A) one-one but not onto
B) onto but not one-one
C) both one-one and onto
D) neither one-one nor onto
Correct Answer: B
Solution :
Given,\[f(x)=(x-1)(x-2)(x-3)\] and \[f(1)=f(2)=f(3)=0\] \[\Rightarrow f(x)\]is not one-one. For each\[y\in R\], there exists\[x\in R\], such that\[f(x)=y\]. Therefore, \[f\] is onto. Hence, \[f:R\to R\]is onto but not one-one.
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