Mathematics of inequality problem

A) (2x-1) 2; 0 with the opening upward, the function has a minimum value, and the minimum point is X=4/(2*(4-A))=2/(4-A).

And f (x) = (4-a) (x-2/(4-a)) 2-4/(4-a)+1= (4-a) (x-2/(4-a)) 2+a/(4-a).

So f (x) < 0.

The range is 2/(4-a)-√ (a/(4-a)).

Because there are exactly three integers in the solution set.

So there must be) 3

That is, there are three integers in the middle of the difference, that is, 3/2 < √ (a/(4-a)) <; 2

9/4 & lt； A/(4-A)& lt； 4 = & gt36/ 13 & lt； A & lt 16/5

You 4-a > 0 = = A<4 synthesis, so the value range of A is 36/ 13.

B)f(x)=x^2+|x-a|+ 1

1)

Because f (0) >; 1 is not equal to 0, so F(X) must not be odd function.

And f (-x) = (-x) 2+|-x-a |+1= x 2+| x+a |+1.

So when |X-A|=|X+A|, that is, A=0 F(X) is an even function.

When a < > when F(X) is not an even function, it is 0.

2). Because x2 >；; =0 |X-A| >=0. So when x 2, | x-a | are all equal to 0, that is, when x =0 a = 0, F(X) has a minimum value, which is equal to 1.

C) the function f(x) defined on r satisfies f(x)=f(4-x), and when x > 2, f(x) monotonically increases.

Let A & gtB& gt；; Then F(A)>F(B)

F(A)= F(4-A)F(B)= F(4-B)= = & gt； F(4-A) > female (4-B)

Corresponding function f (x-4) a >: b >; 2, then 4-a

So when its function F(X) monotonically decreases, x < 2.

If x 1, x2 satisfies x 1 < x2, x 1+x2 < 4, (x 1-2) (x2-2) < 0.

X 1 belongs to

Because (x) satisfies f(x)=f(4-x),

So f (x 1)-f (x2) >: 0.

So choose a.