实数XY满足1≤X＾2+Y＾2≤4.求X＾2+2XY—Y＾2的取值范围
问题描述：
实数XY满足1≤X＾2+Y＾2≤4.求X＾2+2XY—Y＾2的取值范围
答
进行坐标变换,x=rcosθ,y=rsinθ
∵1≤x²+y²≤4
∴1≤(rcosθ)²+(rsinθ)²≤4
∴1≤r²≤4
∴1≤r≤2,θ∈[0,2π)
∴x²+2xy-y²
=(rcosθ)²+2r²cosθsinθ-(rsinθ)²
=r²cos²θ-r²sin²θ+2r²cosθsinθ
=r²cos2θ+r²sin2θ
=√2r²sin(2θ+π/4)
∵1≤r≤2,θ∈[0,2π)
∴x²+2xy-y²的取值范围是[-4√2,4√2]