A) one-one and onto
B) one-one but not onto
C) not one-one but onto
D) neither one-one nor onto
Correct Answer: B
Solution :
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.
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