习题课--函数y=Asin(ωx+φ)的综合应用

1.下列函数中,在[π/4 "," π/2]上是减少的,且周期为π的是0(　　)

A.y=sin(2x+π/2) B.y=cos(2x+π/2)

C.y=sin(x+π/2) D.y=cos(x+π/2)

解析C,D中函数周期为2π,所以错误.当x∈[π/4 "," π/2]时,2x+π/2∈[π"," 3π/2],函数y=sin(2x+π/2)为减少的,而函数y=cos(2x+π/2)为增加的.

答案A

2.已知函数f(x)=2sin ωx(ω>0)在区间["-" π/3 "," π/4]上的最小值是-2,则ω的最小值等于(　　)

A.2/3 B.3/2 C.2 D.3

解析∵ω>0,-π/3≤x≤π/4,∴-ωπ/3≤ωx≤ωπ/4.

　　由已知条件知-ωπ/3≤-π/2,∴ω≥3/2.

答案B

3.将函数y=2sin(2x+π/6)的图像向右平移1/4个周期后,所得图像对应的函数为f(x),则函数f(x)的单调递增区间是(　　)

A.[kπ"-" π/12 "," kπ+5π/12](k∈Z)

B.[kπ+5π/12 "," kπ+11π/12](k∈Z)

C.[kπ"-" 5π/24 "," kπ+7π/24](k∈Z)

D.[kπ+7π/24 "," kπ+19π/24](k∈Z)

解析∵函数y=2sin(2x+π/6)的周期T=2π/2=π,

　　∴将函数y=2sin(2x+π/6)的图像向右平移1/4个周期后,所得图像对应的函数为f(x)=2sin[2(x"-" π/4)+π/6]=2sin(2x"-" π/3),

　　∴令2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z,可得kπ-π/12≤x≤kπ+5π/12,k∈Z,

　　∴函数f(x)的单调递增区间是[kπ"-" π/12 "," kπ+5π/12],k∈Z.故选A.

答案A

4.函数f(x)=sin(2x+φ)("|" φ"|" <π/2)的图像向左平移π/6个单位长度后关于原点对称,则函数f(x)在[0"," π/2]上的最小值为(　　)

A.-√3/2 B.-1/2 C.1/2 D.√3/2

解析函数f(x)=sin(2x+φ)的图像向左平移π/6个单位长度得y=sin[2(x+π/6)+φ]=sin(2x+π/3+φ)的图像.

　　又其为奇函数,则π/3+φ=kπ,k∈Z,

　　解得φ=kπ-π/3.

　　又|φ|<π/2,令k=0,得φ=-π/3,

　　∴f(x)=sin(2x"-" π/3).

　　又∵x∈[0"," π/2],

　　∴sin(2x"-" π/3)∈["-" √3/2 "," 1],

　　即当x=0时,f(x)min=-√3/2,故选A.

答案A