【】$A.(e^{-\frac{2}{e}},e^{\frac{2}{e}})B.(0,e^{-\frac{2}{e}})C.(0,e^{-\frac{2}{e}})\cup(e^{\frac{2}{e}},+\infty)D.(e^{-\frac{2}{e}},1)\cup(1,e^{\frac{2}{e}})$analysis ： There's only one intersection between the images of two functions , That's the equation $$x^2=a^x$$ Only one root , Law 1： Using the image of two functions , In especial $$y=a^x$$ Dynamic graphics to illustrate the problem ; Curves are tangent to curves ; When $$a>1$$ when , function $$y=x^2$$ And functions $$y=a^x$$ The critical position of two intersections is in the case that the first quadrant is tangent , As shown in the figure below ; The following focuses on solving the parameters of tangency $$a$$ Value ; Let the tangent point of two curves be $$P(x_0,y_0)$$, Then there are $$\left\{\begin{array}{l}{2x_0=a^{x_0}\cdot lna ① }\\{y_0=x_0^2 ② }\\{y_0=a^{x_0} ③ }\end{array}\right.$$ from ②③ You know ,$x_02=a{x_0} ④ $, Plug in ① obtain ,$$2x_0=x_0^2\cdot lna$$, Simplify to get $$2=x_0\cdot lna ⑤$$, Again by ④ Take logarithms on both sides to get ,$$2lnx_0=x_0\cdot lna⑥$$, from ⑤⑥ obtain ,$$2lnx_0=2$$, Solution $$x_0=e$$, Plug in ② obtain $$y_0=e^2$$, Re substitution ③ obtain ,$$e^2=a^e$$, Take logarithms on both sides to get ,$$lna=\cfrac{2}{e}$$, be $$a=e^{\frac{2}{e}}$$, When two curves are tangent $$a=e^{\frac{2}{e}}$$, be $$a>e^{\frac{2}{e}}$$ when , There must be only one intersection between two curves . When $$0<a<1$$ when , function $$y=x^2$$ And functions $$y=a^x$$ The critical position of two intersections is in the case that the second quadrant is tangent , As shown in the figure below ; The following focuses on solving the parameters of tangency $$a$$ Value ; Let the tangent point of two curves be $$P(x_0,y_0)$$, Then there are $$\left\{\begin{array}{l}{2x_0=a^{x_0}\cdot lna ① }\\{y_0=x_0^2 ② }\\{y_0=a^{x_0} ③ }\end{array}\right.$$ from ②③ You know ,$x_02=a{x_0} ④ \$, Plug in ① obtain ,$$2x_0=x_0^2\cdot lna$$, Simplify to get $$2=x_0\cdot lna ⑤$$,

Again by ④ Take logarithms on both sides to get ,$$2ln|x_0|=x_0\cdot lna⑥$$, from ⑤⑥ obtain ,$$2ln|x_0|=2$$, Solution $$x_0=-e$$, Plug in ② obtain $$y_0=e^2$$,

Re substitution ③ obtain ,$$e^2=a^{-e}$$, Take logarithms on both sides to get ,$$-lna=\cfrac{2}{e}$$, be $$a=e^{-\frac{2}{e}}$$,

When two curves are tangent $$a=e^{-\frac{2}{e}}$$, be $$0<a<e^{-\frac{2}{e}}$$ when , There must be only one intersection between two curves .

in summary ,$$a\in(0,e^{-\frac{2}{e}})$$, perhaps $$a\in (e^{\frac{2}{e}},+\infty)$$, So choose $$C$$.

Law 2： The separation parameters are obtained ,$$lnx^2=xlna$$, And then it's transformed into $$lna=\cfrac{2ln|x|}{x}$$, Make $$h(x)=\cfrac{2ln|x|}{x}$$, The point is to make an image of it ;

because $$h(x)$$ It's an odd function , Therefore, when $$x>0$$ when ,$$h(x)=\cfrac{2lnx}{x}$$, Let's study its monotonicity with derivatives ;

$$h'(x)=\cdots=\cfrac{2(1-lnx)}{x^2}$$, be $$x\in (0,e)$$ when ,$$h'(x)>0$$,$$h(x)$$ Monotone increasing ; be $$x\in (e,+\infty)$$ when ,$$h'(x)<0$$,$$h(x)$$ Monotonic decline ; also $$h(e)=\cfrac{2}{e}$$, So we can make $$x>0$$ At the time of the $$h(x)$$ Images and $$x<0$$ At the time of the $$h(x)$$ Images , As shown in the figure below ;

It can be seen from the picture that ,$$lna>\cfrac{2}{e}$$ or $$lna<-\cfrac{2}{e}$$ when , There is only one intersection between two function images ,

Solution $$a\in(0,e^{-\frac{2}{e}})$$, perhaps $$a\in (e^{\frac{2}{e}},+\infty)$$, So choose $$C$$.

posted @ 2019-03-14 15:11  Jingyazhai mathematics   read ( 274)  Comment on ( 0 edit   Collection

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