二次函数f(x)=px^2+qx+r中实数p、q、r满足p/(m+2)+q/(m+1)+r/m=0，其中m>0。证:方程f(x)=0在（0，1）恒有

Question

如题所述

Answer 1

∵p/(m+2)+q/(m+1)+r/m=0
∴(p^2)m/(m+2)+pqm/(m+1)+pr=0
∴pqm/(m+1)+pr=-(p^2)m/(m+2)

pf(m/(m+1))
=p[p[m/(m+1)]^2+q[m/(m+1)]+r]
=[pm/(m+1)]^2+pqm/(m+1)+pr
=[pm/(m+1)]^2-(p^2)m/(m+2) {代入上式结论}
=(p^2)m[m/(m+1)^2-1/(m+2)]
=(p^2)m{[m(m+2)-(m+1)^2]/(m+2)(m+1)^2}
=(p^2)m[(-1)/(m+2)(m+1)^2]
=-(p^2)m[1/(m+2)(m+1)^2]<0