题目内容
设点P是曲线![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_ST/0.png)
A.
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_ST/1.png)
B.
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_ST/2.png)
C.
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_ST/3.png)
D.
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_ST/4.png)
【答案】分析:求出曲线解析式的导函数,根据完全平方式大于等于0求出导函数的最小值,由曲线在P点切线的斜率为导函数的值,且直线的斜率等于其倾斜角的正切值,从而得到tanα的范围,由α的范围,求出α的范围即可.
解答:解:∵y′=3x2-
≥-
,∴tanα≥-
,
又∵0≤α≤π,
∴0≤α<
或
.
则角α的取值范围是[0,
)∪[
,π).
故选C.
点评:考查学生会利用导数求曲线上过某点切线方程的斜率,会利用切线的斜率与倾斜角之间的关系k=tanα进行求解.
解答:解:∵y′=3x2-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/0.png)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/1.png)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/2.png)
又∵0≤α≤π,
∴0≤α<
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/3.png)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/4.png)
则角α的取值范围是[0,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/5.png)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131103174348125094390/SYS201311031743481250943010_DA/6.png)
故选C.
点评:考查学生会利用导数求曲线上过某点切线方程的斜率,会利用切线的斜率与倾斜角之间的关系k=tanα进行求解.
![](http://thumb.zyjl.cn/images/loading.gif)
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