We know-that [x] is continuous on R ~ I where I denotes, the set of integers and sin (pi/[x + 1]) is discontinuous for [x + 1] = 0
Thus. the function is defined in the interval
Therefore the points where f can be possibly discontinuous are ..... -3, -.2, -1, 0, 1, 2...... But for
Therefore, f(x) = 0 for 0 ≤ x < 1. Also, f(x) is nor defined on - 1 ≤ x < 0 but f(x) is continuous at x = 0 because RHL at 0 and f(0) is defined and equal. However, LHL does nor exist.
Therefore, set of points of discontinuties of f is I - {0}.