中国移动无线路由器:高1数学题

f(x)=2(sinx)^2+3sinxcosx+cos(x+1/4*Pi)^2+a.
2(sinx)^2=1-cos2x.
3sinxcosx=(3/2)*sin2x.
cos(x+1/4*Pi)^2
=[(cosx-sinx)^2]/2=(1/2)-sinxcosx=(1/2)-(1/2)sin2x.
所以:
f(x)=1-cos2x+(3/2)sin2x+(1/2)-(1/2)*sin2x+a
=sin2x-cos2x+(a+3/2). (引进辅助角)
=(根号2)*sin(2x-Pi/4)+(a+3/2).
所以:
(1)
减区间就是
Pi/2+k*2*Pi<=2x-Pi/4<=3*Pi/2+k*2*Pi.
即
k*Pi+(3*Pi/8)<=x<=k*Pi+(3*Pi/8).

(2)
在(0,Pi/2)上最小值应该是x=0的时候,为
(根号2)*(-sin (Pi/4))+a+3/2=a+1/2.
所以a=3/2.

注:其实我不是抄上面的,我是第一个回答问题的,
只是我不小心做错了,修改了一下......
可恶的抄袭者...............

楼上的不太准确
(1)
cos(x+0.25pi)^2
=(cosxcos0.25pi-sinxsin0.25pi)^2
=(cosx-sinx)^2/2
=(1-sin2x)/2

f(x)=2(sinx)^2+3sinxcosx+cos(x+0.25pi)^2+a
=1-cos2x+(3/2)sin2x+(1-sin2x)/2+a
=1-cos2x+1/2+sin2x+a
=sin2x-cos2x+(a+3/2)
=√2sin(2x-0.25pi)+(a+3/2)

f(x)的减区间为[3pi/8+kpi，7pi/8+kpi]

(2)
0<x<0.5pi，f(x)min=√2*(-√2/2)+(a+3/2)=2
a=3/2

一楼cos(x+1/4*Pi)^2
=(cosx-sinx)^2=1-2sinxcosx=1-sin2x化错了
(1)f(x)=根号2sin(2x-pi/4)+3/2+a
减区间[3pi/8+kpi,7pi/8+kpi] k为整数
(2)f(x)min=根号2（-2根号2）+3/2+a=2
a=3/2

f(x)=2(sinx)^2+3sinxcosx+cos(x+1/4*Pi)^2+a.
2(sinx)^2=1-cos2x.
3sinxcosx=(3/2)*sin2x.
cos(x+1/4*Pi)^2
=(cosx-sinx)^2=1-2sinxcosx=1-sin2x.
所以:
f(x)=1-cos2x+(3/2)sin2x+1-sin2x+a
=(1/2)*sin2x-cos2x+(a+2). (引进辅助角)
=(根号5/2)*sin(2x-t)+(a+2).
其中cost=(1/根号5), sint=(2/根号5).
所以:
(1)
减区间就是
Pi/2+k*2*Pi<=2x+t<=3*Pi/2+k*2*Pi.
即
k*Pi+(Pi/2-t)/2<=x<=k*Pi+(3*Pi/2-t)/2.
(2)
在(0,Pi/2)上最小值应该是x=0的时候,为
(根号5/2)*(-sint)+a+2=a+1.
所以a=1.

f(x)=2(sinx)^2+3sinxcosx+cos(x+1/4*Pi)^2+a.
2(sinx)^2=1-cos2x.
3sinxcosx=(3/2)*sin2x.
cos(x+1/4*Pi)^2
=[(cosx-sinx)^2]/2=(1/2)-sinxcosx=(1/2)-(1/2)sin2x.
所以:
f(x)=1-cos2x+(3/2)sin2x+(1/2)-(1/2)*sin2x+a
=sin2x-cos2x+(a+3/2).
=(根号2)*sin(2x-Pi/4)+(a+3/2).
所以:
(1)
减区间就是
Pi/2+k*2*Pi<=2x-Pi/4<=3*Pi/2+k*2*Pi.
即
k*Pi+(3*Pi/8)<=x<=k*Pi+(3*Pi/8).

(2)
在(0,Pi/2)上最小值应该是x=0的时候,为
(根号2)*(-sin (Pi/4))+a+3/2=a+1/2.
所以a=3/2.

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