2001 AMC 10 第 2 题

Question

一个数

x

比它的倒数与它的加法逆元的乘积多

2

。这个数位于哪个区间？

A number

x

is

2

more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

-4 \le x \le -2

-2 \lt x \le 0

0 \lt x \le 2

2 \lt x \le 4

4 \lt x \le 6

x

的倒数是

\dfrac1x

，加法逆元是

-x

。

The reciprocal of

x

is

\dfrac1x

and the additive inverse is

-x

它们的乘积是

\dfrac1x\cdot(-x)=-1

。

Their product is

\dfrac1x\cdot(-x)=-1

x

的倒数是

\dfrac1x

，加法逆元是

-x

，它们的乘积为

\dfrac1x\cdot(-x)=-1

。

因此

x=2+(-1)=1

，位于区间

0\lt x\le2

。

所以正确答案是 C。

The reciprocal of

x

is

\dfrac1x

and its additive inverse is

-x.

Their product is

\dfrac1x\cdot(-x)=-1.

So

x=2+(-1)=1,

which lies in the interval

0\lt x\le2.

Thus, the correct answer is C.

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