JEE MATHEMATICS MCQ: SET THEORY

1 Easy

1 of 1

Analyzing +4 | -1

Let

S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}}}\theta + {8^{2{{\cos }^2}\theta }} = 16} \right\}.

Then

n(S)+

\sum\limits_{\theta \in S} {\left( {\sec \left( {\frac{\pi }{4} + 2\theta } \right)\cos ec\left( {\frac{\pi }{4} + 2\theta } \right)} \right)}

is equal to :

View answer and explanation

Correct option: C

-4

Option C (-4) is correct because after finding the values of

\theta

that satisfy the given equation and substituting them into the expression, the result is -4.

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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