已知函数f(x)=2cos²x+2√3sinxcosx+a(a∈R) (1)若x∈R,求f(x)的单调增区间

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  • f(x)=2cos²x+2√3sinxcosx+a

    =cos2x+1+√3sin2x+a

    =2×[(1/2)×cos2x+(√3/2)×sin2x]+a+1

    =2sin(2x+π/6)+(a+1)

    (1)

    2kπ-π/2≤2x+π/6≤2kπ+π/2时,f(x)单调递增

    解得,kπ-π/3≤x≤kπ+π/6

    2kπ+π/2≤2x+π/6≤2kπ+3π/2时,f(x)单调递减

    解得,kπ+π/6≤x≤kπ+2π/3

    所以,f(x)的单调增区间为[kπ-π/3,kπ+π/6]

    单调减区间为[kπ+π/6,kπ+2π/3]

    (其中,k为整数)

    (2)

    0≤x≤π/2时

    π/6≤2x+π/6≤7π/6

    当 2x+π/6=π/2,

    即 x=π/6时

    f(x)有最大值=2+(a+1)=3+a

    因为f(x)的最大值为4

    即,a+3=4

    解得,a=1

    所以,a的值为1