JEE MATHEMATICS MCQ: SET THEORY

1 Easy

1 of 1

Analyzing +4 | -1

A=\left\{\theta: 2 \cos ^2 \theta+\sin \theta \leq 2\right\}

and

B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}

, then

A \cap B

is equal to

View answer and explanation

Correct option: C

\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup\{\theta: \pi \leq \theta \leq 3 \pi / 2\}

Option C (

\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup \{\theta: \pi \leq \theta \leq 3 \pi / 2\}

) is correct because it represents the intersection of the solution sets for the trigonometric inequality and the given interval.

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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