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 ⇒ ae2 = a ⇒ e2 = 1 ⇒ e = ±1
e*a = a ⇒ ea2 = a ⇒ e = \(\frac{1}{a}\)
Since e is not unique, this operation has no identity element.