2001 AMC 10 Problem 2

1. Name the pieces. The reciprocal of $x$ is $\frac{1}{x}$ (for example, the reciprocal of $5$ is $\frac15$). The additive inverse of $x$ is $-x$ (the opposite, so for $5$ it is $-5$).
2. Try an example to see what their product does. For $x = 5$: $\frac15 \cdot (-5) = -1$. For $x = 3$: $\frac13 \cdot (-3) = -1$. In general
$$\frac{1}{x} \cdot (-x) = -1$$
for every nonzero $x$, because $x$ on top cancels $x$ on the bottom.
3. "$x$ is $2$ more than the product" means $x = (\text{product}) + 2$. So
$$x = -1 + 2 = 1.$$
4. Check: the reciprocal of $1$ is $1$, its opposite is $-1$, and their product is $-1$. Two more than $-1$ is $1$. It works.
5. The number $1$ satisfies $0 < 1 \le 2$, which is interval (C).

The answer is $\boxed{\textbf{(C)}\ 0 < x \le 2}$.

A number times its own reciprocal is always $1$, so a number times the reciprocal of its opposite is always $-1$. The variable disappears from the product, and the "equation" gives the answer right away. When a word problem sounds tangled, write each phrase as symbols one at a time, then simplify before doing anything else.

The reciprocal of $x$ is $\frac{1}{x}$ and its additive inverse is $-x$. Their product is
$$\frac{1}{x}\cdot(-x) = -1,$$
no matter what nonzero $x$ is. The sentence says $x$ is $2$ more than this product:
$$x = -1 + 2 = 1.$$
Since $0 < 1 \le 2$, the number lies in interval (C).

The answer is $\boxed{\textbf{(C)}\ 0 < x \le 2}$.