Let f : A → A where A = { x : - 1 ≤ x ≤ 1 } . Find whether the function is bijective. f x = x - sin x

Given a function f x = x - sin x We can see horizontal lines parallel to x - axis cutting the graph at one and only one point. Thus, by Horizontal Line Test, the function f x is one-one. Now, finding boundary values : f - 1 = - 1 - sin - 1 = - 1 + sin 1 > - 1 ∵ sin - θ = sin θ and f 1 = 1 - sin 1 = 1 - sin 1 < 1 From graph and boundary values, we can say that range of f x is - 1 + sin 1 , 1 - sin 1 . ∵ Co-domain of f x is - 1 , 1 . Here, range is a subset of co-domain. So, f x is into. Hence, f x is not bijective.

Let f : A → A where A = { x : - 1 ≤ x ≤ 1 } . Find whether the function is bijective. f x = x - sin x

Given a function f x = x - sin x We can see horizontal lines parallel to x - axis cutting the graph at one and only one point. Thus, by Horizontal Line Test, the function f x is one-one. Now, finding boundary values : f - 1 = - 1 - sin - 1 = - 1 + sin 1 > - 1 ∵ sin - θ = sin θ and f 1 = 1 - sin 1 = 1 - sin 1 < 1 From graph and boundary values, we can say that range of f x is - 1 + sin 1 , 1 - sin 1 . ∵ Co-domain of f x is - 1 , 1 . Here, range is a subset of co-domain. So, f x is into. Hence, f x is not bijective.

Let f : A → A where A = { x : - 1 ≤ x ≤ 1 } . Find whether the function is bijective. f x = x x

Given a function f x = x x ; - 1 ≤ x ≤ 1 By definition of modulus function, x = x , x ≥ 0 - x , x < 0 ⇒ f x = x x = x 2 , 0 ≤ x ≤ 1 - x 2 , - 1 ≤ x < 0 Injectivity check : f x 1 = f x 2 ⇒ x 1 = x 2 or not. Case I : 0 < x 1 ≤ 1 , 0 < x 2 ≤ 1 f x 1 = f x 2 ⇒ x 1 2 = x 2 2 ⇒ x 1 2 - x 2 2 = 0 ⇒ x 1 - x 2 x 1 + x 2 = 0 ⇒ x 1 - x 2 = 0 ⇒ x 1 = x 2 ∵ x 1 + x 2 > 0 ∀ x 1 , x 2 ∈ ( 0 , 1 ] Case II : - 1 ≤ x 1 < 0 , - 1 ≤ x 2 < 0 f x 1 = f x 2 ⇒ - x 1 2 = - x 2 2 ⇒ x 1 2 - x 2 2 = 0 ⇒ x 1 - x 2 x 1 + x 2 = 0 ⇒ x 1 - x 2 = 0 ⇒ x 1 = x 2 ∵ x 1 + x 2 < 0 ∀ x 1 , x 2 ∈ [ - 1 , 0 ) Also, f 0 = 0 . Since, x 1 has unique image for each value of x ∈ - 1 , 1 . Thus, f x is one-one i.e. injective. Surjectivity check : Now let y = f x ⇒ y = x 2 , 0 ≤ x ≤ 1 - x 2 , - 1 ≤ x < 0 Every positive real number y = r is hit by x = r and every negative value of y = r is hit by x = - r in its domain. Hence, range of f x is - 1 , 1 . Also, co-domain of f x is - 1 , 1 . Since, range equals co-domain. Thus, f x is onto i.e surjective . ∴ f x is bijective.

Let f : A → A where A = { x : - 1 ≤ x ≤ 1 } . Find whether the function is bijective. f x = tan π x 4

Given a function f x = tan π x 4 Injectivity check : Drawing graph of f x in - 1 , 1 We can see in the graph that y = 1 and y = - 1 cut the graph at only one point. Hence, horizontal line parallel to x - axis cuts the graph at exactly one point. Thus, by Horizontal Line Test, the function f x is one-one. Surjectivity check : Also from graph, f x can attain values - 1 , 1 , i.e. there is at least one pre-image of f x in its domain so, range of f x is - 1 , 1 . Since, range equals co-domain. ∴ f x is onto. Hence, f x is bijective.

Let f : A → A where A = { x : - 1 ≤ x ≤ 1 } . Find whether the function is bijective. f x = x 4

Given a function f x = x 4 ; - 1 ≤ x ≤ 1 Checking f x 1 = f x 2 ⇒ x 1 4 = x 2 4 ⇒ x 1 4 - x 2 4 = 0 ⇒ x 1 - x 2 x 1 + x 2 x 1 2 + x 2 2 = 0 ⇒ x 1 = x 2 or x 1 = - x 2 ∵ x 1 2 + x 2 2 > 0 Since, x 1 does not have unique image. Thus, f x is many-one. Example: f 1 = 1 4 = 1 and f - 1 = - 1 4 = 1 ∵ f 1 = f - 1 but 1 ≠ - 1 ⇒ f x is many-one. For bijection, function should be one-one and onto both. But here function is many-one. Hence, f x is not bijective.