1.已知函数f(x)=ax^2+2ax+4(0＜a＜3）.若x1＜x2,x1+x2=1-a,则f(x1)和f(x2)的大

f(x1)-f(x2)=ax1^2+2ax1+4-ax2^2-2ax2-4
=a(x1^2-x2^2)+2a(x1-x2)
=a(x1+x2)(x1-x2)+2a(x1-x2)
=a(x1-x2)(x1+x2+2)
因x1+x2=1-a 带入上式得
a(x1-x2)(1-a+2)
=a(x1-x2)(3-a)
因0