已知函数f(x)=cos^2(x+12/π),g(x)=1+sin2x/2

1个回答

  • (1)由题设知:

    f(x)=cos²(x+π/12)=1/2[1+cos(2x+π/6)]

    ∵x=x0是函数y=f(x)图像的一条对称轴

    ∴2x0=kπ-π/6(k∈Z)

    ∴g(x0)=1+1/2sin2x0=1+1/2sin(kπ-π/6)

    当k为偶数时,g(x0)=1+1/2sin(-π/6)=1-1/4=3/4

    当k为奇数时,g(x0)=1+1/2sinπ/6=1+1/4=5/4

    (2)

    h(x)=f(x)+g(x)=1/2[1+cos(2x+π/6)]+1+1/2sin2x

    =1/2[cos(2x+π/6)+sin2x]+3/2=1/2(√3/2 cos2x+1/2 sin2x)+3/2

    =1/2sin(2x+π/3)+3/2

    当2kπ-π/2≤2x+π/3≤2kπ+π/2

    即:kπ-5π/12≤x≤kπ+π/12(k∈Z)时,

    函数h(x)=1/2sin(2x+π/3)+3/2是增函数

    当2kπ+π/2≤2x+π/3≤2kπ+3π/2

    即:kπ+π/12≤x≤kπ+7π/12(k∈Z)时,

    函数h(x)=1/2sin(2x+π/3)+3/2是减函数

    故函数h(x)=1/2sin(2x+π/3)+3/2的单调递增区间是:[kπ-5π/12,kπ+π/12] (k∈Z)

    函数h(x)=1/2sin(2x+π/3)+3/2的单调递减区间是:[kπ+π/12,kπ+7π/12] (k∈Z)