The left hand derivative of f(x)=[x]sin⁡(πx)  at  x=k,k an integer, is

The left hand limit L at x=k is  given  by                                     L=limh→0k−hsinπk−h−h                                          =limh→0k−1sinπkcosπh−cosπksinπhas k−h  when  k is an is an integer, is equal to (k−1).Now,   sinπk=0  and  cosπk=−1k∴                              L=limh→0(k−1)0−−1ksinπh−h                                     = (k−1).π.Hence (1) is the correct answer.