f(x)=sin(2x+φ) φ是 实数f(x)≤f(π/6)的绝对值x属于R恒成立且f(π/2)>f(π)则f(x)单调递增区间
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![f(x)=sin(2x+φ) φ是 实数f(x)≤f(π/6)的绝对值x属于R恒成立且f(π/2)>f(π)则f(x)单调递增区间](/uploads/image/z/10752133-13-3.jpg?t=f%28x%29%3Dsin%282x%2B%CF%86%29+%CF%86%E6%98%AF+%E5%AE%9E%E6%95%B0f%28x%29%E2%89%A4f%28%CF%80%2F6%29%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BCx%E5%B1%9E%E4%BA%8ER%E6%81%92%E6%88%90%E7%AB%8B%E4%B8%94f%28%CF%80%2F2%29%3Ef%28%CF%80%29%E5%88%99f%28x%29%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=sin(2x+φ) φ是 实数f(x)≤f(π/6)的绝对值x属于R恒成立且f(π/2)>f(π)则f(x)单调递增区间
f(x)=sin(2x+φ) φ是 实数f(x)≤f(π/6)的绝对值x属于R恒成立且f(π/2)>f(π)则f(x)单调递增区间
f(x)=sin(2x+φ) φ是 实数f(x)≤f(π/6)的绝对值x属于R恒成立且f(π/2)>f(π)则f(x)单调递增区间
f(x)≤|f(π/6)|,
∴sin(π/3+φ)=土1,
π/3+φ=(k+1/2)π,k∈Z.
φ=(k+1/6)π,
由f(π/2)>f(π)得
sin[(k+7/6)π]-sin[(k+13/6)π]>0,
-2sin(π/2)cos[(k+5/3)π>0,取k=-1,
f(x)=sin(2x-5π/6),单调递增区间由
(2k-1/2)π
f(x)≤|f(π/6)|,
∴sin(π/3+φ)=土1,
π/3+φ=(k+1/2)π,k∈Z.
φ=(k+1/6)π,
由f(π/2)>f(π)得
-2sin(π/2)cos[(k+5/3)π>0,取k=-1,
f(x)=sin(2x-5π/6),单调递增区间由
(2k-1/2)π<2x-5π/6<(2k+1/2)π确定,
(2k+1/3)π<2x<(2k+4/3)π,
∴(k+1/6)π
f(x)=sin(2x+φ), φ是 实数,f(x)≤︱f(π/6)︱,当x∈R恒成立,且f(π/2)>f(π)则f(x)单调递增区间是
∵对任何x,不等式f(x)≤︱f(π/6)︱恒成立,∴必有︱f(π/6)︱=︱sin(π/3+φ)︱=1,于是有
π/3+φ=±π/2 +2kπ,即φ=±π/2-π/3+2kπ=π/6+2kπ或-5π/6+2kπ,k∈Z;
又f(π/2)...
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f(x)=sin(2x+φ), φ是 实数,f(x)≤︱f(π/6)︱,当x∈R恒成立,且f(π/2)>f(π)则f(x)单调递增区间是
∵对任何x,不等式f(x)≤︱f(π/6)︱恒成立,∴必有︱f(π/6)︱=︱sin(π/3+φ)︱=1,于是有
π/3+φ=±π/2 +2kπ,即φ=±π/2-π/3+2kπ=π/6+2kπ或-5π/6+2kπ,k∈Z;
又f(π/2)=sin(π+φ)=-sinφ>f(π)=sin(2π+φ)=sinφ,即有2sinφ<0,也就是有sinφ<0,故φ=-5π/6;
∴f(x)=sin(2x-5π/6).
由-π/2+2kπ≦2x-5π/6≦π/2+2kπ,得π/3+2kπ≦2x≦5π/6+2kπ,故单增区间为:
π/6+kπ≦x≦5π/12+kπ,k∈Z;
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