Question

which of the following sets of data represent valid functions?
m = {(-5, 0), (2, 1), (2, 6), (8, 7), (10, 14)}
z = {(0, -4), (3, 2), (5, 4), (7, 9), (12, 10)}
r = {(-4, -5), (1, 2), (5, 4), (7, 9), (16, 13)}
t = {(-5, -1), (1, 3), (5, 6), (-5, -1), (10, 13)}

Explanation:

Step1: Recall function definition

A function is a relation where each input (x - value) has exactly one output (y - value). So we check each set for repeated x - values with different y - values.

Step2: Analyze set M

For \( M=\{(- 5,0),(2,1),(2,6),(8,7),(10,14)\} \), the x - value \( 2 \) is paired with \( 1 \) and \( 6 \). So \( M \) is not a function.

Step3: Analyze set Z

For \( Z =\{(0,-4),(3,2),(5,4),(7,9),(12,10)\} \), each x - value (\( 0,3,5,7,12 \)) has a unique y - value. So \( Z \) is a function.

Step4: Analyze set H

For \( H=\{(-4,-5),(1,2),(5,4),(7,9),(16,13)\} \), each x - value (\( - 4,1,5,7,16 \)) has a unique y - value. So \( H \) is a function.

Step5: Analyze set T

For \( T=\{(-5,-1),(1,3),(5,6),(-5,-1),(10,13)\} \), the x - value \( - 5 \) is paired with \( - 1 \) (repeated, but same y - value) and \( 1 \) is paired with \( 3 \), \( 5 \) with \( 6 \), \( 10 \) with \( 13 \). Wait, actually, in \( T \), the x - value \( - 5 \) has only one unique y - value (\( - 1 \), even though it's repeated), and other x - values are unique. Wait, no, let's re - check. Wait, the definition is each input has exactly one output. So if an x - value is repeated, it must have the same y - value. In \( T \), \( (-5,-1) \) is repeated, but the y - value is the same. But wait, let's check all x - values: \( - 5 \) (output \( - 1 \)), \( 1 \) (output \( 3 \)), \( 5 \) (output \( 6 \)), \( - 5 \) (output \( - 1 \)), \( 10 \) (output \( 13 \)). So \( - 5 \) has only one output (\( - 1 \)), \( 1 \) has one, \( 5 \) has one, \( 10 \) has one. Wait, but earlier for \( M \), the x - value \( 2 \) had two different outputs. But in \( T \), the repeated \( - 5 \) has the same output. Wait, but let's check the original sets again. Wait, the set \( T \) is \( \{(-5,-1),(1,3),(5,6),(-5,-1),(10,13)\} \). So the x - values are \( - 5,1,5,-5,10 \). The x - value \( - 5 \) is mapped to \( - 1 \) both times (same output), \( 1 \) to \( 3 \), \( 5 \) to \( 6 \), \( 10 \) to \( 13 \). So technically, it is a function? Wait, no, wait the definition of a function (in the context of discrete sets) is that for a relation \( f:X ightarrow Y \), each \( x\in X \) has exactly one \( y\in Y \) such that \( (x,y)\in f \). So if an x - value is repeated, it must have the same y - value. In \( M \), \( x = 2 \) has two different y - values (\( 1 \) and \( 6 \)), so it's not a function. In \( Z \), all x - values are unique, so it's a function. In \( H \), all x - values are unique, so it's a function. In \( T \), the x - value \( - 5 \) is repeated but with the same y - value, and other x - values are unique. Wait, but maybe the problem considers the set of ordered pairs where each x - value (even if repeated) has the same y - value. But let's check the options again. Wait, the original question:

Wait, let's re - evaluate:

- For \( M \): \( x = 2 \) has \( y = 1 \) and \( y = 6 \) → not a function.

- For \( Z \): All x - values \( 0,3,5,7,12 \) are unique, so each has one y - value → function.

- For \( H \): All x - values \( - 4,1,5,7,16 \) are unique, so each has one y - value → function.

- For \( T \): \( x=-5 \) is repeated, but \( y = - 1 \) both times. But in the set of ordered pairs, when we define a function, the domain is the set of all x - values (with no repetition in terms of the mapping). Wait, actually, in a function, the domain is the set of input values, and each input has exactly one output. So if we have a relation where an input is repeated, but with the same output, it's still a function (because the input still has only one output). But wait, in the set \( T \), the ordered pairs \( (-5,-1) \) are repeated, but as a set, the set \( T \) is equal to \( \{(-5,-1),(1,3),(5,6),(10,13)\} \) (since sets don't have duplicate elements). Wait, no, the problem is presenting them as lists (maybe tuples), not sets. So if it's a list of ordered pairs (a relation), then for \( T \), the x - value \( - 5 \) is paired with \( - 1 \) twice, but it's still the same output. However, in the case of \( M \), the x - value \( 2 \) is paired with two different outputs. So \( Z \) and \( H \) and \( T \)? Wait, no, let's check again:

Wait, the definition of a function (from a set of ordered pairs) is that for the relation \( R \), if \( (x,y_1)\in R \) and \( (x,y_2)\in R \), then \( y_1 = y_2 \).

So:

- \( M \): \( (2,1) \) and \( (2,6) \) → \( 1 eq6 \) → not a function.

