If f ( x ) = x 2 and g ( x ) = x + 3 , x ∈ R , then find the value of ( g o f ) x , ( f o g ) x · ( f o g ) 2 .

Given that, f ( x ) = x 2 and g ( x ) = x + 3 , x ∈ R . We know that, g o f x = g f x and f o g x = f g x ⇒ g o f x = g f x = x 2 + 3 Now, consider f o g x = f g x = x + 3 2 Now, f o g 2 = f g 2 = 2 + 3 2 = 25 .

Show that : R × R → R defined by a * b = a + 2 b is neither commutative nor associative.

let a = 1 , b = 0 ∈ R a * b = 1 + 2 0 = 1 And b * a = b + 2 a = 0 + 2 = 2 Here, a * b ≠ b * a Hence * is not commutative. Let c = 3 . a * b * c = 1 * c = 1 + 2 3 = 6 a * b * c = a * 2 = 1 * 2 = 1 + 2 2 = 5 Here, a * b * c ≠ a * b * c Hence * is not associative.

Show that : R × R → R defined by a * b = a + 2 b is neither commutative nor associative.

let a = 1 , b = 0 ∈ R a * b = 1 + 2 0 = 1 And b * a = b + 2 a = 0 + 2 = 2 Here, a * b ≠ b * a Hence * is not commutative. Let c = 3 . a * b * c = 1 * c = 1 + 2 3 = 6 a * b * c = a * 2 = 1 * 2 = 1 + 2 2 = 5 Here, a * b * c ≠ a * b * c Hence * is not associative.

If f ( x ) = x 2 and g ( x ) = x + 3 , x ∈ R , then find the value of ( g o f ) x , ( f o g ) x · ( f o g ) 2 .

Given that, f ( x ) = x 2 and g ( x ) = x + 3 , x ∈ R . We know that, g o f x = g f x and f o g x = f g x ⇒ g o f x = g f x = x 2 + 3 Now, consider f o g x = f g x = x + 3 2 Now, f o g 2 = f g 2 = 2 + 3 2 = 25 .