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The application of critical state in Mathematics

2021-07-21 15:04:43 wanghai2018

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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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