 Hello and welcome to this session. Let us understand the following problem today. Check the injectivity and sensitivity of the function given to us as a function f from r to r given by fx is equal to x square. Now let us write the solution. Now first let us check for 1, 1. We observe that for x1, x2 belongs to r, f of x1 is equal to f of x2 which implies x1 square is equal to x2 square which implies x1 is equal to plus minus x2. But x1 is not equal to minus x2. For this to understand let us consider 1 and minus 1 here f of 1 is equal to f of minus 1 but 1 is not equal to minus 1. Therefore x1 is not equal to minus of x2. Hence f is not 1, 1. Now let us check for on 2. Since f of x assumes only non-negative value therefore non-negative real number in co-domain has its pre-image not 1, 1 and f is not on 2 therefore function is neither injective nor surjective. Hope you understood the problem. Bye and have a nice day.