The function f:R to R defined by f(x) = (x - | vedclass

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(B) We have $f(x) = (x - 1)(x - 2)(x - 3)$.First,check for one-one property:$f(1) = (1 - 1)(1 - 2)(1 - 3) = 0$$f(2) = (2 - 1)(2 - 2)(2 - 3) = 0$$f(3)

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Onto but not one-one

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(B) We have $f(x) = (x - 1)(x - 2)(x - 3)$.First,check for one-one property:$f(1) = (1 - 1)(1 - 2)(1 - 3) = 0$$f(2) = (2 - 1)(2 - 2)(2 - 3) = 0$$f(3) = (3 - 1)(3 - 2)(3 - 3) = 0$Since $f(1) = f(2) = f(3) = 0$,the function is not one-one because distinct elements have the same image.Next,check for onto property:$f(x)$ is a polynomial of degree $3$. The range of any odd-degree polynomial function $f: R 	o R$ is $(-\infty, \infty)$,which is the codomain $R$.Therefore,for every $y \in R$,there exists at least one $x \in R$ such that $f(x) = y$.Thus,the function is onto.Conclusion: The function is onto but not one-one.

The function $f:R \to R$ defined by $f(x) = (x - 1)(x - 2)(x - 3)$ is

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