Domain-1

- 1 & lt； =x^2-x

x^2-x+ 1>； =0

Delta & lt0 must be established.

x^2-x<； = 1

x^2-x- 1<； =0

( 1-√5)/2 & lt； = x & lt=( 1+√5)/2

Y=arccosx is the subtraction function.

Then when x 2-x monotonically decreases,

The monotonic increase of y = arccos (x 2-x)

Y = x 2-x- 1 the symmetry axis is x= 1/2.

So (1-√ 5)/2

X 2-x monotonically decreases.

Therefore, the monotonically increasing interval is [( 1-√5)/2, 1/2].

2.f(x)= xz+4ax+2 in (-∞, 6) to find a by decreasing.

3. The monotonic decreasing interval of f (x) = xz+4ax+2 is (-∞, 6) to find a..

4 Find the monotone subtraction interval of y=√xz+3x+3.

5 f(x) satisfies that f(-x)= -f(x) monotonically increases on (-2,2), and f (2+a)+f (1-2a) > 0 to find the value of a.

2.f(x)=(x+2a)^2+2-4a^2

Axis of symmetry x=-2a

The left side of the symmetry axis decreases.

So the symmetry axis is on the right side of the decreasing interval.

so-2a >； =6

a & lt=-3

3.f(x)=(x+2a)^2+2-4a^2

Axis of symmetry x=-2a

The monotone decreasing interval is (-∞, 6)

So x=6 is the axis of symmetry.

So -2a=6

a=-3

4. Let u = x 2+3x+3.

Domain of definition

x^2+3x+3=(x+3/2)^2+3/4>； 0

So it's R.

And the opening of (x+3/2) 2+3/4 is upward, and the symmetry axis is x=-3/2.

So x < -3/2, (x+3/2) 2+3/4 decreases.

X & gt-3/2, (x+3/2) 2+3/4 increasing.

Y = √ u when u > =0 increment

So x 2+3x+3 is the same as y monotone interval.

So the monotone subtraction interval is (-∞, -3/2).

5.f(2+a)+ f( 1-2a)>0

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

f(-x)=-f(x)

So f (2+a) >; f(2a- 1)

Monotonically increasing

2+a & lt； 2a- 1

A> III

Defined domain

-2 & lt； = 2+a & lt； =2，-4 & lt； = a & lt=0

-2 & lt； = 1-2a & lt； =2，-2 & lt； = 2a- 1 & lt； =2，- 1/2 & lt； = a & lt=3/2

And a>3 contradiction

There is no solution to this problem.