solve $\frac{{1 - \left| x \right|}}{{2 - \left| x \right|}} \ge 0$

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

Let $f(x)=2 x^{2}-x-1$ and $S =\{n \in Z :|f(n)| \leq 800\}$ . Then value of $\sum_{n \in S} f(n)$ is . . . .  .

If $P(S)$ denotes the set of all subsets of a given set $S, $ then the number of one-to-one functions from the set $S = \{ 1, 2, 3\}$ to the set $P(S)$ is

If $a+\alpha=1, b+\beta=2$ and $\operatorname{af}(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}, x \neq 0,$ then the value of expression $\frac{ f ( x )+ f \left(\frac{1}{ x }\right)}{ x +\frac{1}{ x }}$ is ..... .

Let $f :R \to R$ be defined by $f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R.$ Then the range of $f$ is