Question

If [x] denotes the integral part of x and f(x)=[n+psin x], 0<x<π , n belongs to integers and p is a prime number, then the number of points , where f(x) is not differentiable is

a) p-1

b) p

c) 2p-1

d)2p+1

Answer 1

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)
correct answer is option C