8.44 The number of real roots of the equation $\frac{x}{100}=\sin x$ is
(A) 91.
(B) 62.
(C) 63.
(D) 64.
(E) 65.
(32nd American High School Mathematics Examination, 1981)
[Solution] Since $\frac{-x}{100}=\sin (-x)$, the equation has as many positive solutions as negative solutions. And by direct verification, we know that $x=0$ is one of its solutions.
Also, because $|\sin x| \leqslant 1$, we have $|x|=100|\sin x| \leqslant 100$,
which means: the absolute value of the solution is no greater than 100.
In the Cartesian coordinate system, draw the graphs of $\frac{x}{100}$ and $\sin x$. Since $15.5<\frac{100}{2 \pi}<16$, i.e.,
$15.5 \cdot 2 \pi<100<16 \cdot 2 \pi$.
From the graph, we can intuitively see that the equation has one solution between 0 and $\pi$, and there are two solutions in each interval from $(2 k-1) \pi$ to $(2 k+1) \pi$, where $1 \leqslant k \leqslant 15$.
Therefore, the number of solutions to the equation is $1+2(1+2 \cdot 15)=63$. Hence, the answer is $(C)$.