解析由函数y=sin x的图像(图略)可知,函数y=sin x在[0"," π/2]上是增加的,

　　∴[0,a]⊆[0"," π/2],∴0

答案(0"," π/2]

5.导学号93774019对于函数f(x)=xsin x,给出下列三个命题:

①f(x)是偶函数;

②f(x)是周期函数;

③f(x)在区间[0"," π/2]上的最大值为π/2.

其中正确的命题是　　　　(写出所有正确命题的序号).

解析∵f(x)的定义域为R,且f(-x)=-xsin(-x)=xsin x=f(x),

　　∴f(x)是偶函数,故①正确;

　　虽然函数y=sin x是周期函数,但f(x)=xsin x不具有周期性,故②错误;

　　易知f(x)在区间[0"," π/2]上是增加的,∴f(x)在π/2处取得最大值,最大值为π/2sinπ/2=π/2,故③正确.

答案①③

6.若函数y=a-bsin x的最大值为3/2,最小值为-1/2,试求函数y=-4asin bx的最值及周期.

解设t=sin x∈[-1,1],则y=a-bt.

　　①当b>0时,a-b≤a-bt≤a+b.

　　∴{■(a+b=3/2 "," @a"-" b="-" 1/2 "," )┤

　　∴{■(a=1/2 "," @b=1"." )┤

　　∴所求函数为y=-2sin x.

　　②当b<0时,同理可得{■(a"-" b=3/2 "," @a+b="-" 1/2 "," )┤

　　∴{■(a=1/2 "," @b="-" 1"." )┤

　　∴所求函数为y=-2sin(-x)=2sin x.

　　∴综合①②得,所求函数为y=±2sin x,其最小值为-2,最大值为2,周期为2π.

7.导学号93774020已知函数f(x)=log_(1/2)|sin x|.

(1)求其定义域和值域;

(2)判断奇偶性;

(3)判断周期性,若是周期函数,求其最小正周期;

(4)写出单调区间.

解(1)由|sin x|>0,得sin x≠0,∴x≠kπ(k∈Z).

　　∴函数的定义域为{x|x≠kπ,k∈Z}.

　　∵0<|sin x|≤1,

　　∴log_(1/2)|sin x|≥0.

　　∴函数的值域为{y|y≥0}.

　　(2)∵函数定义域为{x|x≠kπ,k∈Z},关于原点对称,f(-x)=log_(1/2)|sin(-x)|=log_(1/2)|sin x|=f(x),

　　∴函数f(x)是偶函数.

　　(3)∵f(x+π)=log_(1/2)|sin(x+π)|=log_(1/2)|sin x|=f(x),∴函数f(x)是周期函数,且最小正周期是π.

　　(4)当x∈(kπ"," kπ+π/2]时,t=|sin x|是增加的;

　　当x∈[kπ"-" π/2 "," kπ)时,t=|sin x|是减少的.又函数y=log_(1/2)t为减函数,

　　∴函数f(x)的单调递增区间为[kπ"-" π/2 "," kπ)(k∈Z);单调递减区间为(kπ"," kπ+π/2](k∈Z).