Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 184938 by mnjuly1970 last updated on 14/Jan/23

       If  ,   { ((  f  :  [ (√2) , +∞ ) → R                        )),((   f (x ) = x^( 2)   + ⌊ (( 1)/(1 − ⌊ x^( 2)  ⌋)) ⌋               )) :}                                     ⇒    ⌊  f^( −1)  ( π ) ⌋ = ?

$$ \\ $$$$\:\:\:\:\:\mathrm{If}\:\:,\:\:\begin{cases}{\:\:{f}\:\::\:\:\left[\:\sqrt{\mathrm{2}}\:,\:+\infty\:\right)\:\rightarrow\:\mathbb{R}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{\:\:\:{f}\:\left({x}\:\right)\:=\:{x}^{\:\mathrm{2}} \:\:+\:\lfloor\:\frac{\:\mathrm{1}}{\mathrm{1}\:−\:\lfloor\:{x}^{\:\mathrm{2}} \:\rfloor}\:\rfloor\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\lfloor\:\:{f}^{\:−\mathrm{1}} \:\left(\:\pi\:\right)\:\rfloor\:=\:? \\ $$$$ \\ $$$$ \\ $$

Answered by Frix last updated on 14/Jan/23

(√(π+1))

$$\sqrt{\pi+\mathrm{1}} \\ $$

Commented by mnjuly1970 last updated on 14/Jan/23

thanks alot...

$${thanks}\:{alot}... \\ $$

Commented by Frix last updated on 14/Jan/23

Sorry for the missing solution, it′s just the  result of my playing around with it

$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{the}\:\mathrm{missing}\:\mathrm{solution},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{the} \\ $$$$\mathrm{result}\:\mathrm{of}\:\mathrm{my}\:\mathrm{playing}\:\mathrm{around}\:\mathrm{with}\:\mathrm{it} \\ $$

Answered by mnjuly1970 last updated on 14/Jan/23

     my solution :      x≥(√2) ⇒ x^( 2)  ≥ 2 ⇒ ⌊x^2  ⌋ ≥ 2     ⇒   1−⌊ x^( 2)  ⌋ ≤ −1<0       ⇒ 0>  (1/(1−⌊x^( 2)  ⌋)) ≥−1         f (x)= x^( 2) −1          f^( −1) (x) =(√( x+1))          ⌊ f^( −1) (π )⌋= ⌊ (√(π+1)) ⌋= 2

$$\:\:\:\:\:{my}\:{solution}\:: \\ $$$$\:\:\:\:{x}\geqslant\sqrt{\mathrm{2}}\:\Rightarrow\:{x}^{\:\mathrm{2}} \:\geqslant\:\mathrm{2}\:\Rightarrow\:\lfloor{x}\:^{\mathrm{2}} \:\rfloor\:\geqslant\:\mathrm{2} \\ $$$$\:\:\:\Rightarrow\:\:\:\mathrm{1}−\lfloor\:{x}^{\:\mathrm{2}} \:\rfloor\:\leqslant\:−\mathrm{1}<\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{0}>\:\:\frac{\mathrm{1}}{\mathrm{1}−\lfloor{x}^{\:\mathrm{2}} \:\rfloor}\:\geqslant−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{f}\:\left({x}\right)=\:{x}^{\:\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{f}^{\:−\mathrm{1}} \left({x}\right)\:=\sqrt{\:{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\lfloor\:{f}^{\:−\mathrm{1}} \left(\pi\:\right)\rfloor=\:\lfloor\:\sqrt{\pi+\mathrm{1}}\:\rfloor=\:\mathrm{2} \\ $$$$\:\:\:\: \\ $$

Commented by aba last updated on 14/Jan/23

good

$$\mathrm{good} \\ $$

Commented by Mastermind last updated on 15/Jan/23

Is this a solution to my question  pls, could you please make it easier  than this?

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{my}\:\mathrm{question} \\ $$$$\mathrm{pls},\:\mathrm{could}\:\mathrm{you}\:\mathrm{please}\:\mathrm{make}\:\mathrm{it}\:\mathrm{easier} \\ $$$$\mathrm{than}\:\mathrm{this}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com