What are the topics discussed in Chapter 2 Relations and Functions class 11 Mathematics?

0 8 Views | Posted 6 months ago
Asked by Nishtha Datta

  • 1 Answer

  • S

    Answered by

    Satyendra Dhyani

    6 months ago

    Class 11 Maths Chapter 2 Relations and Functions explores how elements from one set can be connected or related to elements from another.  Students get to learn many concepts, can be checked below;

    • Concept of the Cartesian product of sets
    • Concept of Relations, and how to represent those connections using different methods like arrow diagrams or set notation.
    • Idea of a function, and different types of functions such as one-one, onto, bijective, constant, and identity functions.
    • Real-valued functions, like linear, quadratic, modulus, and the greatest integer function.

    Students can use NCERT Solutions for Chapter 2 Relations and Func

    ...more

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,

...more
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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