# CLASS-11RELATION & FUNCTION-TYPES OF FUNCTION-INJECTIVE FUNCTION (ONE-ONE)

One-One Function (Injective Function)

1.) A function f : A → B is called a one-one or injective function if distinct elements of A have distinct images in B, i.e., if a₁, a₂ ∈ A and a₁ ≠ a₂

=> f(a₁) ≠ f(a₂)

Equivalently, we say f : A → B is one-one if and only if for all a₁ a₂ ∈ A, f(a₁) = f(a₂)

=>  a₁ = a₂

In terms of picture, we say that in one – one function, no more than one arrow terminates at any given element of B. Illustration -

1) If A = {4, 5, 6} and B = {a, b, c, d} and if f : A → B such that f = {(4, a), (5, b), (6, c)}, then f is 1 – 1 2) The mapping f : R → R such that f(x) = x² is not a one-one function since f(-2) = 4 and f(2) = 4, that is, two distinct elements – 2 and 2 have the same image 4.

Example.1) Find whether the following functions are one-one or not.

(i) f : R → R, defined by f(x) = x³, x ∈ R

(ii) f : Z → Z, defined by f(x) = x² + 5 for all x ∈ Z

5x + 7

(iii) f  : R – {3} → R, defined by f(x) = ----------, x ∈ R – {3}

x – 3

Method of check whether a function is one-one or not

Step.1) Take two arbitrary elements a₁, a₂ in the domain of f

Step.2) Put f(a₁) = f(a₂) and solve the equation

Step.3) If it yields a₁ = a₂, only then f : A → B is a one-one function otherwise not.

Remark. Let f : A → B and let a₁, a₂ ∈ A. Then a₁ = a₂ => f(a₁) = f(a₂) is always true from the definition, but f(a₁) = f(a₂) => a₁ = a₂ is true only when f is one-one.

Sol. (i) Let a₁, a₂ be two arbitrary elements of domain f, (a₁, a₂ ∈ R) such that f(a₁) = f(a₂)

Then f(a₁) = f(a₂) => a₁³ = a₂³ => a₁ = a₂

Hence, f is a one-one function.

(ii) Let a₁, a₂ be two arbitrary elements of Z such that f(a₁) = f(a₂) => a₁² + 5 = a₂² + 5

=> a₁² = a₂²

=> a₁ = ± a₂

Since f(a₁) = f(a₂) does not yield a unique answer a₁ = a₂ but gives a₁ = ± a₂, so f is not a one-one function.

In fact, we have f(2) = 2² + 5 = 9 and f(- 2) = (- 2)² + 5 = 9.

Thus two distinct elements 2 and – 2 have the same image.

5a₁ + 7

(iii) We have f(a₁) =  ----------, x  ∈ R – {3}

a₁ - 3

let, a₁, a₂ be two arbitrary elements of domain f(a₁, a₂) ∈ R – {3} such that f(a₁) = f(a₂)

5a₁ + 7          5a₂ + 7

Then, f(a₁) = f(a₂) => ----------- = -----------

a₁ - 3            a₂ - 3

=>  5a₁a₂ + 7a₂ - 15a₁ - 21 = 5a₁a₂ + 7a₁ - 15a₂ - 21

=>  - 15 a₁ - 7 a₁ = - 15 a₂ - 7 a₂

=>        - 22 a₁ = - 22 a₂

=>             a₁ =  a₂

So, f is one-one

Example.2) Show that the modulus function f : R → R, given by f(x) = ǀxǀ, is not one function

Ans.) given => f(x) = ǀxǀ

=> f(1) = ǀ1ǀ = 1  and f(- 1) = ǀ- 1ǀ = 1 Thus,  - 1 ≠ 1 but f(- 1) = f(1)

Hence, f : A → B is not one-one function.

It is a many one function