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Question 98 of 104

Q.Traffic flows from 𝐷 to 𝐸 and 𝐷 to 𝐢. The department wants to represent and analyze this data using relations and functions. Use the given data to answer the following questions: I. Is the traffic flow reflexive? Justify. [1] II. Is the traffic flow transitive? Justify. [1] III
(A) Represent the relation describing the traffic flow as a set of ordered pairs. Also state the domain and range of the relation.

(OR)
III
(B) Does the traffic flow represent a function? Justify your answer. [2] 4
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Part (a)Concept understanding β€” Relation Properties

Properties of a Relation

A relation RR on a set AA pairs elements of AA with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 β€” reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.

Reflexive β€” everything relates to itself

RR is reflexive if a R aa\,R\,a for every a∈Aa\in A. "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on {1,2,3}\{1,2,3\}, {(1,1),(2,2)}\{(1,1),(2,2)\} is not reflexive because (3,3)(3,3) is absent.

Symmetric β€” the relation runs both ways

RR is symmetric if a R bβ€…β€ŠβŸΉβ€…β€Šb R aa\,R\,b \implies b\,R\,a. "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related β€” only that any pair which appears also appears reversed. So {(1,2),(2,1),(3,3)}\{(1,2),(2,1),(3,3)\} is symmetric, but {(1,2),(2,1),(1,3)}\{(1,2),(2,1),(1,3)\} is not, since (3,1)(3,1) is missing.

Transitive β€” relations chain

RR is transitive if a R ba\,R\,b and b R cb\,R\,c together force a R ca\,R\,c. "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: {(1,2),(2,3)}\{(1,2),(2,3)\} is not transitive because (1,3)(1,3) is missing.

Antisymmetric β€” two-way ties force equality

RR is antisymmetric if a R ba\,R\,b and b R ab\,R\,a together force a=ba=b. The order relation ≀\le is antisymmetric: a≀ba\le b and b≀ab\le a give a=ba=b. It does not ban self-pairs like (1,1)(1,1); it only rules out distinct elements related both ways.

Tip

Test the properties in order of ease β€” reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.

PropertyCondition
Reflexiveβˆ€a,Β a R a\forall a,\ a\,R\,a

Part (b)Concept understanding β€” Relation Properties

Properties of a Relation

A relation RR on a set AA pairs elements of AA with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 β€” reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.

Reflexive β€” everything relates to itself

RR is reflexive if a R aa\,R\,a for every a∈Aa\in A. "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on {1,2,3}\{1,2,3\}, {(1,1),(2,2)}\{(1,1),(2,2)\} is not reflexive because (3,3)(3,3) is absent.

Symmetric β€” the relation runs both ways

RR is symmetric if a R bβ€…β€ŠβŸΉβ€…β€Šb R aa\,R\,b \implies b\,R\,a. "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related β€” only that any pair which appears also appears reversed. So {(1,2),(2,1),(3,3)}\{(1,2),(2,1),(3,3)\} is symmetric, but {(1,2),(2,1),(1,3)}\{(1,2),(2,1),(1,3)\} is not, since (3,1)(3,1) is missing.

Transitive β€” relations chain

RR is transitive if a R ba\,R\,b and b R cb\,R\,c together force a R ca\,R\,c. "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: {(1,2),(2,3)}\{(1,2),(2,3)\} is not transitive because (1,3)(1,3) is missing.

Antisymmetric β€” two-way ties force equality

RR is antisymmetric if a R ba\,R\,b and b R ab\,R\,a together force a=ba=b. The order relation ≀\le is antisymmetric: a≀ba\le b and b≀ab\le a give a=ba=b. It does not ban self-pairs like (1,1)(1,1); it only rules out distinct elements related both ways.

Tip

Test the properties in order of ease β€” reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.

PropertyCondition
Reflexiveβˆ€a,Β a R a\forall a,\ a\,R\,a

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