7・101 If $g(x)=1-x^{2}$, and when $x \neq 0$, $f[g(x)]=\frac{1-x^{2}}{x^{2}}$, then $f\left(\frac{1}{2}\right)$ equals
(A) $\frac{3}{4}$.
(B) 1 .
(C) 3 .
(D) $\frac{\sqrt{2}}{2}$.
(E) $\sqrt{2}$.
(25th American High School Mathematics Examination, 1974)
[Solution] From $g(x)=\frac{1}{2}$, i.e., $1-x^{2}=\frac{1}{2}$, we get $x= \pm \sqrt{\frac{1}{2}}$. Therefore, $f\left(\frac{1}{2}\right)=f\left[g\left( \pm \sqrt{\frac{1}{2}}\right)\right]=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1$. Hence, the answer is (B).