The inverse of the function f (x) = _ a (x + x^ 2 + 1 ) (where, a <...

Let $f\left(x\right) = y$, then

$a^{y} = x + \sqrt{x^{2} + 1}$

$⇒ a^{- y} = \dfrac{1}{x + \sqrt{x^{2} + 1}}$

$⇒ a^{- y} = \dfrac{x - \sqrt{x^{2} + 1}}{- 1}$ [rationalising]

$∴ a^{y} - a^{- y} = 2x$

$⇒ x = \dfrac{1}{2}$\left(a^{y} - a^{- y}\right)$</p><p>$∴ f^{- 1} \left(y\right) = \dfrac{1}{2}\left(a^{y} - a^{- y}\right)& \left[∵ f \left(x\right) = y ⇒ x = f^{- 1}\left(y\right)\right]

$⇒ f^{- 1} \left(x\right) =\dfrac{1}{2} \left(a^{x} - a^{- x}\right) [replacing$y$by$x$]</p><p>Hence, option (a) is correct.</p>$