Let A 1 2 3 4 5 6 B 2 4 6 8 Find A B A N D B A

Solved 1 Let A 1 2 3 4 5 6 7 B 1 3 5 8 9 C 1 4 6 8 Chegg Example 18 let a = { 1, 2, 3, 4, 5, 6}, b = { 2, 4, 6, 8 }. find a – b and b – a. a – b = a – (a ∩ b) a ∩ b = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6, 8} = {2, 4, 6} a – b = a – (a ∩ b) = {1, 2, 3, 4, 5, 6} – {2, 4, 6} = {1, 3, 5} b – a = b – (b ∩ a) = b – (a ∩ b) = {2,. Then the sets a∪b and a∩b are. if a = {1,2,3,4,5,6} and b = {2,4,6,8}, then a b = ? let a = {2,4,6,8} and b ={6,8,10,12}. find a∪b.

Solved Let U 1 2 3 4 5 6 A 1 2 4 And B 3 4 5 6 Write Chegg To ask unlimited maths doubts download doubtnut from goo.gl 9wzjcw let a ={ 1, 2, 3, 4, 5, 6 }, b= { 2 ,4 ,6 ,8 }. find `a b` and `b a`. Solve your math problems using our free math solver with step by step solutions. our math solver supports basic math, pre algebra, algebra, trigonometry, calculus and more. Let a = {1, 2, 3, 4, 5} and b = {1, 2, 3, 4, 5, 6}. then the number of functions f : a → b satisfying f (1) f (2) = f (4) – 1 is equal to. One to one function: a function f:a → b is one to one if every element of the range b corresponds to exactly one element of the domain a of f. important: for a function f:a → b to be one to one, the range must have more number of elements than the domain. calculation:.

8 Let A 1 2 3 4 5 B 3 4 5 6 7 And C Chegg Let a = {1, 2, 3, 4, 5} and b = {1, 2, 3, 4, 5, 6}. then the number of functions f : a → b satisfying f (1) f (2) = f (4) – 1 is equal to. One to one function: a function f:a → b is one to one if every element of the range b corresponds to exactly one element of the domain a of f. important: for a function f:a → b to be one to one, the range must have more number of elements than the domain. calculation:. Given a = {1, 2, 3}, b = {3, 4}, c ={4, 5, 6}, find (a × b) ∩ (b × c ). if a = {1, 2, 3}, b = {4}, c = {5}, then verify that: (i) a × ( b ∪ c ) = ( a × b ) ∪ ( a × c ). Let r be the relation on a defined by {(a, b): a, b ∈ a, b is exactly divisible by a}. (iii) find the range of r r = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)} range of r = set of second elements of relation = {1, 2, 3, 4, 6}. Let u = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, a = { 1, 2, 3, 4}, b = { 2, 4, 6, 8 } and c = { 3, 4, 5, 6 }. find (i) a′ (ii) b′ (iii) (a ∪ c)′ (iv) (a ∪ b)′ (v) (a′)′ (vi) (b – c)′. grade. A = {1, 2, 3, 5} and b = {4, 6, 9}. define a relation r from a to b by r = {(x, y): the difference between x and y is odd; x ∈ a, y ∈ b}. write r in roster form. let a = {1, 2, 3, 4, 6}. let r be the relation on a defined by {(a, b): a, b ∈ a, b is exactly divisible by a}. write r in roster form; find the domain of r; find the range of r.