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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.