题目内容
(本小题满分13分)
已知函数
(I)若曲线
在点
处的切线与直线
垂直,求a的值;
(II)求函数
的单调区间;
已知函数

(I)若曲线



(II)求函数

(1) a="1"
(2) 当
时,即
上是增函数.
当
当
单调递增;
当
单调递减
(2) 当


当


当

试题分析:解:(I)函数


又曲线


所以

(II)由于

当


即

当

当

当

点评:解决的关键是能利用导数的几何意义求解切线方程,以及结合导数的符号求解单调性,属于基础题。

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