解:由[f(x1)+f(x2)]/2>f[(x1+x2)/2]可知,抛物线开口向上,故a>0.又f(x+2)为偶函数,即f(-x+2)=f(x+2).===>-b/(2a)=2.即抛物线的对称轴为直线x=2.(1).a的取值范围是(0,+**)。(2).因a>0.b=-4a.f(x)=ax^2-4ax+1.故当a≤1时,f(x)max=f(a)=a^3-4a^2+1.f(x)min=f(2)=1-4a.值域为[1-4a,a^3-4a^2+1].当1<a<2时,值域为[f(2),f(a+2)]=[1-4a,a^3-4a+1].当a≥2时,值域为[f(a),f(a+2)]=[a^2-4a^2+1,a^3-4a+1].(3).显然,当a=2时,在[2,3]上,值域为[-7,-5].长度为2=10-2^3. |