Question

The function $ f:[0,\,\,\infty )\to [0,\,\,\infty ) $ defined by $ f\,(x)=\frac{2x}{1+2x} $ is

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

Answer 1

Answer: (B)

Given, $ f:\,\,[0,\,\infty )\to \,[0,\infty ) $
and $ f(x)=\frac{2x}{1+2x} $
Let, $ y=\frac{2x}{1+2x} $
$ \Rightarrow $ $ y+2xy=2x $
$ \Rightarrow $ $ y=2x(1-y) $
$ \Rightarrow $ $ x=\frac{y}{2(1-y)} $
Here, $ y\in R\tilde{\ }\{1\}. $
So, the range of $ f(x) $ is $ R\tilde{\ }\{\,1\,\}. $
Since, Range $ \ne $ codomain
So, the given function is on to. For one-one,
$ f({{x}_{1}})=f({{x}_{2}}) $
$ \Rightarrow $ $ \frac{2{{x}_{1}}}{1+{{x}_{1}}}=\frac{2{{x}_{2}}}{1+{{x}_{2}}} $
$ \Rightarrow $ $ {{x}_{1}}+{{x}_{1}}{{x}_{2}}={{x}_{2}}+{{x}_{1}}{{x}_{2}} $
$ \Rightarrow $ $ {{x}_{1}}={{x}_{2}} $
So, the above function is one-one.