Question

The function \[f: R \to R\] defined by \[f(x) = (x - 1)(x - 2)(x - 3)\] is
A) one-one but not onto
B) onto but not one-one
C) both one-one and onto
D) neither one-one nor onto

Answer 1

Hint: we can find the correct solution for the given problem by verifying the conditions for the functions to be one-one and onto that is a function f is said to be one to one (or injective function), if the images of distinct elements of X under f are distinct, i.e., for every \[{x_1},{x_2} \in X,f({x_1}) = f({x_2}) \Rightarrow {x_1} = {x_2}\]. To show that f is an onto function, set \[y = f(x)\], and solve for x, or show that we can always express x in terms of y.

Complete step by step answer:
Now let us consider the given function
\[f(x) = (x - 1)(x - 2)(x - 3)\]
Replace x by 1 we get
\[ \Rightarrow f(1) = (1 - 1)(1 - 2)(1 - 3) = 0\]
Replace x by 2 we get
\[ \Rightarrow f(2) = (2 - 1)(2 - 2)(2 - 3) = 0\]
Replace x by 3 we get
\[ \Rightarrow f(3) = (3 - 1)(3 - 1)(3 - 3) = 0\]
We note that
\[ \Rightarrow f(1) = f(2) = f(3) = 0\]
Since the images of 1, 2, 3 are the same, that is 0.
Therefore, f is not one-one.
Now to check onto:
Let y be an element in the co-domain R, such that \[y = f(x)\],
\[ \Rightarrow y = (x - 1)(x - 2)(x - 3)\]
Since, \[y \in R\] and \[x \in R\]
Therefore, $f$ is onto function. So, Option (B) is correct.

Note: We can note that an easy way to determine whether a function is a one-to-one function is to use the horizontal line test on the graph of the function. To do this, draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.
The function $f$ is surjective (i.e., onto) if and only if its graph intersects any horizontal line at least once. f is bijective if and only if any horizontal line will intersect the graph exactly once.