Q. 105.0( 1 Vote )

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Answer :

Given that f: [0, ∞) R where


Let f(x) = f(y)



xy + x = xy + y


x = y


So, f is one-one.


Now, y = f(x)



xy + y = x


y = x – xy



Here, y≠1 i.e. y ϵ R.


So, f is not onto.

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