Functional (JEE)

$\text{GIVEN: }\color{#3D99F6}g(f(x)) = x$ $h(g(g(x)))=x \implies h(g(\textcolor{#3D99F6} {g(f(x)})))=h(g(\textcolor{#3D99F6} {x}))=f(x) \implies h(\textcolor{#3D99F6}{g(f(x))}) = h(\textcolor{#3D99F6} {x}) = f(f(x))$ .

Given: (i): $h(x) = f(x) +x \implies f(f(x))=f(x) +x$

Let $f(x) = x$ in equation (i) to get: (i): $f(x)=x +f^{-1}(x) \implies x = f(x) - f^{-1}(x)$

Substitute this value of x in the LHS of equation (ii) to get (ii): $f(f(x)-(f(x) - f^{-1}(x))) = f(f^{-1}(x)) = f(x) +2\cos(a)x \implies x = f(x)+2\cos(a)x$

Let $2\cos(a) = t \implies f(x) = x(1-t)$

Substitute f(x) in equation (i) to get: $x(1-t)^2 = x(1-t)+x \implies t = \dfrac{1 \pm \sqrt{5}}{2} \implies \cos(a) = \dfrac{1 \pm \sqrt{5}}{4}$

$a = \{\dfrac{\pi}{5}, \dfrac{3\pi}{5}\} \implies \text{SUM= } \dfrac{4\pi}{5}$