Let f: R - \ (α)/(6)\ → R be defined by f(x) = ((5x + 3)/(6x - α))...

We have,

$f\left(x\right) = y = \left(\dfrac{5x + 3}{6x - \alpha}\right) \ldots \left(i\right)$

$⇒ 5x + 3 = 6xy - \alphay$

$⇒ x ( 6y - 5 ) = \alphay + 3$

$⇒ x = \dfrac{\alphay + 3}{6y - 5}$

Hence,

${f}^{- 1}$\left(x\right) = $\left(\dfrac{\alphax + 3}{6x - 5}\right) \ldots \left(ii\right)$

Now,

\left(fof\right)\left(x\right) = x$</p><p>$f ( x ) = {f}^{- 1} ( x )$</p><p>From equation \left(i\right)$ and \left(ii\right)$, we have</p><p>\left(\dfrac{5x + 3}{6x - \alpha}\right) = \left(\dfrac{\alphax + 3}{6x - 5}\right)$

Clearly, $\alpha = 5$