Question

Which of the following functions from Z to itself are bijections?
A.$f\left( x \right) = {x^3}$
B.$f\left( x \right) = x + 2$
C.$f\left( x \right) = 2x + 1$
D.$f\left( x \right) = {x^2} + x$

Answer 1

Hint: We can find the inverse of each of the given functions. Then we can check whether the inverse function is also defined on the given set and determine whether it is onto. If the function is onto, we can check whether the function is one-one. If a function is both one-one and onto, we can say that it is a bijection.

Complete step-by-step answer:
We know that for a function to be bijection, it must be both one-one and onto.
So, we can check whether each of the given functions is onto and if it is onto, we can check whether it is one-one.
Consider option A, $f\left( x \right) = {x^3}$
Let $y = f\left( x \right) = {x^3}$
Now, we can change the equation in terms of y. So, we can take a cube root on both sides.
$ \Rightarrow x = \sqrt[3]{y}$
If the function is bijective, the inverse must be defined for all values of the range. When $y = 2$ , we get $x = \sqrt[3]{2}$ which is not an integer. So, it is not a bijective function.
Consider option B, $f\left( x \right) = x + 2$
Let $y = f\left( x \right) = x + 2$
Now, we can change the equation in terms of y. So, we can subtract 2 on both sides.
$ \Rightarrow x = y - 2$
If the function is bijective, the inverse must be defined for all values of the range. For any integer value of y, we get the value if x integer. So, it is a onto function.
Now we can check whether it is one-one.
Let $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$
$ \Rightarrow {x_1} + 2 = {x_2} + 2$
On cancelling the common terms, we get,
$ \Rightarrow {x_1} = {x_2}$
So, the function is one-one. As the function is both one-one and onto, it is a bijective function
Consider option C, $f\left( x \right) = 2x + 1$
Let $y = f\left( x \right) = 2x + 1$
Now we can change the equation in terms of y. So, we subtract 1 on both sides and the we can divide both sides with 2.
$ \Rightarrow 2x = y - 1$
On dividing the equation by 2 we get,
$ \Rightarrow x = \dfrac{{y - 1}}{2}$
If the function is bijective, the inverse must be defined for all values of the range.
When $y = 2$ , we get
$ \Rightarrow x = \dfrac{{2 - 1}}{2}$
On simplification, we get
$ \Rightarrow x = \dfrac{1}{2}$ which is not an integer. So, it is not a bijective function.
Consider option D, $f\left( x \right) = {x^2} + x$
Let $y = f\left( x \right) = {x^2} + x$
Here, we cannot find the inverse of the function. So, we can check whether it's one-one.
When $x = 0$ , $f\left( 0 \right) = {0^2} + 0 = 0$ and
When $x = - 1$ , \[f\left( { - 1} \right) = {\left( { - 1} \right)^2} + \left( { - 1} \right) = 1 - 1 = 0\]
As the image is the same for more than one value of x, the function is not one-one.
So, we can conclude that only function which is a bijection from Z to itself is $f\left( x \right) = x + 2$
Therefore, the correct answer is option B.

Note: A function must be both one-one and onto to be bijective. So, we must check both the conditions. If one of the conditions fails, we need not to check the other one. We can also check each of the options for one-one first and then onto if it is one-one. As most of the functions are one-one, this will take more time.