Let A be the set of all 30 students of class in a school. Let f : A → N be a function defined by f ( x ) = Roll number of the student x . Show that f is one-one but not onto.

Given A be the set of all 30 students of class in a school and f : A → N be a function defined by f ( x ) = Roll number of the student x . We have ot show that f is one-one but not onto. Without any loss of generality, let us assume that the the roll number of students are from 1 to 30 . Since, no two different students of the class can have same roll number. Therefore, f is one-one. Also 31 ∈ N , the co-domain, but 31 is not the roll number of any student of the class. Thus, 31 is not the image of any element of A under f . Hence, f is not onto. Or Range of f i . e . , the set of roll numbers of 30 students = { 1 , 2 , 3 , . . . . . . . . 30 } is not equal to co-domaiin N . ∴ f is not onto.

Let A be the set of all 30 students of class in a school. Let f : A → N be a function defined by f ( x ) = Roll number of the student x . Show that f is one-one but not onto.

Given A be the set of all 30 students of class in a school and f : A → N be a function defined by f ( x ) = Roll number of the student x . We have ot show that f is one-one but not onto. Without any loss of generality, let us assume that the the roll number of students are from 1 to 30 . Since, no two different students of the class can have same roll number. Therefore, f is one-one. Also 31 ∈ N , the co-domain, but 31 is not the roll number of any student of the class. Thus, 31 is not the image of any element of A under f . Hence, f is not onto. Or Range of f i . e . , the set of roll numbers of 30 students = { 1 , 2 , 3 , . . . . . . . . 30 } is not equal to co-domaiin N . ∴ f is not onto.

Prove that the function f : R → R , given by f x = 2 x is one-one and onto.

Given f : R → R and f x = 2 x We know that A function f : X → Y is one-one, for every x , y ∈ X , if f x = f y then x = y Take any two real numbers x 1 and x 2 f x 1 = f x 2 ⇒ 2 x 1 = 2 x 2 ⇒ x 1 = x 2 ∴ f is one-one We know that A function f : X → Y , for every y ∈ Y there exists an element x ∈ X such that f ( x ) = y Take y = f x ⇒ y = 2 x ⇒ x = y 2 Here for every real number y , there exists a real number x ∴ f is an onto function ∴ The given function is one-one and onto Hence Proved

Let A = x ∈ R : - 1 ≤ x ≤ 1 and f : A → A be a mapping defined by f x = x x . Then f is

neither injective nor surjective

Show that the function f : ℝ → ℝ defined by f ( x ) = x 3 is one-one and onto.

Given that the function f : ℝ → ℝ is defined by f ( x ) = x 3 We need to show that it is one-one, and onto function. Let f x 1 = f x 2 for all x 1 , x 2 ∈ R ⇒ x 1 3 = x 2 3 ⇒ x 1 3 - x 2 3 = 0 ⇒ x 1 - x 2 x 1 2 + x 1 x 2 + x 2 2 = 0 ⇒ x 1 - x 2 = 0 Observe that x 1 2 + x 1 x 2 + x 2 2 = 0 ⇒ x 1 2 + x 2 2 = - x 1 x 2 which is not possible for all x 1 , x 2 ∈ R Hence, f x 1 = f x 2 ⇒ x 1 = x 2 ∀ x 1 , x 2 ∈ R Thus, the given function is one-one. Now, let y = f x ⇒ y = x 3 ⇒ x = y 1 3 Therefore, every image y ∈ R has a unique pre-image y 1 3 in R . ⇒ f is onto function. Hence, Proved.

Let f x = sin π 6 sin π 2 sin x , for all x ∈ R and g x = π 2 sin x , for all x ∈ R . Let f o g x denote f g x and g o f x denote g f x . Then which of the following is (are) true?

lim x → 0 f ( x ) g ( x ) = π 6

There is an x ∈ R such that g o f x = 1

Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.

Show that the function f : R → R defined by f x = x x 2 + 1 , ∀ x ∈ ℝ is neither one-one nor onto. Also, if g : R → R is defined as g ( x ) = 2 x - 1 , find fog ( x ) .

Given, y = x x 2 + 1 ⇒ y x 2 − x + y = 0 Here a = y , b = - 1 and c = y ∴ x = − ( − 1 ) ± 1 − 4 y 2 2 y Clearly, for every value of y , x will have two different values . So the function is many-one not one - one. Since 1 − 4 y 2 ≥ 0 ⇒ ( 1 + 2 y ) ( 1 − 2 y ) ≥ 0 ⇒ − 1 2 ≤ y ≤ 1 2 That means no matter what is x , y belongs to the interval − 1 2 , 1 2 . Hence, the function is not onto. Now, f o g x = f g x = f 2 x - 1 = 2 x - 1 2 x - 1 2 + 1 = 2 x + 1 4 x 2 - 4 x + 1 + 1 = 2 x + 1 2 2 x 2 - 2 x + 1