f(x) = ln(sin x), g(x) = sin –1 (e –x ) f(g(x)) = ln(sin(sin –1 e –x )) = -x f(g($\alpha$)) = – $\alpha$ = b As f(g(x)) = – x $\therefore$ (f(g(x)))' = – 1 $\Rightarrow$ (f(g($\alpha$)))' = – 1 = a $\therefore$ b = – $\alpha$, a = – 1 $\therefore$ a$\alpha$ 2 - b$\alpha$ - a = - $\alpha$ 2 + $\alpha$ 2 + 1 = 1