1. Let $a, b$ be unequal real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ .
Solve for the axis of symmetry of $f(x)=x^{2}+a x+b$ which is $x=-\frac{a}{2}=\frac{a+b}{2} \Rightarrow b=-2 a$, thus $f(x)=x^{2}+a x-2 a \Rightarrow f(2)=4+2 a-2 a=4$.