题目内容

((本题满分15分)设函数,且的极值点. (Ⅰ) 若的极大值点,求的单调区间(用表示); (Ⅱ)若恰有1解,求实数的取值范围.

 

【答案】

解:

因为的极值点,所以

所以 ……………3分

(I)因为的极大值点,所以

时,;当时,;当时,

所以的递增区间为;递减区间为.…………6分


(II)若,则上递减,在上递增

恰有1解,则,即,所以;…………9分

,则

因为,则

,从而恰有一解; ……………12分

,则

,从而恰有一解;                      

所以所求的范围为.   ……………15分

 

【解析】略

 

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