设函数f(x)=sin(xω+ψ)+cos(ωx+ψ)(ω>0,ψ绝对值<2分之π)的最小值正周期为π,且f(-x)=f(x)则 (A) f(x)在(0,2分之π)单间递减 (B) f(x)在(四分之π,4分之3π)单调递减 (C) f(x)在(0,2分之
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![设函数f(x)=sin(xω+ψ)+cos(ωx+ψ)(ω>0,ψ绝对值<2分之π)的最小值正周期为π,且f(-x)=f(x)则 (A) f(x)在(0,2分之π)单间递减 (B) f(x)在(四分之π,4分之3π)单调递减 (C) f(x)在(0,2分之](/uploads/image/z/981825-33-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x%CF%89%2B%CF%88%29%2Bcos%28%CF%89x%2B%CF%88%29%EF%BC%88%CF%89%EF%BC%9E0%2C%CF%88%E7%BB%9D%E5%AF%B9%E5%80%BC%EF%BC%9C2%E5%88%86%E4%B9%8B%CF%80%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80%2C%E4%B8%94f%28-x%29%3Df%28x%29%E5%88%99+%28A%29+f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%880%2C2%E5%88%86%E4%B9%8B%CF%80%EF%BC%89%E5%8D%95%E9%97%B4%E9%80%92%E5%87%8F+%EF%BC%88B%EF%BC%89+f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%88%E5%9B%9B%E5%88%86%E4%B9%8B%CF%80%2C4%E5%88%86%E4%B9%8B3%CF%80%EF%BC%89%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F+%EF%BC%88C%EF%BC%89+f%28x%29%E5%9C%A8%EF%BC%880%2C2%E5%88%86%E4%B9%8B)
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设函数f(x)=sin(xω+ψ)+cos(ωx+ψ)(ω>0,ψ绝对值<2分之π)的最小值正周期为π,且f(-x)=f(x)则 (A) f(x)在(0,2分之π)单间递减 (B) f(x)在(四分之π,4分之3π)单调递减 (C) f(x)在(0,2分之
设函数f(x)=sin(xω+ψ)+cos(ωx+ψ)(ω>0,ψ绝对值<2分之π)的最小值正周期为π,且f(-x)=f(x)
则 (A) f(x)在(0,2分之π)单间递减 (B) f(x)在(四分之π,4分之3π)单调递减 (C) f(x)在(0,2分之π)单调递增 (c)f(x)(4分之π,4分之3π)单调递增
设函数f(x)=sin(xω+ψ)+cos(ωx+ψ)(ω>0,ψ绝对值<2分之π)的最小值正周期为π,且f(-x)=f(x)则 (A) f(x)在(0,2分之π)单间递减 (B) f(x)在(四分之π,4分之3π)单调递减 (C) f(x)在(0,2分之
f(x)=sin(xω+ψ)+cos(ωx+ψ)
=√2sin(xω+ψ+π/4)
f(-x)=√2sin(-xω+ψ+π/4)
所以sin(xω+ψ+π/4)=sin(-xω+ψ+π/4)
2π/ω=π
ω=2
sin(2x+ψ+π/4)=sin(-2x+ψ+π/4)
sin(2x+ψ+π/4)+sin(2x-ψ-π/4)=0
2sin(2x)cos(2x+ψ+π/4)=0
cos(2x+ψ+π/4)=0
2x+ψ+π/4=kπ+π/2
2x+ψ=kπ+π/4
所以
f(x)=√2sin(xω+ψ+π/4)
=√2sin(kπ+π/2)
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