If `f(x)=[x] sin ((pi)/([x+1]))`, where [.] denotes the greatest integer function, then the set of point of discontiuity of f in its domain isA. ZB. `Z-{-1,0}`C. `R-[-1,0)`D. none of these

If `f(x)=[x] sin ((pi)/([x+1]))`, where [.] denotes the greatest integ

1.	If `f(x)=[x] sin ((pi)/([x+1]))`, where [.] denotes the greatest integer function, then the set of point of discontiuity of f in its domain isA. ZB. `Z-{-1,0}`C. `R-[-1,0)`D. none of these
Answer» Correct Answer - B Clearly , f(x) is defined for all `x in R -[-1,0)`. For any integer `k ne -1`, we have `f(x)=k sin (pi)/(k+1), k le x lt k +1` So, f(x) being a constant function is continuous at all points other than integers in its domain. Also, `underset(x to k)lim f(x)=underset(h to 0)lim f(k-h)` `Rightarrow underset(x to k^(-))lim f(x)=underset(h to 0)lim [k-h] "sin" (pi)/([k+h+1])=k sin ((pi)/(k+1)), k ne -1` `therefore underset(x to k^(-))lim f(x) ne underset(x to k^(+))(lim f(x), k ne 0` Thus, f(x) is a discontinous at all non-zero integer points in its domain. When k=0, we have `f(x)=0 sin pi=0"for all " x in [0,1)` `Rightarrow underset(x to 0^(+))lim f(x)=f(0)` So, f(x) is right continous at x=0 Hence, the set of points of discontinuity in its domain is `Z-(-1,0)`