题目内容
求函数f(x)="sinx+cosx+sinxcosx." x∈﹝0,
﹞的最大值并求出相应的x值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747371413.png)
x=
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747387396.png)
试题分析:利用sinx与cosx的平方关系,令sinx+cosx=t,通过换元,将三角函数转化为二次函数,求出对称轴,利用二次函数的单调性求出最值.
设t=sinx+cosx=sin(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747387396.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747371413.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240007474341151.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747465602.png)
∴函数f(x)=sinx+cosx+sinxcosx=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747481860.png)
∴函数f(x)在(1,)单调递增,∴当t=,t=sinx+cosx=sin(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747387396.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747512338.png)
此时,t=sinx+cosx=sin(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747387396.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000747387396.png)
点评:本小题主要是利用两角和公式的化简求值,二次函数的性质.此题是用换元法,转化思想.但要注意在换元时变量的取值范围.
![](http://thumb2018.1010pic.com/images/loading.gif)
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