If f(x) = ax + b, a and b are positive real numbers and if f(f(x)) = 9x + 8, then the value of a + b is:

$$f(x) = ax+b$$.
$$f(f(x)) = f(ax+b) = a(ax+b)+b = a^2x + ab+b$$
We have been given that $$f(f(x)=9x+8$$
$$9x+8=a^2x+ab+b$$
Equating the co-efficient of $$x$$, we get, $$a^2=9$$. Therefore, $$a$$ can be $$3$$ or $$-3$$. But, it has been given that $$a$$ is a positive real number. 
Equating the constants, we get, $$ab+b=8$$
If we substitute $$a=3$$, we get, $$3b+b=8$$
$$4b=8$$
$$b=2$$
Therefore, $$a+b=3+2=5$$
Therefore, option C is the right answer.