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 $$A = \{ x \in R: - 1 \le x \le 1\} $$ and $$f:A \to A$$ be a mapping defined by $$f(x) = x\left| x \right|$$. Then f is