Let $$f:( - 1,1) \to B$$, be a function defined by
$$f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}$$,
then $$f$$ is both one-one and onto when B is the interval

A real valued function f(x) satisfies the functional equation

f(x - y) = f(x)f(y) - f(a - x)f(a + y)

where a is given constant and f(0) = 1, f(2a - x) is equal to

The domain of the function
$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$

The graph of the function y = f(x) is symmetrical about the line x = 2, then