Let `A={x:xepsilonR-}f` is defined from `ArarrR` as `f(x)=(2x)/(x-1)` then `f(x)` is (a) Surjective but nor injective (b) injective but nor surjective (c) neither injective surjective (d) injectiveA. injective but not surjectiveB. not injectiveC. surjective but not injectiveD. neither injective nor surjective

Let `A={x:xepsilonR-}f` is defined from `ArarrR` as `f(x)=(2x)/(x-1)`

1.	Let `A={x:xepsilonR-}f` is defined from `ArarrR` as `f(x)=(2x)/(x-1)` then `f(x)` is (a) Surjective but nor injective (b) injective but nor surjective (c) neither injective surjective (d) injectiveA. injective but not surjectiveB. not injectiveC. surjective but not injectiveD. neither injective nor surjective
Answer» Correct Answer - A We have a function `f:A to R "defined as, " f(x)=(2x)/(x-1)` One-one let `x_(1),x_(2) in A` such that `f(x_(1))=f(x_(2))` `rArr (2x_(1))/(x_(1)-1)=(2x_(2))/(x_(2)-1)` `rArr 2x_(1)x_(2)-2x_(1)=2x_(1)x_(2)-2x_(2)` `rArr x_(1)=x_(2)` Thus, `f(x_(1))=f(x_(2))` has only one solution `x_(1)=x_(2)` ` therefore` f(x) is one-one (injective) Onto Let `x = 2 " then " f(2)=(2xx2)/(2-1)=4` But x = 2 is not in the domain, and f(x) is one-one function ` therefore f(x)` can never be 4. similarly, f(x) can not take many values. Hence, f(x) is into (not surjective). `therefore f(x)` is injective but not surjective.