CA Foundation Exam > CA Foundation Questions > {(x, y)|x<y} isa)not a functionb)a functi...
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{(x, y)|x<y} is
- a)not a function
- b)a function
- c)one-one mapping
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
{(x, y)|x<y} isa)not a functionb)a functionc)one-one mappingd)none...
Given set {(x, y)|xy}, we need to determine whether it represents a function or not.
Explanation:
A function is a relation between two sets in which each element of the first set (called the domain) is paired with exactly one element of the second set (called the range). In other words, if (x, y) is a part of a function, then there cannot be another element (x, z) in the same function where z is not equal to y.
Let's consider the given set {(x, y)|xy}. Here, the element (x, y) is included in the set if and only if xy is true (i.e., not equal to zero).
However, this set does not satisfy the definition of a function because for some values of x, there can be multiple values of y that satisfy the condition xy ≠ 0. For example, if x = 2, then both (2, 1) and (2, -1) are included in the set, since 2*1 ≠ 0 and 2*(-1) ≠ 0. Therefore, there is no unique value of y corresponding to each value of x, which violates the definition of a function.
Hence, the correct answer is option 'A': not a function.
Explanation:
A function is a relation between two sets in which each element of the first set (called the domain) is paired with exactly one element of the second set (called the range). In other words, if (x, y) is a part of a function, then there cannot be another element (x, z) in the same function where z is not equal to y.
Let's consider the given set {(x, y)|xy}. Here, the element (x, y) is included in the set if and only if xy is true (i.e., not equal to zero).
However, this set does not satisfy the definition of a function because for some values of x, there can be multiple values of y that satisfy the condition xy ≠ 0. For example, if x = 2, then both (2, 1) and (2, -1) are included in the set, since 2*1 ≠ 0 and 2*(-1) ≠ 0. Therefore, there is no unique value of y corresponding to each value of x, which violates the definition of a function.
Hence, the correct answer is option 'A': not a function.
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Community Answer
{(x, y)|x<y} isa)not a functionb)a functionc)one-one mappingd)none...
Some relations are not functions
ex : x=y^2 here is function of y
but y is not function of x
ex : x=y^2 here is function of y
but y is not function of x
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{(x, y)|x<y} isa)not a functionb)a functionc)one-one mappingd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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