 Hello and welcome to the session. Let us discuss the following problem tree. In the following case state whether the function is 1, 1 onto a bijective. Justify the answer. We have function f from r to r defined by fx is equal to 1 plus x square. Now let us write the solution. Given to us is function f from r to r defined by fx is equal to 1 plus x square. Now let us check for one one. Let x, y be two arbitrary elements in r. Then fx is equal to fy which implies 1 plus x square is equal to 1 plus y square. This 1, 1 gets cancelled. So we are left with x square is equal to y square which implies x is equal to y and x is equal to minus y. x is equal to minus y is not possible. Therefore f is not one one. Now let us check for onto. Now f of x is equal to 1 plus x square which is greater than equal to 1 for all x belongs to r. So negative real numbers in r of co-domain do not have their edges in r of domain. Therefore f is not onto. Hence we can see that f is not onto and f is not one one. Therefore f is not one one and onto. I hope you understood this problem. Bye and have a nice day.