Then x=2 or x=-3.

Then from the symmetry of unary quadratic function

The symmetry axis is x=(2-3)/2=-0.5.

Because of a man

rule

The image opening of unary quadratic function is downward.

Then the monotonically increasing interval is [-∞, -0.5]

Your question is not clearly written, so the answer is different.

2. Let 3/π-2X=t

Then the increment interval of sint is

[-π/2+2*Kπ，π/2+2*Kπ]

that is

-π/2+2 * kπ= & lt； t & lt=π/2+2*Kπ

So -π/2+2 * kπ =

rule

kπ+(5/ 12)π= & lt； x & lt=Kπ+( 1 1/ 12)π

Then: the increasing interval of the function Y=sin(3/π-2X) is

[Kπ+(5/ 12)π，Kπ+( 1 1/ 12)π]

3.

Because the function F(X)=3|X-b|+2

The symmetry axis is x = b.

The image is also due to 3>0, which first decreases and then increases.

And because the function F(X)=3|X-b|+2 is increasing function in [0, positive infinity].

Then b < 0

The range of is (-∞, 0)

4. Because

F(X)=X^2-2X+3

=X^2-2X+ 1+3- 1

=(X- 1)^2+2

The image opening of function F(X) is upward.

The lowest point is (1, 2)

And F(0)=3.

According to the symmetry of the image

F(2)=3

On the closed interval [0.a], the minimum value of F(X)=X2-2X+3 is 2 and the maximum value is 3'.

1 & lt； = a & lt+2 rule

The range of a is 1, 2.