The solution set of |x-1| + |x+1|

Question

(C) We are given the inequality $|x-1| + |x+1| < 2$. Case $1$: If $x < -1$,then $|x-1| = -(x-1) = -x+1$ and $|x+1| = -(x+1) = -x-1$. The inequality be

Vedclass · Accepted Answer

$\emptyset$

Vedclass · Answer

(C) We are given the inequality $|x-1| + |x+1| Case $1$: If $x The inequality becomes $(-x+1) + (-x-1)  -1$. Since this contradicts our assumption $x Case $2$: If $-1 \le x The inequality becomes $(-x+1) + (x+1) This is a contradiction,so there is no solution in this interval. Case $3$: If $x \ge 1$,then $|x-1| = x-1$ and $|x+1| = x+1$. The inequality becomes $(x-1) + (x+1) Since this contradicts our assumption $x \ge 1$,there is no solution in this interval. Therefore,the solution set is the empty set,denoted by $\emptyset$.

The solution set of $|x-1| + |x+1| < 2$ is...

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