Mathematics - Functions Question with Solution | TestHub

For (P), we need $\frac{x^2-1}{x^2-4}\ge 0$, which implies $x\in (-\infty ,-2)\cup [-1,1]\cup (2,\infty )$. So (P)-(1). For (Q), we need $\frac{1-|x|}{1+|x|}>0$. Since $1+|x|>0$, we need $1-|x|>0⟹|x|<1⟹x\in (-1,1)$. So (Q)-(2). For (R), we need $-1\le \frac{1+x^2}{2x}\le 1$. This inequality holds only for $x=1$ or $x=-1$ because $\frac{1+x^2}{2|x|}\ge 1$ (AM-GM or $(|x|-1{)}^2\ge 0$). Thus $x\in \{-1,1\}$. So (R)-(3). For (S), we need $[x{]}^2-4\ne 0⟹[x]\ne 2$ and $[x]\ne -2$. This means $x\notin [2,3)$ and $x\notin [-2,-1)$. So $x\in \mathbb{R}\setminus ([-2,-1)\cup [2,3))$. So (S)-(4).