Knowing that f x is an even function, the domain defined is , , , and is a subtraction function on 0

p [3 4,+ f(x) is an even function, and on [0,+ is a subtraction function.

So f(x) increases monotonically on (0), and . f(-3/4)=.f(3 4), choose c

f(x) is an even function defined on r and subtracts on [0,+.

So f(p)=f(a 2-a+1)=f[(a-1 2) 2+3 4] f(3 4).

Because of the even function, f(-3 4) = f(3 4) is c

Answer: f(x) is an even function.

f(log2 x)=f(-|log2 x|),f(1/2)=f(-1/2)

f(log2 x)>0

i.e. f(-|log2 x|> f(-1 2) f(x) is a subtractive function on (-0).

log2 x|<-1/2

log2(x)|>1/2

log2(x)>1 2 or log2(x)<-1 2log2(x)>log2(2) or log2(x) 2 or 0 i.e. the solution set is (0, 2 2)u( 2,+

f(x) is an even function, f(-3 2)=f(3 2) is a subtraction function on [0,+, e.g. x 3 2 , then: f(x) f(3 2).

x= a²+2a+5/2 = (a+1）² 3/2 ≥ 3/2

f(x)=f(a²+2a+5/2) ≤f(3/2) = f(-3/2)

First of all, since f is an even function, it should be a subtraction function on (-1,0).

The inequality is then deformed f(3-2a)<-f(a-2) =f(a-2) (property of even function).

If you draw a picture, you can see that it should be |a-2|>|3-2a|Then the squares of both sides (both are positive and squared) and then we get a unary quadratic inequality 3a 2-8a+5<0, and the answer should be (1,5, 3).

On [0,1) is the increment function.

then 0<=a0

So f(x) at (-1,0) is a subtractive function.

Define domains-1a, 2-4

A 2-a-2<0, -10, a 2-4>0 multiplication function a-20

a>2,a<-1

So 2a 2-4

a^2+a-6<0

34-a^2

a^2+a-6>0

a<-3,a>2

So 2 so 3 [Welcome to ask, thank you for adopting!] o(∩_o~】

Solve the inequality and get the result.

There are many even functions that define the domain as (-1,1), and different even functions should get different a.

Hello, the function f(x) is an even function f(-1 2)=f(1 2)a +a+1=(a +a+1 4)+3 4=(a+1 2) +3 4 3 4 1 2

So f(a +a+1) f(1 2).

f(-1/2)＞f(a²+a+1)

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a-2, 4-a are all on the defined domain.

13 - 50 a -4>0, the function is an increasing function on [0,1), so a-20 (a-2)(a+1) >0

a>2 or a<-1, and 2a -4

a²-a-2<0 (a-2)(a+1)<0-1 is obtained

1 All f(x) are even functions and are subtracting functions on [0,+, so are increasing functions on (- 0].

f(2a+1)>f(-1).

It is only necessary to solve the absolute value of 2a+1 <-1 (you can see the image of the even function, if you don't understand it, you can discuss it by category).

1

(1) Let x1-x2, and -x1, -x2 (-0] because f(x) is a subtraction function in (- 0), so f(-x1)f(1).

f(x) is increased on [0,+, so.

a-1|>1, i.e., a-1>1 or a-1<-1, gives a>2 or a<0

f(x) is an even function, f(-x)=f(x)Let -x be at (-o) i.e. x 0, ]f -x at (0) subtract function, with respect to y-axis symmetry, f(x) in... Increment function.

From the question, it can be seen that <-1 a<-2 to sum up a>2 or a<-2