A is the domain of fx=1x−x and B is the domain of gx=1−x then A∩B=

fx=1x−x is defined iff x−x>0⇒ x>x ∀ x∈R−⇒ x<0A=x/x<0gx=1−x is defined iff 1−x≥0⇒ x≤1⇒ −1≤x≤1B=x/−1≤x≤1∴ A∩B=−1, 0

A is the domain of fx=1x−x and B is the domain of gx=1−x then A∩B=

fx=1x−x is defined iff x−x>0⇒ x>x ∀ x∈R−⇒ x<0A=x/x<0gx=1−x is defined iff 1−x≥0⇒ x≤1⇒ −1≤x≤1B=x/−1≤x≤1∴ A∩B=−1, 0

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$A i s t h e d o m a i n o f f (x) = \frac{1}{\sqrt{|x| - x}} a n d B i s t h e d o m a i n o f g (x) = \sqrt{1 - |x|} t h e n A \cap B =$

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$(- 1, 0)$

$[- 1, 0)$

$(- 1, 1)$

$[- 1, 1]$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{\sqrt{|x| - x}} i s d e f i n e d i f f |x| - x > 0 \Rightarrow |x| > x \forall x \in R^{-} \Rightarrow x < 0 \\ A = \{x / x < 0\} \\ g (x) = \sqrt{1 - |x|} i s d e f i n e d i f f 1 - |x| \geq 0 \Rightarrow |x| \leq 1 \Rightarrow - 1 \leq x \leq 1 \\ B = \{x / - 1 \leq x \leq 1\} \\ ∴ A \cap B = [- 1, 0) \end{array}$