If left|x+frac{1}{x}right| geq 2,then x in | vedclass

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(A) We are given the inequality $\left|x+\frac{1}{x}\right| \geq 2$. Case $1$: If $x > 0$,then $x + \frac{1}{x} \geq 2$. By the Arithmetic Mean-Geomet

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$R - \{0\}$

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(A) We are given the inequality $\left|x+\frac{1}{x}\right| \geq 2$. Case $1$: If $x > 0$,then $x + \frac{1}{x} \geq 2$. By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for $x > 0$,$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1$,which implies $x + \frac{1}{x} \geq 2$. This holds for all $x > 0$. Case $2$: If $x  0$. Then $\left|-y - \frac{1}{y}\right| = \left|-(y + \frac{1}{y})\right| = y + \frac{1}{y} \geq 2$. This also holds for all $y > 0$,meaning it holds for all $x Since $x$ cannot be $0$ (as $\frac{1}{x}$ is undefined),the inequality holds for all $x \in R - \{0\}$.

If $\left|x+\frac{1}{x}\right| \geq 2$,then $x \in$

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