Question

設(shè)二次函數(shù)f(x)=mx²+nx+t的圖像過原點(diǎn),g(x)=ax³+bx?3(x>0),f(x), g(x)的導(dǎo)函數(shù)為,g¢(x),且="0," =?2,f(1)="g(1)," =g¢(1).
(Ⅰ)求函數(shù)f(x),g(x)的解析式；
(Ⅱ)求F(x)=f(x)?g(x)的極小值；
(Ⅲ)是否存在實(shí)常數(shù)k和m,使得f(x)³kx+m和g(x)£kx+m成立?若存在,求出k和m的值；若不存在,說明理由.

Answer 1

(Ⅰ)由已知得t=0,=2mx+n,
則="n=0," =?2m+n=?2,從而n="0," m=1,
∴f(x)=x². 
則="2x, " g¢(x)=3ax²+b.
由f(1)="g(1)," =g¢(1)得a+b?3=2,3a+b=2,解得a=?1,b=5,
∴g(x)=?x³+5x?3(x>0) ……4分
(Ⅱ)∵F(x)=f(x)?g(x)=x³+x²?5x+3(x>0),
求導(dǎo)數(shù)得F¢(x)=3x²+2x?5=(x?1)(3x+5)
∴F(x)在(0,1)單調(diào)遞減,在(1,+¥)單調(diào)遞增,從而F(x)的極小值為F(1)="0. " ……8分
(Ⅲ)因 f(x)與g(x)有一個(gè)公共點(diǎn)(1,1),而函數(shù)f(x)在點(diǎn)(1,1)的切線方程為y=2x?1.
下面驗(yàn)證都成立即可.
由(x?1)²=x²?2x+1³0,得x²³2x?1,知f(x)³2x?1恒成立.
設(shè)h(x)=?x³+5x?3?(2x?1)= ?x³+3x?2(x)>0,
求導(dǎo)數(shù)得h¢(x)=?3x²+3=?3(x+1)(x?1)(x>0),
∴h(x)在(0,1)上單調(diào)遞增,在(1,+¥)上單調(diào)遞減,所以h(x)=?x³+5x?3?(2x?1)的最大值為h(1)=0,
所以?x³+5x?3£2x?1,即g(x)£2x?1恒成立.
故存在這樣的實(shí)常數(shù)k和m,且k=2,m=?1

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