已知函数f(x)=cos²﹙x+π/12﹚,g﹙x﹚=1+1/2sin2x 求:

1个回答

  • (1)

    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+π/6=kπ

    即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+1/2sin(π/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+1/2sin2x)+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

    故函数h(x)的单调增区间是[kπ-5π/12,kπ+π/12](k∈Z)