The domain of definition of the function f(x)

Solution:

(c) ϕ

$f(x)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$

For $\mathrm{f}(\mathrm{x})$ to be defined,

$x+2 \neq 0$

$\Rightarrow x \neq-2 \ldots(1)$

And $1+x \neq 0$

$\Rightarrow \mathrm{x} \neq-1 \ldots(2)$

Also, $\frac{x-2}{x+2} \geq 0$

$\Rightarrow \frac{(x-2)(x+2)}{(x+2)^{2}} \geq 0$

$\Rightarrow(x-2)(x+2) \geq 0$

$\Rightarrow \mathrm{x} \in(-\infty,-2) \cup[2, \infty) \ldots(3)$

And $\frac{1-\mathrm{x}}{1+\mathrm{x}} \geq 0$

$\Rightarrow \frac{(1-\mathrm{x})(1+\mathrm{x})}{(1+\mathrm{x})^{2}} \geq 0$

$\Rightarrow(1-\mathrm{x})(1+\mathrm{x}) \geq 0$

$\Rightarrow \mathrm{x} \in(-\infty,-1) \cup[1, \infty) \ldots(4)$

From $(1),(2),(3)$ and $(4)$, we get,

$\mathrm{x} \in \phi$Thus, $\operatorname{dom}(\mathrm{f}(\mathrm{x}))=\phi$