The correct option is
C 2p−1f(x)=[n+psinx]=n+[psinx] ( n is an integer )
[x] is not differentiable at those points where x is an integer.
So, f(x) is not differentiable at those points where value of psinx is an integer, on the interval (0,π)
That is,
sinx=1p,2p,3p............p−1p,pp
sinx=1p,2p,3p............p−1p,1
From the above values, sinx attains the values 1p,2p,3p............p−1p twice on the interval (0,π)
But sinx=1 at only one point, x=π2
So, total no. of points =2(p−1)+1=2p−1