1.设f(x)=ax2+bx+c,则f(x)>-2x可化为ax2+(b+2)x+c>0由题意得方程ax2+(b+2)x+c=0的两根是1和3所以1+3=-(b+2)/a,1×3=c/a即b=-4a-2,c=3af(x)=ax2-(4a+2)x+3a(1)方程f(x)+6a=0可化为ax2-(4a+2)x+9a=0Δ=(4a+2)2-36a2=0a=1,b=-6,c=3或a=-1/5,b=-6/5,c=-3/5(2)由题意知a<0最大值为[12a2-(4a+2)2]/4a>0解得a<-2-√3或a>-2+√32.(...
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