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Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is
- a)Surjective
- b)Not a function
- c)Injective
- d)Bijective
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, ...
Here, f(1) = 8, f(2) = 7, f(3) = 6.
Since, different points of domain have the different f-image in the range, therefore f is a one-one function.
Most Upvoted Answer
Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, ...
Solution:
To determine whether f is surjective, injective, or bijective, we need to understand what these terms mean.
- Surjective: A function f from set A to set B is surjective (onto) if for every element y in set B, there is an element x in set A such that f(x) = y. In other words, every element in set B has at least one preimage in set A.
- Injective: A function f from set A to set B is injective (one-to-one) if for every pair of distinct elements x₁ and x₂ in set A, f(x₁) and f(x₂) are distinct in set B. In other words, no two distinct elements in set A have the same image in set B.
- Bijective: A function f from set A to set B is bijective if it is both surjective and injective. In other words, every element in set B has exactly one preimage in set A, and no two distinct elements in set A have the same image in set B.
Now, let's look at the given function f(x) = {(1, 8), (2, 7), (3, 6)}.
- f is not surjective because set B has elements 5 and 9 that do not have any preimages in set A.
- f is not a function if there is an element in set A that has more than one image in set B. However, this is not the case for f, so it is a function.
- f is injective because no two distinct elements in set A have the same image in set B. For example, f(1) = 8, f(2) = 7, and f(3) = 6, so no two distinct elements in set A have the same image in set B.
- Since f is injective but not surjective, it is not bijective.
Therefore, the correct answer is option 'C' - f is injective.
To determine whether f is surjective, injective, or bijective, we need to understand what these terms mean.
- Surjective: A function f from set A to set B is surjective (onto) if for every element y in set B, there is an element x in set A such that f(x) = y. In other words, every element in set B has at least one preimage in set A.
- Injective: A function f from set A to set B is injective (one-to-one) if for every pair of distinct elements x₁ and x₂ in set A, f(x₁) and f(x₂) are distinct in set B. In other words, no two distinct elements in set A have the same image in set B.
- Bijective: A function f from set A to set B is bijective if it is both surjective and injective. In other words, every element in set B has exactly one preimage in set A, and no two distinct elements in set A have the same image in set B.
Now, let's look at the given function f(x) = {(1, 8), (2, 7), (3, 6)}.
- f is not surjective because set B has elements 5 and 9 that do not have any preimages in set A.
- f is not a function if there is an element in set A that has more than one image in set B. However, this is not the case for f, so it is a function.
- f is injective because no two distinct elements in set A have the same image in set B. For example, f(1) = 8, f(2) = 7, and f(3) = 6, so no two distinct elements in set A have the same image in set B.
- Since f is injective but not surjective, it is not bijective.
Therefore, the correct answer is option 'C' - f is injective.
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Community Answer
Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, ...
Injective because A has only one element in B.
not surjective because B does not have for every element In B as preimage in A.
not a bijective because it is not both surjective and injective.
And itz a function ,so not b option...
not surjective because B does not have for every element In B as preimage in A.
not a bijective because it is not both surjective and injective.
And itz a function ,so not b option...
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Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f isa)Surjectiveb)Not a functionc)Injectived)BijectiveCorrect answer is option 'C'. Can you explain this answer?
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