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Sagot :
To determine which of the given sets of pairs is not a function, we need to understand the definition of a function. A function from a set [tex]\( A \)[/tex] to a set [tex]\( B \)[/tex] assigns each element [tex]\( a \)[/tex] in [tex]\( A \)[/tex] exactly one element [tex]\( b \)[/tex] in [tex]\( B \)[/tex]. In other words, for each input [tex]\( x \)[/tex] in the domain, there should be exactly one corresponding output [tex]\( y \)[/tex] in the codomain.
Let's analyze each set of pairs one by one:
### Set 1:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline 2 & 5 \\ \hline 3 & 9 \\ \hline \end{tabular} \][/tex]
In this set, the input [tex]\( x = 2 \)[/tex] is associated with two different outputs ([tex]\( y = 4 \)[/tex] and [tex]\( y = 5 \)[/tex]). This violates the definition of a function because a single input [tex]\( x \)[/tex] cannot map to more than one output [tex]\( y \)[/tex]. Therefore, set 1 is not a function.
### Set 2:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -2 & 0 \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline 3 & 7 \\ \hline \end{tabular} \][/tex]
In this set, each input [tex]\( x \)[/tex] is associated with exactly one output [tex]\( y \)[/tex]:
- [tex]\( x = -2 \rightarrow y = 0 \)[/tex]
- [tex]\( x = 1 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 2 \rightarrow y = 4 \)[/tex]
- [tex]\( x = 3 \rightarrow y = 7 \)[/tex]
Therefore, set 2 satisfies the conditions of a function.
### [tex]\( y = 3x^2 - 6x + 4 \)[/tex]:
This is an equation defining a quadratic function. For each input [tex]\( x \)[/tex], there will be exactly one output [tex]\( y \)[/tex] because it represents a parabola. Hence, this is a function.
### Set 3:
[tex]\[ \{ (3, 4), (6, 5), (7, 9), (9, 15) \} \][/tex]
In this set, each input [tex]\( x \)[/tex] is associated with exactly one output [tex]\( y \)[/tex]:
- [tex]\( x = 3 \rightarrow y = 4 \)[/tex]
- [tex]\( x = 6 \rightarrow y = 5 \)[/tex]
- [tex]\( x = 7 \rightarrow y = 9 \)[/tex]
- [tex]\( x = 9 \rightarrow y = 15 \)[/tex]
Therefore, set 3 satisfies the conditions of a function.
### Conclusion:
Among the sets provided, set 1 is identified as not being a function since it maps the input [tex]\( x = 2 \)[/tex] to two different outputs.
Therefore, the answer is 1.
Let's analyze each set of pairs one by one:
### Set 1:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline 2 & 5 \\ \hline 3 & 9 \\ \hline \end{tabular} \][/tex]
In this set, the input [tex]\( x = 2 \)[/tex] is associated with two different outputs ([tex]\( y = 4 \)[/tex] and [tex]\( y = 5 \)[/tex]). This violates the definition of a function because a single input [tex]\( x \)[/tex] cannot map to more than one output [tex]\( y \)[/tex]. Therefore, set 1 is not a function.
### Set 2:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -2 & 0 \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline 3 & 7 \\ \hline \end{tabular} \][/tex]
In this set, each input [tex]\( x \)[/tex] is associated with exactly one output [tex]\( y \)[/tex]:
- [tex]\( x = -2 \rightarrow y = 0 \)[/tex]
- [tex]\( x = 1 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 2 \rightarrow y = 4 \)[/tex]
- [tex]\( x = 3 \rightarrow y = 7 \)[/tex]
Therefore, set 2 satisfies the conditions of a function.
### [tex]\( y = 3x^2 - 6x + 4 \)[/tex]:
This is an equation defining a quadratic function. For each input [tex]\( x \)[/tex], there will be exactly one output [tex]\( y \)[/tex] because it represents a parabola. Hence, this is a function.
### Set 3:
[tex]\[ \{ (3, 4), (6, 5), (7, 9), (9, 15) \} \][/tex]
In this set, each input [tex]\( x \)[/tex] is associated with exactly one output [tex]\( y \)[/tex]:
- [tex]\( x = 3 \rightarrow y = 4 \)[/tex]
- [tex]\( x = 6 \rightarrow y = 5 \)[/tex]
- [tex]\( x = 7 \rightarrow y = 9 \)[/tex]
- [tex]\( x = 9 \rightarrow y = 15 \)[/tex]
Therefore, set 3 satisfies the conditions of a function.
### Conclusion:
Among the sets provided, set 1 is identified as not being a function since it maps the input [tex]\( x = 2 \)[/tex] to two different outputs.
Therefore, the answer is 1.
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