 Hello and welcome to this session. Let us discuss the following problem today. Show that the function f from r to where x belongs to r such that minus 1 is less than x less than 1 defined by fx is equal to x by 1 plus mod of x, so that x belongs to r is 1, 1 and on to function. Now let us write the solution. Now first let x is greater than equal to 0 then fx is equal to x by 1 plus x let f of x1 is equal to f of x2 for any x1, x2 which implies x1 by 1 plus x, x1 is equal to x2 by 1 plus x2 which implies x1 into 1 plus x2 is equal to x2 into 1 plus x1 which implies x1 plus x1 x2 is equal to x2 plus x1 x2 which implies x1 is equal to x2. Now for x is less than 0 fx is equal to x by 1 minus x and let fx1 is equal to fx2 for any x1, x2 which implies x1 by 1 minus x1 is equal to x2 by 1 minus x2 which implies x1 into 1 minus x2 is equal to x2 into 1 minus x1 which implies x1 minus x1 x2 is equal to x2 minus x1 x2 which implies x1 is equal to x2 thus f is 1 1 for all x. Now let us check for on to first for minus 1 is less than x less than 1 fx is equal to x by 1 plus mod of x lies between minus half and half let y be an arbitrary element in the core domain then x is greater than equal to 0 let y is equal to x by 1 plus x therefore y into 1 plus x is equal to x or y plus yx is equal to x or y is equal to x into 1 minus y or x is equal to y by 1 minus y. Now when x is less than 0 fx is equal to x by 1 minus x or y is equal to x by 1 minus x which implies y into 1 minus x is equal to x which implies y minus yx is equal to x which implies y is equal to x into 1 plus y which implies x is equal to y by 1 plus y thus we find that in both cases for x greater than equal to 0 and x less than 0 for y belongs to core domain there exist a unique value in its domain f is on to f is 1 1 and on to I hope you enjoyed the session bye and have a nice day