Let f : A → B be a function defined by f x = x 2 - 1 , where A = { 2 , 4 , 6 , 10 , 12 } , B = { 0 , 1 , 2 , 4 , 5 , 9 } . Represent f by an arrow diagram.

Given f : A → B , A = { 2 , 4 , 6 , 10 , 12 } and B = { 0 , 1 , 2 , 4 , 5 , 9 } Here, f x = x 2 - 1 f 2 = 2 2 - 1 = 1 - 1 = 0 f 4 = 4 2 - 1 = 2 - 1 = 1 f 6 = 6 2 - 1 = 3 - 1 = 2 f 10 = 10 2 - 1 = 5 - 1 = 4 f 12 = 12 2 - 1 = 6 - 1 = 5 The representation of the values in arrow diagram is

Let f : A → B be a function defined by f x = x 2 - 1 , where A = { 2 , 4 , 6 , 10 , 12 } , B = { 0 , 1 , 2 , 4 , 5 , 9 } . Represent f by an arrow diagram.

Given f : A → B , A = { 2 , 4 , 6 , 10 , 12 } and B = { 0 , 1 , 2 , 4 , 5 , 9 } Here, f x = x 2 - 1 f 2 = 2 2 - 1 = 1 - 1 = 0 f 4 = 4 2 - 1 = 2 - 1 = 1 f 6 = 6 2 - 1 = 3 - 1 = 2 f 10 = 10 2 - 1 = 5 - 1 = 4 f 12 = 12 2 - 1 = 6 - 1 = 5 The representation of the values in arrow diagram is

Let f : A → B be a function defined by f x = x 2 - 1 , where A = { 2 , 4 , 6 , 10 , 12 } , B = { 0 , 1 , 2 , 4 , 5 , 9 } . Represent f by A graph

Given f : A → B , A = { 2 , 4 , 6 , 10 , 12 } and B = { 0 , 1 , 2 , 4 , 5 , 9 } The graph of a function f is the set of ordered pairs ( x , y ) , where f ( x ) = y . In the common case where x and f ( x ) are real numbers, these pairs are Cartesian coordinates of points in two-dimensional space and thus form a subset of this plane. f x = x 2 - 1 f 2 = 2 2 - 1 = 1 - 1 = 0 f 4 = 4 2 - 1 = 2 - 1 = 1 f 6 = 6 2 - 1 = 3 - 1 = 2 f 10 = 10 2 - 1 = 5 - 1 = 4 f 12 = 12 2 - 1 = 6 - 1 = 5 The points to be plotted are 2 , 0 , 4 , 1 , 6 , 2 , 8 , 3 , 10 , 4 , 12 , 5 The representation of the values in graph is

Show that the function f : N → N defined by f ( x ) = 2 x - 1 is one-one but not onto.

Given f : N → N and f x = 2 x - 1 ∵ N = 1 , 2 , 3 , 4 , 5 , … f ( 1 ) = 2 ( 1 ) - 1 = 1 f ( 2 ) = 2 ( 2 ) - 1 = 3 f ( 3 ) = 2 ( 3 ) - 1 = 5 f ( 4 ) = 2 ( 4 ) - 1 = 7 f ( 5 ) = 2 ( 5 ) - 1 = 9 In the figure, for different elements in x , there are different images in f ( x ) Hence, f : N → N is a one-one function. A function f : N → N is said to be onto function if the range of f is equal to the co-domain of f Range = { 1 , 3 , 5 , 7 , 9 , … } As given co-domain is natural numbers = { 1 , 2 , 3 , … } But here the range is not equal to co-domain. Therefore, the given function is one-one but not onto function.

Show that the function f : N → N defined by f ( m ) = m 2 + m + 3 is one-one function.

Given f : N → N and f ( m ) = m 2 + m + 3 A function f : A → B is called one-one function if distinct element of A have distinct images in B . N = { 1 , 2 , 3 , 4 , 5 … . . } m ∈ N ∵ f ( m ) = m 2 + m + 3 ∴ f ( 1 ) = 1 2 + 1 + 3 = 5 f ( 2 ) = 2 2 + 2 + 3 = 9 f ( 3 ) = 3 2 + 3 + 3 = 15 f ( 4 ) = 4 2 + 4 + 3 = 23 From the figure, from different elements in the X domain, there are different images in f ( x ) . Thus, f : N → N is a one-one function. Hence, proved.

Let A = { 1 , 2 , 3 , 4 } and B = N . Let f : A → B be defined by f ( x ) = x 3 then find the range of f .

Given A = { 1 , 2 , 3 , 4 } and B = N ∴ B = { 1 , 2 , 3 , 4 , 5 , … . } Now, f ( x ) = x 3 f ( 1 ) = 1 3 = 1 f ( 2 ) = 2 3 = 8 f ( 3 ) = 3 3 = 27 f ( 4 ) = 4 3 = 64 If f : A → B then the range of any function is defined by the set of all images of the elements of X under f . Thus, the required range = { 1 , 8 , 27 , 64 }

Let A = { 1 , 2 , 3 , 4 } and B = ℕ . Let f : A → B be defined by f ( x ) = x 3 then identify the type of function.

Given A = { 1 , 2 , 3 , 4 } and B = N ∴ B = { 1 , 2 , 3 , 4 , 5 , … . } Now, f ( x ) = x 3 f ( 1 ) = 1 3 = 1 f ( 2 ) = 2 3 = 8 f ( 3 ) = 3 3 = 27 f ( 4 ) = 4 3 = 64 Here for all the elements of A there are distinct images. So, the function is one-one function. But for some images of B there is no elements in A . So, the function is into function Thus, the given function is one-one and into function.