It is known that A is a set of functions f x defined on 2,4 and satisfying the following conditions

I did all the math homework books, and I couldn't find them when I looked them up, so I made a lot of effort to do them myself. I don't have an answer, I come back and praise a few words when I do it right, and if I do it wrong, I should go back to school to pay for it. Proof of the following:

Assuming that such x0 is not unique, there is x1, x2 to make the equation hold.

then x1= (2 x1).

x2=ψ（2 x2） ②

Gotta |x1-x2|=|ψ（2 x1）| 2 x2）|Anyway, the equation is equal on both sides, and the addition of the absolute value does not affect, in order for the form to be consistent with condition 2).

and l|x1-x2|≥|2 x1）| 2 x2）|

Solution 1 l

Contradicts 0 l 1.

The assumption is not true.

Such x0 is unique.

In addition, the argument of the function is 2x, and for any x it belongs to [1,2], which means that it defines the domain[2,4], so there is nothing wrong with that.

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Well, no matter how you look at it, it feels very contradictory Since f(x) belongs to a, then he should be in the middle of the interval 2 to 4, so no matter what x is, he should belong to this interval, that is, when x is equal to 2x, he will still be in the 2 to 4 interval, how can there be a f(2x) belongs to the false interval of 1 to 2 tangled mathematics.

Yes! The first defined field is [2,4], and then how can x be on [1,2].

1) f1(49) =2-sqr49 =-5 does not belong to (1,4]f1(x) does not call for respect in set a.

x 0, 0 (1 2) x 1

0 3 (1 2) 2 3 thus 1 1+3 (1 2) x 4f2(x) (1,4] and f2(x)=1+3 (1 2) x on [0,+ and ashwagandha as a subtraction function.

f2(x)=1+3(1 2) x in set a.

2) When x 0, f(x)+f(x+2)=2+15 4 (1 2) x 23 4

Again it is known that f(x)+f(x+2) k is always true for any x 0, k 23 4

Therefore, the range of the value of the real number k is [23 4 ,+

If a a, then 1 1-a a, 1-1 a a, and a, 1 1-a, 1-1 a are not equal to each other, that is, if there is element a in a, then there must be 2 other elements in a 1 1-a, 1-1 a, so the elements in a are all in a group of 3 at the same time in a, so the number of elements in a a may be 3, 6, 9 ,..That is, the number of elements in a must be an integer multiple of 3.

Because 4 (x-3) is an integer, x-3 is the divisor of 4, and the divisor of 4 is -4, -2, -1, 1, 2, 4, so x-3=-4, -2, -1, 1, 2, 4

The solution gives x=-1,1,2,4,5,7, and x is a natural number, so a={1,2,4,5,7}

Proof: ( For any <>

So <>

to arbitrary <>

So <>

Ling <> "So<>

Counter-evidence: There are two <>

Make <>

then by <>

Therefore, l 1, contradictory, so the conclusion is valid.

So <>

1) f1(49)=2 49= 5 does not belong to (1,4) f1(x) is not wise in set a.

x 0, 0 (1 2) x 1

6＜-6(1/2)^x≤0

2 4-6(1 2) x object bend 4

f2(x)∈(2，4]

and f2(x) is a subtraction function on [0,+.

f2(x)=4-6(1 2) x in set a.

2 When x 0, f(x)+f(x+2)=2+[15(1 2)x] 4 23 4

2f(x+1)≥23/4＞0

F(x)+f(x+2) 2f(x+1) is always true for any x 0.

1) F1(x) does not change the accompaniment in set A, because when X>36, F1(X) >4; f2(x) belongs to the set a, because 6(1 2) x at x>0, decreasing and (0,1], so f2(x) increments at x>0 and f2(x) [2,4].

2) For f(x)=f2(x)=4-6(1 2) x(x 0), there is a scuff [f(x)+f(x+2)] 2=4-6 2 (x+1)-6 2 (x+2)<4-6 2 (x+1)=f(x+1), so the inequality f(x)+f(x+2) 2f(x+1) holds for any x-total nucleus.

f1(x) is not f1(x) is certainly greater than 4 when x is large

1 2) x when (x is greater than or equal to 0 belongs to 0, 1 left open and right closed, so f2 (x) belongs to [bored -2, destroy the cover with 4) and he is a subtraction function plus fiber with a negative sign is an increase function so f2 (x) is in a.

Bring in 6*(1 2) (x+2) greater than 0 to be constant.

1) f1(49) =2- 49 =-5 does not belong to (1,4]f1(x) is not in set a.

x 0, 0 (1 2) x 1

Object 0 3 (1 2) 2 3 thus 1 1 + 3 (1 2) x 4

f2(x) (1,4] and f2(x)=1+3 (1 2) x is the subtraction function of the Changju on [0,+.

f2(x)=1+3(1 2) x in set a.

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1.Substitute 1 for 6 (1+x) to get 3, and then beat 3 for Leaky Tong to lose into Kitan for 3 2

So 3 2 belongs to A

2.Let x=6 (1+x) give x=2 or x=-3 (rounded off because x 0).