设函数f(x)=sinx+3cosx+1.(1)求函数f(x)在[0,π2]的最大值与最小值;(2)若实数a,b,c使得af(
设函数f(x)=sinx+3cosx+1.(1)求函数f(x)在[0,π2]的最大值与最小值;(2)若实数a,b,c使得af(x)+bf(x-c)=1对任意x∈R恒成立,...
设函数f(x)=sinx+3cosx+1.(1)求函数f(x)在[0,π2]的最大值与最小值;(2)若实数a,b,c使得af(x)+bf(x-c)=1对任意x∈R恒成立,求bcosca的值.
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(1)f(x)=sinx+
cosx+1
=2(
sinx+
cosx)+1
=2sin(x+
)+1
∵x∈[0,
],∴x+
∈[
,
]
∴
≤sin(x+
)≤1,∴2≤2sin(x+
)+1≤3
∴函数f(x)在[0,
]的最大值为3;最小值为2.
(2)af(x)+bf(x-c)=a[2sin(x+
)+1]+b[2sin(x+
-c)+1]=1
2asin(x+
)+2bsin(x+
-c)=1-a-b
2asin(x+
)+2bsin(x+
)cosc-2bcos(x+
)sinc=1-a-b
(2a+2bcosc)sin(x+
)-(cos(x+
)=1-a-b
sin(x+
+φ)=1-a-b
因为上式对一切的x恒成立,所以
=0
∴
∴由2a+2bcosc=0得:
=-1.
3 |
=2(
1 |
2 |
| ||
2 |
=2sin(x+
π |
3 |
∵x∈[0,
π |
2 |
π |
3 |
π |
3 |
5π |
6 |
∴
1 |
2 |
π |
3 |
π |
3 |
∴函数f(x)在[0,
π |
2 |
(2)af(x)+bf(x-c)=a[2sin(x+
π |
3 |
π |
3 |
2asin(x+
π |
3 |
π |
3 |
2asin(x+
π |
3 |
π |
3 |
π |
3 |
(2a+2bcosc)sin(x+
π |
3 |
π |
3 |
(2a+2bcosc)2+(2bsinc)2 |
π |
3 |
因为上式对一切的x恒成立,所以
(2a+2bcosc)2+(2bsinc)2 |
∴
|
∴由2a+2bcosc=0得:
bcosc |
a |
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