If f : R → R is defined by f x = x x 2 + 1 , find f ( f ( 2 ) ) .

We have, f x = x x 2 + 1 ⇒ f f x = f x f x 2 + 1 ⇒ f f x = x x 2 + 1 x x 2 + 1 2 + 1 ⇒ f f x = x x 2 + 1 x 2 + x 2 + 1 2 Put x = 2 , then f f 2 = 2 2 2 + 1 2 2 + 2 2 + 1 2 ⇒ f f 2 = 10 29

Let f ( x ) = 3 x - 4 for all x ∈ R . Find g : R → R such that g o f = I R .

We have, f x = 3 x - 4 Now, g o f x = I R ⇒ g f x = I R Hence, f x is the inverse of g x or we can say g x is the inverse of f x . f x is a linear function, hence it is one-one and onto. Now, y = f x = 3 x - 4 ⇒ x = y + 4 3 Hence, f - 1 x = g x = x + 4 3 .

Let f x = a x + b for all x ∈ R , where a , b ∈ R and a ≠ 0 , then find f - 1 x .

We have, f x = a x + b Since, f x is a linear function, hence it is one-one and onto. Now, y = f x = a x + b ⇒ x = y - b a Hence, f - 1 x = x - b a .

Let f x = 1 , if x ∈ Q - 1 , if x ∉ Q and g x = - 1 , if x ∈ Q 1 , if x ∉ Q , then for all x ∈ R , find f ( g ( x ) ) .

We have, f x = 1 , if x ∈ Q - 1 , if x ∉ Q g x = - 1 , if x ∈ Q 1 , if x ∉ Q Then, f g x = 1 , if x ∈ Q 1 , if x ∉ Q

Check whether the relation R defined in the set 1 , 2 , 3 , 4 , 5 , 6 as R = a , b : b = a + 1 is reflexive, symmetric or transitive.

Suppose A = 1 , 2 , 3 , 4 , 5 , 6 . A relation R is defined on set A as: R = { ( a , b ) : b = a + 1 } ∴ R = { ( 1, 2 ) , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.