A math finale problem with a detailed solution

Since f(x+1) and f(x-1) are both odd functions, f(x)=f(x+4)t>4 t=4l (l is a positive integer).

t<=4 t=4 k (k is a positive integer).

The domain of f(x) is r, f(1)=f(-1)=0, if t=4l, zero, xo=2l·n+1 is odd (untitled, rounded), if t=4 k, xo=2 k·n+1

k=1 is not true.

k=2 Yes.

K>2 xo will appear as a score, no.

So if and only if x is an integer point, it is a zero point.

1+2011 2012 is an integer.

The number of zeros is 2012-(-2011)+1=4024, and g(x)=a(x-x1)(x-x2) is set to the zero distribution

ax^2-a(x1+x2)+ax1x2

y max = -[a(x1-x2) 2] 4

The minimum absolute value of the two intersections x1- x2 is obviously 1

Make the absolute value of the maximum value of y minimum.

a≠0a = -1 or 1

Let a=1 x1=0 x2=1

For g(x) at [0,1] integral.

3) For:

Equivalent to TN <=3N28(n-6).

tn²>=(5n/(4n+3))²1

tn >=3 4-3n 16(n-5) against and in order to make tn 2 as large as possible.

an, as small as possible.

1.Obviously f(x) is a periodic function, and 2 is one of the periods: f(1)=f(-1)=0f(x-1+2011 2012) has 2012 zeros on [-2011, 2012].

The answer on the fourth floor is purely copied, or is it copied randomly without looking at the question. In that case, I had to ask again.

This is compulsory four, I'll study it.

A math finale, and hopefully a detailed solution to the one, I would like to revise the containers that keep happening and threaten to get into everything.

**Can't see clearly, and the direction is also problematic. As for you saying that the teacher doesn't talk about it, I probably don't think it's necessary.

What the hell, are you sharing it for us to do?

These mathematicians all mention them specifically in **.

1.Rectangular piece of paper ABCD, where CE intersects GF at O and CE perpendicular to FG

CG=eg=cf, and when g coincides with d, eg=3

So 0ag so e is not on the ad edge, on the ab edge, ae = eg -ag

ae = 1 if g is on the cd edge, cg = 2

eg=2de=²=eg²-dg²=3

de=√3ae=4-√3

So the chances of winning for 100 people are the same, 1 4

Let p(6,(8 root number), by the angle between the and x axis is 30 degrees, we can know that the slope is tan30=(root number 3) 3, and the equation for the straight line can be found as: y=(root number 3) 3x+(2 root number 3) 3, and because it is tangent to the circle m, it can be seen that the distance from the center of the circle (2,0) to the straight line is 2, at this time, you can find it directly!

It's a hassle! Also draw pictures, good luck! Actually, I feel like you should be able to make it!

Solution: 1) Flipping the trapezoid twice is denoted as A''b''c''d''，b''n=2bc-nc=2*5-8=2cm；

Obviously, because DCB is 60°, the overlapping part is also an equilateral triangle, and the side length = b''n=2, so the area = (1 2)*2*2*sin 60°=root number 3cm 2;(It shouldn't be difficult to find the area of an equilateral triangle knowing the edge length!) ）

2) Flipping the trapezoid three times is denoted as a'''b'''c'''d''', by calculating ad=2, ab=3, root number 3

To make the area of the overlap of the right-angled trapezoidal and the equilateral triangle equal to the area of the right-angled trapezoidal ABCD.

That is, the trapezoid is completely within the equilateral triangle, and the drawing can be seen, putting a'''d'''Extending the intersection of pn and e, apparently en=d'''c'''=6

pe=a'''E (easy to prove triangle PD'''e is an equilateral triangle), a'''e=a'''d'''+d'''e=ad+c'''n=2+2=4()

Therefore, the side length pn=pe+en=4+6=10, which is the minimum side length, so a is at least 10cm;

3) From the meaning of the title, it is not difficult to know that the PM ratio is pn at point F, the overlapping area is the trapezoidal area, and the trapezoidal area is s=1 2*(2+5)*3 root number 3=21 root number 3 2; Half is 21 roots numbered 3 4, that is, equilateral triangle FMN=21 roots numbered 3 4, so that there is an edge length mc'''= root number 21, and c'''n=2, so mn=mc'''+c'''n = root number 21 + 2

It is not difficult to draw a good picture in this question, attached: equilateral triangle area = side length * root number 3 4

1、a²=4,b²=3

c²=4-3=1

c=1, so x=a, c=4

2. x=4, substitute x 4-y 3=1

So p(4,3).

a1(-2,0),a2(2,0)

then PA1 is x-2y+2=0

PA2 is 3x-2y-6=0

pa1:x=2y-2

Substitute x 4 + y 3 = 1

2y²-3y=0

y=3/2,x=1

So q(1,3,2).

pa1:y=3x/2-3

Substitute x 4 + y 3 = 1

2y²-3y=0

y=3/2,x=1

So q(1,3,2).

x=1,y=-3/2

r(1,-3/2)

So qr=3 2-(-3 2)=3

(1) x 2-x-6=0 (x 2)(x-3)=0 x1=-2,x2=3 The parabola y=ax 2 bx c intersects the positive semi-axis of the y-axis at c, then c 0 s abc=|x1-x2|·c/2=5c/2=15/2