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 

 

पर्याय

• (a) (−∞, −2] ∪ [2, ∞)

• (b) [−1, 1]

• (c) ϕ

• (d) None of these

   

Answer 1

(c) ϕ 

\[f\left( x \right) = \sqrt{\frac{x - 2}{x + 2}} + \sqrt{\frac{1 - x}{1 + x}}\]

\[\text{ For f(x) to be defined,}  \]
\[x + 2 \neq 0\]
\[ \Rightarrow x \neq - 2 . . . (1)\]
\[\text{ And }  1 + x \neq 0\]
\[ \Rightarrow x \neq - 1 . . . . (2)\]
\[\text{ Also } , \frac{x - 2}{x + 2} \geq 0\]
\[ \Rightarrow \frac{(x - 2)(x + 2)}{(x + 2 )^2} \geq 0\]
\[ \Rightarrow (x - 2)(x + 2) \geq 0\]
\[ \Rightarrow x \in ( - \infty , - 2) \cup [2, \infty ) . . . (3)\]
\[\text{ And } \frac{1 - x}{1 + x} \geq 0\]
\[ \Rightarrow \frac{(1 - x)(1 + x)}{(1 + x )^2} \geq 0\]
\[ \Rightarrow (1 - x)(1 + x) \geq 0\]
\[ \Rightarrow x \in ( - \infty , - 1) \cup [1, \infty ) . . . (4)\]
\[\text{ From (1), (2), (3) and (4), we get,}  \]
\[x \in \phi . \]
\[\text{ Thus, dom }  (f(x)) = \phi . \]