For | x - 1 | x + 2 1 , x lies in the interval

Question

For $\frac{| x - 1 |}{x + 2} < 1$ , $x$ lies in the interval

Smart Achievers · Accepted Answer

(1)

$Case I:$ $| x - 1 |$ $= x - 1,$ $x \geq 1$

$∴$ $\frac{| x - 1 |}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} - 1 < 0 \Rightarrow \frac{- 3}{x + 2} < 0$

$\Rightarrow$ $\frac{3}{x + 2} > 0$ $which is possible when x > - 2$

$But it is true for x \geq 1, so common solution is x \geq 1 or x \in [1, \infty)$ ...(i)

$Case II:$ $| x - 1 |$ $= - x + 1,$ $x < 1$

$∴$ $\frac{| x - 1 |}{x + 2} < 1 \Rightarrow \frac{- x + 1}{x + 2} < 1 \Rightarrow \frac{x - 1}{x + 2} + 1 > 0 \Rightarrow \frac{2 x + 1}{x + 2} > 0$

$\Rightarrow$ $x < - 2 or x > - \frac{1}{2}$

$But x < 1, so x \in (- \infty, - 2) \cup (- \frac{1}{2}, 1)$ ...(ii)

$∴$ $From (i) and (ii), x lies in the interval$

$(- \infty, - 2) \cup (- \frac{1}{2}, 1) \cup [1, \infty)$ $or x \in (- \infty, - 2) \cup (- \frac{1}{2}, \infty)$

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