 Hi and welcome to the session. Let us discuss the following question. Question says, find the values of k so that the function f is continuous at indicated point. fx is equal to kx plus 1 if x is less than equal to 5. fx is equal to 3x minus 5 if x is greater than 5. And x is equal to 5. First of all let us understand that if function f is continuous at x is equal to a then limit of x tending to a minus fx is equal to limit of x tending to a plus fx is equal to fa. That is left hand side limit of the function is equal to right 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 are given fx is equal to kx plus 1 if x is less than equal to 5 and fx is equal to 3x minus 5. If x is greater than 5 we are also given that function is continuous at x is equal to 5. Now let us find out left hand side limit of the function at x is equal to 5 so we can write limit of x tending to 5 minus fx is equal to limit of x tending to 5 minus kx plus 1 this is equal to 5k plus 1 so we get left hand side limit of the function at x is equal to 5 equal to 5k plus 1. Now let us find right hand side limit of the function at x is equal to 5 so we can write limit of x tending to 5 plus fx is equal to limit of x tending to 5 plus 3x minus 5 this is equal to 3 multiplied by 5 minus 5 which is further equal to 15 minus 5 or we can say right hand side limit of the function at x is equal to 5 is equal to then we know given function is continuous at x is equal to 5 so left hand side limit must be equal to right hand side limit so we can write limit of x tending to 5 minus fx is equal to limit of x tending to 5 plus fx substituting the corresponding values of the limits we get 5k plus 1 is equal to 10 this implies 5k is equal to 10 minus 1 this further implies 5k is equal to 9 or we can say k is equal to 9 upon 5 squared value of k is equal to 9 upon 5 this completes the session hope you understood the session goodbye