Show that the function f : Q → Q defined by f ( x ) = 3 x - 2 is one-one.

Given f : Q → Q be defined by f ( x ) = 3 x - 2 We know that, a function f : A → B is said to be a one-one function or an injection, if different elements of A have different images in B . ∀ x , y ∈ Q , f ( x ) = f ( y ) ⇒ 3 x - 2 = 3 y - 2 ⇒ 3 x - 3 y ⇒ x = y ∴ f is one-one.

Show that the function f : N → N defined by f ( x ) = 2 x - 1 is not onto.

Given function be f : N → N defined by f ( x ) = 2 x - 1 We know that, an onto function is such that for every element in the codomain there exists an element in domain which maps to it. Clearly 2 ∈ N . Then f ( x ) = 2 ⇒ 2 x - 1 = 2 ⇒ x = 3 2 ∉ N Thus element 2 has no pre-image in N (domain of f ). ∴ f is not onto.

Show that the function f : N → N defined by f ( n ) = n 2 is injective.

Given function be f : N → N defined by f ( n ) = n 2 We know that, a function f : A → B is said to be a one-one function or an injection, if different elements of A have different images in B . ∀ n , m ∈ N , f ( n ) = f ( m ) ⇒ n 2 = m 2 ⇒ ( n - m ) ( n + m ) = 0 ⇒ n = m [ ∵ n + m > 0 ∀ n , m ∈ N ] ∴ f is one-one or injective.

In the function f : N → N defined by f ( m ) = m 2 + m + 2 one-one? Justify your answer.

Given that, f : N → N defined by f ( m ) = m 2 + m + 2 We know that, a function f : A → B is said to be a one-one function or an injection, if different elements of A have different images in B . ∀ n , m ∀ N , f ( n ) = f ( m ) ⇒ n 2 + m + 2 = m 2 + m + 2 ⇒ n 2 - m 2 + n - m = 0 ⇒ ( n - m ) ( n + m + 1 ) = 0 ⇒ n - m ) 0 [ ∵ n + m + 1 > 0 ∀ n , m ∈ N ] ⇒ n = m Hence f is injective. ∴ f is one-one.

What is the range of the function f ( x ) = | x - 1 | x - 1 ?

Let y = f ( x ) = | x - 1 | x - 1 , x ≠ 1 We know that, a modulus function is a function which gives the absolute value of a number or variable.When x > 0 then f x = x and when x < 0 then f x = - x . For x - 1 > 0 ⇒ x - 1 = x - 1 f ( x ) = x - 1 x + 1 = 1 ∵ x ≠ 1 For x - 1 < 0 ⇒ | x - 1 | = - ( x - 1 ) ∴ f ( x ) = - ( x - 1 ) x - 1 = - 1 [ ∵ x ≠ 1 ] Thus, R f = { - 1 , 1 }

Find the domain of the function f ( x ) = 1 5 - 3 sin x .

Given f ( x ) = 1 5 - 3 sin x For D f : f ( x ) must be a real number. We know that, domain of sin ( x ) is all real numbers ⇒ 1 5 - 3 sin x must be a real number. ⇒ 5 - 3 sin x ≠ 0 which is true ∀ x ∈ R [ ∵ - 1 ≤ sin x ≤ 1 ⇒ 3 ≥ - 3 sin x ≥ - 3 ⇒ 5 + 3 ≥ 5 - 3 sin x ≥ 5 - 3 ⇒ 2 ≤ 5 - 3 sin x ≤ 8 ] Hence, D f = R

Find the range of the function f ( x ) = 1 2 - cos x

Let y = f ( x ) = 1 2 - cos x We know that, the range of a cos x function is - 1 , 1 . Now, - 1 ≤ cos x ≤ 1 ⇒ 1 ≥ - cos x ≥ - 1 ⇒ 2 + 1 ≥ 2 - cos x ≥ 2 - 1 ⇒ 1 ≤ 2 - cos x ≤ 3 On dividing both sides the equality sign changes. ⇒ 1 ≥ 1 2 - cos x ≥ 1 3 ⇒ 1 3 ≤ y ≤ 1 ∴ R f = 1 3 , 1 is the range of a function.