f of (x) = f(f(x)) = \(\frac{2f(x) + 1}{3}\)

= \(\frac{2(\frac{2x+1}{3})+1}{3}\) = \(\frac{4x + 5}{9}\)

f(x1) = f(x2) ⇒ \(\frac{2x_1 + 1}{3
}\) = \(\frac{2x_2 + 1}{3
}\)

⇒ 3y = 2x +1

⇒ 3y -1 = 2x ⇒ x = \(\frac{3y-1}{2}\) ∈ R

Therefore f is onto. Hence f is bijective and invertible.

(b) let ‘e’ be the identity element, then

a*e = a ⇒ ae^{2} = a ⇒ e^{2} = 1 ⇒ e = ±1

e*a = a ⇒ ea^{2} = a ⇒ e = \(\frac{1}{a}\)

Since e is not unique, this operation has no identity element.