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Question Number 97206 by O Predador last updated on 07/Jun/20

Answered by john santu last updated on 07/Jun/20

$$\mathrm{let}\sqrt{\mathrm{x}+3}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{t}}^2-3$$$$\mathrm{P}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^2-6}{{\mathrm{t}}^2-3}.\Rightarrow \mathrm{P}\left(\mathrm{Q}\left(\mathrm{x}\right)\right)=\frac{\mathrm{Q}{\left(\mathrm{x}\right)}^2-6}{\mathrm{Q}{\left(\mathrm{x}\right)}^2-3}$$$$\iff \frac{\mathrm{Q}{\left(\mathrm{x}\right)}^2-6}{\mathrm{Q}{\left(\mathrm{x}\right)}^2-3}=\sqrt{5\mathrm{x}+11}$$$$\mathrm{Q}{\left(1\right)}^2-6=(\mathrm{Q}{\left(1\right)}^2-3)\left(4\right)$$$$\mathrm{Q}{\left(1\right)}^2-6=4\mathrm{Q}{\left(1\right)}^2-12$$$$3\mathrm{Q}{\left(1\right)}^2=6\Rightarrow \mathrm{Q}\left(1\right)=\pm \sqrt{2}$$

Commented by O Predador last updated on 07/Jun/20

$$\mathbf{Thank}\mathbf{you}!$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{poq}\left(\mathrm{x}\right)=\sqrt{5\mathrm{x}+11}\Rightarrow \mathrm{q}\left(\mathrm{x}\right)={\mathrm{p}}^{-1}\left(\sqrt{5\mathrm{x}+11}\right)\Rightarrow \mathrm{q}\left(1\right)={\mathrm{p}}^{-1}\left(4\right)$$$$\mathrm{let}\sqrt{\mathrm{x}+3}=\mathrm{t}\Rightarrow \mathrm{x}+3={\mathrm{t}}^2\Rightarrow \mathrm{x}={\mathrm{t}}^2-3\Rightarrow \mathrm{p}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^2-6}{{\mathrm{t}}^2-3}$$$$\mathrm{p}\left(\mathrm{t}\right)=\mathrm{y}\Rightarrow \frac{{\mathrm{t}}^2-6}{{\mathrm{t}}^2-3}=\mathrm{y}\Rightarrow {\mathrm{t}}^2-6={\mathrm{yt}}^2-3\mathrm{y}\Rightarrow (1-\mathrm{y}){\mathrm{t}}^2=6-3\mathrm{y}\Rightarrow$$$${\mathrm{t}}^2=\frac{6-3\mathrm{y}}{1-\mathrm{y}}\Rightarrow \mathrm{t}=\stackrel{-}{+}\sqrt{\frac{6-3\mathrm{y}}{1-\mathrm{y}}}=\stackrel{-}{+}\sqrt{\frac{3\mathrm{y}-6}{\mathrm{y}-1}}={\mathrm{p}}^{-1}\left(\mathrm{y}\right)\Rightarrow {\mathrm{p}}^{-1}\left(4\right)=\stackrel{-}{+}\sqrt{\frac{12-6}{4-1}}$$$$=\stackrel{-}{+}\sqrt{\frac{6}{3}}=\stackrel{-}{+}\sqrt{2}\mathrm{so}\mathrm{the}\mathrm{answer}\mathrm{is}\left(\mathrm{d}\right)$$

Commented by bobhans last updated on 07/Jun/20

$$\mathrm{typo}.\mathrm{C}$$

Commented by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{yes}.$$

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