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If $\mathrm{S}=\{\mathrm{a} \in \mathrm{R}:|2 \mathrm{a}-1|=3[\mathrm{a}]+2\{\mathrm{a}\}\}$, where $[\mathrm{t}]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$, then $72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}$ is equal to....................
$18$
$16$
$13$
$75$
Solution
$ |2 \mathrm{a}-1|=3[\mathrm{a}]+2\{\mathrm{a}\} $
$ |2 \mathrm{a}-1|=[\mathrm{a}]+2 \mathrm{a}$
Case $-1$ : $\mathrm{a}>\frac{1}{2} $
$ 2 \mathrm{a}-1=[\mathrm{a}]+2 \mathrm{a} $
$ {[\mathrm{a}]=-1 \quad \therefore \mathrm{a} \in[-1,0) \text { Reject }} $
Case-$2$: $\mathrm{a}<\frac{1}{2} $
$ -2 \mathrm{a}+1=[\mathrm{a}]+2 \mathrm{a} $
$ \mathrm{a}=\mathrm{I}+\mathrm{f} $
$ -2(\mathrm{I}+\mathrm{f})+1=\mathrm{I}+2 \mathrm{I}+2 \mathrm{f} $
$ \mathrm{I}=0, \mathrm{f}=\frac{1}{4} \quad \therefore \mathrm{a}=\frac{1}{4} $
Hence $ \mathrm{a}=\frac{1}{4} $
$ 72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}=72 \times \frac{1}{4}=18$
