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Question Number 208149 by mnjuly1970 last updated on 06/Jun/24

Answered by A5T last updated on 06/Jun/24

⌊2x^2 ⌋>2x^2 −1⇒x−⌊2x^2 ⌋<1−2x^2 +x Suppose D_f ,R_f ⊆R 1−2x^2 +x<0⇒x>1 or x<−(1/2) If R_f ⊆R, then f(x) is undefined for x<((−1)/2)∪x>1 when x=((−1)/2),1;f(x) is undefined in R ((−1)/2)<x<0⇒∣x∣<(1/2)⇒0<x^2 <(1/4)⇒0<2x^2 <(1/2) ⇒⌊2x^2 ⌋=0⇒f(x)=(√(x−0))=f(x)=(√x) which is undefined for ((−1)/2)<x<0 when x=0;f(x)=0 when 0<x<(1/( (√2))); ⌊2x^2 ⌋=0⇒f(x)=(√x)∈(0,((8)^(1/4) /2)) when x=(1/( (√2)));f(x) is undefined in R when (1/( (√2)))<x<1, ⌊2x^2 ⌋=1⇒f(x)=(√(x−1)) which is undefined in R since (1/( (√2)))<x<1. So,for D_f ,R_f ⊆R, R_f =[0,((8)^(1/4) /2)),D_f =[0,(1/( (√2))))

$$\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor>\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\Rightarrow{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor<\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +{x} \\ $$$${Suppose}\:{D}_{{f}} ,{R}_{{f}} \subseteq\mathbb{R} \\ $$$$\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +{x}<\mathrm{0}\Rightarrow{x}>\mathrm{1}\:{or}\:{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${If}\:{R}_{{f}} \subseteq\mathbb{R},\:{then}\:{f}\left({x}\right)\:{is}\:{undefined}\:{for}\:{x}<\frac{−\mathrm{1}}{\mathrm{2}}\cup{x}>\mathrm{1} \\ $$$${when}\:{x}=\frac{−\mathrm{1}}{\mathrm{2}},\mathrm{1};{f}\left({x}\right)\:{is}\:{undefined}\:{in}\:\mathbb{R} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}<{x}<\mathrm{0}\Rightarrow\mid{x}\mid<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{0}<{x}^{\mathrm{2}} <\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\mathrm{0}<\mathrm{2}{x}^{\mathrm{2}} <\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor=\mathrm{0}\Rightarrow{f}\left({x}\right)=\sqrt{{x}−\mathrm{0}}={f}\left({x}\right)=\sqrt{{x}} \\ $$$${which}\:{is}\:{undefined}\:{for}\:\frac{−\mathrm{1}}{\mathrm{2}}<{x}<\mathrm{0}\:\: \\ $$$${when}\:{x}=\mathrm{0};{f}\left({x}\right)=\mathrm{0} \\ $$$${when}\:\mathrm{0}<{x}<\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}};\:\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor=\mathrm{0}\Rightarrow{f}\left({x}\right)=\sqrt{{x}}\in\left(\mathrm{0},\frac{\sqrt[{\mathrm{4}}]{\mathrm{8}}}{\mathrm{2}}\right) \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}};{f}\left({x}\right)\:{is}\:{undefined}\:{in}\:\mathbb{R} \\ $$$${when}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}<{x}<\mathrm{1},\:\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor=\mathrm{1}\Rightarrow{f}\left({x}\right)=\sqrt{{x}−\mathrm{1}} \\ $$$${which}\:{is}\:{undefined}\:{in}\:\mathbb{R}\:{since}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}<{x}<\mathrm{1}. \\ $$$${So},{for}\:{D}_{{f}} ,{R}_{{f}} \subseteq\mathbb{R},\:{R}_{{f}} =\left[\mathrm{0},\frac{\sqrt[{\mathrm{4}}]{\mathrm{8}}}{\mathrm{2}}\right),{D}_{{f}} =\left[\mathrm{0},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$

Commented by mnjuly1970 last updated on 06/Jun/24

$$\: \\ $$

Commented by A5T last updated on 06/Jun/24

Do you have the answer? Is it correct?

$${Do}\:{you}\:{have}\:{the}\:{answer}?\:{Is}\:{it}\:{correct}? \\ $$

Commented by mnjuly1970 last updated on 07/Jun/24

yes sir A5T

$${yes}\:{sir}\:{A}\mathrm{5}{T} \\ $$

Answered by MM42 last updated on 06/Jun/24

D=[0 , (1/( (√2)))) & R=[0 , (1/( (2)^(1/4) )))

$${D}=\left[\mathrm{0}\:,\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\:\:\&\:\:{R}=\left[\mathrm{0}\:,\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by MM42 last updated on 06/Jun/24

