Mark the correct alternative in each of the following:
Let f : R –{n} → R be a function defined by where m ≠ n. Then,
Given that f : R –{n} → R where
Let f(x) = f(y)
⇒ (x-m)(y-n)=(x-n)(y-m)
⇒ xy – xn – my + mn = xy – xm – ny + mn
⇒ x = y
So, f is one-one.
⇒ y(x-n)=(x-m)
⇒ xy – ny = x – m
⇒ x(y-1) = ny – m
For y = 1 , no x is defined.
So, f is into.