Find-the-range-of-i-f-x-x-x-ii-f-x-x-x-x-

Question Number 205297 by mnjuly1970 last updated on 15/Mar/24

$$ \\ $$$$\:\:\:{Find}\:\:{the}''\:{range}\:''\:{of}\:\:: \\ $$$$ \\ $$$$\:\:\:{i}\::\:\:\:{f}\:\left({x}\right)\:=\lfloor\:\frac{\:{x}}{\:\lfloor\:{x}\:\rfloor}\:\rfloor \\ $$$$\:\:\:{ii}:\:{f}\left({x}\right)\:=\:\frac{\:{x}}{\lfloor\:{x}\:\rfloor\:+\:\lfloor\:−{x}\:\rfloor} \\ $$$$\:\: \\ $$

Answered by A5T last updated on 15/Mar/24

(i)f is undefined when ⌊x⌋=0 (0≤x<1) Otherwise x=⌊x⌋+{x},where 0≤{x}<1 ⇒f(x)=⌊1+(({x})/(⌊x⌋))⌋, when x≥1,{x}<⌊x⌋,⇒f(x)=1 when x<0 f(x)=⌊1−(({x})/(∣⌊x⌋∣))⌋; {x}<∣⌊x⌋∣ ⇒0<1−(({x})/(∣⌊x⌋∣))≤1⇒ f(x)_(x<0) = 0 or 1 ⇒Range: {0,1}

$$\left({i}\right){f}\:{is}\:{undefined}\:{when}\:\lfloor{x}\rfloor=\mathrm{0}\:\left(\mathrm{0}\leqslant{x}<\mathrm{1}\right) \\ $$$${Otherwise}\:{x}=\lfloor{x}\rfloor+\left\{{x}\right\},{where}\:\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)=\lfloor\mathrm{1}+\frac{\left\{{x}\right\}}{\lfloor{x}\rfloor}\rfloor,\:{when}\:{x}\geqslant\mathrm{1},\left\{{x}\right\}<\lfloor{x}\rfloor,\Rightarrow{f}\left({x}\right)=\mathrm{1} \\ $$$${when}\:{x}<\mathrm{0}\:{f}\left({x}\right)=\lfloor\mathrm{1}−\frac{\left\{{x}\right\}}{\mid\lfloor{x}\rfloor\mid}\rfloor;\:\left\{{x}\right\}<\mid\lfloor{x}\rfloor\mid \\ $$$$\Rightarrow\mathrm{0}<\mathrm{1}−\frac{\left\{{x}\right\}}{\mid\lfloor{x}\rfloor\mid}\leqslant\mathrm{1}\Rightarrow\:{f}\left({x}\right)_{{x}<\mathrm{0}} =\:\mathrm{0}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{Range}:\:\left\{\mathrm{0},\mathrm{1}\right\} \\ $$

Commented by mnjuly1970 last updated on 15/Mar/24

$${thanks}\:{a}\:{lot}\:{ali} \\ $$

Commented by Frix last updated on 15/Mar/24

Yes. Just to clarify: f(x)= { ((1; x∈Z^− )),((0; x∈R^− \Z)),((undefined; 0≤x<1)),((1; x≥1)) :}

$$\mathrm{Yes}.\:\mathrm{Just}\:\mathrm{to}\:\mathrm{clarify}: \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{1};\:{x}\in\mathbb{Z}^{−} }\\{\mathrm{0};\:{x}\in\mathbb{R}^{−} \backslash\mathbb{Z}}\\{\mathrm{undefined};\:\mathrm{0}\leqslant{x}<\mathrm{1}}\\{\mathrm{1};\:{x}\geqslant\mathrm{1}}\end{cases} \\ $$

Answered by A5T last updated on 15/Mar/24

(ii) when x∈Z,f(x) is undefined when x>0, ⌊−x⌋=−⌊x⌋−1⇒f(x)=−x when x<0, ⌊−x⌋=−⌊x⌋−1⇒f(x)=−x ⇒f(x)∈R\Z

$$\left({ii}\right)\:{when}\:{x}\in\mathbb{Z},{f}\left({x}\right)\:{is}\:{undefined} \\ $$$${when}\:{x}>\mathrm{0},\:\lfloor−{x}\rfloor=−\lfloor{x}\rfloor−\mathrm{1}\Rightarrow{f}\left({x}\right)=−{x} \\ $$$${when}\:{x}<\mathrm{0},\:\lfloor−{x}\rfloor=−\lfloor{x}\rfloor−\mathrm{1}\Rightarrow{f}\left({x}\right)=−{x} \\ $$$$\Rightarrow{f}\left({x}\right)\in\mathbb{R}\backslash\mathbb{Z} \\ $$

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