题目内容

(本题满分15分) 已知函数f (x)=x3ax2bxa , bR

(Ⅰ) 曲线C:yf (x) 经过点P (1,2),且曲线C在点P处的切线平行于直线y=2x+1,求ab的值;

(Ⅱ) 已知f (x)在区间 (1,2) 内存在两个极值点,求证:0<ab<2

 

【答案】

(Ⅰ)解:

由题设知:  解得     

(Ⅱ)解:因为在区间内存在两个极值点 ,

所以,即内有两个不等的实根.

由 (1)+(3)得.

由(4)得

,故,从而.

所以.                       

【解析】略

 

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