Let's address each part of the question step by step.
### Part (i): Is the function [tex]\( f \)[/tex] onto?
A function is considered onto (surjective) if every element in the codomain has at least one pre-image in the domain. Here, the function is defined as [tex]\( f: \mathbb{N} \rightarrow \mathbb{N} \)[/tex] with the rule [tex]\( f(x) = 2x - 1 \)[/tex].
- For [tex]\( f(x) = 2x - 1 \)[/tex], the output will always be an odd number because doubling any natural number [tex]\( x \)[/tex] and then subtracting 1 results in an odd number.
- Since the codomain [tex]\(\mathbb{N}\)[/tex] contains both odd and even numbers and not all natural numbers are odd, it implies that not every element in the codomain (which includes even numbers) has a pre-image.
- Therefore, the function [tex]\( f(x) = 2x - 1 \)[/tex] is not onto.
### Part (ii): What is the image of 8?
To find the image of 8 under the function [tex]\( f \)[/tex]:
- Use the function rule [tex]\( f(x) = 2x - 1 \)[/tex].
- Substitute [tex]\( x = 8 \)[/tex]:
[tex]\[
f(8) = 2 \cdot 8 - 1 = 16 - 1 = 15
\][/tex]
So, the image of 8 is 15.
### Part (iii): What is the pre-image of 31?
To find the pre-image of 31, we need to solve the equation [tex]\( f(x) = 31 \)[/tex]:
- Given the function [tex]\( f(x) = 2x - 1 \)[/tex], set it equal to 31:
[tex]\[
2x - 1 = 31
\][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = 31 + 1
\][/tex]
[tex]\[
2x = 32
\][/tex]
[tex]\[
x = 16
\][/tex]
So, the pre-image of 31 is 16.
### Part (iv): If [tex]\( (k, 4k - 5) \)[/tex] lies in the function, find the value of [tex]\( k \)[/tex].
We need to find [tex]\( k \)[/tex] such that [tex]\( 4k - 5 = f(k) \)[/tex]:
- The function [tex]\( f \)[/tex] is given by [tex]\( f(x) = 2x - 1 \)[/tex].
- So, substitute [tex]\( f(k) = 4k - 5 \)[/tex]:
[tex]\[
4k - 5 = 2k - 1
\][/tex]
- Solve for [tex]\( k \)[/tex]:
[tex]\[
4k - 2k = -1 + 5
\][/tex]
[tex]\[
2k = 4
\][/tex]
[tex]\[
k = 2
\][/tex]
So, the value of [tex]\( k \)[/tex] is 2.
### Summary of all parts:
(i) The function [tex]\( f \)[/tex] is not onto.
(ii) The image of 8 is 15.
(iii) The pre-image of 31 is 16.
(iv) The value of [tex]\( k \)[/tex] such that [tex]\( (k, 4k - 5) \)[/tex] lies in the function is 2.
