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If-for-nonzero-x-2f-x-2-3f-1-x-2-x-2-1-then-f-x-2-

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Question Number 96222 by bobhans last updated on 30/May/20
If for nonzero x ; 2f (x^2 )+3f ((1/x^2 )) = x^2 −1 then f (x^2 ) = ?
$$\mathrm{If}\:\mathrm{for}\:\mathrm{nonzero}\:{x}\:;\:\mathrm{2}{f}\:\left({x}^{\mathrm{2}} \right)+\mathrm{3}{f}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:=\:{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${then}\:{f}\:\left({x}^{\mathrm{2}} \right)\:=\:? \\ $$
Commented by bobhans last updated on 31/May/20
we have 2 f(x^2 ) +3 f((1/x^2 )) = x^2 −1 ⇒4 f(x^2 ) + 6 f((1/x^2 )) = 2x^2 −2 ...(i) also 2 f((1/x^2 )) + 3 f(x^2 ) = (1/x^2 )−1...(ii) 3×(ii)−(i) ⇒ 5 f(x^2 ) = ((3−3x^2 )/x^2 )−(2x^2 −2) f(x^2 ) = ((3−3x^2 −2x^4 +2x^2 )/(5x^2 )) = ((3−x^2 −2x^4 )/(5x^2 ))
$$\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:{f}\left({x}^{\mathrm{2}} \right)\:+\mathrm{3}\:{f}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:=\:{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}\:{f}\left({x}^{\mathrm{2}} \right)\:+\:\mathrm{6}\:{f}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\:\:…\left({i}\right) \\ $$$${also}\:\mathrm{2}\:{f}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:+\:\mathrm{3}\:{f}\left({x}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}…\left({ii}\right) \\ $$$$\mathrm{3}×\left({ii}\right)−\left({i}\right)\:\Rightarrow\:\mathrm{5}\:{f}\left({x}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\right) \\ $$$${f}\left({x}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{5}{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{3}−{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{5}{x}^{\mathrm{2}} }\: \\ $$
Commented by PRITHWISH SEN 2 last updated on 31/May/20
put x=(1/x) 2f((1/x^2 ))+3f(x^2 )=((1−x^2 )/x^2 ) .......(ii) (i)×2 −(ii)×3 given equ. is (i) f(x^2 )= ((3−x^2 −2x^4 )/(5x^2 ))
$$\mathrm{put}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{3f}\left(\mathrm{x}^{\mathrm{2}} \right)=\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:…….\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)×\mathrm{2}\:−\left(\boldsymbol{\mathrm{ii}}\right)×\mathrm{3}\:\:\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{equ}}.\:\boldsymbol{\mathrm{is}}\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)=\:\frac{\mathrm{3}−\mathrm{x}^{\mathrm{2}} −\mathrm{2x}^{\mathrm{4}} }{\mathrm{5x}^{\mathrm{2}} } \\ $$
Answered by mathmax by abdo last updated on 01/Jun/20
we have 2f(x^2 )+3f((1/x^2 )) =x^2 −1 ⇒2 f((1/x^2 ))+3f(x^2 ) =(1/x^2 )−1 (change x by (1/x))⇒ we get the system { ((2f(x^2 )+3f((1/x^2 )) =x^2 −1)),((3f(x^2 )+2f((1/x^2 )) =((1−x^2 )/x^2 ))) :} Δ_s = determinant (((2 3)),((3 2)))=4−9 =−5 Δ_(f(x^2 )) = determinant (((x^2 −1 3)),((((1−x^2 )/x^2 ) 2))) =2x^2 −2 −((3(1−x^2 ))/x^2 ) =((2x^4 −2x^2 −3+3x^2 )/x^2 )=((2x^4 +x^2 −3)/x^2 ) ⇒f(x^2 ) =((2x^4 +x^3 −3)/(−5x^2 )) =((−2x^4 −x^3 +3)/(5x^2 ))
$$\mathrm{we}\:\mathrm{have}\:\mathrm{2f}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{3f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\:=\mathrm{x}^{\mathrm{2}} \:−\mathrm{1}\:\Rightarrow\mathrm{2}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{3f}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\:\left(\mathrm{change}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}}\right)\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{\mathrm{2f}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{3f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\:=\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\\{\mathrm{3f}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\:=\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }}\end{cases} \\ $$$$\Delta_{\mathrm{s}} =\begin{vmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\mathrm{4}−\mathrm{9}\:=−\mathrm{5} \\ $$$$\Delta_{\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)} \:=\begin{vmatrix}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}\:=\mathrm{2x}^{\mathrm{2}} −\mathrm{2}\:−\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }\:=\frac{\mathrm{2x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{3}+\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\frac{\mathrm{2x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{3}} −\mathrm{3}}{−\mathrm{5x}^{\mathrm{2}} }\:=\frac{−\mathrm{2x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} \:+\mathrm{3}}{\mathrm{5x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 01/Jun/20
error of typo f(x^2 ) =((2x^4 +x^2 −3)/(−5x^2 )) =((−2x^4 −x^2 +3)/(5x^2 ))
$$\mathrm{error}\:\mathrm{of}\:\mathrm{typo}\:\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\frac{\mathrm{2x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} −\mathrm{3}}{−\mathrm{5x}^{\mathrm{2}} }\:=\frac{−\mathrm{2x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{5x}^{\mathrm{2}} } \\ $$

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