Find the values of a for which the function f x = 1 + a x , a ≠ 0 is the inverse to itself.

Given: f x = 1 + a x Since, f x is a linear function, hence it is one-one and onto. Now, if f x is self invertible, then f x = f - 1 x ⇒ f f x = f f - 1 x ⇒ 1 + a f x = x ⇒ 1 + a 1 + a x = x ⇒ 1 + a = 1 - a 2 x ⇒ x = 1 + a 1 - a 2 ⇒ x = 1 1 - a , a ≠ 1 Hence, function is self invertible for a ∈ R - 1 .

Let f x = x 1 - x 2 , x ≠ ± 1 , then find ( f o f o f ) ( x )

We have, f x = x 1 - x 2 Now, f o f x = f x 1 - f x 2 ⇒ f o f x = x 1 - x 2 1 - x 2 1 - x 2 ⇒ f o f x = x 1 - 2 x 2 , x ≠ ± 1 2 Now, f o f o f x = f f o f x ⇒ f o f o f x = x 1 - 2 x 2 1 - x 2 1 - 2 x 2 ⇒ f o f o f x = x 1 - 3 x 2 , x ≠ ± 1 3

Let f , g , h : R → R , given by f x = 2 x + 1 , g x = 1 - 4 x and h x = x 2 + 1 . Find f o g o h and g o f o h .

We have, f x = 2 x + 1 g x = 1 - 4 x h x = x 2 + 1 Now, f o g o h x = f g h x ⇒ f o g o h x = f g x 2 + 1 ⇒ f o g o h x = f 1 - 4 x 2 + 1 ⇒ f o g o h x = f - 3 - 4 x 2 ⇒ f o g o h x = 2 - 3 - 4 x 2 + 1 ⇒ f o g o h x = - 8 x 2 - 5 And, g o f o h x = g f h x ⇒ g o f o h x = g f x 2 + 1 ⇒ g o f o h x = g 2 x 2 + 1 + 1 ⇒ g o f o h x = g 2 x 2 + 3 ⇒ g o f o h x = 1 - 4 2 x 2 + 3 ⇒ g o f o h x = - 11 - 8 x 2

Let N m be the set of all ordered m - tuples of natural numbers. If x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m where x i , y i ∈ N , i = 1 , 2 , 3 , … . , m and an operation ' + ' is defined by x + y = x 1 + y 1 , x 2 + y 2 , … … x m + y m , then prove that given operation is commutative as well as associative.

We have, x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m Now, x + y = x 1 + y 1 , x 2 + y 2 , … … , x m + y m y + x = y 1 + x 1 , y 2 + x 2 , … … , y m + x m But addition is commutative, hence x 1 + y 1 = y 1 + x 1 , x 2 + y 2 = y 2 + x 2 , . . . . , x m + y m = y m + x m Therefore, x + y = y + x , so " + " is commutative. Also, let z = z 1 , z 2 , z 3 , … . , z m , then x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m But addition is associative, hence x + y + z = x + y + z Therefore, " + " is associative.

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.