解ï¼
f'(x)=x²+ax+2b=0
æä¸¤ä¸ªæ ¹x1åx2ï¼å
¶ä¸0<x1<1ï¼1<x2<2
fââ(x1)=2x1+a<0,fââ(x2)=2x2+a>0
ç±a<-2x1ï¼-2<-2x1<0ï¼å¾aâ¤-2ï¼
ç±a>-2x2ï¼-4<2x2<-2ï¼å¾aâ¥-2
æ
a=-2
ç±0<x1<1ï¼1<x2<2ï¼å¾ï¼x1<x1x2<2x1ï¼æ
0<x1x2<2ï¼ç±x1x2=2bå¾ï¼0<b<1
å³-2<b-2<-1
æ
æï¼1/3<(b-2)/(a-1)<2/3
ä½ ççæ¡å¯¹ä¸å¯¹å§â¦â¦
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