AMC10 2001 A

AMC10 2001 A · Q2

AMC10 2001 A · Q2. It mainly tests Linear equations, Absolute value.

A number $x$ is 2 more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

一个数 $x$ 比它的倒数与其加法逆元的乘积多 2。它位于哪个区间？

(A) $ -4 \le x \le -2 $ $ -4 \le x \le -2 $

(B) $ -2 < x \le 0 $ $ -2 < x \le 0 $

(D) $ 2 < x \le 4 $ $ 2 < x \le 4 $

(E) $ 4 < x \le 6 $ $ 4 < x \le 6 $

Answer

Correct choice: (C)

正确答案：（C）

Solution

The reciprocal of $x$ is $\frac{1}{x}$, and the additive inverse of $x$ is $-x$. The product of these is $\left(\frac{1}{x}\right) \cdot (-x) = -1$. So $x = -1 + 2 = 1$, which is in the interval $0 < x \le 2$.

$x$ 的倒数是 $\frac{1}{x}$，$x$ 的加法逆元是 $-x$。它们的乘积是 $\left(\frac{1}{x}\right) \cdot (-x) = -1$。所以 $x = -1 + 2 = 1$，位于区间 $0 < x \le 2$。

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