(A) Given the inequality $|x-2| \leq 1$. We know that the property $|x| \leq a$ is equivalent to $-a \leq x \leq a$. Applying this property to the giv

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(A) Given the inequality $|x-2| \leq 1$. We know that the property $|x| \leq a$ is equivalent to $-a \leq x \leq a$. Applying this property to the given inequality,we get $-1 \leq x-2 \leq 1$. Adding $2$ to all parts of the inequality,we get $-1 + 2 \leq x \leq 1 + 2$. This simplifies to $1 \leq x \leq 3$. Therefore,the solution set is $x \in [1, 3]$.