Question

The domain of definition of the function $f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$is
(a) (−∞, −2] ∪ [2, ∞)
(b) [−1, 1]
(c) ϕ
(d) None of these

Answer 1

(c) ϕ

$f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$
$$\begin{gathered} \mathrm{For}\mathrm{f}\left(\mathrm{x}\right)\mathrm{to}\mathrm{be}\mathrm{defined}, \\ x+2\ne 0 \\ \Rightarrow x\ne -2...\left(1\right) \\ \text{And}1+x\ne 0 \\ \Rightarrow \mathrm{x}\ne -1....\left(2\right) \\ \mathrm{Also},\frac{\mathrm{x}-2}{\mathrm{x}+2}\ge 0 \\ \Rightarrow \frac{(\mathrm{x}-2)(\mathrm{x}+2)}{(\mathrm{x}+2{)}^2}\ge 0 \\ \Rightarrow (\mathrm{x}-2)(\mathrm{x}+2)\ge 0 \\ \Rightarrow \mathrm{x}\in (-\infty ,-2)\cup [2,\infty )...\left(3\right) \\ \text{And}\frac{1-\mathrm{x}}{1+\mathrm{x}}\ge 0 \\ \Rightarrow \frac{(1-\mathrm{x})(1+\mathrm{x})}{(1+\mathrm{x}{)}^2}\ge 0 \\ \Rightarrow (1-\mathrm{x})(1+\mathrm{x})\ge 0 \\ \Rightarrow \mathrm{x}\in (-\infty ,-1)\cup [1,\infty )...\left(4\right) \\ \mathrm{From}\left(1\right),\left(2\right),\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}, \\ \mathrm{x}\in \mathrm{\varphi }. \\ \mathrm{Thus},\mathrm{dom}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\mathrm{\varphi }. \end{gathered}$$