已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b
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![已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b](/uploads/image/z/2691635-59-5.jpg?t=%E5%B7%B2%E7%9F%A5a%EF%BC%9E0%2C%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9D-2asin%282x%EF%BC%8B%CF%80%2F6%29%EF%BC%8B2a%EF%BC%8Bb%2C%E5%BD%93x%E2%88%88%EF%BC%BB0%2C%CF%80%2F2%EF%BC%BD%E6%97%B6%2C-5%E2%89%A4f%28x%29%E2%89%A41+%281%29%E6%B1%82%E5%B8%B8%E6%95%B0a%2Cb%E7%9A%84%E5%80%BC+%282%29%E8%AE%BEg%28x%29%EF%BC%9Df%28x%EF%BC%8B%CF%80%2F2%29%E4%B8%94lgg%28x%29%EF%BC%9E0%2C%E6%B1%82g%28x%29%E7%9A%84%E5%8D%95%E8%B0%83+%E6%88%91%E7%AE%97%E4%BA%86%E5%A5%BD%E5%87%A0%E9%81%8D%E7%AC%AC%E4%B8%80%E9%97%AE%2Ca%2Bb)
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b
(1)因为,x∈[0,π/2],
2x+π/6∈[π/6,7π/6],
sin(2x+π/6)∈[-1/2,1],
又 a>0
所以, -2a+2a+b=-5
a+2a+b=1
解得: a=2, b=-5
(2) 由(1)知,f(x)=-4sin(2x+π/6)-1
由题意 g(x)=f(x+π/2)
=-4sin(2x+π+π/6)-1
=4sin(2x+π/6)-1>1
即 sin(2x+π/6)>1/2
所以 2x+π/6∈(2kπ+π/6,2kπ+5π/6)
单调增区间满足 2x+π/6∈(2kπ+π/6,2kπ+π/2]
单调减区间满足 2x+π/6∈[2kπ+π/2,2kπ+5π/6)
解得 g(x)的单调增区间为 (kπ,kπ+π/6]
单调减区间为 [kπ+π/6,kπ+π/3]
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