1

IIT-JEE 2009

MCQ (More than One Correct Answer)
For $$0 < \theta < {\pi \over 2},$$ the solution (s) of $$$\sum\limits_{m = 1}^6 {\cos ec\,\left( {\theta + {{\left( {m - 1} \right)\pi } \over 4}} \right)\,\cos ec\,\left( {\theta + {{m\pi } \over 4}} \right) = 4\sqrt 2 } $$$ is (are)
A
$$\,{\pi \over 4}$$
B
$$\,{\pi \over 6 }$$
C
$$\,{\pi \over 12}$$
D
$$\,{5\pi \over 12}$$
2

IIT-JEE 1999

MCQ (More than One Correct Answer)
For a positive integer $$\,n$$, let
$${f_n}\left( \theta \right) = \left( {\tan {\theta \over 2}} \right)\,\left( {1 + \sec \theta } \right)\,\left( {1 + \sec 2\theta } \right)\,\left( {1 + \sec 4\theta } \right).....\left( {1 + \sec {2^n}\theta } \right).$$ Then
A
$${f_2}\left( {{\pi \over {16}}} \right) = 1$$
B
$${f_3}\left( {{\pi \over {32}}} \right) = 1$$
C
$${f_4}\left( {{\pi \over {64}}} \right) = 1$$
D
$${f_5}\left( {{\pi \over {128}}} \right) = 1$$
3

IIT-JEE 1988

MCQ (More than One Correct Answer)
The values of $$\theta $$ lying between $$\theta = \theta $$ and $$\theta = \pi /2$$ and satisfying the equation

$$\left| {\matrix{ {1 + {{\sin }^2}\theta } & {{{\cos }^2}\theta } & {4\sin 4\theta } \cr {{{\sin }^2}\theta } & {1 + {{\cos }^2}\theta } & {4\sin 4\theta } \cr {{{\sin }^2}\theta } & {{{\cos }^2}\theta } & {1 + 4\sin 4\theta } \cr } } \right| = 0$$ are

A
$$7\pi /24$$
B
$$5\pi /24$$
C
$$11\pi /24$$
D
$$\pi /24$$

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