for X belongs to [0,pi] sin x lies in [0,1]

[sinx] is discontinuous at pi/2 , so not differentiable at it

and now considering [n + sin x] where sin x is just shifted by n units along Y axes , but still the values lie in ( n , n+1) and here too the curve will be discontinues/not differentiable at 1 point i.e., at pi/2

now taking p( a prime no.) into consideration,

for p=1 , we have [sinx] with 1 discontinues/not differentiable points

for p=2 , we have [2.sinx] with 3 discontinues/not differentiable points

for p=3 , we have [3.sinx] with 5 discontinues/not differentiable points

.

...

for p = p ,it follows to (2p-1) discontinues/not differentiable points

option is D

[n +  p sin x] = n + [p sin x].

The points of discontinuity and hence of non-differentiability are the points where p sin x is an integer. There are no other points of non-differentiability.

So, the points of discontinuity are when sin x = 1/p or 2/p,..., (p-1)/p each of which have two corresponding values of x and sin x = 1, which has a unique solution in the given interval.

That makes 2(p-1)+1 = 2p-1 solutions