Q. 55.0( 2 Votes )

Mark the correct

Answer :

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.

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