5个回答
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解:(1)依题意,得
f(x)=根号3 m sinx cosx - m cosx cosx
=(二分之根号3 m sin2x -m/2 cos2x) -m/2
=msin(2x-π/6)-m/2
(2)依题意,得
x∈(0,π/2)2x-π/6∈(-π/6,5π/6)
2x-π/6=-π/6 f(x)min=-m=-4
∴m=4
2x-π/6=π/2时f(x)max=m/2=2,x=π/3
f(x)=根号3 m sinx cosx - m cosx cosx
=(二分之根号3 m sin2x -m/2 cos2x) -m/2
=msin(2x-π/6)-m/2
(2)依题意,得
x∈(0,π/2)2x-π/6∈(-π/6,5π/6)
2x-π/6=-π/6 f(x)min=-m=-4
∴m=4
2x-π/6=π/2时f(x)max=m/2=2,x=π/3
追问
最后不应该是加m/2么。。。因为是a-b所以负负为正应该是加m cos x平方
2014-01-18
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a=(根号3msinx,cosx) b=(cosx,-mcosx)
f(x)=ab=√3msinxcosx-mcosxcosx
=(√3/2)msin2x-(m/2)[2(cosx)^2-1+1]
=msin2xcosπ/6-mcos2xsinπ/6-m/2
=msin(2x-π/6)-m/2
x∈[0,π/2]时, 2x-π/6∈[-π/6,5π/6],sin(2x-π/6)∈[-1/2,1]
m>0,f(x)最小值=-m/2-m/2=-m=-4 m=4
最大值=m-m/2=m/2=2,2x-π/6=π/2 , x=π/3
m<0时,f(x)最小值=m-m/2=m/2=-4 m=-8
最大值=-m/2-m/2=-m=8, 2x-π/6=-π/6, x=0
f(x)=ab=√3msinxcosx-mcosxcosx
=(√3/2)msin2x-(m/2)[2(cosx)^2-1+1]
=msin2xcosπ/6-mcos2xsinπ/6-m/2
=msin(2x-π/6)-m/2
x∈[0,π/2]时, 2x-π/6∈[-π/6,5π/6],sin(2x-π/6)∈[-1/2,1]
m>0,f(x)最小值=-m/2-m/2=-m=-4 m=4
最大值=m-m/2=m/2=2,2x-π/6=π/2 , x=π/3
m<0时,f(x)最小值=m-m/2=m/2=-4 m=-8
最大值=-m/2-m/2=-m=8, 2x-π/6=-π/6, x=0
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