一道函数题二次函数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,求证（1)p*f(m/(m+1))

Question

一道函数题
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,求证（1)p*f(m/(m+1))

Answer 1

二次函数f(x)=px²+qx+r中,p,q,r∈R,且p/(m+2)+q/(m+1)+r/m=0其中,m>0,求证:
1、pf[m/(m+1)]＜0；2.f(x)=0在(0,1)内恒有解
(1)∵p/(m+2)+q/(m+1)+r/m=0
--->p²m/(m+2)+pqm/(m+1)+pr=0
--->pqm/(m+1)+pr=-p²m/(m+2)
--->pf[m/(m+1)]
= p[p[m/(m+1)]²+pq[m/(m+1)]+pr
= [pm/(m+1)]² - p²m/(m+2)
= p²m[m/(m+1)²-1/(m+2)]
= p²m[m(m+2)-(m+1)²]/[(m+2)(m+1)²]
= -p²m/[(m+2)(m+1)²]
＜0
(2) ∵f(0)=r,f(1)=p+q+r,令m/(m+1)=t∈(0,1)
--->p＞0时,由(1)--->f(t)＜0
若r＞0--->f(0)＞0,f(t)＜0--->f(x)=0在(0,t)内有解；
若r≤0--->f(t)＜0
f(1)=p+q+r
=p-(m+1)[p/(m+2)+r/m]+r
=p/(m+2)-r/m≥0--->f(x)=0在[t,1)内有解
综上：f(x)=0在(0,1)内恒有解
p＜0时,同理可证.