Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{n+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{6}(6)+f^{7}(7)=$

Let $A$ denote the set of all real numbers $x$ such that $x^3-[x]^3=\left(x-[x]^3\right)$, where $[x]$ is the greatest integer less than or equal to $x$. Then,

For a suitably chosen real constant $a$, let a function, $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x} .$ Further suppose that for any real number $x \neq- a$ and $f( x ) \neq- a ,( fof )( x )= x .$ Then $f\left(-\frac{1}{2}\right)$ is equal to

The range of function $f : R \rightarrow  R$, $f(x) = \frac{{{{(x\, + \,1)}^4}}}{{{x^4} + \,1}}$ is

The domain of the function

$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$

Let $f : R \rightarrow R$ be a function such that $f(x)=\frac{x^2+2 x+1}{x^2+1}$. Then