Let $$f(a) = {{a - 1} \over {a + 1}}$$ consider the following :

1. $$f(2a) = f(a) + 1$$

2. $$f\left( {{1 \over a}} \right) = - f(a)$$

Which of the above is/are correct?

Let $$f(x):\left\{ {\matrix{ {x,} & {x\,is\,rational} \cr {0,} & {x\,is\,irrational} \cr } } \right.$$

and $$g(x):\left\{ {\matrix{ {x,} & {x\,is\,rational} \cr {0,} & {x\,is\,irrational} \cr } } \right.$$

if $$f:R \to R$$ and $$g:R \to R$$, then (f $$-$$ g) is

If $$f(x) = {{\sqrt {x - 1} } \over {x - 4}}$$, defines a function on R, then what is its domain?

When one of the following is correct in respect of the function f : R $$\to$$ R⁺ defined as f(x) = |x + 1| ?