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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Functions

If S=\ a ∈ R:|2 a-1|=3[a]+2\ a\ \ , where [t] denotes the greatest...

If S={a∈R:∣2a−1∣=3[a]+2{a}}S=\{a \in \mathbf{R}:|2 a-1|=3[a]+2\{a\}\}, where [t][t] denotes the greatest integer less than or equal to tt and {t}\{t\} represents the fractional part of tt, then 72∑a∈Sa72 \sum_{a \in S} a is equal to ______

Correct answer

0

Step-by-step solution

∣2a−1∣=3[a]+2{a}∣2a−1∣=[a]+2a\begin{aligned} & |2 \mathrm{a}-1|=3[\mathrm{a}]+2\{\mathrm{a}\} \\ & |^{2} \mathrm{a}-1|=[\mathrm{a}]+2 \mathrm{a}\end{aligned} Case-1 :a>122a−1=[a]+2a[a]=−1∴a∈[−1,0) Reject  Case-2 :a<12−2a+1=[a]+2aa=I+f−2(I+f)+1=I+2I+2fI=0,f=14∴a=14 Hence a=1472∑a∈Sa=72×14=18\begin{aligned} & \text{ Case-1 }: \mathrm{a}>\frac{1}{2} \\ & 2 a-1=[a]+2 a \\ & {[a]=-1 \quad \therefore a \in[-1,0) \text{ Reject }} \\ & \text{ 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} \\ & \text{ Hence } \mathrm{a}=\frac{1}{4} \\ & 72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}=72 \times \frac{1}{4}=18\end{aligned}$
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