已知f(x)=sin²x+2sinxcosx+3cos²x,x∈R,求
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解:
f(x)=sin²x+2sinxcosx+3cos²x
=sin²x+cos²x+2cos²x+sin2x
=1+2cos²x+sin2x
=1+con2x+1+sin2x
=2+√2sin(π/4+2x)
∵|sin(π/4+2x)|<=1
∴f(x)=2+√2sin(π/4+2x) >0
∵正弦函数的周期是2π
∴sin(π/4+2x)的周期是π
因为f(x)>0
∴最小正周期是π
(2)
2nπ-π/2<= π/4+2x <=2nπ+π/2
2nπ-π3/4<= 2x <=2nπ+π/4
nπ-π3/8<= x <=nπ+π/8
f(x)的单调增区间为:[nπ-π3/8 , nπ+π/8] (n为整数)
(3)
π/4+2x=2nπ+π/2
X=nπ+π/8时 f(x)最大=2+√2
f(x)=sin²x+2sinxcosx+3cos²x
=sin²x+cos²x+2cos²x+sin2x
=1+2cos²x+sin2x
=1+con2x+1+sin2x
=2+√2sin(π/4+2x)
∵|sin(π/4+2x)|<=1
∴f(x)=2+√2sin(π/4+2x) >0
∵正弦函数的周期是2π
∴sin(π/4+2x)的周期是π
因为f(x)>0
∴最小正周期是π
(2)
2nπ-π/2<= π/4+2x <=2nπ+π/2
2nπ-π3/4<= 2x <=2nπ+π/4
nπ-π3/8<= x <=nπ+π/8
f(x)的单调增区间为:[nπ-π3/8 , nπ+π/8] (n为整数)
(3)
π/4+2x=2nπ+π/2
X=nπ+π/8时 f(x)最大=2+√2
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解:y=sin2x+cos2x+2=√ 2sin(2x+ π4)+2;
①,T= 2π/2=π;函数的最小正周期为:π
②,当x=kπ+ π/8(kÎZ)时,ymax=2+ √2;函数的最大值为:2√ +2;
③,因为y=sinx的单调增区间为:[2kπ- π/2,2kπ+ π/2]k∈Z,所以2x+ π/4∈[2kπ- π/2,2kπ+ π/2]
解得x∈[kπ- 3π/8,kπ+ π/8],k∈Z就是函数的单调增区间.
①,T= 2π/2=π;函数的最小正周期为:π
②,当x=kπ+ π/8(kÎZ)时,ymax=2+ √2;函数的最大值为:2√ +2;
③,因为y=sinx的单调增区间为:[2kπ- π/2,2kπ+ π/2]k∈Z,所以2x+ π/4∈[2kπ- π/2,2kπ+ π/2]
解得x∈[kπ- 3π/8,kπ+ π/8],k∈Z就是函数的单调增区间.
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这个(+2)怎么算出来的,其他的我都算出来了
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