The function `f:[0,\ oo)->R` given by `f(x)=x/(x+1)` is (a) one-one and onto (b) one-one but not onto (c) onto but not one-one (d) neither one-one nor onto

Options

A. one-one and into
B. one-one but not onto
C. onto but not one-one
D. neither one-one nor onto

Answer

B

Explanation

We have,
`f(x)=(x)/(x+1)=1-(1)/(1+x)`
Let `x, y in R` be such that `f(x)=f(y)`. Then, f(x)=f(y)
`Rightarrow 1-(1)/(1+x)=1-(1)/(1+y)Rightarrow 1+x=1+y Rightarrow x=y`
`therefore ` f is one-one
Clearly, f is not onto as f takes only values less than 1 i.e. Range `f=(-oo, 1) ne` co-domain of f.