f(1-m)+f(1-m^2)<0f(1-m)<-f(1-m^2)=f(m^2-1)1-m>m^2-1m^2+m-2<0(m+2)(m-1)<0-2而:-1<裂差冲1-m<1, ==> 0-1<1-m^2<1, ==> -根号2联立(1),(2),(3),得:0
奇函数,减函数f(1-m)<-f(1-m^2)=f(m^2-1)1>1-m>m^2-1>-1