Mark the correct alternative in each of the following:
Let the function f : R – {–b} → R – {1} be defined by then
Given that f: R – {–b} → R – {1} where
Here, f(x) = f(y) only when x=y.
Hence, it is one-one.
Now, f(x) = y
⇒ x + a = y(x + b)
⇒ x – yx = yb – a
So, x ϵ R – {1}
Hence, it is onto.