Question

求|(k-2)/√1+k²|的最大值

y=sinθ+cosθ+ksinθcosθ,θ∈R,其中k为实常数
（1）求y的最大值Mk和最小值mk（2）求Mk的最小值和mk的最大值

Answer 1

设 x = sinθ+cosθ = √2 sin(θ+ π/4) ∈[－√2,√2]
x² = sin²θ+cos²θ+2sinθcosθ = 1 + 2sinθcosθ
sinθcosθ = (x²－1)/2
y = x + k(x²－1)/2
当k=0时,y = x ,Mk = √2,mk = ﹣√2
当k≠0时,配方得
y = k/2 (x +1/k )² －(1+k²)/(2k)
对称轴 x = - 1/k
①当 -1/k < －√2 即 0 < k < 1/√2 时,开口向上
x=√2取最大值,Mk = √2 + k/2
x=－√2取最小值,mk = －√2 + k/2
②当 －√2 ≤ -1/k < 0 即 k ≥ 1/√2 时,开口向上
x= √2取最大值,Mk = √2 + k/2
x = - 1/k 取最小值,mk = －(1+k²)/(2k)
③当0 < - 1/k ≤ √2 即 k ≤ - 1/√2 时,开口向下
x = - 1/k 取最大值,Mk = －(1+k²)/(2k)
x=－√2取最小值,mk = －√2 + k/2
④当 - 1/k > √2 即 －1/√2 < k < 0 时,开口向下
x= √2 取最大值,Mk = √2 + k/2
x=－√2 取最小值,mk= －√2 + k/2
综上,
Mk = √2 + k/2 ,k > －1/√2
－(1+k²)/(2k),k ≤ ﹣1/√2
mk = －√2 + k/2 ,k < 1/√2
－(1+k²)/(2k),k ≥ 1/√2
(2) 只是简单的求值域问题
不难求得 Mk最小值为1,在 k = - 1时取到
mk 最大值为 - 1,在 k = 1 时取到