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①f(x)=sin2x + √3(2cos²x - 1)
=sin2x + √3cos2x
=2[(1/2)sin2x + (√3/2)cos2x]
=2sin(2x + π/3)
②∵正弦函数的值域是[-1,1]
∴f(x)的最小值是-2
此时2x + π/3=2kπ - π/2
2x=2kπ - 5π/6
∴x=kπ - 5π/12,k∈Z
③递增区间是:
2kπ - π/2≤2x + π/3≤2kπ + π/2
2kπ - 5π/6≤2x≤2kπ + π/6
kπ - 5π/12≤x≤kπ + π/12,k∈Z
④∵x∈R
∴-1≤sin(2x + π/3)≤1
∴-2≤f(x)≤2
=sin2x + √3cos2x
=2[(1/2)sin2x + (√3/2)cos2x]
=2sin(2x + π/3)
②∵正弦函数的值域是[-1,1]
∴f(x)的最小值是-2
此时2x + π/3=2kπ - π/2
2x=2kπ - 5π/6
∴x=kπ - 5π/12,k∈Z
③递增区间是:
2kπ - π/2≤2x + π/3≤2kπ + π/2
2kπ - 5π/6≤2x≤2kπ + π/6
kπ - 5π/12≤x≤kπ + π/12,k∈Z
④∵x∈R
∴-1≤sin(2x + π/3)≤1
∴-2≤f(x)≤2
追答
⑤∵-π/12≤x≤π/6
∴-π/6≤2x≤π/3
则π/6≤2x + π/3≤2π/3
∴1/2≤sin(2x + π/3)≤1
则1≤f(x)≤2
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