已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.
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![已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.](/uploads/image/z/1078109-53-9.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%2C%E5%87%BD%E6%95%B0f%28x%29%3D-2acos%282x%2B%CF%80%2F6%29%2B2a%2Bb%2C%E5%BD%93x%E2%88%88%28-%CF%80%2F4%2C%CF%80%2F4%29%E6%97%B6%2C-5%E2%89%A4f%EF%BC%88x%EF%BC%89%E2%89%A41.%EF%BC%881%EF%BC%89%E6%B1%82%E5%B8%B8%E6%95%B0a%2Cb%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E8%AE%BEg%EF%BC%88x%EF%BC%89%3Df%EF%BC%88x%2B%CF%80%2F4%EF%BC%89%2C%E4%B8%94%E6%BB%A1%E8%B6%B3lg%E3%80%90g%EF%BC%88x%EF%BC%89%E3%80%91%3E0%2C%E6%B1%82g%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4.)
已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.
已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值
(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.
已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.
(1)因为x∈(-π/4,π/4),所以2X+π/6∈(-2π/3,2π/3)
所以得出2acos(2x+π/6)∈(-1/2,1],
a>0,当2acos(2x+π/6)=-1/2时,f(x)max=1=-2a*(-1/2)+2a+b,得出:a+b=1
当2acos(2x+π/6)=1时,f(x)min=-5=-2a*1+2a+b,得出:b=-5
a+b=1,b=-5 ,所以a=6
(2)g(x)=f(x+π/4)=-2*6cos(2(x+π/4)+π/6)+2*6-5=-12cos(2X+2π/3)+7
因为x∈(-π/4,π/4),所以2X+2π/3∈(π/6,7π/6)
g(x)在 (π/6 ,π]上单调递减,在[π,7π/6)上单调递增.
因为考虑要满足lg【g(x)】>0=lg1,根据lgX在定义域范围内是单调递增的,
所以得出:g(x)>1,-12cos(2X+2π/3)+7>1,cos(2X+2π/3)