(1) f(x) is the derivative of increasing function's statement f(x) (-2x 2+2ax+4)/(x 2+2) 2 >; =0 stays in the interval [- 1, 1].

That is, -2x 2+2ax+4 >: =0 holds in the interval [- 1, 1].

Then f (- 1) >: =0 f( 1)>=0 means-1

(2) arrange f(x)= 1/x into a quadratic function in the form of x 2-ax-2 = 0, two of which are x 1, x 2.

Then x 1+x2=a x 1x2=-2, so there is

|x 1-x2i^2=(x 1+x2)^2-4x 1x2=a^2+8

So m2+TM+1>; =|x 1-x2| is m 2+TM+1> = √ (a 2+8).

m2+TM+ 1 >； = √ (A 2+8) holds for any A belonging to [- 1, 1] and any T belonging to [- 1, 1].

Then you need the left f (t) of the above formula = mt+m 2+1,and the minimum value must be >; = Maximum value of f(a) on the right.

And f(t) is a linear function, which needs to be discussed.

When m>0, the minimum value f (t) = f (-1) = m 2-m+1> m > = 3; =2

When the minimum value at m<0 is f (t) = f (1) = m 2+m+1> m.

When m=0, it is obviously not true.

So the range of m is m & gt=2 or m.

∵| x | = x(x≥0)-x(x & lt； 0)

∴ 1- 1|x|dx=0- 1|x|dx+0 1|x|dx

=0- 1(-x)dx+0 1xdx, so C.

4. Let f (x) = x2 (0 ≤ x

a . 34a b . 45

C.56 d. does not exist.

[answer] c

[resolution] 02f (x) dx = 01x2dx+12 (2-x) dx

Let F 1(x)= 13x3, F2(x)=2x- 12x2,

Then F' 1(x)=x2, and f' 2 (x) = 2-x.

∴02f(x)dx=f 1( 1)-f 1(0)+f2(2)-f2( 1)

=13-0+2× 2-12× 22-2×1-12×12 = 56.

5 . abf′(3x)dx =()

Afaxi (b)- Afaxi (a) Bafaxi (3b)- Afaxi (3a)

c . 13[f(3b)-f(3a)]d . 3[f(3b)-f(3a)]