Given that f : S $$\to$$ R is said to have a fixed point at c of S if f(c) = c. Let f : [1, $$\infty$$) $$\to$$ R be defined by f(x) = 1 + $$\sqrt x $$. Then

Let $$f(x) = 1 - \sqrt {({x^2})} $$, where the square root is to be taken positive, then

The domain of $$f(x) = \sqrt {\left( {{1 \over {\sqrt x }} - \sqrt {x + 1} } \right)} $$ is

Let $$f(x) = \sqrt {{x^2} - 3x + 2} $$ and $$g(x) = \sqrt x $$ be two given functions. If S be the domain of fog and T be the domain of gof, then