Let $A = [x ∶ x \in R, | x | < 1];$ $B = [x ∶ x \in R, | x - 1 | \geq 1]$ and $A \cup B = R - D$ , then the set $D$ is

[x : 1 < x \leq 2]

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[x : 1 \leq x < 2]

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[x : 1 \leq x \leq 2]

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None of these

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Solution

The correct option is B $[x : 1 \leq x < 2]$
$A = {x : x \in R, | x | < 1} B = {x : x \in R, | x - 1 | \geq 1} A \cup B = R - D | x | < 1 \Rightarrow - 1 < x < 1 ∴ A \in (- 1, 1) | x - 1 | \geq 1 \Rightarrow (x - 1) \geq 1 or (x - 1) \leq - 1 \Rightarrow x \geq 2 or x \leq 0 B \equiv (- \infty, 0] \cup [2, \infty) A \cup B \in (- \infty, 1) \cup [2, \infty) \Rightarrow A \cup B \in R - [1, 2)$

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Similar questions

Let $A = {x : x \in R, | x | < 1}$ ;
$B = {x : x \in R, | x - 1 | \geq 1}$ and

$A \cup B = R - D$ , then the set $D$ is

A = {x : x \in R, | x | < 1}

B = {x : x \in R, | x - 1 | \geq 1}

A \cup B = R - D

D

A = [x ∶ x \in R, | x | < 1];

B = [x ∶ x \in R, | x - 1 | \geq 1]

A \cup B = R - D

D

Let A = { x : x ∈ R, |x| < 1}; B = {x : x ∈ R, |x-1| ≥ 1} and

A ∪ B = R - D, then the set D is

A = {x : - 1 < x < 1; x \in R}

B = {x : x \leq 0 or x \geq 2; x \in R}

A \cup B = R - C

C

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