The solution set of |x-1|+|x-2|

Question

(A) To solve the inequality $|x-1|+|x-2| < 3$,we consider the critical points $x=1$ and $x=2$.
Case $1$: If $x < 1$,then $-(x-1) - (x-2) < 3

Vedclass · Accepted Answer

$(0,3)$

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(A) To solve the inequality $|x-1|+|x-2| < 3$,we consider the critical points $x=1$ and $x=2$.
Case $1$: If $x < 1$,then $-(x-1) - (x-2) < 3 \implies -x+1-x+2 < 3 \implies -2x+3 < 3 \implies -2x < 0 \implies x > 0$. So,$0 < x < 1$.
Case $2$: If $1 \le x < 2$,then $(x-1) - (x-2) < 3 \implies x-1-x+2 < 3 \implies 1 < 3$,which is always true for $1 \le x < 2$.
Case $3$: If $x \ge 2$,then $(x-1) + (x-2) < 3 \implies 2x-3 < 3 \implies 2x < 6 \implies x < 3$. So,$2 \le x < 3$.
Combining all cases,the solution set is $(0, 1) \cup [1, 2) \cup [2, 3) = (0, 3)$.

The solution set of $|x-1|+|x-2| < 3$ is

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