Explanation

(a)
Given, $\frac{|x - 1|}{x + 2} - 1 < 0$
Case I When $x < 1,\ |x - 1| = 1 - x$
$\therefore\ \frac{1 - x}{x + 2} - 1 < 0\ \Rightarrow \ \frac{- 2x - 1}{x + 2} < 0$
$\Rightarrow \ \ \frac{2x + 1}{x + 2} > 0\ \Rightarrow x < - 2$ or $x > - \frac{1}{2}$
But $x < 1$
$\therefore x \in ( - \infty,\ - 2) \cup \left( - \frac{1}{2},\ 1 \right)$
Case II When $x \geq 1,\ |x - 1| = x - 1$
$\therefore\ \frac{x - 1}{x + 2} - 1 < 0\ \Rightarrow \ - \frac{3}{x + 2} < 0$
$\Rightarrow \frac{3}{x + 2} > 0$
$\Rightarrow \ \ x > - 2$
But $x \geq 1$
$\therefore\ \ x \geq 1,\ ie,\ x \in \lbrack 1,\ \infty)$ ...(iii)
$\therefore$ From Eqs. (i) and (ii), we get
$x \in ( - \infty,\ - 2) \cup \left( - \frac{1}{2},\ \infty \right)$