QUESTION IMAGE
Question
- which of the following is a function?
a. {(2, 6), (2, 10), (8, 6), (5, 1), (4, 6), (3, 9)}
b. {(3, 7), (4, 8), (5, 9), (6, 10), (7, 11)}
c. {(1, 5), (2, 5), (3, 5), (4, 5)}
d. {(0, 0), (1, 1), (1, 2), (2, 3)}
To determine if a relation is a function, we use the definition: a function is a relation where each input (x - value) has exactly one output (y - value).
Step 1: Analyze Option A
For the relation \( A=\{(2, 6),(2, 10),(8, 6),(5, 1),(4, 6),(3, 9)\} \), the input \( x = 2 \) is paired with two different outputs (\( y=6 \) and \( y = 10 \)). So, this is not a function.
Step 2: Analyze Option B
For the relation \( B=\{(3, 7),(4, 8),(5, 9),(6, 10),(7, 11)\} \), each \( x \) - value (3, 4, 5, 6, 7) is paired with exactly one \( y \) - value (7, 8, 9, 10, 11) respectively.
Step 3: Analyze Option C
Wait, actually, let's correct. Option C: \( C=\{(1, 5),(2, 5),(3, 5),(4, 5)\} \) is also a function (each \( x \) has one \( y \)), but let's check the original problem. Wait, the user's image shows option B is circled, but let's re - check. Wait, no, let's follow the definition. Wait, in option B, all \( x \) values are unique and each has one \( y \). In option C, all \( x \) values (1,2,3,4) have the same \( y = 5 \), but since each \( x \) has only one \( y \), it is a function. But wait, the original problem's options:
Wait, let's re - evaluate:
- Option A: \( x = 2 \) has two \( y \) values (\( 6 \) and \( 10 \)) → not a function.
- Option B: Each \( x \) (3,4,5,6,7) has exactly one \( y \) (7,8,9,10,11) → function.
- Option C: Each \( x \) (1,2,3,4) has exactly one \( y \) (5) → function.
- Option D: \( x = 1 \) has two \( y \) values (\( 1 \) and \( 2 \)) → not a function.
But maybe there is a mistake in the problem or maybe I misread. Wait, the original problem's option B is \( \{(3, 7),(4, 8),(5, 9),(6, 10),(7, 11)\} \) (assuming the typo in the original image, maybe a missing curly brace, but as a set of ordered pairs, each \( x \) is unique and has one \( y \)). And option C is also a function. But maybe in the context of the question, both B and C are functions? Wait, no, let's check the definition again. A function allows multiple inputs to have the same output (many - to - one is allowed), but not one input to have multiple outputs (one - to - many is not allowed).
So:
- Option A: One - to - many (x = 2 → two y's) → not a function.
- Option B: Each x has one y (x's are 3,4,5,6,7; y's are 7,8,9,10,11) → function.
- Option C: Each x (1,2,3,4) has one y (5) → function (many - to - one).
- Option D: One - to - many (x = 1 → two y's) → not a function.
But maybe the question has a typo, or maybe I misread option B. Wait, the original image shows option B as \( \{(3, 7),(4, 8),(5, 9),(6, 10),(7, 11)\} \) (assuming the correct set of ordered pairs), and option C as \( \{(1, 5),(2, 5),(3, 5),(4, 5)\} \). Both B and C are functions. But maybe in the problem's context, the intended answer is B (and C is also a function, but maybe there was a mistake). But according to the analysis of option B:
Each \( x \) - value in B is unique and paired with exactly one \( y \) - value. So, following the function definition, B is a function. Also, C is a function. But let's check the original problem's options again.
Wait, the user's image:
Option A: \( \{(2, 6),(2, 10),(8, 6),(5, 1),(4, 6),(3, 9)\} \)
Option B: \( \{(3, 7),(4, 8),(5, 9),(6, 10),(7, 11)\} \) (assuming the correct set, maybe a missing '{' at the start, but as a set of ordered pairs, each x has one y)
Option C: \( \{(1, 5),(2, 5),(3, 5),(4, 5)\} \)
Option D: \( \{(0, 0),(1, 1),(1, 2),(2, 3)\} \)
So, for option B:
- \( x = 3 \) → \( y = 7 \)
- \( x = 4 \) → \( y = 8 \)
- \( x = 5 \) → \( y = 9 \)
- \( x = 6 \) → \( y = 10 \)
- \( x = 7 \) → \( y = 11 \)
Each \( x \)…
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B. \(\{(3, 7), (4, 8), (5, 9), (6, 10), (7, 11)\}\), C. \(\{(1, 5), (2, 5), (3, 5), (4, 5)\}\) (But based on the circled option in the image, the intended answer is B. \(\{(3, 7), (4, 8), (5, 9), (6, 10), (7, 11)\}\))