What is the difference between a relation and a function?

0 3 Views | Posted 6 months ago

  • 1 Answer

  • A

    Answered by

    Aayush Kumari

    6 months ago

    A relation deals with, how elements from any set  (A) are related to elements of another set (B). there are various form of relations, such as Reflexive, Symatric, Transitive and Equivalance. On the other hand, Functions are special case of Relations where every element in Set A is related to exactly one element in Set B. Functions works like mathematics formulas, for a given set of input the out put will always remain the same. Students will be abel to check the type of relations, and application in class 11 Math Relation and Function chapter. We have provided Class 11 Math Relations and Functions NCERT Solutions on shiskha 

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Similar Questions for you

A
alok kumar singh

Reflexive :for (a, b) R (a, b)

-> ab– ab = 0 is divisible by 5.

So (a, b) R (a, b) " a, b Î Z

R is reflexive

Symmetric:

For (a, b) R (c, d)

If ad – bc is divisible by 5.

Then bc – ad is also divisible by 5.

-> (c, d) R (a, b) "a, b, c, dÎZ

R is symmetric

Transitive:

If (a, b) R (c, d) ->ad –bc divisible by 5 and (c, d) R (e, f) Þcf – de divisible by 5

ad – bc = 5k1               k1 and k2 are integers

cf– de = 5k2

afd – bcf = 5k1f

bcf – bde = 5k2b

afd – bde = 5 (k1f + k2b)

d (af– be) = 5 (k1f + k2b)

-> af – be is not divisible by 5 for every a, b,

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V
Vishal Baghel

x = n = 0 c o s 2 n θ = c o s 0 θ + c o s 2 θ + c o s 4 θ + . . . . . = 1 + c o s 2 θ + c o s 4 θ + . . . . .

a = 1, r = cos2

x = S = a 1 r = 1 1 c o s 2 θ = 1 s i n 2 θ

Similarly, y = 1 c o s 2 θ 1 y = c o s 2 θ

z = x y x y 1 x y z z = x y . . . . . . . . . . . ( i )

Also, 1 x + 1 y = 1 x + y = x y . . . . . . . . . . ( i i )

(i) & (ii) xyz = xy + z (x + y) z = xy + z

V
Vishal Baghel

Total number of possible relation = 2 n 2 = 2 4 = 1 6

Favourable relations = ? , { ( x , x ) } , { ( y , y ) }

{ ( x , x ) , ( y , y ) }

{ ( x , x ) , ( y , y ) , ( x , y ) , ( y , x ) }

Probability = 5 1 6

V
Vishal Baghel

A = {1, 2, 3, 4}. As (4, 4)  R so not reflexive

Also not transitive

V
Vishal Baghel

3 2 x 2 1 x 0 p u t 2 x = t

3 t 2 t 0 t 2 3 t + 2 0

( t 1 ) ( t 2 ) 0 t [ 1 , 2 ]

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