Question Number 114764 by dw last updated on 21/Sep/20
$${Find}\:{period}\:{of}\:{f}\left({x}\right)={e}^{{cos}^{\mathrm{4}} \left(\pi{x}\right)+{x}−\lfloor{x}\rfloor+{cos}^{\mathrm{2}} \left(\pi{x}\right)} \\ $$$$ \\ $$
Answered by PRITHWISH SEN 2 last updated on 22/Sep/20
$$\mathrm{period}\:\mathrm{of}\:\mathrm{cos}\:^{\mathrm{4}} \left(\pi\mathrm{x}\right)=\:\mathrm{1} \\ $$$$\mathrm{period}\:\mathrm{of}\:\mathrm{cos}\:^{\mathrm{2}} \left(\pi\mathrm{x}\right)=\:\mathrm{1} \\ $$$$\mathrm{period}\:\mathrm{of}\:\mathrm{x}−\lfloor\mathrm{x}\rfloor\:=\:\mathrm{1} \\ $$$$\mathrm{period}\:\mathrm{of}\:\:\mathrm{cos}\:^{\mathrm{4}} \pi\mathrm{x}\:+\mathrm{x}\:−\lfloor\mathrm{x}\rfloor+\mathrm{cos}\:^{\mathrm{2}} \pi\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{L}.\mathrm{C}.\mathrm{M}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)=\mathrm{1} \\ $$$$\therefore\:\mathrm{period}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1} \\ $$
Commented by dw last updated on 23/Sep/20
$${Thank}'{s}\:{Sir} \\ $$
Commented by PRITHWISH SEN 2 last updated on 22/Sep/20
$$\mathrm{yes}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$