Let $\mathrm{A}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R},-1<\mathrm{x}<1\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R}, \mathrm{x} \leq 0$ or $\mathrm{x} \geq 2\}$ and $\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D}$, then the set $\mathrm{D}$ is

JEE MATHEMATICS MCQ: SET THEORY

1 Easy

1 of 1

Application +4 | -1

Let

\mathrm{A}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R},-1<\mathrm{x}<1\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R}, \mathrm{x} \leq 0

\mathrm{x} \geq 2\}

and

\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D}

, then the set

\mathrm{D}

View answer and explanation

Correct option: B

\{\mathrm{x}: 1 \leq \mathrm{x}<2\}

Option B is correct because it represents the set D, which is the complement of the union of sets A and B with respect to the set of real numbers.

Practice All Chapter MCQs →

S

2 a_1 a_2 a_3 \ldots . . a_{2006}

a_1, a_2, a_3, \ldots ., a_{2006}

S

a N=\{a n: n \in N: b N=\{n b: n \in N\}

c N=\{c N: n \in N\}

b N \cap c N=d N

a, b, c \in N

b, c

A=\{x: x

\}

B=\{x: x

A-B

F(n)

n

[\mathrm{F}(20) \cap \mathrm{F}(16)] \subseteq \mathrm{F}(\mathrm{y})

40 \%

A, 20 \%

10 \%

C

A

C

4 \%

A

C

2 \%

A