已知向量a=(cos3/2x,sin3/2x),b=(cos1/2x,sin1/2x),x属于[0,π/2] 求f(x)
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已知向量a=(cos3/2x,sin3/2x),b=(cos1/2x,sin1/2x),x属于[0,π/2] 求f(x)=m|a+b|-a*b(m∈R)的最大值求大
向量a =(cos(3x/2),sin(3x/2)),b =(cos(x/2),sin(x/2)),所以a·b =cos(3x/2)cos(x/2) + sin(3x/2)sin(x/2) = cos(3x/2 – x/2) = cosx ; |a + b| =|cos(3x/2) + cos(x/2),sin(3x/2) + sin(x/2)| = √[(cos(3x/2) + cos(x/2)) 2 + (sin(3x/2) + sin(x/2)) 2 ] = √[(cos 2 (3x/2) + cos 2 (x/2)+ 2cos(3x/2)cos(x/2)) + (sin 2 (3x/2) + sin 2 (x/2) +2sin(3x/2)sin(x/2))] = √[2 + 2cos(3x/2)cos(x/2) + 2sin(3x/2)sin(x/2)]= √(2 + 2cosx) = √[2(cos 2 (x/2) + sin 2 (x/2))+ 2(cos 2 (x/2) – sin 2 (x/2))] = √[4(cos 2 (x/2)] = 2cos(x/2) ; 所以f(x) = m|a +b| - a·b =2mcos(x/2) – cosx = 2mcos(x/2) – [2cos 2 (x/2) – 1] = -2cos 2 (x/2)+ 2mcos(x/2) + 1 = -2[cos(x/2) – m/2] 2 + 1 + m 2 /2,对称轴为cos(x/2) – m/2 = 0,即cos(x/2) = m/2,因为x∈[0,π/2],所以(x/2)∈[0,π/4] => cos(x/2)∈[√2/2,1],记g(t) = -2(t – m/2) 2 + 1 + m 2 /2,对m的值分类讨论可得: 1)如果m < √2,那么m/2 < √2/2,函数g(t) = -2(t – m/2) 2 + 1 + m 2 /2在t∈[√2/2,1]上单调递减,当t = √2/2,即x = π/2时,最大值是f(π/2) = √2m ; 2)如果√2 ≤ m ≤ 2,那么√2/2 ≤ m/2 ≤ 1,当cos(x/2) =m/2时,即x =2arccos(m/2)时,f(x)的最大值是 1 + m 2 /2 ; 3)如果m > 2,那么m/2 > 1,函数g(t) = -2(t – m/2) 2 + 1 + m 2 /2在t∈[√2/2,1]上单调递增,当t = 1,即x = 0时,最大值是f(π/2) = 2m – 1 ; 综上所述,当 m < √2 ,f(x)的最大值是 √2m ,(此时x = π/2) ; 当 √2 ≤ m ≤ 2 ,f(x)的最大值是 1 + m 2 /2 ,(x = 2arccos(m/2)); 当 m > 2 ,f(x)的最大值是 2m– 1 ,(x = 0) .
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