JEE MATHEMATICS MCQ: SET THEORY

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Which one of the following is correct ?

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Correct option: B

\mathrm{A}-(\mathrm{B} \cup \mathrm{C})=\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right) \cap \mathrm{C}^{\prime}

Option B (

A-(B \cup C)=(A \cap B^{\prime}) \cap C^{\prime}

) is correct because it represents the set of elements in A that are not in B and not in C, which is equivalent to the intersection of A with the complements of B and C.

More Questions from SET THEORY

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Let

S

be the set of all those 2007-place decimal integers

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

which contain odd number of digit 9 in each sequence

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

. The cardinal number of

S

is If

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

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

and

b N \cap c N=d N

, where

a, b, c \in N

and

b, c

are coprime, then Which of the following cannot be the number of elements in the power set of any finite set? If

A=\{x: x

is a multiple of 3

\}

and

B=\{x: x

is a multiple of 5}, then

A-B

is equal to If

F(n)

denotes the set of all divisors of

n

except 1 , what is the least value of y satisfying

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

? In a town of 10000 families, it was found that

40 \%

families buy newspaper

A, 20 \%

families buy newspaper B and

10 \%

families buy newspaper

C

, 5% families buy newspapers

A

and B, 3% buy newspapers B and

C

and

4 \%

buy newspapers

A

and

C

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2 \%

families buy all the three newspapers, then number of families which buy

A

only is