(2010•通州区一模)已知函数f(x)=2cos2x+2sinxcosx.
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(2010•通州区一模)已知函数f(x)=2cos2x+2sinxcosx.
(I)求f(x)的最小正周期;
(II)若x∈[0,
]
(I)求f(x)的最小正周期;
(II)若x∈[0,
π |
2 |
(Ⅰ)f(x)=2cos2x+2sinxcosx
=2cos2x-1+2sinxcosx+1
=cos2x+1+sin2x
=
2sin(2x+
π
4)+1,
∴函数f(x)的最小正周期 T=
2π
2=π.
(Ⅱ)当x∈[0,
π
2]时,2x+
π
4∈[
π
4,
5π
4],
当2x+
π
4=
π
2,即x=
π
8时,
f(x)取得最大值f(
π
8) =
2+1;
当2x+
π
4=
5π
4,即x=
π
2时,
f(x)取得最小值f(
π
2) =
2(−
2
2) +1=0.
∴当x∈[0,
π
2]时,
f(x)最大值与最小值的和为f(
π
8) +f(
=2cos2x-1+2sinxcosx+1
=cos2x+1+sin2x
=
2sin(2x+
π
4)+1,
∴函数f(x)的最小正周期 T=
2π
2=π.
(Ⅱ)当x∈[0,
π
2]时,2x+
π
4∈[
π
4,
5π
4],
当2x+
π
4=
π
2,即x=
π
8时,
f(x)取得最大值f(
π
8) =
2+1;
当2x+
π
4=
5π
4,即x=
π
2时,
f(x)取得最小值f(
π
2) =
2(−
2
2) +1=0.
∴当x∈[0,
π
2]时,
f(x)最大值与最小值的和为f(
π
8) +f(
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