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 :
We have \[f(x)=(x-1)(x-2)(x-3)\] and \[f(1)=f(2)=f(3)=0\] Þ \[f(x)\] is not one-one. For each x\[y\in R\], 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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