 Hi and welcome to the session. Let us discuss the following question. Question says, find all points of discontinuity of f, where f is defined by fx is equal to x raised to the power 10 minus 1. If x is less than equal to 1, fx is equal to x square if x is greater than 1. First of all, let us understand that function f is discontinuous at point x is equal to a. The function is not equal to right hand side limit of the function at x is equal to a. This is the key idea to solve the given question. Let us now start the solution. We are given fx is equal to x raised to the power 10 minus 1 if x is less than equal to 1, fx is equal to x square if x is greater than 1. We are given fx is equal to x raised to the power 10 minus 1. If x is less than 1, now this is a polynomial function and we know polynomial function is continuous at every real number. So, this again implies that function f is continuous at all the real numbers less than 1. Now, we are given fx is equal to x square if x is greater than 1. Now, this is again a polynomial function and we know polynomial function is continuous at every real number. So, this again implies that function f is continuous at all the real numbers greater than 1. Now, let us take the continuity of the function at x is equal to 1. Clearly, we can see that the function is defined at x is equal to 1. So, we can write x is equal to 1, function f is defined. Now, let us find out left hand side limit of the function at x is equal to 1. So, we can write limit of x tending to 1 minus fx is equal to limit of x tending to 1 minus x raised to the power 10 minus 1. Now, this is equal to 1 raised to the power 10 minus 1 which is further equal to 1 minus 1. But we can say it is equal to 0 and therefore we get limit of x tending to 1 minus fx equal to 0. Let us now find out right hand side limit of the function at x is equal to 1. So, we can write limit of x tending to 1 plus fx is equal to limit of x tending to 1 plus x y. Now, this is further equal to minus 1 which is equal to 1. So, we get limit of x tending to 1 plus fx is equal to 1. Clearly, we can see left hand side limit and right hand side limit do not coincide each other. So, we can write limit of x tending to 1 minus fx is not equal to limit of x tending to 1 plus fx. This implies function at x is continuous at x is equal to 1. So, this is our required answer. This completes the session. Hope you understood the session. Goodbye.