Solve frac{1 - |x|}{2 - |x|} ge 0. | vedclass

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(D) Let $t = |x|$. Since $|x| \ge 0$,we have $t \ge 0$. The inequality becomes $\frac{1 - t}{2 - t} \ge 0$. Multiplying by $-1$ in the numerator and d

Vedclass · Accepted Answer

$(-\infty, -2) \cup [-1, 1] \cup (2, \infty)$

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(D) Let $t = |x|$. Since $|x| \ge 0$,we have $t \ge 0$. The inequality becomes $\frac{1 - t}{2 - t} \ge 0$. Multiplying by $-1$ in the numerator and denominator,we get $\frac{t - 1}{t - 2} \ge 0$. Using the wavy curve method for $t$,the critical points are $t = 1$ and $t = 2$. The inequality holds for $t \in [0, 1] \cup (2, \infty)$. Substituting $t = |x|$ back,we have $|x| \in [0, 1] \cup (2, \infty)$. Case $1$: $0 \le |x| \le 1 \implies x \in [-1, 1]$. Case $2$: $|x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$. Combining these,the solution is $x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

Solve $\frac{1 - |x|}{2 - |x|} \ge 0$.

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