Explanation

(d)
$\left| |x| - 1 \right| < |1 - x|$
Case I $x \geq 0$
$\therefore$ Inequality (i) becomes $|x - 1| < x - 1\$or$\ |1 - x| < 1 - x$ which is not satisfied by any $x,\$because
$|a| \geq \forall a \in R$
Case II$\ - 1 \leq x < 0$
$\therefore$ Inequality (i) becomes $| - 1 - x| < 1 - x\$or$\ |x + 1| < 1 - x$
Or $x + 1 < 1 - x\$or$\ x < 0$
Thus, inequality (i) is satisfied for $- 1 \leq x < 0$
Case III $x < - 1$
Inequality (i) becomes $| - 1 - x| < 1 - x \Rightarrow |1 + x| < 1 - x$
$\Rightarrow - (1 + x) < 1 - x \Rightarrow - 2 < 0,\$which is true
So, solution set is $( - \infty,0)$