Solve each of the following in equations and represent the solution set on the number line.
$\frac{1}{|x|-3} \leq \frac{1}{2}, x \in \mathbf{R}$
Given
$\frac{1}{|x|-3} \leq \frac{1}{2}, x \in R$
Intervals of $|x|$ :
$|x|=-x, x<0$
$|x|=x, x \geq 0$
Domain of $\frac{1}{|x|-3} \leq \frac{1}{2}$
$|x|+3=0$
$x=-3$ or $x=3$
Therefore,
$-3 Now, combining intervals with domain: $x<-3,-3 For $x<-3$ $\frac{1}{|x|-3} \leq \frac{1}{2} \rightarrow \frac{1}{-x-3} \leq \frac{1}{2}$ Now, subtracting $\frac{1}{2}$ from both the sides $\frac{1}{-x-3}-\frac{1}{2} \leq \frac{1}{2}-\frac{1}{2}$ $\frac{2-(-x-3)}{2(-x-3)} \leq 0$ $\frac{x+5}{-2 x-6} \leq 0$ Signs of x + 5: $x+5=0 \rightarrow x=-5$ (Subtracting 5 from both the sides) $x+5>0 \rightarrow x>-5$ (Subtracting 5 from both the sides) $x+5<0 \rightarrow x<-5$ (Subtracting 5 from both the sides) Signs of -2x - 6: $-2 x-6=0 \rightarrow x=-3$ (Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2) $-2 x-6>0 \rightarrow x<-3$ (Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2) $-2 x-6<0 \rightarrow x>-3$ (Adding 6 on both the sides, then multiplying both the sides by -1 and then dividing both the sides by 2) Intervals satisfying the required condition: ≤ 0 $x<-5, x=-5, x>-3$ Or $x \leq-5$ or $x>-3$ Similarly, for $-3 $x \leq-5$ or $x>-3$ Merging overlapping intervals: $-3 For, $0 \leq x<3$ $\frac{1}{|x|-3} \leq \frac{1}{2} \rightarrow \frac{1}{x-3} \leq \frac{1}{2}$ Subtracting $\frac{1}{2}$ from both the sides $\frac{1}{x-3}-\frac{1}{2} \leq \frac{1}{2}-\frac{1}{2}$ $\frac{2-(x-3)}{2(x-3)} \leq 0$ $\frac{5-x}{2(x-3)} \leq 0$ Multiplying both the sides by 2 $\frac{2(5-x)}{2(x-3)} \leq 0(2)$ $\frac{(5-x)}{(x-3)} \leq 0$ Signs of 5 – x: $5-x=0 \rightarrow x=5$ (Subtracting 5 from both the sides and then dividing both sides by -1) $5-x>0 \rightarrow x<5$ (Subtracting 5 from both the sides and then multiplying both sides by -1) $5-x<0 \rightarrow x>5$ (Subtracting 5 from both the sides and then multiplying both sides by -1) Signs of x – 3 $x-3=0 \rightarrow x=3$ (Adding 3 to both the sides) $5-x>0 \rightarrow x>3$ (Adding 3 to both the sides) $5-x<0 \rightarrow x<3$ (Adding 3 to both the sides) Intervals satisfying the condition: x ≤ 0 x < 3 or x = 5 or x > 5 Or x <3 and x ≥ 5 Similarly, for 0 ≤ x < 3: x <3 and x ≥ 5 Merging overlapping intervals: 0 ≤ x < 3 Now, combining all the intervals satisfying condition: ≤ 0 x ≤ -5 or -3 < x < 0 or 0 ≤ x < 3 or x ≥ 5 Therefore $x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty)$
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