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Question Number 173240 by mnjuly1970 last updated on 08/Jul/22

Commented by a.lgnaoui last updated on 09/Jul/22

Commented by a.lgnaoui last updated on 09/Jul/22

$$m+n=\frac{i}{2}-1and-\frac{i}{2}-i$$

Answered by mr W last updated on 09/Jul/22

$$f^{-1}\left(x\right)=\frac{x}{\sqrt{1-2x}}$$$$f^{-1}(mx+\frac{1}{2})=\frac{mx+\frac{1}{2}}{\sqrt{-2mx}}$$$$f^{-1}(nx+\frac{1}{2})=\frac{nx+\frac{1}{2}}{\sqrt{-2nx}}$$$$-\frac{-mx-\frac{1}{2}}{\sqrt{-2mx}}+\frac{-nx-\frac{1}{2}}{\sqrt{-2nx}}=\sqrt{x}+\frac{1}{\sqrt{x}}$$$$-\sqrt{-mx}+\frac{1}{2\sqrt{-mx}}+\sqrt{-nx}-\frac{1}{2\sqrt{-nx}}=(\sqrt{x}+\frac{1}{\sqrt{x}})\sqrt{2}$$$$letM=\sqrt{-m},N=\sqrt{-n}$$$$N-M=\sqrt{2}$$$$\frac{1}{M}-\frac{1}{N}=2\sqrt{2}$$$$\frac{\sqrt{2}}{MN}=2\sqrt{2}$$$$MN=\frac{1}{2}$$$$\Rightarrow M=1-\frac{\sqrt{2}}{2},N=1+\frac{\sqrt{2}}{2}$$$$\Rightarrow m=-M^2=-(\frac{3}{2}-\sqrt{2})$$$$\Rightarrow n=-N^2=-(\frac{3}{2}+\sqrt{2})$$$$\Rightarrow n+m=-3✓$$

Commented by mnjuly1970 last updated on 09/Jul/22

$$\mathrm{thanks}\mathrm{alot}\mathrm{sir}\mathrm{W}$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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