 Hi and welcome to the session. My name is Shashi and I am going to help you with the following question. Question is, find the values of a so that the function f is continuous at the indicated point. fx is equal to kx square if x is less than equal to 2, fx is equal to 3 if x is greater than 2 and x is equal to 2. First of all let us understand that if function f is continuous and x is equal to a then limit of x tending to a plus fx is equal to limit of x tending to a minus fx is equal to f a. Or we can say right hand side limit of the function is equal to fth hand side limit of the function is equal to value 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 have given fx is equal to kx square if x is less than equal to 2 and fx is equal to 3 if x is greater than 2. We are given that given function f is continuous at x is equal to 2. First of all let us find out left hand side limit of the function at x is equal to 2. So we can write limit of x tending to 2 minus fx is equal to limit of x tending to 2 minus kx square which is equal to 4. So we get 2 minus fx is equal to 4 and find out right hand side limit of the function at x is equal to 2. So we can write limit of x tending to 2 plus fx is equal to limit of x tending to 2 plus 3 is equal to 3. So we get right hand side limit of the function at x is equal to 2 is equal to 3. Now given function is continuous so left hand side limit must be equal to right hand side limit limit of x tending to 2 minus fx is equal to limit of x tending to 2 plus fx. Now we know left hand side limit is equal to 4k and right hand side limit is equal to 3. So we get 4k is equal to 3. Now this implies k is equal to 3 upon 4. So required value of k is equal to 3 upon 4. This completes the session. Hope you understood the session. Good bye.