A high school math problem, everyone helps, urgent, about functions

He's an increment function, right, so let x1>x2 have f(x1)>f(x2).

Original = (x+a+a-2) (x+a) =1+(a-2) (x+a).

f(x1)-f(x2)=1+(a-2)/(x1+a)-1-(a-2)(x2+a)=(x1-x2)(2-a)/(x1+a)(x2+a) >0

x1-x2>0 is known, and (2-a) can be discussed categorically when a>2f(x1)-f(x2)>0 doesn't work, right, then a can only be less than 2

When it is less than 2, it is necessary to consider (x1+a)(x2+a)>0, so that a should be greater than 0 regardless of adding x1 or x2, so a should be greater than -1 from here -10 (a-2) (x+a)>-1

Since he is an additive function, when x = 1 and his value is = 0 then it can be determined that the values of (1, infinite) are all greater than 0

So bring x=1 in.

a-2)/(1+a)>=-1

The above gives a>-1, so a-2>=-1-a gives a>=

Combine the above, <=a<2 so a can only be equal to 1, and that's it.

Solution: Order 10

1. a>-1 a>-n n+2a-2<0 1 20 -1 Because a is an integer, a=1

1．f(x)=ax^5-bx+2

f(-x)=a(-x)^5-b(-x)+2= -ax^5+bx+2，f(x)+f(-x)=4

f(-3)=1，∴f(3)=3；

2 f(x) is an odd function defined on (-1,1), f(1-a)+f(2a-1)<0 can be reduced to.

f(1-a)< f(2a-1)

f(1-a)< f(-2a+1), and f(x) is a subtractive function on (-1,1), -1<-2a+1<1-a<1, the solution is 00, when x>0, f(x)=x(1+x(1 3)), x(1 3) is the cube root of x).

f(-x)=(-x)[1+(-x) (1 3)]= -x[1-x(1 3)], and f(x) is an odd function, f(x) = -f(-x)= x[1-x (1 3)], so when x<0, f(x) = x[1-x (1 3)].

The axis of symmetry of u=x +2x-3=(x+1) -4 is x=-1, obviously it decreases monotonically at (negative infinity, -1), but it should be noted that the number of squares should not be less than zero at the same time, that is, (x+1) -4 0 gives x -1 or x -3, and the comprehensive answer is (negative infinity, -3).

The domain of f(x) is [a,b], and the domain of f(x) is a<=-x<=b.

i.e. the domain [-b,-a],a b 0 b>-a

a>0.

a>-a

g(x) is an empty set.

a>0.

a<=-a

g(x) [a,-a],

y＝－x²＋2x=-（x-1）²+1≤1

Therefore, the value range of the elements in b is a real number less than or equal to 1, and for the real number k b, if there is no element corresponding to k in the set a, then k 1

x^2+3x-4≥0

x^2+4x+12≥0

Solving the above set of inequalities gives its definition domain: 1 x 6

Solution: By the question, there is.

x^2+3x-4≥0，①

x^2+4x+12≥0，②

From , the solution is x (-4] [1,+

From , the solution is x[-2,6].

Two sets of solutions are intersected, and there is.

x∈[1,6].

Reference).

1<=x<=6

Look at your answer, right.

1. Just bring in the expression.

2，f（1）=1/2,f(1/3)+f(3)=1,f(1/2)+f(2）=1

You have already proved in the first question that f(a)+f(1 a)=1, then f(3)+f(1 3)=1, the same f(2)+f(1 2)=1, f(1)+f(1 1)=1, so.

f(1)=1/2.So f(1 3) + f(1 2) + f(1) + f(2) + f(3) =