8.55 The function $f(x)$ is undefined at $x=0$, but for all non-zero real numbers $x$, $f(x)+2 f\left(\frac{1}{x}\right)=3 x$. The solution situation of the equation $f(x)=f(-x)$ is
(A) Exactly one real solution.
(B) Exactly two real solutions.
(C) No real solutions.
(D) Infinitely many, but not all non-zero real numbers.
(E) All non-zero real numbers.
(32nd American High School Mathematics Examination, 1981)
[Solution] Substituting $-\frac{1}{x}$ into the equation $f(x)+2 f\left(\frac{1}{x}\right)=3 x$, we get
$$
f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x} \text {. }
$$
Multiplying (2) by 2 and subtracting (1) gives $\quad f(x)=\frac{2-x^{2}}{x}$.

Therefore, $f(x)=f(-x)$ if and only if $\frac{2-x^{2}}{x}=\frac{2-(-x)^{2}}{-x}$, which simplifies to $x^{2}=2$.

Solving the above equation yields $x= \pm \sqrt{2}$. Hence, the answer is $(B)$.