Consider f: given by Show that f is invertible with

Hence, find

(i)

(ii) y if

Where is the set of all non-negative real numbers.

OR

Discuss the commutativity and associativity of binary operation ‘*’ defined on by the rule for all Also find the identity element of * in A and hence find the invertible elements of A.

Let x, y ϵ N; f(x) = f(y)

⇒ 9x² + 6x − 5 = 9y² + 6y − 5

⇒ 9(x² − y²) + 6(x − y) = 0

⇒ (x − y) (9x + 9y + 6) = 0

⇒ x − y = 0 or x = y

This means the function is one-one, and we can see from the range, it is also onto.

Hence, f(x) is invertible and f_.f⁻¹(x) = x

⇒ 9[f⁻¹(x)]² + 6[f⁻¹(x)] − 5 = x

OR

Given: ‘*’ is a binary operation on A = Q − {1} defined by

a*b = a − b + ab

Commutativity:

Let any a, b ϵ A,

a*b = a − b + ab ⇒ b*a = b − a + ab

∵ a − b + ab ≠ b − a + ab

∴ a*b ≠ b*a

∴ ‘*’ is not Commutative on A

Associativity:

Let any a, b, c ϵ A

(a*b)*c = (a − b + ab)*c

= (a − b + ab) − c + (a − b + ab)c

= a − b − c + ab + ac − bc + abc

a*(b*c) = a*(b − c + bc)

= a − (b − c + bc) + a(b − c + bc)

= a − b + c + ab − bc − ac + abc

(a*b)*c ≠ a*(b*c)

∴ ‘*’ is not Associative on A.

Let e be the identity element in A.

⇒ a*e = a = e*a

⇒ a − e + ae = e − a + ae

⇒ (a − 1)e = 0

∴ e = 0 is the identity element in A.

Let b be the inverse of a

a*b = e = b*a

⇒ a − b + ab = 0

∴ b = a ÷ (1 − a)

∴ Every element of A is invertible.