If ${A}=\{x \in {R} / {f}(x)=0\}$ and ${B}=\{x \in {R} / {g}(x)=0\}$ then ${A} \cap {B}=$

$\{x \in {R} / {f}(x)+{g}(x)=0\}$

JEE MATHEMATICS MCQ: SET THEORY

1 Easy

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Understanding +4 | -1

{A}=\{x \in {R} / {f}(x)=0\}

and

{B}=\{x \in {R} / {g}(x)=0\}

then

{A} \cap {B}=

View answer and explanation

Correct option: D

\left\{x \in {R} / {f}(x)^2+{g}(x)^2=0\right\}

Option D is correct because A ∩ B represents the intersection of sets A and B, meaning elements that satisfy both f(x) = 0 and g(x) = 0, which is equivalent to f(x)² + g(x)² = 0.

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}

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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

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a, b, c \in N

and

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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

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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

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A, 20 \%

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10 \%

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C

, 5% families buy newspapers

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