Let f and g be two functions defined by $$f(x) = \mid x + \mid x \mid \mid$$ and $$g(x) = \frac{1}{x}$$ for $$x \neq 0$$. If $$f(a) + g(f(a)) = \frac{13}{6}$$ for some real a, then the maximum possible value of $$f(g(a))$$ is:

Given that f(a) + g(f(a)) = $$\dfrac{\ 13}{6}$$

f(a) + $$\dfrac{\ 1}{f\left(a\right)}$$ = $$\dfrac{\ 13}{6}$$ 

$$6(f(a))^2-13\left(f\left(a\right)\right)+6\ =\ 0$$

f(a) = $$\dfrac{13\pm\sqrt{13^2-24\left(6\right)\ }\ }{2\times6}$$ 

this implies , f(a) = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 

Given that f(a) = |a+|a||
if $$a\le0\ $$, f(a) = |a-a| = 0
else if $$a\ge\ 0$$ , then f(a) = 2a.

Therefore , 2a = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 
that is, a = $$\ \dfrac{\ 3}{4}\ (or)\ \dfrac{\ 1}{3}$$

g($$\dfrac{\ 3}{4}$$) = $$\dfrac{\ 4}{3}$$

g($$\dfrac{\ 1}{3}$$) = $$\dfrac{\ 3}{1}$$

Therefore f(g(a)) = $$\dfrac{\ 8}{3}$$ or 6.

Maximum value is 6.