(2012•通州区一模)已知函数f(x)=2sinxcosx+2cos2x+1.
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(2012•通州区一模)已知函数f(x)=2sinxcosx+2cos2x+1.
(Ⅰ)求f(x)的最小正周期;
(Ⅱ)求f(x)在区间[-
(Ⅰ)求f(x)的最小正周期;
(Ⅱ)求f(x)在区间[-
π |
2 |
(Ⅰ)f(x)=2sinxcosx+2cos2x+1
=2sinxcosx+2cos2x-1+2=sin2x+cos2x+2=
2sin(2x+
π
4)+2,
∵ω=2,∴T=
2π
2=π,
则函数f(x)的最小正周期为π;
(Ⅱ)∵x∈[-
π
2,0]时,∴2x+
π
4∈[-
3π
4,
π
4],
当2x+
π
4=
π
4,即x=0时,sin(2x+
π
4)=sin
π
4=
2
2,
∴f(x)取得最大值3;
当2x+
π
4=-
π
2,即x=-
3π
8时,sin(2x+
π
4)=sin(-
π
2)=-1,
∴f(x)取得最小值2-
2.
=2sinxcosx+2cos2x-1+2=sin2x+cos2x+2=
2sin(2x+
π
4)+2,
∵ω=2,∴T=
2π
2=π,
则函数f(x)的最小正周期为π;
(Ⅱ)∵x∈[-
π
2,0]时,∴2x+
π
4∈[-
3π
4,
π
4],
当2x+
π
4=
π
4,即x=0时,sin(2x+
π
4)=sin
π
4=
2
2,
∴f(x)取得最大值3;
当2x+
π
4=-
π
2,即x=-
3π
8时,sin(2x+
π
4)=sin(-
π
2)=-1,
∴f(x)取得最小值2-
2.
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