Question-208149

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

Answered by A5T last updated on 06/Jun/24

⌊ 2 x^{2} ⌋ > 2 x^{2} - 1 \Rightarrow x - ⌊ 2 x^{2} ⌋ < 1 - 2 x^{2} + x

S u p p o s e D_{f}, R_{f} \subseteq R

1 - 2 x^{2} + x < 0 \Rightarrow x > 1 o r x < - \frac{1}{2}

I f R_{f} \subseteq R, t h e n f (x) i s u n d e f i n e d f o r x < \frac{- 1}{2} \cup x > 1

w h e n x = \frac{- 1}{2}, 1; f (x) i s u n d e f i n e d i n R

\frac{- 1}{2} < x < 0 \Rightarrow∣ x ∣< \frac{1}{2} \Rightarrow 0 < x^{2} < \frac{1}{4} \Rightarrow 0 < 2 x^{2} < \frac{1}{2}

\Rightarrow ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x - 0} = f (x) = \sqrt{x}

w h i c h i s u n d e f i n e d f o r \frac{- 1}{2} < x < 0

w h e n x = 0; f (x) = 0

w h e n 0 < x < \frac{1}{\sqrt{2}}; ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x} \in (0, \frac{\sqrt[4]{8}}{2})

w h e n x = \frac{1}{\sqrt{2}}; f (x) i s u n d e f i n e d i n R

w h e n \frac{1}{\sqrt{2}} < x < 1, ⌊ 2 x^{2} ⌋ = 1 \Rightarrow f (x) = \sqrt{x - 1}

w h i c h i s u n d e f i n e d i n R s i n c e \frac{1}{\sqrt{2}} < x < 1 .

S o, f o r D_{f}, R_{f} \subseteq R, R_{f} = [0, \frac{\sqrt[4]{8}}{2}), D_{f} = [0, \frac{1}{\sqrt{2}})

Commented by mnjuly1970 last updated on 06/Jun/24

Commented by A5T last updated on 06/Jun/24

D o y o u h a v e t h e a n s w e r ? I s i t c o r r e c t ?

Commented by mnjuly1970 last updated on 07/Jun/24

y e s s i r A 5 T

Answered by MM42 last updated on 06/Jun/24

D = [0, \frac{1}{\sqrt{2}}) & R = [0, \frac{1}{\sqrt[4]{2}})

Commented by MM42 last updated on 06/Jun/24

i f x < 0 \Rightarrow x - ⌊ 2 x^{2} ⌋ < 0 \Rightarrow\notin D_{f}

i f \sqrt{\frac{n}{2}} ⩽ x < \sqrt{\frac{n + 1}{2}} \Rightarrow x < n; \forall n \in N

\Rightarrow n ⩽ 2 x^{2} < n + 1 \Rightarrow ⌊ 2 x^{2} ⌋ = n

\Rightarrow x - ⌊ 2 x^{2} ⌋ < 0 \Rightarrow\notin D_{f}

f (0) = 0 & {l i m}_{x \to \frac{1}{\sqrt{2}}^{}} f (x) = \frac{1}{\sqrt[4]{2}}

\Rightarrow R_{f} = [0, \frac{1}{\sqrt[4]{2}})

Answered by mnjuly1970 last updated on 07/Jun/24

x - ⌊ 2 x^{2} ⌋ ⩾ 0 \Rightarrow x ⩾ ⌊ 2 x^{2} ⌋ ⩾ 0 (★)

- - - - - -

⌊ 2 x^{2} ⌋ = k \overset{k \in Z^{⩾ 0}}{\Rightarrow} k ⩽ 2 x^{2} < k + 1

f r o m (★) : x ⩾ k \overset{x ⩾ 0}{\Rightarrow} 2 x^{2} ⩾ 2 k^{2} (★ ★)

- - - - - -

f r o m (★), (★ ★) \overset{★ \cap ★ ★ = {}}{\Rightarrow} : 2 k^{2} ⩾ k + 1

k ⩾ 1 ()

c o m p l e m e n t o f () i s o u r p u r p o s e (★ ★ ★)

k = 0 \Rightarrow 0 ⩽ 2 x^{2} < 1 \Rightarrow {\begin{cases} D_{f} = 0 ⩽ x < \frac{1}{\sqrt{2}} \\ ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x} \end{cases}

\Rightarrow {\begin{cases} 0 ⩽ x < \frac{1}{\sqrt{2}} \\ f (x) = \sqrt{x} \Rightarrow R_{f} = [0, \frac{1}{\sqrt[4]{2}}) \end{cases}