已知a∈R,函数f(X)=-2aSin(2x+π/6)+2a+b当x∈〔0,π/2〕,-5≤f(X)≤1. (1)求常数a,b的值
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![已知a∈R,函数f(X)=-2aSin(2x+π/6)+2a+b当x∈〔0,π/2〕,-5≤f(X)≤1. (1)求常数a,b的值](/uploads/image/z/1078110-54-0.jpg?t=%E5%B7%B2%E7%9F%A5a%E2%88%88R%2C%E5%87%BD%E6%95%B0f%28X%29%3D-2aSin%282x%2B%CF%80%2F6%EF%BC%89%2B2a%2Bb%E5%BD%93x%E2%88%88%E3%80%940%2C%CF%80%2F2%E3%80%95%2C-5%E2%89%A4f%28X%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)
已知a∈R,函数f(X)=-2aSin(2x+π/6)+2a+b当x∈〔0,π/2〕,-5≤f(X)≤1. (1)求常数a,b的值
已知a∈R,函数f(X)=-2aSin(2x+π/6)+2a+b当x∈〔0,π/2〕,-5≤f(X)≤1. (1)求常数a,b的值
已知a∈R,函数f(X)=-2aSin(2x+π/6)+2a+b当x∈〔0,π/2〕,-5≤f(X)≤1. (1)求常数a,b的值
由x∈[0,π/2],可知2x+π/6∈[π/6,7π/6],Sin(2x+π/6)∈[-1/2,1],
(1)a>0时,有a+2a+b=1,-2a+2a+b=-5,解得a=2,b=-5
(2)a<0时,有,-2a+2a+b=1,a+2a+b=-5解得a=-2,b=1
综上,a=2,b=-5或
a=-2,b=1
x∈〔0,π/2)
2x+π/6∈〔π/6,7π/6)
Sin(2x+π/6)∈[-1/2,1]
(1)-2a(-1/2)+2a+b=-5且-2a+2a+b=1
得a=-2,b=1
(2)-2a(-1/2)+2a+b=1且-2a+2a+b=-5
得a=2,b=-5
所以a=-2,b=1或a=2,b=-5
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