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```        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}.
```
8 years ago

147 Points
```										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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.  All the best Sanchit gupta.  Regards, Askiitians Experts Badiuddin

```
8 years ago
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