Question

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x^2

Answer 1

(i)  it is given that f : R → R defined by f (x) = 3 – 4x
Let x₁, x₂∈ R such that f(x₁) = f(x₂)
⇒ 3 – 4x₁ = 3 – 4x₂
⇒ -4x₁ = -4x₂
⇒ x₁ = x₂
⇒ f is one- one

we know on thing if a function f(x) is inversible then f(x) is definitely a bijective function. means, f(x) will be one - one and onto.
Let's try to do innverse of f(x) = 3 - 4x
y = 3 - 4x
y - 3 = 4x => x = (y - 3)/4
f⁻¹(x) = (x - 3)/4
hence, f(x) is inversible .
so, f(x) is one - one and onto function.
hence, f(x) is bijective function [ if any function is one -one and onto then it is also known as bijective function.]

(ii) it is given that f :R→R defined by f(x) = 1 +x²
Let x₁, x₂∈ R such that f(x₁) = f(x₂)
1 + x₁² = 1 + x₂²
x₁² = x₂²
taking square root both sides ,
x₁ = ±x₂
now, f(1) = f(-1) = 2
so, f is not one - one function.

also for all real value of x , f(x) is always greater than 1 . so, range of f(x) ∈ [1,∞)
but co-domain ∈ R
e.g., Co - domain ≠ range
so, f is not onto function.
also f is not bijective function.