question_answer

Assertion (A): The function  is continuous everywhere.

Reason (R): f(x) is periodic function.

 

A) Both A and R are individually true and R is the correct explanation of A.

B) Both A and R are individually true and R is not the correct explanation of A.

C) 'A' is true but 'R' is false

D) 'A' is false but 'R' is true

E) Both A and R are false.

Correct Answer: C

Solution :

L.H. Lt = \[\underset{x\to {{0}^{-}}}{\mathop{Lt}}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{Lt}}\,\sin x=0\] R.H. Lt \[\therefore \,\,\,L.H.\,Lt=R.H.\,\,Lt=0\] \[\Rightarrow \,\,f\left( x \right)\]is continuous at x = 0 \[\therefore \]Assertion (A) is true Also the graph of   is Clearly given function is not periodic for every x \[\therefore \]Reason (R) is not true Have option [C] is the correct option.