= max 0 ⩽ θ < 2 π ( a − 1 2 cos 2 θ + 1 2 sin 2 θ + a + 1 2 ) = max 0 ⩽ θ < 2 π ( ( a − 1 2 ) 2 + 1 4 sin ( 2 θ + a ) + a + 1 2 ) = ( a − 1 2 ) 2 + 1 4 + a + 1 2 .
\begin{array}{l}
=\max _{0 \leqslant \theta<2 \pi}\left(\frac{a-1}{2} \cos 2 \theta+\frac{1}{2} \sin 2 \theta+\frac{a+1}{2}\right) \\
=\max _{0 \leqslant \theta<2 \pi}\left(\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}} \sin (2 \theta+a)+\frac{a+1}{2}\right) \\
=\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}}+\frac{a+1}{2} .
\end{array}
= max 0 ⩽ θ < 2 π ( 2 a − 1 cos 2 θ + 2 1 sin 2 θ + 2 a + 1 ) = max 0 ⩽ θ < 2 π ( ( 2 a − 1 ) 2 + 4 1 sin ( 2 θ + a ) + 2 a + 1 ) = ( 2 a − 1 ) 2 + 4 1 + 2 a + 1 .
Similarly, max x 2 + y 2 y 0 f ( x , y ) = − ( a − 1 2 ) 2 + 1 4 + a + 1 2 \max _{x^{2}+y^{2} y_{0}} f(x, y)=-\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}}+\frac{a+1}{2} max x 2 + y 2 y 0 f ( x , y ) = − ( 2 a − 1 ) 2 + 4 1 + 2 a + 1 max x 2 + y 2 + 0 f ( x , y ) − min x 2 + y 2 ≠ 0 f ( x , y ) \max _{x^{2}+y^{2}+0} f(x, y)-\min _{x^{2}+y^{2} \neq 0} f(x, y) max x 2 + y 2 + 0 f ( x , y ) − min x 2 + y 2 = 0 f ( x , y ) = 2 ( a − 1 2 ) 2 + 1 4 .
=2 \sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}} \text {. }
= 2 ( 2 a − 1 ) 2 + 4 1 .
Thus, ( a − 1 ) 2 = 3 (a-1)^{2}=3 ( a − 1 ) 2 = 3 a = 1 ± 3 a=1 \pm \sqrt{3} a = 1 ± 3 f ( k x , k y ) = f ( x , y ) f(k x, k y)=f(x, y) f ( k x , k y ) = f ( x , y ) max x 2 + y 2 f ( x , y ) = max x 2 + y 2 = 1 f ( x , y ) = max 0 < θ < 2 π ( a cos 2 θ + cos θ ⋅ sin θ + sin 2 θ ) = max 0 < θ < 2 π ( ( a − 1 ) cos 2 θ + cos θ ⋅ sin θ + 1 )
\begin{array}{l}
\max _{x^{2}+y^{2}} f(x, y)=\max _{x^{2}+y^{2}=1} f(x, y) \\
=\max _{0<\theta<2 \pi}\left(a \cos ^{2} \theta+\cos \theta \cdot \sin \theta+\sin ^{2} \theta\right) \\
=\max _{0<\theta<2 \pi}\left((a-1) \cos ^{2} \theta+\cos \theta \cdot \sin \theta+1\right)
\end{array}
max x 2 + y 2 f ( x , y ) = max x 2 + y 2 = 1 f ( x , y ) = max 0 < θ < 2 π ( a cos 2 θ + cos θ ⋅ sin θ + sin 2 θ ) = max 0 < θ < 2 π ( ( a − 1 ) cos 2 θ + cos θ ⋅ sin θ + 1 )