If-a-function-f-R-R-satisfies-the-relation-f-x-1-f-x-1-3-f-x-for-all-x-R-then-a-period-of-f-is-A-10-B-12-C-6-D-4-

Question Number 119635 by Ar Brandon last updated on 25/Oct/20

$$\mathrm{If}\:\mathrm{a}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:{f}\left(\mathrm{x}+\mathrm{1}\right)+{f}\left(\mathrm{x}−\mathrm{1}\right)=\sqrt{\mathrm{3}}{f}\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Answered by Olaf last updated on 26/Oct/20

$${f}\left({x}+\mathrm{1}\right)+{f}\left({x}−\mathrm{1}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}+\mathrm{1}\right) \\ $$$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}\right)\:=\:\sqrt{\mathrm{3}}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right] \\ $$$${f}\left({x}+\mathrm{2}\right)\:=\:\mathrm{2}{f}\left({x}\right)−\sqrt{\mathrm{3}}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\mathrm{2}{f}\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\mathrm{2}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right]−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}\right)−\mathrm{2}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}+\mathrm{1}\right)−\mathrm{2}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:\sqrt{\mathrm{3}}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right]−\mathrm{2}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:{f}\left({x}\right)−\sqrt{\mathrm{3}}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:{f}\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:{f}\left({x}+\mathrm{1}\right)−\left[{f}\left({x}+\mathrm{1}\right)+{f}\left({x}−\mathrm{1}\right)\right] \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:−{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{6}\right)\:=\:−{f}\left({x}\right) \\ $$$$\Rightarrow\:{f}\left({x}+\mathrm{12}\right)\:=\:−{f}\left({x}+\mathrm{6}\right) \\ $$$${f}\left({x}+\mathrm{12}\right)\:=\:−\left[−{f}\left({x}\right)\right] \\ $$$${f}\left({x}+\mathrm{12}\right)\:=\:{f}\left({x}\right) \\ $$$$ \\ $$$$\mathrm{Period}\:\mathrm{is}\:\mathrm{12}. \\ $$

Commented by Ar Brandon last updated on 26/Oct/20

Thank you Sir

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