A) both one-one and onto
B) one-one but not onto
C) onto but not one-one
D) neither one-one nor onto.
Correct Answer: C
Solution :
\[f(x)=\frac{\left| x \right|-1}{\left| x \right|+1}\]for one-one function if \[f({{x}_{1}})=f({{x}_{2}})\] then \[{{x}_{1}}\]must be equal to \[{{x}_{2}}\] Let \[f({{x}_{1}})=f({{x}_{2}})\] \[\frac{\left| {{x}_{1}} \right|-1}{\left| {{x}_{1}} \right|+1}=\frac{\left| {{x}_{2}} \right|-1}{\left| {{x}_{2}} \right|+1}\] \[\left| {{x}_{1}} \right|\left| {{x}_{2}} \right|+\left| {{x}_{1}} \right|-\left| {{x}_{2}} \right|-1=\left| {{x}_{1}} \right|\left| {{x}_{2}} \right|-\left| {{x}_{1}} \right|+\left| {{x}_{2}} \right|-1\] \[\Rightarrow \]\[\left| {{x}_{1}} \right|-\left| {{x}_{2}} \right|=\left| {{x}_{2}} \right|-\left| {{x}_{1}} \right|\] \[2\left| {{x}_{1}} \right|=2\left| {{x}_{2}} \right|\] \[\left| {{x}_{1}} \right|=\left| {{x}_{2}} \right|\] \[{{x}_{1}}={{x}_{2}},{{x}_{1}}=-{{x}_{2}}\] here \[{{x}_{1}}\]has two values therefore function is many one function. For onto : \[f(x)=\frac{\left| x \right|-1}{\left| x \right|+1}\]for every value of f (x) there is a value of x in domain set. If f (x) is negative then x = 0 for all positive value of f (x), domain contain atleast one element. Hence f (x) is onto function.
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