Let f (x) = _ e ( x), (0 < x < π) and g(x) = ^ - 1 ( e ^ - x ), ( x...

Given $f$\left(x\right) = \log_{e}$\left(\sin x\right), \left(0 < x < \pi\right)$ and $g\left(x\right) = \sin^{- 1}( {e}^{- x} ), ( x \ge 0 ) ,$ then we have

$fog\left(x\right) = f\left(g\left(x\right)\right)$

$= f\left(\sin^{- 1}\left({e}^{- x}\right)\right)$

$= \log_{e}\left(\sin \left(\sin^{- 1}\left({e}^{- x}\right)\right)\right)$

$= \log_{e}\left({e}^{- x}\right) = - x$

$⇒ fog\left(x\right) = - x$

$⇒ b = f\left(g\left(\alpha\right)\right) = - \alpha$

Now ${( f\left(g ( x )\right) )}^{'} = - 1$

$⇒ a = {\left(f\left(g\left(\alpha\right)\right)\right)}^{'} = - 1$

$∴ a{\alpha}^{2} - b\alpha - a = \left(- 1\right){\alpha}^{2} - \left(- \alpha\right)\alpha - \left(- 1\right)$

$= - {\alpha}^{2} + {\alpha}^{2} + 1 = 1 .$