I can t do the function problems in the first year of high school, so let s help, I can t do the mat

Then there are: |x-1|=kx （*

You may wish to set y=|x-1| ；z=kx .

Make y=|x-1|images.

It is an image with a straight line x=1 as the axis of symmetry, the absolute value of the slope is 1, and the V-shaped opening is pointing upward.

z=kx If there is an intersection with the absolute value of the image, then at this point the y function and the z function of the image.

The abscissa x is equal, the ordinate y is equal (equivalent to the (*) equation is true), and the x at this point is the solution of the equation.

Obviously, when k 1 or k = 0 is it possible to have only 1 solution.

When 00 ensures that the intersection is in the first quadrant of the Cartesian coordinate system, x>0 so the value range of k is: k=0 or k 1

If x>=1, then (1-k)x=1, x=1 (1-k)>=1, so k belongs to [0,1].

If x<1, then 0=0

By |x-1|-kx=0 kx=|x-1|0 is conditionally known, x>0

So k 0 then discusses the relationship between x and 1.

When x>1, the equation is sorted to (k-1)x=-1

When k = 1, there is no solution to the equation and it is rounded.

When k! =1, x=1 (1-k) >0

Solution: k<1

When x (0,1), the equation is sorted as (k+1)x=1, where x=1 (k+1).

0<1 (k+1)<1.

k>0 can be summarized to get the value range of k.

When x (0,1], k 0

When x (1, ). k＜1

Let l:y=kx

Then when l intersects the line y=x-1 or y=1-x above the x-axis, the x-value corresponding to the intersection satisfies |x-1|-kx=0

Geometrically, if l has only one intersection point with these two lines at x>0, k>=1 (0=1 or k=0.)

The first question mark:

f(√x-1)=x+2

x) 2-2 x + 1 + 2 x - 2 + 3=( x-1) 2 + 2( x-1) +3, so f(x)=x 2+2x+3

Second question mark:

Let the two equations y=x+2 and y=x 2

Then these two upper curves intersect at two points (-1,1) and (2,4) and for all real numbers x, f(x) has the following equation:

x<-1,f(x)=x+2

1≤x≤2，f(x)=x^2

x>2, f(x)=x+2

The third question mark:

If the function f(x) is known to be odd and defined at x equals 0, then f(0) is equal to 0.

The fourth question mark:

It is known that f(x/x) is equal to one minus x-squared plus 1 f(x)f(1 x)=(x 2+1) (1-x 2) Let 1 x=m, then x=1 m

f(m)=[1/m)^2+1]/[1-(1/m)^2]=(m^2+1)/(m^2-1)

Therefore f(x)=(x 2+1) (x 2-1).

1、sin60=√3/2

sin45=√2/2

The answer is 3 2,45°

sets, set A maps to B, ,0 C,1, 0 B,1

0 a, 1 other, the same goes for it.

f: a --b , x --y=sinx ∈ b=(0,1) ;x∈ (0,π/2)

60 --y=sin 60 degrees = 3 22 2 = y =sinx --x=45 degrees [Let the set A have n elements and B has m elements, then there are m n mappings from A to B] A to B has 2 3=8 mappings.

I said it myself, it's too long)

(1) Knowing from the meaning of the title. Let the element in b be f(x), then f(x)=sinx

So f(60°) = root number 3 2, and sinx = 2 2, x = 45°

2) 6 in total, a-0, b-0, c-0, a-1, b-1, c-1

In addition to the learning of function knowledge points separately mentioned upstairs: 1. Understand the meaning of functions, that is, the correspondence between function values and independent variables, and the function values change with the change of independent variablesWhen solving some function transformation problems, all equivalent transformations should be carried out around independent variables 2. Be familiar with the properties of common elementary functions, including exponential functions, logarithmic functions, power functions, etc. Define domain, value range, periodicity, symmetry, monotonicity, parity 3, use images, numbers and shapes to solve problems Jiedeng Education High School Group teachers also suggest that high school students understand the characteristics of high school learning, so as to find their own learning methods and improve faster:

1. Compared with junior high school mathematics, high school mathematics is more difficult [coping methods] It is necessary to thoroughly understand the content supplemented by the teacher in books and classrooms, sometimes to think repeatedly and study repeatedly, to be able to draw inferences from one another on the basis of understanding, and to ask questions on the basis of diligence. 2. Caused by the different requirements for mathematics at different learning stages of junior high and high school. 【Correct method】To see the student's performance can not only look at the score value, the key is to look at the relative position of the class or grade, but also to look at the position of the student's school in the city.

3. The inadaptability of learning methods [coping methods] In class, it is necessary not only to understand, but also to write down the content supplemented by the teacher appropriately, and it is best to digest the learned content after class before doing homework, and do not read notes or formulas while doing questions. After class, try to choose some relevant questions to practice so that you can bypass the class. 4. Relax your mind.

Coping method] The course content of the first year of high school must not be slack, the knowledge of functions runs through the whole of high school mathematics, and the idea of function is a powerful tool to solve many problems. At the beginning of high school, it is very important to develop a diligent and assiduous study attitude, rigorous and serious study habits and methods. There are more than a dozen chapters of high school mathematics, and the first year of high school mathematics is mainly about functions, and some students do not learn functions very well, but they can learn solid geometry and analytic geometry well in the second year of high school, so they must treat students with a changing point of view.

Encouragement and self-confidence are the magic weapons of education that will never fail.

The element x in m maps to the set n and remains x

Description: m=n, then b a=0 a=1

i.e. b = 0 a = 1

a+b=1

For the question of Figure 2:

The solution set is r, which means that for all x belongs to r, fx 2a is true. a<0, indicating that the function image opening is downward, as shown in the following figure.

For Figure 3: If you don't understand, you can ask.

When you encounter this kind of problem, you have to judge it every time.

a=0, a<0, a>0, and then judge that 4ac must be greater than or equal to 0 so that this is feasible, and everything else is not feasible.

I'll do this question, what do you want to ask about that small step?

Let me tell you about the second question, because the coefficient of the quadratic term is indefinite, so we must discuss it first, when a is not equal to 0, the parabola opening is downward, and the discriminant formula must be less than 0 to conform to this parabola is always below the x-axis.