f'(x)=x？ -3ax？ +4

And x+2y-3=0, and the tangent slope is 2.

So f'( 1)= 1-3a+4=2.

a= 1

2、

f'(x)=x？ -3ax？ +4

X & gt=0 is an increasing function.

So x & gt=0 and f(x)>0.

Let g(x)=f'(x)=x? -3ax？ +4

g'(x)=3x？ -6ax = x(3x-6a)= 1

x=0，x=2a

a & gt0

So x < 0, x & gt2a, g'(x)>0, and g(x) is increasing function.

0<x & lt2a, g(x) is a decreasing function.

So x=0 is the maximum point and x=2a is the minimum point.

X> when =0, x=2a has the minimum value.

So as long as g (2a) >; 0 is enough.

8a？ - 12a？ +4 & gt； 0

Answer? & lt 1

0 & lta & lt 1