 Hi and welcome to the session, I am Shashi and I am going to help you with the following question. Question says, find all points of discontinuity of function f where fx is equal to sin x upon x if x is less than 0, fx is equal to x plus 1 if x is greater than equal to 0. First of all let us understand that function f is discontinuous at x equal to a if left hand side limit of the function is not equal to right hand side limit of the function at x equal to a. Then left hand side limit of the function do not coincide right hand side limit of the function at x equal to a then function is discontinuous at x equal to a. This is the key idea to solve the given question. Now let us start the solution. We are given function f given by fx equal to sin x upon x if x is less than 0, fx is equal to x plus 1 if x is greater than equal to 0. Now clearly we can see given function is defined at every real number. Now let us discuss continuity of the function at every point less than 0. We are given fx is equal to sin x upon x if x is less than 0, right function discontinuous at every real number and this is a polynomial function it is also continuous at every real number. Now we also know that quotient obtained by dividing two continuous functions is also continuous. So we can write function f is continuous at every real number less than 0. Now let us discuss continuity of the function at all the points greater than 0. We are given function f given by fx equal to x plus 1 if x is greater than 0. Now this is a polynomial function and we know polynomial function is continuous at every real number. So even function f is continuous at every real number greater than 0. Now let us discuss continuity of the function at x equal to 0, at x equal to 0, limit of x tending to 0 minus fx. First of all we will find out left hand side limit of the function this is equal to limit of x tending to 0 minus sin x upon x this is equal to 1, 0, limit of theta tending to 0 sin theta upon theta this is equal to 1, here we have used this limit. Now we get left hand side limit of the function at x equal to 0 as 1, now let us find out right hand side limit of the function at x equal to 0 this is equal to limit of x tending to 0 plus x plus 1, now this is equal to 0 plus 1 which is equal to 1 only, so we get right hand side limit as 1, here we can see left hand side limit of the function is equal to right hand side limit of the function at x equal to 0, let us now find out value of the function at x equal to 0, f0 is equal to 0 plus 1 which is equal to 1 only, now we get limit of x tending to 0 minus fx is equal to limit of x tending to 0 plus fx is equal to f0 is equal to 1, this implies given function f is continuous at x equal to 0, since left hand side limit is equal to right hand side limit is equal to value of the function at x equal to 0, so given function f is continuous at x equal to 0, now we know function f is continuous at all x greater than 0, all x less than 0 and x equal to 0, this implies function f is continuous at every remember, so there is no point of discontinuity this is our required answer, this completes the session, I hope you understood the session, take care and have a nice day.