The period of the function 3sin2⁡πx+x−[x]+sin4⁡πx, where [⋅] denotes the greatest integer function is 503kthen k=

sin2⁡πx=1−cos⁡2πx2Since cos 2πx is a periodic function with period 2π2π=1, therefore sin2πx is periodic with period 1.    (1)x – [x] is a periodic function with period 1.    (2)sin4⁡πx=sin2⁡πx2=14(1−cos⁡2πx)2 =141+cos2⁡2πx−2cos⁡2πx=18(3+cos⁡4πx−4cos⁡2πx)Since, cos 4πx is a periodic function with period 2π4π= 1/2and cos 2πx is a periodic function with period2π2π=1, therefore, period of sin4πx is equal to L.C.M. 1,12= L.C.M. (1,1) H.C.F. (1,2)=11=1From Eq. (1), (2) and (3), we getPeriod of 3sin2⁡πx+x−[x]+sin4⁡πx=1