A math homework problem in high school during the summer vacation, a math problem in the first year

For the sake of simplicity, write m 1 2 [f(x1)+f(x2)] and take a new function g(x)=f(x)-m, then the equation f(x) m< >g(x) 0

Therefore, it is only necessary to prove that g(x) 0 has a real root between x1 and x2.

This is equivalent to proving g(x1)g(x2)<0Whereas.

g(x1)g(x2)

f(x1)-m][f(x2)-m]

f(x1)f(x2)-m[f(x1)+f(x2)]+m 2f(x1)f(x2)-(1 2)[f(x1)+f(x2)] 2+(1 4))[f(x1)+f(x2)] 2[bring m back].

f(x1)f(x2)-(1/4))[f(x1)+f(x2)]^2-(1/4)[f(x1)-f(x2)]^2f(x1)≠f(x2)

1/4)[f(x1)-f(x2)]^2<0g(x1)g(x2)<0.

f(x)=1 2[f(x1)+f(x2)] has a real root between x1 and x2.

The function range is y

d=max，e=min=

a=(d,e) y, then f=[f(x) f(x)] 2 a y because f(x) ≠ f(x), so c does not obtain an extremum.

Because e< f

This is easy to understand by drawing! First of all, f(x) is a continuous quadratic function, for f(x)=(f(x1)+f(x2)) 2 has: assuming that f(x1) is based on this, we can determine that there is a fixed one between (x1, x2)!

lg(x - y) +lg(x + 2y)= lg[(x - y)(x + 2y)]lg^2 + lgx + lgy

lg(2xy)

So: (x - y) (x + 2y) = 2xyx 2 + 2xy - xy - 2y 2 = 2xyx 2 - 2y 2 = xy

x/y - 2y/x = 1

Let a = x y then there is.

a^2 - a - 2 = 0

a - 2 )(a + 1) = 0

a = 2,-1 ..According to the meaning of the question, so: x y = 2

lg(x-y)+lg(x+2y)=lg2+lgx+lgylg(x-y)(x+2y)=lg2xy

x-y)(x+2y)=2xy

x^2+2xy-xy-2y^2=2xy

x^2-xy-2y^2=0

x-2y)(x+y)=0

x=2y or x+y=0

Because x>0 y>0

So x+y=0 (rounded).

So x y=2

Because g is the center of gravity of the triangle abc, the vector ga + vector gb + vector gc = vector 0

Set the center of gravity g (xo, yo).

Then (xo+2,yo)reed with +(xo-2,yo)+(xo-x,yo-y)=(0,0).

i.e. 3xo-x=0, 3yo-y=0

And because vertex c moves on the curve x + y = 4, 9xo +9yo = 4

That is, the trajectory equation for the center of gravity g of the triangle ABC is x +y = 4 9

p(4,2) is a point in the circle c:x 2+y 2-24x-28y-36=0, and the moving points a,b on the circle satisfy apb=90°

q(x,y)

2x=xa+xb,2y=ya+yb

4x^2=(xa+xb)^2

4y^2=(ya^2+yb)^2

x^2+y^2-24x-28y-36=0

xa)^2+(ya)^2-24xa-28ya-36=0

xb)^2+(yb)^2-24xb-28yb-36=0

xa)^2+(ya)^2-24xa-28ya-36]+[xb)^2+(yb)^2-24xb-28yb-36]=0

xa)^2+(xb)^2+(ya)^2+(yb)^2-24*(xa+xb)-28*(ya+yb)-72=0

xa+xb)^2-2xa*xb+(ya+yb)^2-2ya*yb-24*2x-28*2y-72=0

4x^2+4y^2-48x-56y-72=2(xa*xb+ya*yb)

pa⊥pb(ya-2)/(xa-4)]*yb-2)/(xb-4)]=1

xa-4)*(xb-4)+(ya-2)*(yb-2)=0

xa*xb+ya*yb=4(xa+xb)+2(ya+yb)-20

xa*xb+ya*yb=4*2x+2*2y-20

16x+8y-40=2(xa*xb+ya*yb)

4x^2+4y^2-48x-56y-72=16x+8y-40

x^2+y^2-16x-16y-8=0

The orbital finch carrying equation for point q in AB is garden: (x-8) 2+(y-8) 2=136

15.Solution: Let the vertex c coordinate state trillion be (x0,y0), the center of gravity g coordinate is (x,y), and x=(-2+0+x0) 3,y=(0+(-2)+y0) 3, then x0=3x+2, y0=3y+2, substitute the equation to obtain, (3x+2) 2+(3y+2) 2=4, finish, the trajectory equation of the center of gravity g is 9x 2+9y 2+12x+12y+4=0

Let the point q be the point of symmetry q with respect to l'(a,b), then b (a-2)=1 ; 2+a)/2+b/2=4 ②

From , we get a=4 and b=2 q'(4,2)

Set the point q with respect to the y-axis symmetry point q'', easy to get q''The straight line where (-2,0)ef is q'q''The straight line, i.e., x-3y+2=0

Find the symmetry point of Q with respect to the y-axis and Q with respect to the line L, and the line connecting these two points is the straight line EF.

The method of finding the symmetry point b of a point a with respect to a straight line:

Let the symmetry point be b(a,b), and list two equations to solve in the search condition, according to my experience, the conditions for listing the simplest equation are (1) ab is on a symmetrical line (2) ab is perpendicular to a symmetrical line (the product of the two slopes is equal to -1).

For example, let q(2,0) be the symmetry point of the straight line l as p(m,n) then:

2+m) 2 + n+0 2 =4 (condition 1) (n-0) (m-2) *1 =-1 (condition 2) to get m=4, n=2, i.e. p(4,2) (the straight line in this question is special, you can use a simpler method, you can draw a picture to get the law).

Obviously, the symmetry point of q(2,0) on the y-axis is (-2,0) Therefore, the equation for the straight line EF obtained from the two-point equation is: x-3y+2=0 Comments: This problem is a typical combination of physics and mathematics.