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Question Number 12403 by sin (x) last updated on 21/Apr/17
f(x)=⌊x−(3/e)⌋+⌊x+(3/e)⌋⇒f(1)=?
$${f}\left({x}\right)=\lfloor{x}−\frac{\mathrm{3}}{{e}}\rfloor+\lfloor{x}+\frac{\mathrm{3}}{{e}}\rfloor\Rightarrow{f}\left(\mathrm{1}\right)=? \\ $$
Answered by FilupS last updated on 21/Apr/17
⌊x⌋ = floor(x)  f(1)=⌊1−(3/e)⌋+⌊1+(3/e)⌋  2e>3>e  ⇒   2>(3/e)>1  ∴−1<1−(3/e)<0   ⇒   ⌊1−(3/e)⌋=−1  ∴2<1+(3/e)<3   ⇒   ⌊1+(3/e)⌋=2     ∴f(1)=−1+2  f(1)=1
$$\lfloor{x}\rfloor\:=\:\mathrm{floor}\left({x}\right) \\ $$$${f}\left(\mathrm{1}\right)=\lfloor\mathrm{1}−\frac{\mathrm{3}}{{e}}\rfloor+\lfloor\mathrm{1}+\frac{\mathrm{3}}{{e}}\rfloor \\ $$$$\mathrm{2}{e}>\mathrm{3}>{e}\:\:\Rightarrow\:\:\:\mathrm{2}>\frac{\mathrm{3}}{{e}}>\mathrm{1} \\ $$$$\therefore−\mathrm{1}<\mathrm{1}−\frac{\mathrm{3}}{{e}}<\mathrm{0}\:\:\:\Rightarrow\:\:\:\lfloor\mathrm{1}−\frac{\mathrm{3}}{{e}}\rfloor=−\mathrm{1} \\ $$$$\therefore\mathrm{2}<\mathrm{1}+\frac{\mathrm{3}}{{e}}<\mathrm{3}\:\:\:\Rightarrow\:\:\:\lfloor\mathrm{1}+\frac{\mathrm{3}}{{e}}\rfloor=\mathrm{2} \\ $$$$\: \\ $$$$\therefore{f}\left(\mathrm{1}\right)=−\mathrm{1}+\mathrm{2} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$

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