题目内容

(本题满分15分 )已知函数.

(1)求函数的最大值;

(2)若,不等式恒成立,求实数的取值范围;

(3)若,求证:.

 

【答案】

(1)在处取得最大值,且最大值为0.(2). (3)见解析。

【解析】(1)先求出 ,然后求导确定单调区间,极值,最值即可.

(2) 本小题转化为在上恒成立,进一步转化为,然后构造函数,利用导数研究出h(x)的最大值,再利用基础不等式可知,从而可知a的取值范围.

(1),则.…………2分

当时,,则在上单调递增;

当时,,则在上单调递减,

所以,在处取得最大值,且最大值为0.     ………………………4分

(2)由条件得在上恒成立.           ………………………6分

设,则.

当时,;当时,,所以,.

要使恒成立,必须.                  ………………………8分

另一方面,当时,,要使恒成立,必须.

所以,满足条件的的取值范围是.            ………………………10分

(3)当时,不等式等价于.……12

令,设,则,

在上单调递增,,

所以,原不等式成立.          ………………15分

 

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