A high school question, a high school question

Solution: Let 2l=20cm, l=10cm

The geometric relationships in the diagram are known. When the rod is under pressure on x and y at any moment, the trajectory of the rod's centroid o is an arc with bo as the radius.

again ob = 1 2 ac = l

The force analysis yields n1, n2, mg

Get the equation: ..."()"Indicates a corner marker.

n(2)lsin -n(1)lcos =0---1 gives n(2) n(1)=tan from 1

.Here comes a conclusion. The resultant force of the two supporting forces is in the radius direction along the centroid trajectory.

Then it was found that when A descended some distance. n(1)=0 the rod will break away from the wall y, but since n(1)=0 and n(2)≠0, then the n(2) rod will rotate infinitely. It's impossible. Not true.

So both support forces are 0

Obviously. Equivalent treatment can be done here. It resembles a ball rolling down a semicircle.

In the process before detaching from the hemisphere, when the analysis is just out of the hemisphere, there is 2 indicating that the quadratic mgsin = m[v(0) 2 l]--2mg[l-h]=[1 2]mv(0) 2---3 is solved to obtain h=[2 3]l; v(0)=√

It is clear to everyone that the ball will be thrown diagonally.

There is v = v(0)cos; v⊥=v(0)sinα--4sinα=h/l=2/3

cosα=[√5]/3

Write the kinematic equations in both vertical and horizontal directions.

There is h=v t+[1 2]gt 2---5

s(1)=v‖t---6

Simultaneous 1-5 gives t1 = rounded)

t2 = and then substitute 6 to get s(1) =

The distance traveled by o is s=d+s(1)=

Just get it done.

In fact, if such a problem is done with advanced mathematics knowledge in college, it is much easier to do, because the upper and lower regions are infinitely 0 at the same time, so the derivative is found up and down at the same time: the derivative above is 2x and the bottom is 1, so 2x when x is 5, you can get the answer: 10

If it is a high school method, then the numerator can only be written as a square difference formula, and then x-5 is reduced up and down, leaving x+5, x can be brought in, and it can only be simplified.

It's so simple.

tana=sina cosa tana<0, cosa>0, then sina<0

So sina = -4 5

f（x1x2）=f（x1）+f（x2），f（4）=1

f（x+6）+f（x）=f[(x+6）x]2=f（4）+f(4)=f(16)

f（x+6）+f（x）＞2

f[(x+6）x]>f(16)

The function f(x) on r+ is an increasing function.

x+6）x>16

x^2+6x-16>0

x^2+6x-16=0

x1=-8,x2=2

x<-8 or x>2

x>2 can be a logarithmic function with a base of 4, because it satisfies all the known conditions, so that the problem can be easily solved by turning it into concrete, no longer abstract.

f(x+6)+f(x)2 is equivalent to.

log[4]（x+6）+log[4]x>log[4]16

x∈(0,2)

You can put this function as a special bit logarithmic function, and it can be easily solved.

For formulas with absolute values, the absolute value symbol should be removed first;

To remove the absolute value symbol, we must first find the "zero point" of the absolute value formula

How to find the zero point: let each absolute value formula 0, for the problem, the zero point is found: let x-1 0, x=1 is the zero point; Then make x+1 0, x -1 is the zero point!

So 2 zeros can divide the number line from left to right into three segments: (1)x<-1;（2）-1≤x＜1；（3）x≥1

When x takes the values in these three paragraphs, according to the positive and negative values in the two absolute value formulas, remove the absolute value symbol, and add "+" in front of the formula

For example: (1) when x<-1: x-1 =1-x; ｜x+1｜＝-x-1；

2) -1 x 1: x-1 =1-x; ｜x+1｜＝x+1；

3) x 1: x-1 = x-1; x+1 x+1 So when solving the problem, it should be divided into the above three situations.

According to the formula = 0 in the absolute value symbol, the absolute value symbol is removed from the range, and the calculated ...... is simplified

The absolute value is 0, then |x-1|≥0，|x,+1|≥0。Then for X-1, there are these cases, ; 1 ；<1.For x+1, there is -1; >1 ；<1.

You have learned, small take the smallest, big take the biggest.

The three ranges correspond to two absolute values, i.e., x-1 0 and x+1 0 respectively, then we get x -1

x-1 0, x+1 0, then -1 x 1 x-1 0, x+1 0, then x 1

Then correspond to each range and solve the absolute value formula.

However, it should be -2x at x -1.

The first question is that f(x)-a x and g(x) are inverse functions of each other, so it is easy to find f(x) (x-2) (x+1)+a x(x≠ 1).

The second question proves that f(x) 0 has no negative real root, that is, (x-2) (x+1) a x=0 has no negative real root, and (x-2) (x+1)=1-3 (x+1), that is, 1-3 (x+1)+a x=0 has no negative real root, that is, 3 (x+1)=a x+1 has no negative real root. Then divide a>1 and 0

You wait, I'll report to you the teacher to go.

(1) The image of y=f(x)-a x and y=g(x) is symmetrical with respect to the straight line y=x.

f(x)-a x g 1(x) inverse function g(x) x 1 3) (x 1) 1 3 (x 1)g(x) 1 3 ( x 1).

x＋1=3/(g(x)＋1)

x 1 3 (g(x) 1) (g(x) 2) (g(x) 1) g 1(x) inverse function (x 2) (x 1) f(x) (x 2) (x 1) a x(x≠ 1)(2) The title is.

Try to prove that the equation f(x=0) f(x) 0 does not have a negative real root?

y=(x-a)^2-a^2-1

The opening is upward, and the axis of symmetry x=a

If a<0, then y is to the right of the axis of symmetry and is an increasing function.

So x=0 and y min=-1

x = 2, y max = 3-4a

If 0<=a<=1

Then x = a and y minimum = -a 2-1

2 to 0 farther away from the axis of symmetry.

So x = 2 and y max = 3-4a

If 1<=a<=2

Then x = a and y minimum = -a 2-1

From 0 to 2 farther away from the axis of symmetry.

So x=0, y max=-1

If a>2, then y is to the left of the axis of symmetry and is a subtractive function.

So x = 2 and y min = 3-4a

x = 0, y max = -1

1) At the beginning of the n point, the electric field force gravity is decomposed into along the rod and perpendicular to the rod, and then the two components along the rod and the Coulomb force between Mn are combined to form the resultant force, f=ma.

2) The method is the same as the above method, this time when the resultant force along the direction of the Zheng rod in front of Hui Lasong is equal to zero, that is, when the velocity is the maximum, go to find the Coulomb force solution r.

The ball starts to do an acceleration motion with a gradual decrease in acceleration, and when the velocity is maximum, the acceleration is zero, and then the Coulomb force increases and starts to do a deceleration motion, and the direction perpendicular to the rod is not considered, because the supporting force of the rod is always balanced).