- \( Z \): All x - values are unique, so no two pairs with same x → function.

- \( H \): All x - values are unique, so no two pairs with same x → function.

- \( T \): \( (-5,-1) \) is repeated, but \( y \) - values are same. Also, other x - values are unique. So \( T \) is a function? Wait, but the problem is a multiple - choice (check - box) question. Wait, maybe I made a mistake. Wait, let's check the x - values in \( T \): \( - 5,1,5,-5,10 \). The x - value \( - 5 \) has \( y=-1 \) (twice), \( 1 \) has \( y = 3 \), \( 5 \) has \( y = 6 \), \( 10 \) has \( y = 13 \). So according to the function definition, since \( - 5 \) is mapped to only \( - 1 \) (even though it's written twice), and other x - values are mapped to unique y - values, \( T \) is a function? But wait, maybe the problem considers the list as a set of ordered pairs where duplicates are not allowed in the sense of the relation. Wait, no, the standard definition is that in a function, each input has exactly one output. So if an input is used multiple times, it must have the same output. So \( T \) is a function, \( Z \) is a function, \( H \) is a function, and \( M \) is not. But let's check the original options again.

Wait, the options are:

- \( M=\{(-5,0),(2,1),(2,6),(8,7),(10,14)\} \) → \( x = 2 \) has two y's → not a function.

- \( Z=\{(0,-4),(3,2),(5,4),(7,9),(12,10)\} \) → all x's are unique → function.

- \( H=\{(-4,-5),(1,2),(5,4),(7,9),(16,13)\} \) → all x's are unique → function.

- \( T=\{(-5,-1),(1,3),(5,6),(-5,-1),(10,13)\} \) → \( x=-5 \) has \( y = - 1 \) (twice, same y), others unique → function. But maybe the problem considers \( T \) as a function? Wait, but maybe I misread \( T \). Wait, \( T \) has \( (-5,-1) \) and \( (-5,-1) \) (same pair), so as a relation, it's a function. But let's check the answer options. Wait, the question is "Which of the following sets of data represent valid functions?" So we need to check which ones satisfy the function definition.

So \( Z \): yes, \( H \): yes, \( T \): yes? Wait, no, wait \( T \) has \( (-5,-1) \) twice, but in a function, the key is that each x has one y. So if x is - 5, y is - 1 (even if written twice), so it's a function. But \( M \) has x = 2 with two different y's, so not a function.

But let's check the original problem again. Maybe the options are:

- \( M \): no

- \( Z \): yes

- \( H \): yes

- \( T \): yes? Wait, but maybe the problem considers \( T \) as a function? Wait, no, maybe I made a mistake. Wait, let's check the x - values:

For \( Z \): x - values are 0,3,5,7,12 (all unique) → function.

For \( H \): x - values are - 4,1,5,7,16 (all unique) → function.

For \( T \): x - values are - 5,1,5,-5,10. The x - value - 5 is repeated, but with the same y - value. So according to the function definition, it is a function (because each x has exactly one y). For \( M \): x = 2 has y = 1 and y = 6 → not a function.

So the valid functions are \( Z \), \( H \), and \( T \)? But that seems odd. Wait, maybe the problem has a typo, or I misread \( T \). Wait, \( T=\{(-5,-1),(1,3),(5,6),(-5,-1),(10,13)\} \). So the ordered pairs are \( (-5,-1) \), \( (1,3) \), \( (5,6) \), \( (-5,-1) \), \( (10,13) \). So when we look at the relation, the input \( - 5 \) is associated with output \( - 1 \) (twice, but same output), input \( 1 \) with \( 3 \), input \( 5 \) with \( 6 \), input \( 10 \) with \( 13 \). So it is a function. \( Z \) and \( H \) have all unique x - values, so they are functions. \( M \) has a repeated x - value with different y - values, so not a function.

But let's check the answer options again. The check - boxes are next to each set. So the valid ones are \( Z \), \( H \), and \( T \)? Wait, no, maybe I made a mistake with \( T \). Wait, the definition of a function is that for every input, there is exactly one output. So if an input is used more than once, it must have the same output. So \( T \) is a function, \( Z \) is a function, \( H \) is a function, and \( M \) is not.

But let's confirm with the definition: A function is a relation in which no two ordered pairs have the same first element (input) and different second elements (output). So:

- \( M \): (2,1) and (2,6) have same first element, different second → not a function.

- \( Z \): All first elements are unique → function.

- \( H \): All first elements are unique → function.

- \( T \): The first element - 5 appears twice, but with the same second element (-1), and other first elements are unique → function.

So the valid functions are \( Z \), \( H \), and \( T \). But maybe the problem considers \( T \) as a function, \( Z \) as a function, and \( H \) as a function, and \( M \) as not.

Answer:

The sets that represent valid functions are \( \boldsymbol{Z=\{(0, - 4),(3,2),(5,4),(7,9),(12,10)\}} \), \( \boldsymbol{H=\{(-4,-5),(1,2),(5,4),(7,9),(16,13)\}} \), and \( \boldsymbol{T=\{(-5,-1),(1,3),(5,6),(-5,-1),(10,13)\}} \) (the check - boxes next to \( Z \), \( H \), and \( T \) should be selected, and the check - box next to \( M \) should be unselected).