1. Given the function $f(x)=\frac{2 x^{2}+\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)}{2 x^{2}+\cos x}$, the maximum and minimum values are $a, b$ respectively, then $a+b=$
1. 2 Detailed Explanation: Convert the function to $f(x)=\frac{2 x^{2}+\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)}{2 x^{2}+\cos x}=1+\frac{\sin x}{2 x^{2}+\cos x}$. Let $g(x)=\frac{\sin x}{2 x^{2}+\cos x}$, since it is an odd function, if there is a maximum value $g(m)$, then there is a minimum value $g(-m)$, so $a+b=2$,