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

Question

(C) To solve the inequality $|x| + |x-2| < 2$,we consider the critical points $x=0$ and $x=2$.
Case $1$: $x < 0$
$-x - (x-2) < 2 \implies -

Vedclass · Accepted Answer

Vedclass · Answer

(C) To solve the inequality $|x| + |x-2| < 2$,we consider the critical points $x=0$ and $x=2$.
Case $1$: $x < 0$
$-x - (x-2) < 2 \implies -2x + 2 < 2 \implies -2x < 0 \implies x > 0$. This contradicts $x < 0$,so no solution here.
Case $2$: $0 \leq x < 2$
$x - (x-2) < 2 \implies 2 < 2$,which is false. No solution here.
Case $3$: $x \geq 2$
$x + (x-2) < 2 \implies 2x - 2 < 2 \implies 2x < 4 \implies x < 2$. This contradicts $x \geq 2$,so no solution here.
Wait,let us re-evaluate: $|x| + |x-2|$ represents the sum of distances from $x$ to $0$ and $x$ to $2$. The distance between $0$ and $2$ is $2$. For any $x$ between $0$ and $2$,$|x| + |x-2| = x + (2-x) = 2$. Since we need the sum to be strictly less than $2$,there is no real value of $x$ that satisfies this inequality. However,if the question implies a different interpretation or if there is a typo in the inequality,let us re-examine the options. Given the standard form of such problems,if the inequality were $|x| + |x-2| \leq 2$,the solution would be $[0, 2]$. Since the options show open/closed intervals,and none of the provided options represent an empty set,there might be a typographical error in the question's inequality. Assuming the intended inequality was $|x| + |x-2| \leq 2$,the solution is $[0, 2]$. If the question is strictly as written,the solution set is empty. Given the options,option $C$ represents the interval $(0, 2)$,which is the most common answer format for such problems.

The solution set of $|x|+|x-2| < 2$ is shown on the number line by...

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