Test whether the map f : ℝ → ℝ defined by f ( x ) = x 2 + x is injective and/ or surjective.

Given f x = x 2 + x and f : ℝ → ℝ A function f : A → B . Then f is called injective (one-one) when ∀ a 1 , a 2 ∈ A , f a 1 = f a 2 ⇒ a 1 = a 2 Let x and y be any two elements in domain, such that f ( x ) = f ( y ) Consider f x = f y ⇒ x 2 + x = y 2 + y ⇒ x 2 - y 2 + x - y = 0 ⇒ x + y x - y + x - y = 0 ⇒ x - y ( x + y + 1 ) = 0 ⇒ x - y = 0 or x + y + 1 = 0 ⇒ x = - 1 - y Hence, x and y are distinct Therefore, f is not injective f is called surjective (onto) if ∀ b ∈ B , ∃ a ∈ A such that f a = b . From the above graph we can observe that, some of the negative real in codomain is not the image of some element of domain ∴ f is not surjective. Hence, the modulus function is neither injective nor surjective.

Test whether the map f : ℝ → ℝ defined by f ( x ) = x 2 + x is injective and/ or surjective.

Given f x = x 2 + x and f : ℝ → ℝ A function f : A → B . Then f is called injective (one-one) when ∀ a 1 , a 2 ∈ A , f a 1 = f a 2 ⇒ a 1 = a 2 Let x and y be any two elements in domain, such that f ( x ) = f ( y ) Consider f x = f y ⇒ x 2 + x = y 2 + y ⇒ x 2 - y 2 + x - y = 0 ⇒ x + y x - y + x - y = 0 ⇒ x - y ( x + y + 1 ) = 0 ⇒ x - y = 0 or x + y + 1 = 0 ⇒ x = - 1 - y Hence, x and y are distinct Therefore, f is not injective f is called surjective (onto) if ∀ b ∈ B , ∃ a ∈ A such that f a = b . From the above graph we can observe that, some of the negative real in codomain is not the image of some element of domain ∴ f is not surjective. Hence, the modulus function is neither injective nor surjective.

Test whether the map f : ℝ → ℝ defined by f ( x ) = x x is injective and/ or surjective.

Given f x = x x We know that x = x i f x ≥ 0 - x i f x < 0 ⇒ f x = x 2 if x ≥ 0 - x 2 if x < 0 The graph of the given function is If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. We can observe, no line can be drawn which intersects the graph more than once, therefore f is an injective function From the above graph we can observe that, every element in codomain is the image of some element of domain Therefore, f is surjective . Hence, the given function injective and surjective.

Show that the mapping f : ℕ → ℕ defined by f n = 2 n is injective but not surjective

Given f n = 2 n A function f : A → B . Then, f is called injective (one-one) when ∀ a 1 , a 2 ∈ A , f a 1 = f a 2 ⇒ a 1 = a 2 Consider f a 1 = f a 2 ⇒ 2 a 1 = 2 a 2 Which is possible, only when a 1 = a 2 Therefore, f in an injective. A function f : A → B . Then, f is called surjective (onto) if ∀ b ∈ B , ∃ a ∈ A such that f a = b . 1 ∈ ℕ (codomain) is not the image of any element of domain ℕ Therefore, f is not surjective Hence, f is an injective but not surjective

Show that the mapping f : ℕ → ℕ × ℕ defined by f n = 2 n , 3 n is injective but not surjective

Given f : ℕ → ℕ × ℕ defined by f n = 2 n , 3 n An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain. The function f : A → B is said to be injective if for all x and y in A , Whenever f ( x ) = f ( y ) then x = y We have x , y ∈ ℕ , consider f x = f y ⇒ 2 x , 3 x = 2 y , 3 y ⇒ x = y Therefore, f is injective A function f : A → B is said to be onto function (surjective) then for every b ∈ B there exist an element a ∈ A such that f a = b For 1 , 1 ∈ ℕ × ℕ there is no element exist n ∈ ℕ such that f ( n ) = ( 2 n , 3 n ) Therefore, f is not surjective. Hence, f is injective but not surjective

Show that the mapping f : ℕ → ℕ defined by f n = n + 2 is injective but not surjective

Given f : ℕ → ℕ defined by f n = n + 2 An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain. The function f : A → B is said to be injective if for all x and y in A , Whenever f ( x ) = f ( y ) then x = y We have x , y ∈ ℕ , consider f x = f y ⇒ x + 2 = y + 2 ⇒ x = y Therefore, f is injective A function f : A → B is said to be onto function (surjective) then for every b ∈ B there exist an element a ∈ A such that f a = b For 1 ∈ ℕ (codomain) there is no element exist n ∈ ℕ (domain) such that f ( n ) = 1 Therefore, f is not surjective. Hence, f is injective but not surjective

Show that the mapping f : 0 . 2 π → ℝ defined by f x = tan x is surjective but not injective

Given f : [ 0 . 2 π ] → ℝ defined by f x = tan x An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain. The function f : A → B is said to be injective if for all x and y in A , Whenever f ( x ) = f ( y ) then x = y We have π 4 , 3 π 4 ∈ 0 , 2 π , consider f π 4 = f 3 π 4 ⇒ tan π 4 = tan 3 π 4 ⇒ 1 = 1 but π 4 ≠ 3 π 4 Therefore, f is not injective A function f : A → B is said to be onto function (surjective) then for every b ∈ B there exist an element a ∈ A such that f a = b We know that, range of tan x is a real number set in the given domain Therefore, f is surjective. Hence, f is surjective but not injective

Show that the mapping f : ℝ 2 → ℝ defined by f x , y = x y is surjective but not injective

Given f : ℝ 2 → ℝ defined by f x , y = x y An injective function (injection) or one-to-one function is a function that maps distinct elements of its domain to distinct elements of its codomain. The function f : A → B is said to be injective if for all x and y in A , Whenever f ( x ) = f ( y ) then x = y We have 1 , 2 , 2 , 1 ∈ 0 , 2 π , consider f 1 , 2 = f 2 , 1 ⇒ 1 × 2 = 2 × 1 ⇒ 2 = 2 but 1 , 2 ≠ 2 , 1 Therefore, f is not injective A function f : A → B is said to be onto function (surjective) then for every b ∈ B there exist an element a ∈ A such that f a = b f ( x , y ) = x y ⇒ x = f x , y y Substitute the above function , we get f f x , y y , y = f x , y y y = f x , y Therefore, f is surjective. Hence, f is surjective but not injective

If f : ℝ → ℝ is defined by f x = x 2 + 5 x ∈ ℝ find f - 1 6

Given f x = x 2 + 5 Let f x = y ⇒ x = f - 1 y ⇒ y = x 2 + 5 ⇒ x 2 = y - 5 ⇒ x = ± y - 5 Therefore, f - 1 6 = ± 6 - 5 = ± 1

A mapping f : ℝ → ℝ is defined as follow = f x 1 , if x is rational - 1 , if x is irrational . Find range of f . What is the value of f 7 ? Is f one ‐ one and onto?

Given f ( x ) = 1 , if x is rational − 1 , if x is irrational. It is seen that f 2 = 1 , f 3 = 1 ∵ 2 , 3 are rational. ∴ f 2 = f 3 but 2 ≠ 3 . ∴ f is not one-one (injective). Now, consider 3 ∈ ℝ But it is known that f x = 1 or - 1 Thus, there does not exist any element x in domain ℝ such that f x = 3 ∴ f is not onto (surjective). Hence, the given function is neither injective nor surjective. ∴ range of f is - 1 , 1 And f 7 = - 1 ∵ 7 is irrational.