Find whether following function is One-one or not, f : R → R defined by f ( x ) = x 2 + 5 .

Given function, f ( x ) = x 2 + 5 We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ R , then the function is One-one function. Let x 1 , x 2 ∈ R be such that f x 1 = f x 2 ⇒ x 1 2 + 5 = x 2 2 + 5 ⇒ x 1 2 = x 2 2 ⇒ x 1 = ± x 2 Therefore, by definition f is not One-one function.

Find whether following function is One-one or not, f : R → R defined by f ( x ) = x 2 + 5 .

Given function, f ( x ) = x 2 + 5 We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ R , then the function is One-one function. Let x 1 , x 2 ∈ R be such that f x 1 = f x 2 ⇒ x 1 2 + 5 = x 2 2 + 5 ⇒ x 1 2 = x 2 2 ⇒ x 1 = ± x 2 Therefore, by definition f is not One-one function.

Find whether following function is One-one or not, f : R - { 3 } → R defined by f x = 5 x + 7 x - 3 for x ∈ R - { 3 } .

Given, f : R - { 3 } → R is defined by f x = 5 x + 7 x - 3 We know that, if f a = f b ⇒ a = b , a , b ∈ R , then the function is One-one function. Let a , b ∈ R be such that f a = f b ∴ 5 a + 7 a - 3 = 5 b + 7 b - 3 ⇒ ( 5 a + 7 ) ( b - 3 ) = ( a - 3 ) ( 5 b + 7 ) ⇒ 5 a b - 15 a + 7 b - 21 = 5 a b + 7 a - 15 b - 21 ⇒ - 22 a = - 22 b ⇒ a = b Therefore, by definition, f is One-one function.

Find whether the following function is onto or not, f : Z → Z defined by f ( x ) = 6 x - 7 , for all x ∈ Z .

Consider any y ∈ Z (Co-domain of f ) Let y = f ( x ) = 6 x - 7 ⇒ y + 7 = 6 x ⇒ y + 7 6 = x Thus , for every y ∈ Z (Co-domain of f ), there does not exist x ∈ Z (Domain of f ) Hence, the f is not Onto.

Find whether following function is onto or not, f : R → R defined by f ( x ) = x 2 + 3 , for all x ∈ R .

Given that, f : R → R is defined by ( x ) = x 2 + 3 , for all x ∈ R Consider any y ∈ R (Co-domain of f ) Let y = f x = x 2 + 3 ⇒ y - 3 = x 2 ⇒ x = ± y - 3 We can clearly see, x is valid only when y - 3 ≥ 0 i.e, y ≥ 3 . Thus, for every y ∈ R (Co-domain of f ), there does not exist x ∈ R (Domain of f ). Hence, f is not onto function.

Let f : R → R be a function defined by f ( x ) = 9 x 3 + 8 , for all x ∈ R , show that f is One-one and Onto. Hence find f - 1 .

Given that, f ( x ) = 9 x 3 + 8 Let f ( a ) = f ( b ) ∴ 9 a 3 + 8 = 9 b 3 + 8 ⇒ 9 a 3 = 9 b 3 ⇒ a 3 = b 3 ⇒ a = b By definition of One-one function, f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ R . Hence, f is One -one function. Let y = f ( x ) = 9 x 3 + 8 ⇒ y - 8 = 9 x 3 ⇒ x 3 = y - 8 9 ⇒ x = y - 8 9 3 Thus, for every y ∈ R , there exist x ∈ R , such that f ( x ) = y ∴ f is Onto function. Since the function is One-one and Onto. ∴ f - 1 exists. Now, f - 1 ( y ) = x if and only if f ( x ) = y ∴ f - 1 y = y - 8 9 3 Replacing y with x , we get f - 1 x = x - 8 9 3 , which is the required inverse function.

A function f : R → R defined by f x = 6 x - 7 3 , x ∈ R . Show that f is one-one and onto. Hence find f - 1 .

Given that f x = 6 x - 7 3 Let f ( a ) = f ( b ) ⇒ 6 a - 7 3 = 6 b - 7 3 ⇒ 6 a - 7 = 6 b - 7 ⇒ 6 a = 6 b ⇒ a = b ∴ f ( a ) = f ( b ) ⇒ a = b ∴ f is one-one Let y = f ( x ) = 6 x - 7 3 ∴ 3 y = 6 x - 7 ∴ 3 y + 7 = 6 x ∴ x = 3 y + 7 6 For every y ∈ R , we have x ∈ R such that f ( x ) = y ∴ f is onto ∴ f - 1 exists Now, f - 1 ( y ) = x if and only if f ( x ) = y ∴ f - 1 ( x ) = 3 x + 7 6 , which is the required inverse of a function.

A function f is defined as f ( x ) = 4 x + 5 , for - 4 ≤ x < 0 . Find the values of f ( - 1 ) , f ( - 2 ) , f ( 0 ) , if they exist.

Given that, f ( x ) = 4 x + 5 , - 4 ≤ x < 0 As we clearly seen that, the domain of the function is x ∈ [ - 4 , 0 ) . ∴ f ( - 1 ) = [ 4 x + 5 ] x = - 1 = 4 ( - 1 ) + 5 = 1 (Since - 1 lies in the given domain) And also, f ( - 2 ) = [ 4 x + 5 ] x = - 2 = 4 ( - 2 ) + 5 = - 3 (Since - 2 lies in the given domain) So, f ( 0 ) is not defined because, 0 does not belong to the domain.