Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) = $${{x - {1 \over 2}} \over {x - 1}}$$. Then the composition function f(g(x)) is :

For a suitably chosen real constant a, let a

function, $$f:R - \left\{ { - a} \right\} \to R$$ be defined by

$$f(x) = {{a - x} \over {a + x}}$$. Further suppose that for any real number $$x \ne - a$$ and $$f(x) \ne - a$$,

(fof)(x) = x. Then $$f\left( { - {1 \over 2}} \right)$$ is equal to :

If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$ \in $$ N, where N is the set of all natural number, then the value of $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is :

Let f : R $$ \to $$ R be a function which satisfies
f(x + y) = f(x) + f(y) $$\forall $$ x, y $$ \in $$ R. If f(1) = 2 and
g(n) = $$\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)} $$, n $$ \in $$ N then the value of n, for which g(n) = 20, is :