Let f be the subset of Q × Z defined by f = m n , m : m , n ∈ Z , n ≠ 0 . Is f a function from Q to Z ? Justify your answer.

Given, f = m n , m : m , n ∈ Z , n ≠ 0 ∴ f m n = m , where m , n ∈ Z Now, when m = 1 , n = 2 then f 1 2 = 1 ∈ Z And when m = 2 , n = 4 then f 2 4 = 2 ∈ Z But here 1 2 = 2 4 . So by the definition of function, one element cannot have more than one element as image. Hence, given f is not a function.

Let f be the subset of Q × Z defined by f = m n , m : m , n ∈ Z , n ≠ 0 . Is f a function from Q to Z ? Justify your answer.

Given, f = m n , m : m , n ∈ Z , n ≠ 0 ∴ f m n = m , where m , n ∈ Z Now, when m = 1 , n = 2 then f 1 2 = 1 ∈ Z And when m = 2 , n = 4 then f 2 4 = 2 ∈ Z But here 1 2 = 2 4 . So by the definition of function, one element cannot have more than one element as image. Hence, given f is not a function.

Consider the function f ( x ) = x + 1 x , x ∈ R , x ≠ 0 . Is f One-one ?

Given function, f ( x ) = x + 1 x , x ∈ R , x ≠ 0 Let x = 2 then f 2 = 2 + 1 2 = 5 2 And x = 1 2 then f 1 2 = 1 2 + 1 1 2 = 1 2 + 2 = 5 2 Here different values of x we are getting the same result. By the definition of One-one function different elements should have different images. Hence, f is not One-one function.

Show that the function f : N → N defined by f ( x ) = 3 x - 2 is One-one function but not Onto.

Given function f : N → N defined by f ( x ) = 3 x - 2 . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ N , then its a One-one function. Let x 1 , x 2 ∈ N be such that f x 1 = f x 2 ⇒ 3 x 1 - 2 = 3 x 2 - 2 ⇒ 3 x 1 = 3 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ N (co-domain of f ) Let y = f ( x ) = 3 x - 2 ⇒ y + 2 = 3 x ⇒ y + 2 3 = x Thus , for every y ∈ N (Co-domain of f ), there does not exist x ∈ N (Domain of f ) Hence, the f is not Onto.

Find the domain of the real function f x = 1 x - x .

Given, f x = 1 | x | - x For the domain of a function D f , f x must be a real number. ⇒ 1 | x | - x must be a real number ⇒ | x | - x > 0 ⇒ | x | > x . . . . 1 We know that, x = x for x ≥ 0 - x for x < 0 First case: x = x then inequality one becomes x - x > 0 ⇒ 0 > 0 , which is not possible. Second case: x = - x then inequality one becomes - x - x > 0 ⇒ - 2 x > 0 Therefore, x < 0 . ∴ D f = set of all negative real numbers except at x = 0 Hence, D f = ( - ∞ , 0 ) .

Find the domain and range of the real function f ( x ) = x - 3 .

Let y = f ( x ) = x - 3 Domain : For the domain of a function D f , f x must be a real number. ⇒ x - 3 must be a real number ⇒ x - 3 ≥ 0 (Square root never be negative) ⇒ x ≥ 3 Since we know If x ≥ a then x ∈ [ a , ∞ ) Hence, D f = [ 3 , ∞ ) . Range : Since, x ≥ 3 ⇒ x - 3 ≥ 3 - 3 (Added - 3 on both side of inequality) ⇒ x - 3 ≥ 0 ⇒ x - 3 ≥ 0 (Since square root is always positive) ⇒ y ≥ 0 Given Hence, the range of the function is [ 0 , ∞ ) .