Question

函数题！

函数题 函数f(x)=cosx+kcos(x-兀/3) (k为常数),将f(x)的图象向右平移2兀/3个单位所得图象经过坐标原点O. (1)求k值;. (2)求y=f(x)的单调增区间.

Answer 1

（1) 函数f(x)的图象向右平移2兀/3个单位所得函数
g(x)=cos(x-2兀/3)+kcos(x-兀)=cos(x-2兀/3)-kcos(x)图象
又图象经过坐标原点O，即g（0）=0，所以cos(0-2兀/3)-kcos(0)=0，k=-(1/2)
(2)f(x)=cosx-(1/2)cos(x-兀/3)
=cosx-(1/2)[cosx*cos（兀/3)+sinx*sin（兀/3)]=-(1/2)sin(x-兀/3）
由2nπ+π/2≤(x-π/3）≤2nπ+3π/2,得到：2nπ+5π/6≤x≤2nπ+11π/6
f(x)的单调增区间为[2nπ+5π/6,2nπ+11π/6]

Answer 2

关于函数平移的问题一定要记住口诀：左加右减，上加下减！【如此这类题就简单多了】
（1) 函数f(x)的图象向右平移2π/3个单位所得函数:
g(x)=cos(x-2π/3)+kcos(x-π)
=cos(x-2π/3)-kcos(x)
又图象经过坐标原点O(0,0),即g（0）=0.
可得：cos(0-2π/3)-kcos(0)=0解得：k=-0.5
(2)f(x)=cosx-0.5cos(x-π/3)
=cosx-0.5[cosx*cosπ/3+sinx*sinπ/3]
=(根号3/2)cos(x+π/6）由余弦函数的单调性可知：
2kπ+π≤(x+π/6）≤2kπ+2π (k为整数),
得到：2kπ+5π/6≤x≤2kπ+11π/6
f(x)的单调增区间为[2kπ+5π/6,2kπ+11π/6] (k为整数)。

Answer 3

(1)将f(x)的图象向右平移2兀/3个单位所得函数是
y=f(x-2π/3)=cos（x-2π/3）+kcos(x-π/3-2π/3)=cos（x-2π/3）-kcosx
此函数图象经过坐标原点O，则cos（-2π/3）-k=0，得 k=-1/2.
(2)f(x)=cosx-(1/2)*cos(x-π/3)=(3/4)cosx-(√3/4)sinx
=(√3/2)cos(x+π/6),
f(x)的单调增区间:2nπ+π≤x+π/6≤2nπ+2π (n为整数)
即 2nπ+5π/6≤x≤2nπ+11π/6 (n为整数)