Ask for help in high school math with the problem of the constant establishment of parametric inequa

The known function f(x)=x -2ax+1, g(x)=a x, where a 0, x ≠ 0

1）.For any x [1,2], there is f(x) g(x) constant, and the range of the value of the real number a is obtained.

2）.For any x1 [1,2], x2 [2,4], there is f(x1) g(x2) constant, and the value range of the real number a is obtained.

Solution: (1).For any x [1,2], there is f(x) g(x) constant.

This is to find the minimum value of f(x) on [1,2].

and g(x) at [1,2], apparently g(x) at [1,2] is maximum at g(1)=a

The minimum value of f(x) is discussed, f(x) = (x-a) +1-a

When 0a, the solution is a<2 3

So 02, then the minimum value is f(2)=5-4a

To satisfy 5-4a>a, solve a<1

No solution. When 1 a 2, then the minimum value is f(a) = 1-a

1-a >a needs to be solved to obtain (-1- 5) 2a 2, solution a<4 5

So 02, then the minimum value is f(2)=5-4a

To satisfy 5-4a>a 2, a<10 9 is solved

No solution. When 1 a 2, then the minimum value is f(a) = 1-a

1-a >a 2 needs to be solved to (-1- 17) 4, so 1 a<(-1+ 17) 4

Sum up. That is, the scope of a is.

0

fx>gx

i.e. x -2ax + 1-a x>0

fx=x²-2ax+1-a/x

Because fx>0

So x 3-2ax 2+x-a>0

Let gx=x 3-2ax 2+x-a

Then derive it.

gx'= 3x^2-4ax+1

Now you can discuss the value of a.

The second step is to find the minimum value of fx from 1 to 2 and the maximum value of gx from 2 to 4.

are expressed in the formula containing a, and then the inequality about a can be obtained to solve the range of values of a) The following is the key It's up to you, the general idea has been given, if you want to learn it, you have to do it yourself.

If you are satisfied, I hope you can approve it.

Usage scenario: For types where parameters cannot be separated or it is difficult to find the maximum value after separating parameters.

The second step starts from the study of the properties of the function, which is transformed into the discussion of the monotonicity and extreme value of the function.

Step 3 Draw conclusions.

Example] known function , where .If on the interval , the constant is true, and the range of values for is found.

Solution], order, solution or

1) If , then.

So when, ;

When , .

So at that time, there is a maximum.

So, it is equivalent to.

Solve 2) if , then , then then, ;

When , .

So when , there is a maximum, and when , there is a minimum.

So, it is equivalent to.

solution or , therefore.

Synthesis (1) (2).

Summary] For the problem that the parameters cannot be separated or it is difficult to find the maximum value or the demarcation after separating the parameters, we can organize the inequality with parameters into appropriate forms such as , , etc., and then start from the study of the properties of the function and transform it into the discussion of the monotonicity and extreme value of the function. The following conclusions are often used in the process of problem solving:

1) If there is a minimum value, then the constant imitation spine grip is established, and the constant is established;

2) If there is a maximum, then Heng is established, and Heng Cheng is prepared to be established;

You're a freshman in high school, right? If you do the math yourself, you can also use quadratic functions to calculate!

x²+1＞2ax （x∈[1,2]

x＞0（x²+1]/x＞2a

Treat the above as a mean inequality) because x [1,2] so a (1,5 4).

And because a 0

A (1,5 4).

Quadratic unequal.

Formula ax +bx+c>0 on r.

The sufficient and necessary conditions for a constant formation are a>0 and b -4ac<0 has a solution.

Sufficient and necessary conditions A 0 or A<0 and B -4ac >0

The unsolved sufficient and necessary conditions a<0 and b -4ac 0

2.The sufficient condition for the quadratic non-answer equation ax +bx+c<0 to hold on r is a<0 and b -4ac<0 has a solution, the sufficient condition a 0 or a>0 and b -4ac > has no solution, and the sufficient condition a>0 and b -4ac 0 has no solution

4 questions can imitate 3 questions.

x²+(m-4)x+4-2m>0

then (x-2)m>-(x-2).

and x[-1,1], x-2<0

Therefore, divide the two sides of the above equation by x-2 to obtain.

m<2-x.

Obviously, at -1 x 1, 1 2-x 3

Therefore, when the original formula is established, m<1

m∈(-1)。

Since x +x+1=(x+1 2) +3 4 3 4>0 then 3x +2x+2 m(x +x+1).

3-m)x +(2-m)x+2-m 0 always holds m=3, -x-1 0 x 1 does not hold.

Therefore, only when 3-m>0 is m<3, it is possible to follow the discriminant principle at this time.

2-m)²-4(2-m)(3-m)≤0

m-2)(m-2-4m+12)≤0

m-2)(-3m+10)≤0

m-2)(3m-10)≥0

Solution m2 or m10 3

So m 2

The question should be (3x +2x+2) (x +x+1 ) m (below is done here).

First, analyze the denominator x +x+1=(x+1 2) +3 4 3 4, so the inequality is multiplied by the denominator at the same time, and the unequal sign does not change.

The inequality can be reduced to 3x +2x+2 m(x +x+1 )(3-m)x +(2-m)x+(2-m) 0 From the image of the binomial f(x)=ax +bx+c, it can be seen that in order to make f(x) 0 a constant function function image above the x-axis, only the following two conditions must be satisfied at the same time.

1. a>0 (ensure that the image opening is upward).

2. There is no intersection between the image and the x-axis or only one intersection point, i.e., δ=b -4ac 0, so as long as (3-m)>0 and (2-m) -4(3-m)(2-m) 0 are held at the same time.

Solving the inequality yields the range of values of the natural number m.