The period of the function $3^{\left({\sin }^2\mathrm{\pi x}+\mathrm{x}-\left[\mathrm{x}\right]+{\sin }^4\mathrm{\pi x}\right)},$ where [⋅] denotes the greatest integer function is $\frac{503}{k}$then k=

${\text{sin}}^2\mathrm{\pi x}=\frac{1-\cos 2\mathrm{\pi x}}{2}$
Since cos 2$\mathrm{\pi }$x is a periodic function with period $\frac{2\mathrm{\pi }}{2\mathrm{\pi }}=1,$
therefore ${\sin }^2\pi x$is periodic with period 1.    (1)
x – [x] is a periodic function with period 1.    (2)
$$\begin{gathered} {\sin }^4\mathrm{\pi x}={\left({\sin }^2\mathrm{\pi x}\right)}^2=\frac{1}{4}(1-\cos 2\mathrm{\pi x}{)}^2 \\ \begin{array}{r}=\frac{1}{4}\left(1+{\cos }^{2}2\mathrm{\pi x}-2\cos 2\mathrm{\pi x}\right)\\ =\frac{1}{8}(3+\cos 4\mathrm{\pi x}-4\cos 2\mathrm{\pi x})\end{array} \end{gathered}$$
Since, cos 4$\mathrm{\pi }$x is a periodic function with period $\frac{2\mathrm{\pi }}{4\mathrm{\pi }}$
= 1/2and cos 2$\mathrm{\pi }$x is a periodic function with period
$\frac{2\mathrm{\pi }}{2\mathrm{\pi }}=1,$ therefore, period of sin⁴$\mathrm{\pi }$x is equal to
$\text{ L.C.M. }\left(1,\frac{1}{2}\right)=\frac{\text{ L.C.M. }(1,1)}{\text{ H.C.F. }(1,2)}=\frac{1}{1}=1$
From Eq. (1), (2) and (3), we get
Period of $3^{\left({\sin }^2\mathrm{\pi x}+\mathrm{x}-\left[\mathrm{x}\right]+{\sin }^4\mathrm{\pi x}\right)}=1$