Let a relation be defined on a set of functions defined on R - R such that R -f- g- f-g is an even function then R is

Reflexive, f f = 0 ∈ R Symmetric: f g is even g f is even so (g,f) ∈ R Transitive ∀ (f,g) ∈ R, (g,h) ∈ R ⇒ f g is even, g h is even ⇒ f h is even So ...

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Standard XII

Mathematics

Equivalence Relation

Let a relatio...

Question

Let a relation be defined on a set of functions defined on $R \to R$ such that R ={(f, g): f-g is an even function} then R is

Reflexive but not symmetric

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Not reflexive but symmetric & transitive

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Symmetric but not transitive

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Equivalence relation

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Solution

The correct option is D Equivalence relation
Reflexive, $f - f = 0 \in R$
Symmetric: f - g is even
g - f is even so $(g, f) \in R$
Transitive $\forall (f, g) \in R, (g, h) \in R$
$\Rightarrow f - g$ is even, g - h is even
$\Rightarrow f - h$ is even
So R is equivalence Relation.

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Let a relation be defined on a set of functions defined on R→R such that R ={(f, g): f-g is an even function} then R is

The correct option is D Equivalence relationReflexive, f−f=0∈R Symmetric: f - g is even g - f is even so (g,f)∈R Transitive ∀(f,g)∈R,(g,h)∈R ⇒f−g is even, g - h is even ⇒f−h is even So R is equivalence Relation.

Let a relation be defined on a set of functions defined on $R \to R$ such that R ={(f, g): f-g is an even function} then R is

The correct option is D Equivalence relation
Reflexive, $f - f = 0 \in R$
Symmetric: f - g is even
g - f is even so $(g, f) \in R$
Transitive $\forall (f, g) \in R, (g, h) \in R$
$\Rightarrow f - g$ is even, g - h is even
$\Rightarrow f - h$ is even
So R is equivalence Relation.