äºæ¬¡å½æ°f(x)=px�0�5+qx+rä¸ï¼p,q,râRï¼ä¸p/(m+2)+q/(m+1)+r/m=0å
¶ä¸ï¼m>0ï¼æ±è¯:
1ãpf[m/(m+1)]ï¼0ï¼2.f(x)=0å¨(0,1)å
ææ解
(1)âµp/(m+2)+q/(m+1)+r/m=0
--->p�0�5m/(m+2)+pqm/(m+1)+pr=0
--->pqm/(m+1)+pr=-p�0�5m/(m+2)
--->pf[m/(m+1)]
ã= p[p[m/(m+1)]�0�5+pq[m/(m+1)]+pr
ã= [pm/(m+1)]�0�5 - p�0�5m/(m+2)
ã= p�0�5m[m/(m+1)�0�5-1/(m+2)]
ã= p�0�5m[m(m+2)-(m+1)�0�5]/[(m+2)(m+1)�0�5]
ã= -p�0�5m/[(m+2)(m+1)�0�5]
ãï¼0
(2) âµf(0)=r, f(1)=p+q+rï¼ä»¤m/(m+1)=tâ(0,1)
--->pï¼0æ¶ï¼ç±(1)--->f(t)ï¼0
ããããè¥rï¼0--->f(0)ï¼0,f(t)ï¼0--->f(x)=0å¨(0,t)å
æ解ï¼
ããããè¥râ¤0--->f(t)ï¼0
ãããããããããf(1)=p+q+r
ããããããããããã=p-(m+1)[p/(m+2)+r/m]+r
ããããããããããã=p/(m+2)-r/mâ¥0--->f(x)=0å¨[t,1)å
æ解 综ä¸ï¼f(x)=0å¨(0,1)å
ææ解 pï¼0æ¶ï¼åçå¯è¯ã
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