If the function ƒ : R – {1, –1} $$ \to $$ A defined by
ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to

Let $$\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)} $$ where the function ƒ satisfies
ƒ(x + y) = ƒ(x)ƒ(y) for all natural numbers x, y and ƒ(1) = 2. then the natural number 'a' is

Let ƒ(x) = a^x (a > 0) be written as
ƒ(x) = ƒ₁ (x) + ƒ₂ (x), where ƒ₁ (x) is an even function of ƒ₂ (x) is an odd function.
Then ƒ₁ (x + y) + ƒ₁ (x – y) equals

If $$f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$$, $$\left| x \right| < 1$$ then $$f\left( {{{2x} \over {1 + {x^2}}}} \right)$$ is equal to