设函数f(x)=a*b,其中向量a=(2sin(π/4+x),cos2x),b=(sin(π/4+x),-根3),x属于R,求:1.函数f(x)的单调递增区间2.当x属于[0,π/2]时,求函数f(x)的值域
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设函数f(x)=a*b,其中向量a=(2sin(π/4+x),cos2x),b=(sin(π/4+x),-根3),x属于R,求:1.函数f(x)的单调递增区间2.当x属于[0,π/2]时,求函数f(x)的值域
设函数f(x)=a*b,其中向量a=(2sin(π/4+x),cos2x),b=(sin(π/4+x),-根3),x属于R,
求:1.函数f(x)的单调递增区间
2.当x属于[0,π/2]时,求函数f(x)的值域
设函数f(x)=a*b,其中向量a=(2sin(π/4+x),cos2x),b=(sin(π/4+x),-根3),x属于R,求:1.函数f(x)的单调递增区间2.当x属于[0,π/2]时,求函数f(x)的值域
f(x)=a*b=2sin(x+π/4)cos(x+π/4)-√3cos2x
=sin(2x+π/2)-√3cos2x
=(1-√3)cos2x
1.单增区间为2x∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/4,kπ+π/4]
2.x属于[0,π/2]时,2x∈[0,π]
f(x)的值域为[1-√3,√3-1]
1问,f(x)=a*b=2sin(2x-π/6)+1,单调增区间为(-π/2)+2kπ≤2x-π/6≤π/2+2kπ即有曾区间为【(-π/6)+kπ,π/3+kπ】。2问,当x属于(0,π/2)时,(2x-π/6)属于(-π/6,5π/6),而当k=0时,函数的增区间为(-π/6,π/3),所以函数的值域为(0,√3+1)
这种题目其实是伪向量体,只要按点乘算出函数后,题目就和向量无关了
f(x)=2sin(π/4+x)sin(π/4+x) -根号3 cos2x
= 2[sin(π/4)cosx + cos(π/4) sinx] ^2 -根号3 cos2x
= 1+2sinxcosx -根号3 cos2x
= 1 + sin2x - 根号3 cos2x
=1 + 2[0.5si...
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这种题目其实是伪向量体,只要按点乘算出函数后,题目就和向量无关了
f(x)=2sin(π/4+x)sin(π/4+x) -根号3 cos2x
= 2[sin(π/4)cosx + cos(π/4) sinx] ^2 -根号3 cos2x
= 1+2sinxcosx -根号3 cos2x
= 1 + sin2x - 根号3 cos2x
=1 + 2[0.5sin2x - 0.5根号3 cos2x]
=1+2 sin(2x-π/3)
剩下就是普通三角函数lz应该会了吧
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1.f(x)=a*b=2[sin(π/4+x)]^2-√3cos2x
=1-cos(π/2+2x)-√3cos2x
=1+sin2x-√3cos2x
=1+2sin(2x-π/3),
其增区间由(2k-1/2)π<2x-π/3<(2k+1/2)π,k∈Z确定,
即(k-1/12)π
2x-π...
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1.f(x)=a*b=2[sin(π/4+x)]^2-√3cos2x
=1-cos(π/2+2x)-√3cos2x
=1+sin2x-√3cos2x
=1+2sin(2x-π/3),
其增区间由(2k-1/2)π<2x-π/3<(2k+1/2)π,k∈Z确定,
即(k-1/12)π
2x-π/3∈[-π/3,2π/3],
sin(2x-π/3)∈[(-√3)/2,1],
∴f(x)的值域为[1-√3,3].
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