Let f(x) = sec ^-1 ( [ 1 + sin ² x ] ) ; where [.] denotes the greatest integeral function. Then show that the set of points where f(x) is not continous is { (2n-1) Π /2 , n belonging to integers}.

Dear Sanchit

we know that f(x) = sec ^-1 ( [ 1 + sin ² x ] ) exists when its domain

[ 1 + sin ² x ]>=1     or [ 1 + sin ² x ] <=-1

case 1 :

[ 1 + sin ² x ] <=-1    no solution exists

case 2:

[ 1 + sin ² x ]>=1

[1 + sin ² x =1] =1 for all value of x except for those value of x where sin ² x =1

ie odd multiple of Π /2

so       x=(2n-1) Π /2 wher n is integer are the points where function is discontinous