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.