Q. 255.0( 1 Vote )

# Consider f: <span

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.

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