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(x+1)+f(x−1) = (√3)f(x) f(x+2)+f(x) = (√3)f(x+1) f(x+2)+f(x) = (√3)[(√3)f(x)−f(x−1)] f(x+2) = 2f(x)−(√3)f(x−1) f(x+3) = 2f(x+1)−(√3)f(x) f(x+3) = 2[(√3)f(x)−f(x−1)]−(√3)f(x) f(x+3) = (√3)f(x)−2f(x−1) f(x+4) = (√3)f(x+1)−2f(x) f(x+4) = (√3)[(√3)f(x)−f(x−1)]−2f(x) f(x+4) = f(x)−(√3)f(x−1) f(x+5) = f(x+1)−(√3)f(x) f(x+5) = f(x+1)−[f(x+1)+f(x−1)] f(x+5) = −f(x−1) f(x+6) = −f(x) ⇒ f(x+12) = −f(x+6) f(x+12) = −[−f(x)] f(x+12) = f(x) Period is 12.](Q119660.png)
$${f}\left({x}+\mathrm{1}\right)+{f}\left({x}−\mathrm{1}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}+\mathrm{1}\right) \\ $$$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}\right)\:=\:\sqrt{\mathrm{3}}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right] \\ $$$${f}\left({x}+\mathrm{2}\right)\:=\:\mathrm{2}{f}\left({x}\right)−\sqrt{\mathrm{3}}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\mathrm{2}{f}\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\mathrm{2}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right]−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{3}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}\right)−\mathrm{2}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:\sqrt{\mathrm{3}}{f}\left({x}+\mathrm{1}\right)−\mathrm{2}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:\sqrt{\mathrm{3}}\left[\sqrt{\mathrm{3}}{f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)\right]−\mathrm{2}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{4}\right)\:=\:{f}\left({x}\right)−\sqrt{\mathrm{3}}{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:{f}\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{3}}{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:{f}\left({x}+\mathrm{1}\right)−\left[{f}\left({x}+\mathrm{1}\right)+{f}\left({x}−\mathrm{1}\right)\right] \\ $$$${f}\left({x}+\mathrm{5}\right)\:=\:−{f}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$${f}\left({x}+\mathrm{6}\right)\:=\:−{f}\left({x}\right) \\ $$$$\Rightarrow\:{f}\left({x}+\mathrm{12}\right)\:=\:−{f}\left({x}+\mathrm{6}\right) \\ $$$${f}\left({x}+\mathrm{12}\right)\:=\:−\left[−{f}\left({x}\right)\right] \\ $$$${f}\left({x}+\mathrm{12}\right)\:=\:{f}\left({x}\right) \\ $$$$ \\ $$$$\mathrm{Period}\:\mathrm{is}\:\mathrm{12}. \\ $$
Commented by Ar Brandon last updated on 26/Oct/20
Thank you Sir