Question Number 165550 by mnjuly1970 last updated on 03/Feb/22
Answered by mahdipoor last updated on 03/Feb/22
$${f}\left(\mathrm{1}\right)=\sqrt{\mathrm{11}}\:,\:{f}\left({f}\left(\mathrm{1}\right)\right)=\sqrt{\mathrm{22}}\:,\:{f}\left({f}\left({f}\left(\mathrm{1}\right)\right)\right)=\sqrt{\mathrm{33}} \\ $$$$…\:{f}\left({f}…{f}\left(\mathrm{1}\right)…\right)=\sqrt{\mathrm{10}{n}+\mathrm{1}} \\ $$$$\Rightarrow{f}'\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{10}}} \\ $$$${f}'\left(\mathrm{1}\right)×{f}\left(\sqrt{\mathrm{11}}\right)×{f}\left(\sqrt{\mathrm{22}}\right)×…×{f}\left(\sqrt{\mathrm{10}{n}+\mathrm{1}}\right)= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}×\frac{\sqrt{\mathrm{11}}}{\:\sqrt{\mathrm{22}}}×\frac{\sqrt{\mathrm{22}}}{\:\sqrt{\mathrm{33}}}×…×\frac{\sqrt{\mathrm{10}{n}+\mathrm{1}}}{\:\sqrt{\mathrm{10}{n}+\mathrm{11}}}= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}{n}+\mathrm{11}}}=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow{n}^{\mathrm{2}} −\mathrm{8}{n}−\mathrm{10}=\mathrm{0}\:\Rightarrow \\ $$$${n}=\frac{\mathrm{8}\pm\sqrt{\mathrm{104}}}{\mathrm{2}}\:\notin\mathrm{N}\:\Rightarrow{withuot}\:{answer}\: \\ $$$${but}\:{if}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}{n}+\mathrm{11}}}=\frac{\mathrm{1}}{{n}}\:\Rightarrow{n}^{\mathrm{2}} −\mathrm{10}{n}−\mathrm{11}=\mathrm{0}\: \\ $$$$\Rightarrow{n}=\mathrm{11}\:{or}\:−\mathrm{1}\:\Rightarrow\:{n}=\mathrm{11} \\ $$