设函数fx=向量a*向量b,其中向量a=(m,cos2x),向量b=(1+sin2x,1),x∈R且fx的图像经过点(π
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设函数fx=向量a*向量b,其中向量a=(m,cos2x),向量b=(1+sin2x,1),x∈R且fx的图像经过点(π/4,2) 求函数fx在区间[0,π/2]上的最小值
![设函数fx=向量a*向量b,其中向量a=(m,cos2x),向量b=(1+sin2x,1),x∈R且fx的图像经过点(π](/uploads/image/z/19445100-60-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0fx%3D%E5%90%91%E9%87%8Fa%EF%BC%8A%E5%90%91%E9%87%8Fb%2C%E5%85%B6%E4%B8%AD%E5%90%91%E9%87%8Fa%3D%EF%BC%88m%2Ccos2x%EF%BC%89%2C%E5%90%91%E9%87%8Fb%3D%EF%BC%881%2Bsin2x%2C1%EF%BC%89%2Cx%E2%88%88R%E4%B8%94fx%E7%9A%84%E5%9B%BE%E5%83%8F%E7%BB%8F%E8%BF%87%E7%82%B9%EF%BC%88%CF%80)
f(x) = m(1 + sin2x) + cos2x
f(π/4) = m[1 + sin(π/2)] + cos(π/2) = m(1 + 1) + 0 = 2m = 2
m = 1
f(x) = 1 + sin2x + cos2x = 1 + 2[(sin2x)(√2/2) + (cos2x)(√2/2)]
= 1 + 2[sin2xcos(π/4) + cos2xsin(π/4)]
= 1+ 2sin(2x + π/4)
x ∈ [0,π/2], 2x + π/4 ∈ [π/4,5π/4]
π/4 < 2x+ π/4 < π/2时, f(x)为增函数
π/2 < 2x + π/4< 5π/4时, f(x)为减函数
所以只须比较f(0), f(π/2)
f(0) = 1 + 2sin(π/4) = 1 + √2
f(π/2) = 1 + 2sin(5π/4) = 1 - 2sin(π/4) = 1 - √2 < f(0)
fx在区间[0,π/2]上的最小值 = 1 - √2
f(π/4) = m[1 + sin(π/2)] + cos(π/2) = m(1 + 1) + 0 = 2m = 2
m = 1
f(x) = 1 + sin2x + cos2x = 1 + 2[(sin2x)(√2/2) + (cos2x)(√2/2)]
= 1 + 2[sin2xcos(π/4) + cos2xsin(π/4)]
= 1+ 2sin(2x + π/4)
x ∈ [0,π/2], 2x + π/4 ∈ [π/4,5π/4]
π/4 < 2x+ π/4 < π/2时, f(x)为增函数
π/2 < 2x + π/4< 5π/4时, f(x)为减函数
所以只须比较f(0), f(π/2)
f(0) = 1 + 2sin(π/4) = 1 + √2
f(π/2) = 1 + 2sin(5π/4) = 1 - 2sin(π/4) = 1 - √2 < f(0)
fx在区间[0,π/2]上的最小值 = 1 - √2
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