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Let A={x:xepsilonR-{1}}f is defined from...

Let `A={x:xepsilonR-{1}}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) injective

A

injective but not surjective

B

not injective

C

surjective but not injective

D

neither injective nor surjective

Text Solution

Verified by Experts

The correct Answer is:
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.
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