So f(x) is increasing function.

The maximum value f( 1)=a+b+2=4.

a+b=2

The minimum value of f (-1) =-(a+b)+2-1=-3/2.

1 1.acos^2(x/2)=a/2(cosx+ 1)

f(x)=sinx+a/2cosx+a/2

=√( 1+(a/2)^2)*sin(x+π/4)+a/2

Because 0≤x≤π, π/4≤x+π/4≤5π/4.

According to the image of the evil function:

When x= π/4, f(x) has the maximum value: √2-1.

When x= π, f(x) has a minimum value: -2.

So the range of f(x) is [-2, √2- 1].

12 . x & lt； When =0, f (x) = 2 x, and its value range is (0, 1).

X>0, f(x)=log2 (x), and its value range is R ..

Therefore, x < =0, y = f (2x)-1= log2 (2x)-1= x-1

X>0, y=f(log2(x))- 1, and then discuss in sections:

When x> is in 1, log2 (x) >: 0, y=log2(log2(x))- 1=0, so: log2(x)=2, x=4, which is consistent.

When 0

So y has two zeros: x=4, 1.

13. Because ∠ A = 30,

So in ABC of right triangle, AB = 2, ∠ B = 60.

And because D is the midpoint of AB, DB= 1.

So the triangle DCB is an equilateral triangle.

So CD= 1

Because the angle between vector AB and vector CD is 120.

So vector ab vector CD = | AB || CD | cos120 = 2 *1* (1/2) =-1.

13. according to "the existence of non-zero real number t makes it possible for any x∈M(M? D), with x+t∈D, and f(x+t)≥f(x), then f(x) is said to be a t treble function on m ". For the function whose domain is [- 1, +∞], f(x)=x2 is m high-pitched functions, so it is easy to know that the function f(x) whose domain is r is odd function. When x≥0, f(x)=|x-a2|-a2. Draw the function image and get 4≥3a2-(-a2)? - 1≤a≤ 1。

∫f(- 1)= f( 1), m≥ 1-(- 1), that is, m≥2,

The image of f(x)=|x-a2|-a2 is as shown in the figure, ∴4≥3a2-(-a2)? - 1≤a≤ 1。

Therefore: m ≥ 2; - 1≤a≤ 1