Let f(x) = x $$-$$ 3, g(x) = 4 $$-$$ x. Then the set of values of x for which $$|f(x) + g(x)|\, < \,|f(x)| + |g(x)|$$ is true, is given by :

$$\left\{ {x \in R:{{2x - 1} \over {{x^3} + 4{x^2} + 3x}} \in R} \right\}$$ is equal to

The solution set of $${{|x - 2|\, - 1} \over {|x - 2|\, - 2}} \le 0$$ is

Let $$f(x) = {x \over {\sqrt {1 + {x^2}} }}$$, $$\underbrace {fofofo.....of(x)}_{x\,times}$$ is