Question

设函数f(x)=3ax^2+2bx+c,若a+b+c=0,f(0)f(1)>0,求证

一：-2<-b/a<-1二：设x1,x2是f(x)=0两实根，证根号3/3小于等于|x1-x2|<2/3

Answer 1

1.因为f(0)f(1)>0所以：c(3a+2b+c)>0因为a+b+c=0-(a+b)(2a+b+a+b+c)>0(a+b)(2a+b)<02a^2+3ab+b^2<0(b/a)^2+3(b/a)+2<0 （同除a^2）-2<b/a<-12.x1+x2=-2b/6a=-b/3ax1x2=c/3a(x1-x2)^2=(x1+x2)^2-4x1x2 =(b/3a)^2-4c/3a其中(b/3a)^2=1/9*(b/a)^2,4c/3a=4/3*(c/a)根据1，b/a在(-2,-1)，(b/a)^2为最大，取-2,b=-2aa+b+c=0=-a+c=0,a=c.(b/a)^2为最小，取-1,a=-ba+b+c=0=c,c=0所以i.(b/3a)^2=1/9*(b/a)^2<=4/9 4c/3a=4/3*(c/a)>=4/3ii.(b/3a)^2=1/9*(b/a)^2>=1/9 4c/3a=4/3*(c/a)>=0总上所述1/9<=(x1-x2)^2<4/9即1/3<=|x1-x2|<2/3