Find the sum of all integral values of a such | vedclass

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Question

(C) Let $f(x) = ||x - 2| - |3 - x||$. By the triangle inequality property of absolute values,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$

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$3$

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(C) Let $f(x) = ||x - 2| - |3 - x||$. By the triangle inequality property of absolute values,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$. More simply,the expression $||x - 2| - |3 - x||$ represents the absolute difference between the distances of a point $x$ from $2$ and $3$. For $x For $2 \leq x \leq 3$,$f(x) = |(x - 2) - (3 - x)| = |2x - 5|$. As $x$ goes from $2$ to $3$,$2x - 5$ goes from $-1$ to $1$,so $|2x - 5|$ goes from $1$ to $0$ and back to $1$. For $x > 3$,$f(x) = |(x - 2) - (x - 3)| = |1| = 1$. Thus,the range of $f(x)$ is $[0, 1]$. For the equation $f(x) = 2 - a$ to have a solution,we must have $0 \leq 2 - a \leq 1$. Solving for $a$: $-2 \leq -a \leq -1$,which implies $1 \leq a \leq 2$. The integral values of $a$ are $1$ and $2$. The sum of these values is $1 + 2 = 3$.

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

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