$2 \cdot 89$ Given the equation $|x|=a x+1$ has one negative root and no positive root, the range of values for $a$ is
(A) $a>-1$.
(B) $a=1$.
(C) $a \geqslant 1$.
(D) None of the above.
(China Junior High School Mathematics League, 1987)
[Solution 1] Let $x$ be the negative root of the equation, then $-x = ax + 1$, hence $x = \frac{-1}{a+1}$, where $a > -1$.
Clearly, when $a > -1$, the equation has a negative root.
Assume the equation has a positive root $x$, then $x = ax + 1$.
Thus,
$$
(1-a)x = 1,
$$
which gives $x = \frac{1}{1-a} > 0$, so $a \neq 1$ and $a < 1$, hence $a \geqslant 1$.
Therefore, the answer is $(C)$.
[Solution 2] Draw the graphs of $y = |x|$ and $y = ax + 1$.
Clearly, the line $y = ax + 1$ intersects only the second quadrant bisector (one branch of $y = |x|$) and not the first quadrant bisector (the other branch of $y = |x|$) only when its slope $a \geqslant 1$.
Therefore, the answer is $(C)$.