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Question Number 212518 by efronzo1 last updated on 16/Oct/24

$$\cancel{\underbrace{\downharpoonleft }\underset{―}{}}$$

Answered by golsendro last updated on 16/Oct/24

$$\cancel{\underset{―}{\underbrace{\pm }}}={(\mathrm{x}-\mathrm{a})}^2+\mathrm{a}$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}-\mathrm{a}}+\mathrm{a}$$$$\iff {(\mathrm{x}-\mathrm{a})}^2=\sqrt{\mathrm{x}-\mathrm{a}}$$$$\sqrt{\mathrm{x}-\mathrm{a}}(\sqrt{{(\mathrm{x}-\mathrm{a})}^3}-1)=0$$$$\mathrm{x}=\mathrm{a}\mathrm{or}\mathrm{x}=\mathrm{a}+1$$$$\mathrm{so}\mathrm{if}\mathrm{x}=5049=\mathrm{a}\mathrm{then}\mathrm{the}\mathrm{other}\mathrm{solution}$$$$$$

Answered by mr W last updated on 16/Oct/24

$$x^2-2ax+a(a+1)=x$$$$x^2-(2a+1)x+a(a+1)=0$$$$saytheothersolutionis\beta .$$$$5049+\beta =2a+1$$$$5049\beta =a(a+1)$$$$4\times 5049\beta ={(5049+\beta )}^2-1$$$${(\beta -5049)}^2-1=0$$$$\Rightarrow \beta =5049\pm 1$$$$i.e.theothersolutionis5048or5050.$$$$witha=5048or5049.$$

Commented by mr W last updated on 16/Oct/24

Commented by mr W last updated on 16/Oct/24

Commented by golsendro last updated on 16/Oct/24

$$\mathrm{f}:[\mathrm{a},\mathrm{\infty })\to [\mathrm{a},\mathrm{\infty })\mathrm{sir}$$$$\mathrm{it}\mathrm{follow}\mathrm{that}\beta =5059-1\mathrm{not}$$$$\mathrm{solution}$$

Commented by mr W last updated on 16/Oct/24

$$witha=5048therearetwosolutions:$$$$x=5048,5049$$$$witha=5049therearetwosolutions:$$$$x=5049,5050$$$$whatiswrong?$$

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