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Question-165550




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

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