If S = {a in R : |2a - 1| = 3[a] + 2{a}},wher | vedclass

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(A) Given the equation: $|2a - 1| = 3[a] + 2\{a\}$.Since $a = [a] + \{a\}$,we have $2\{a\} = 2a - 2[a]$.Substituting this into the equation: $|2a - 1|

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$18$

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(A) Given the equation: $|2a - 1| = 3[a] + 2\{a\}$.Since $a = [a] + \{a\}$,we have $2\{a\} = 2a - 2[a]$.Substituting this into the equation: $|2a - 1| = 3[a] + 2a - 2[a] = [a] + 2a$.Case $1$: $a \ge \frac{1}{2}$.Then $2a - 1 = [a] + 2a$,which implies $[a] = -1$.Since $[a] = -1$,$a \in [-1, 0)$. However,this contradicts the condition $a \ge \frac{1}{2}$. Thus,no solution exists in this case.Case $2$: $a Then $-(2a - 1) = [a] + 2a$,which simplifies to $1 - 2a = [a] + 2a$,or $4a = 1 - [a]$.Let $a = I + f$,where $I = [a]$ and $f = \{a\} \in [0, 1)$.Then $4(I + f) = 1 - I$,so $5I + 4f = 1$.Since $0 \le f Thus,$0 \le 1 - 5I If $I = 0$,then $4f = 1 \implies f = \frac{1}{4}$. Thus $a = 0 + \frac{1}{4} = \frac{1}{4}$.If $I = -1$,then $5(-1) + 4f = 1 \implies 4f = 6 \implies f = 1.5$,which is not possible as $f Therefore,the only solution is $a = \frac{1}{4}$.Finally,$72 \sum_{a \in S} a = 72 	imes \frac{1}{4} = 18$.

If $S = \{a \in R : |2a - 1| = 3[a] + 2\{a\}\}$,where $[t]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$,then $72 \sum_{a \in S} a$ is equal to:

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