SAT Exam > SAT Questions > Let A = {1, 2, 3} and B = {4, 5, 6}. Which on...
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Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?
- a)f = {(2, 4),(2, 5), (2, 6)}
- b)f = {(1, 5), (2, 4), (3, 4)}
- c)f = {(1, 4), (1, 5), (1, 6)}
- d)f = {(1, 4), (2, 5), (3, 6)}
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functi...
Explanation:
To determine if a function is bijective, we need to check if it is both injective (one-to-one) and surjective (onto).
Injective:
A function is injective if each element in the domain maps to a unique element in the codomain. In other words, no two different elements in the domain can map to the same element in the codomain.
In option 'A', we have f(2) = 4, f(2) = 5, and f(2) = 6, which violates the injective condition. Therefore, option 'A' is not bijective.
In option 'B', we have f(1) = 5 and f(2) = 4, which are both unique mappings. However, f(3) = 4 violates the injective condition. Therefore, option 'B' is not bijective.
In option 'C', we have f(1) = 4, f(1) = 5, and f(1) = 6, which violates the injective condition. Therefore, option 'C' is not bijective.
In option 'D', we have f(1) = 4, f(2) = 5, and f(3) = 6, which are all unique mappings. No two different elements in the domain map to the same element in the codomain. Therefore, option 'D' is injective.
Surjective:
A function is surjective if every element in the codomain is mapped to by at least one element in the domain. In other words, there are no "missing" elements in the codomain.
In option 'D', every element in the codomain {4, 5, 6} is mapped to by at least one element in the domain {1, 2, 3}. Therefore, option 'D' is surjective.
Since option 'D' satisfies both the injective and surjective conditions, it is bijective.
Conclusion:
The function f = {(1, 4), (2, 5), (3, 6)} is bijective.
To determine if a function is bijective, we need to check if it is both injective (one-to-one) and surjective (onto).
Injective:
A function is injective if each element in the domain maps to a unique element in the codomain. In other words, no two different elements in the domain can map to the same element in the codomain.
In option 'A', we have f(2) = 4, f(2) = 5, and f(2) = 6, which violates the injective condition. Therefore, option 'A' is not bijective.
In option 'B', we have f(1) = 5 and f(2) = 4, which are both unique mappings. However, f(3) = 4 violates the injective condition. Therefore, option 'B' is not bijective.
In option 'C', we have f(1) = 4, f(1) = 5, and f(1) = 6, which violates the injective condition. Therefore, option 'C' is not bijective.
In option 'D', we have f(1) = 4, f(2) = 5, and f(3) = 6, which are all unique mappings. No two different elements in the domain map to the same element in the codomain. Therefore, option 'D' is injective.
Surjective:
A function is surjective if every element in the codomain is mapped to by at least one element in the domain. In other words, there are no "missing" elements in the codomain.
In option 'D', every element in the codomain {4, 5, 6} is mapped to by at least one element in the domain {1, 2, 3}. Therefore, option 'D' is surjective.
Since option 'D' satisfies both the injective and surjective conditions, it is bijective.
Conclusion:
The function f = {(1, 4), (2, 5), (3, 6)} is bijective.
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Community Answer
Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functi...
f = {(1, 4), (2, 5), (3, 6)} is a bijective function.
One-one: It is a one-one function because every element in set A = {1, 2, 3} has a distinct image in set B = {4, 5, 6}.
Onto: It is an onto function as every element in set B = {4, 5, 6} is the image of some element in set A = {1, 2, 3}.
f = {(2, 4), (2, 5), (2, 6)} and f = {(1, 4), (1, 5), (1, 6)} are many-one onto.
f = {(1, 5), (2, 4), (3, 4)} is neither one – one nor onto.
One-one: It is a one-one function because every element in set A = {1, 2, 3} has a distinct image in set B = {4, 5, 6}.
Onto: It is an onto function as every element in set B = {4, 5, 6} is the image of some element in set A = {1, 2, 3}.
f = {(2, 4), (2, 5), (2, 6)} and f = {(1, 4), (1, 5), (1, 6)} are many-one onto.
f = {(1, 5), (2, 4), (3, 4)} is neither one – one nor onto.
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Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?a)f = {(2, 4),(2, 5), (2, 6)}b)f = {(1, 5), (2, 4), (3, 4)}c)f = {(1, 4), (1, 5), (1, 6)}d)f = {(1, 4), (2, 5), (3, 6)}Correct answer is option 'D'. Can you explain this answer?
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Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?a)f = {(2, 4),(2, 5), (2, 6)}b)f = {(1, 5), (2, 4), (3, 4)}c)f = {(1, 4), (1, 5), (1, 6)}d)f = {(1, 4), (2, 5), (3, 6)}Correct answer is option 'D'. Can you explain this answer? for SAT 2025 is part of SAT preparation. The Question and answers have been prepared according to the SAT exam syllabus. Information about Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?a)f = {(2, 4),(2, 5), (2, 6)}b)f = {(1, 5), (2, 4), (3, 4)}c)f = {(1, 4), (1, 5), (1, 6)}d)f = {(1, 4), (2, 5), (3, 6)}Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for SAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?a)f = {(2, 4),(2, 5), (2, 6)}b)f = {(1, 5), (2, 4), (3, 4)}c)f = {(1, 4), (1, 5), (1, 6)}d)f = {(1, 4), (2, 5), (3, 6)}Correct answer is option 'D'. Can you explain this answer?.
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