if x<0⇒x−⌊2x^2 ⌋<0⇒∉ D_f if (√((n/2) ))≤x<(√((n+1)/2)) ⇒x<n ; ∀n∈N ⇒n≤2x^2 <n+1⇒⌊2x^2 ⌋=n ⇒x−⌊2x^2 ⌋<0⇒∉ D_f f(0)=0 & lim_(x→(1/( (√2)))^ ) f(x)=(1/( (2)^(1/4) )) ⇒R_f =[0,(1/( (2)^(1/4) )))

$${if}\:\:{x}<\mathrm{0}\Rightarrow{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor<\mathrm{0}\Rightarrow\notin\:{D}_{{f}} \\ $$$${if}\:\:\:\sqrt{\frac{{n}}{\mathrm{2}}\:}\leqslant{x}<\sqrt{\frac{{n}+\mathrm{1}}{\mathrm{2}}}\:\Rightarrow{x}<{n}\:\:\:;\:\forall{n}\in\mathbb{N} \\ $$$$\Rightarrow{n}\leqslant\mathrm{2}{x}^{\mathrm{2}} <{n}+\mathrm{1}\Rightarrow\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor={n} \\ $$$$\Rightarrow{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor<\mathrm{0}\Rightarrow\notin\:{D}_{{f}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\:\&\:{lim}_{{x}\rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:^{} } {f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}} \\ $$$$\Rightarrow{R}_{{f}} =\left[\mathrm{0},\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\right) \\ $$

Answered by mnjuly1970 last updated on 07/Jun/24

x−⌊2x^2 ⌋≥0 ⇒ x ≥ ⌊ 2x^2 ⌋≥0 (★) −−−−−− ⌊ 2x^2 ⌋=k ⇒^(k∈Z^(≥0) ) k ≤ 2x^2 < k+1 from (★): x ≥ k ⇒^(x≥0) 2x^2 ≥ 2k^2 (★★) −−−−−− from(★),(★★)⇒^(★∩★★={}) : 2k^2 ≥ k+1 k ≥ 1 (■) complement of(■) is our purpose (★★★) k =0 ⇒ 0 ≤ 2x^2 < 1 ⇒ { ((D_f = 0≤x <(1/( (√2))))),(( ⌊2x^2 ⌋=0 ⇒ f(x)=(√x))) :} ⇒ { (( 0≤x <(1/( (√2))))),(( f(x)=(√x) ⇒ R_f = [0 , (1/( (2)^(1/4) )) ))) :}

$$\:\: \\ $$$$\:\:\:\:\:\:\:\:{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor\geqslant\mathrm{0}\:\Rightarrow\:{x}\:\geqslant\:\lfloor\:\mathrm{2}{x}^{\mathrm{2}} \rfloor\geqslant\mathrm{0}\:\left(\bigstar\right) \\ $$$$\:\:\:\:\:\:\:−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\lfloor\:\mathrm{2}{x}^{\mathrm{2}} \rfloor={k}\:\overset{{k}\in\mathbb{Z}^{\geqslant\mathrm{0}} } {\Rightarrow}{k}\:\leqslant\:\mathrm{2}{x}^{\mathrm{2}} <\:{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:{from}\:\left(\bigstar\right):\:\:{x}\:\geqslant\:{k}\:\overset{{x}\geqslant\mathrm{0}} {\Rightarrow}\mathrm{2}{x}^{\mathrm{2}} \geqslant\:\mathrm{2}{k}^{\mathrm{2}} \left(\bigstar\bigstar\right) \\ $$$$\:\:\:\:\:\:−−−−−− \\ $$$$\:\:\:\:{from}\left(\bigstar\right),\left(\bigstar\bigstar\right)\overset{\bigstar\cap\bigstar\bigstar=\left\{\right\}} {\Rightarrow}\::\:\:\:\:\mathrm{2}{k}^{\mathrm{2}} \:\geqslant\:{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:{k}\:\geqslant\:\mathrm{1}\:\:\:\left(\blacksquare\right)\: \\ $$$$\:\:\:\:\:{complement}\:{of}\left(\blacksquare\right)\:{is}\:{our}\:{purpose}\:\:\left(\bigstar\bigstar\bigstar\right) \\ $$$$\:\:\:\:\:\:\:{k}\:=\mathrm{0}\:\Rightarrow\:\:\mathrm{0}\:\leqslant\:\mathrm{2}{x}^{\mathrm{2}} \:<\:\mathrm{1}\:\Rightarrow\:\begin{cases}{{D}_{{f}} =\:\mathrm{0}\leqslant{x}\:<\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{\:\:\:\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor=\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)=\sqrt{{x}}}\end{cases}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:\begin{cases}{\:\:\mathrm{0}\leqslant{x}\:<\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{\:{f}\left({x}\right)=\sqrt{{x}}\:\Rightarrow\:{R}_{{f}} \:=\:\left[\mathrm{0}\:,\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\:\right)}\end{cases} \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$

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