 Hello and welcome to the session. Let us discuss the following problem today. Let a is equal to r minus singleton 3 and b is equal to r minus singleton 1. Consider the function f from a to b defined by fx is equal to x minus 2 by x minus 3. Is f 1 1 and on to justify your answer. Now let us write the solution. Now let us first check for 1 1. Let xy be any two elements of a. Then f of x is equal to f of y. Which implies x minus 2 by x minus 3 is equal to y minus 2 by y minus 3. Which implies x minus 2 into y minus 3 is equal to y minus 2 into x minus 3. Which implies xy minus 3x minus 2y plus 6 is equal to xy minus 3y minus 2x plus 6. Now this gets cancelled and this also gets cancelled. So we are left with minus 3x plus 2x is equal to minus 3y plus 2y. Which implies minus x is equal to minus y. Which implies x is equal to y. Thus f of x is equal to f of y. Which implies x is equal to y. For all x belongs to a. Thus f is 1 1. Now let us check for on 2. Let y be any arbitrary element of b. Then f of x is equal to y. Which implies x minus 2 by x minus 3 is equal to y. Which implies x minus 2 is equal to y into x minus 3. Which implies x minus 2 is equal to xy minus 3y. Which implies x is equal to 2 minus 3y by 1 minus y. Now clearly x is equal to 2 minus 3y by 1 minus y is a real number for all y not equal to 1. Also 2 minus 3y by 1 minus y is not equal to 3 for any y. Because if 2 minus 3y by 1 minus y is equal to 3. Then we have 2 is equal to 3. Which is not possible. Another element in b. How it's premade x in a. Given by this is equal to 2 minus 3y by 1 minus y. Therefore f is on 2. Now f is 1 1 f is on 2. Therefore yes f is 1 1 and on 2. I hope you understood the problem. Bye and have a nice day.