A) a one-one function
B) an onto function
C) a bijection
D) neither one-one nor onto
Correct Answer: C
Solution :
Given, \[f(x)=\,{{x}^{3}}-1\] Let \[{{x}_{1}}\,,\,{{x}_{2}}\,\in \,R.\] Now, \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow \] \[x_{1}^{3}-1=x_{2}^{3}-1\] \[\Rightarrow \] \[x_{1}^{3}=x_{2}^{3}\,\,\,\Rightarrow \,\,\,{{x}_{1}}={{x}_{2}}\] \[\therefore \] \[f(x)\] is one-one. Also, it is onto. Hence, it is a bisection.
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