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1 ,(0,π/2) 2x+π/6~(π/6,7π/6),sin(2x+π/6)~(-1/2,1],
由a》0,所以f(x)~[-2a+2a+b,a+2a+b)=[-5,1),所以a=2,b=-5
2,f(x)=-4sin(2x+π/6)-1带入g(x)化简:
g(x)=4sin(2x+π/6)-1,lgg(x)》0,则g(x)>1,sin(2x+π/6)>1/2
2x+π/6~(2kπ+π/6,2kπ+5π/6),(kπ,kπ+π/3).
对应递增:2x+π/6~(2kπ+π/6,2kπ+π/2],(kπ,kπ+π/6]
递减:(kπ+π/6,kπ+π/3)
注意区间的开闭,题中有一处问题
1年前
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