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If f:[0,oo)vec[0,oo)a n df(x)=x/(1+x),t ...

If `f:[0,oo)vec[0,oo)a n df(x)=x/(1+x),t h e nf` is one-one and onto one-one but not onto onto but not one-one neither on-one nor onto

A

one-one and onto

B

one-one but not onto

C

onto but not one-one

D

neither one-one nor onto

Text Solution

Verified by Experts

The correct Answer is:
B
Given `f:[0,oo)to [0,oo)`
Here, domain is `[0,oo)` and codoamain is `[0,oo)`. Thus, to check one-one
Since, `f(x)=(x)/(1+x) rArrf'(x)=(1)/((1+x)^(2)) gt 0,AA x in [0,oo)`
`therefore f(x)` is increasing in its domain. Thus, f(x) is one-one in its domain. To check onto (we find range)
Again, `y=f(x)=(x)/(1+x)`
`rArr y+yx=x`
`rArr x=(y)/(1-y) rArr (y)/(1-y) ge 0`
Since, `x ge 0` therefore `0 le x lt 1`
i.e Range `ne ` Codomain
`therefore f(x)` is one-one but not onto.
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