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 $1$: $x < 1$.
The inequality becomes $-(x-

Vedclass · Accepted Answer

$(0, 3)$

Vedclass · Answer

(A) To solve the inequality $|x-1|+|x-2| < 3$,we consider the critical points $x=1$ and $x=2$.
Case $1$: $x < 1$.
The inequality becomes $-(x-1) - (x-2) < 3$,which simplifies to $-x+1-x+2 < 3$,or $-2x+3 < 3$. This gives $-2x < 0$,so $x > 0$. Combining with $x < 1$,we get $0 < x < 1$.
Case $2$: $1 \le x < 2$.
The inequality becomes $(x-1) - (x-2) < 3$,which simplifies to $x-1-x+2 < 3$,or $1 < 3$. This is always true for the interval $1 \le x < 2$.
Case $3$: $x \ge 2$.
The inequality becomes $(x-1) + (x-2) < 3$,which simplifies to $2x-3 < 3$,or $2x < 6$,so $x < 3$. Combining with $x \ge 2$,we get $2 \le x < 3$.
Combining all cases,the solution set is $(0, 1) \cup [1, 2) \cup [2, 3)$,which simplifies to $(0, 3)$.

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

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