A is the domain of f(x)=|x|−x and B is the domain of g(x)=11−|x| , then A∩B

|x|−x≥0⇒x∈(−∞,∞)−A 1−|x|>0⇒x∈(−1,1)−B A∩B=(−1,1)

A is the domain of f(x)=|x|−x and B is the domain of g(x)=11−|x| , then A∩B

|x|−x≥0⇒x∈(−∞,∞)−A 1−|x|>0⇒x∈(−1,1)−B A∩B=(−1,1)

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A is the domain of $f (x) = \sqrt{| x | - x}$ and B is the domain of $g (x) = \frac{1}{\sqrt{1 - | x |}}$ , then $A \cap B$

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$(- 1, 1)$

$(- 1, 0)$

$[- 1, 0]$

$(0, 1)$

answer is B.

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Detailed Solution

$| x | - x \geq 0 \Rightarrow x \in (- \infty, \infty) - A$

$1 - | x | > 0 \Rightarrow x \in (- 1, 1) - B$

$A \cap B = (- 1, 1)$

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A is the domain of f(x)=|x|−x and B is the domain of g(x)=11−|x| , then A∩B

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A is the domain of $f (x) = \sqrt{| x | - x}$ and B is the domain of $g (x) = \frac{1}{\sqrt{1 - | x |}}$ , then $A \cap B$