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Question Number 87536 by M±th+et£s last updated on 04/Apr/20

solve   ∣2x−1∣=3⌊x⌋+2{x}

$${solve}\: \\ $$$$\mid\mathrm{2}{x}−\mathrm{1}\mid=\mathrm{3}\lfloor{x}\rfloor+\mathrm{2}\left\{{x}\right\} \\ $$$$ \\ $$

Answered by mr W last updated on 05/Apr/20

x=n+δ  ⌊x⌋=n  {x}=δ  2x−1=2n+2δ−1    if 2x−1≥0, i.e. x≥(1/2):  2n+2δ−1=3n+2δ  ⇒2n−1=3n ⇒n=−1  ⇒δ=any value 0≤δ<0  ⇒−1≤x<0  ⇒no solution    if 2x−1<0, i.e. x<(1/2):  −2n−2δ+1=3n+2δ  5n+4δ=1  4δ must be intger.  δ=0: 5n=1 ⇒no solution  δ=(1/4): 5n=0 ⇒n=0 ⇒x=(1/4) <(1/2)⇒solution  δ=(1/2): 5n=−1 ⇒no solution   δ=(3/4): 5n=−2 ⇒no solution    ⇒only solution is x=(1/4).

$${x}={n}+\delta \\ $$$$\lfloor{x}\rfloor={n} \\ $$$$\left\{{x}\right\}=\delta \\ $$$$\mathrm{2}{x}−\mathrm{1}=\mathrm{2}{n}+\mathrm{2}\delta−\mathrm{1} \\ $$$$ \\ $$$${if}\:\mathrm{2}{x}−\mathrm{1}\geqslant\mathrm{0},\:{i}.{e}.\:{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\mathrm{2}{n}+\mathrm{2}\delta−\mathrm{1}=\mathrm{3}{n}+\mathrm{2}\delta \\ $$$$\Rightarrow\mathrm{2}{n}−\mathrm{1}=\mathrm{3}{n}\:\Rightarrow{n}=−\mathrm{1} \\ $$$$\Rightarrow\delta={any}\:{value}\:\mathrm{0}\leqslant\delta<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{x}<\mathrm{0}\:\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${if}\:\mathrm{2}{x}−\mathrm{1}<\mathrm{0},\:{i}.{e}.\:{x}<\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$−\mathrm{2}{n}−\mathrm{2}\delta+\mathrm{1}=\mathrm{3}{n}+\mathrm{2}\delta \\ $$$$\mathrm{5}{n}+\mathrm{4}\delta=\mathrm{1} \\ $$$$\mathrm{4}\delta\:{must}\:{be}\:{intger}. \\ $$$$\delta=\mathrm{0}:\:\mathrm{5}{n}=\mathrm{1}\:\Rightarrow{no}\:{solution} \\ $$$$\delta=\frac{\mathrm{1}}{\mathrm{4}}:\:\mathrm{5}{n}=\mathrm{0}\:\Rightarrow{n}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}}\:<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{solution} \\ $$$$\delta=\frac{\mathrm{1}}{\mathrm{2}}:\:\mathrm{5}{n}=−\mathrm{1}\:\Rightarrow{no}\:{solution}\: \\ $$$$\delta=\frac{\mathrm{3}}{\mathrm{4}}:\:\mathrm{5}{n}=−\mathrm{2}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$$\Rightarrow{only}\:{solution}\:{is}\:{x}=\frac{\mathrm{1}}{\mathrm{4}}. \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

thank you sir nice solution

$${thank}\:{you}\:{sir}\:{nice}\:{solution} \\ $$

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