If $f (x) = |\begin{matrix} \cos (2 x) & \cos (2 x) & \sin (2 x) \\ - \cos x & \cos x & - \sin x \\ \sin x & \sin x & \cos x \end{matrix}|,$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$f^{'} (x) = 0$ at exactly three points in $(- π, π)$

$f^{'} (x) = 0$ at more than three points in $(- π, π)$

$f (x)$ attains its maximum at x = 0

$f (x)$ attains its minimum at x = 0

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = |\begin{matrix} \cos 2 x & \cos 2 x & \sin 2 x \\ - \cos x & \cos x & - \sin x \\ \sin x & \sin x & \cos x \end{matrix}| = \cos 4 x + \cos 2 x$
Now $f^{'} (x) = - 2 \sin 2 x - 4 \sin 4 x = 0$
$\Rightarrow f^{'} (x) = - 2 \sin 2 x (1 + 4 \cos 2 x) = 0$
$\sin 2 x = 0 o r \cos 2 x = - \frac{1}{4}$
For $\sin 2 x - 0; x - 0, π / 2 - π / 2$
For $\cos 2 x = 1 / 4$ there are four solutions.
$f' (x) = 0$ has more than three solutions.
Again $f'' (x) = - (4 \cos 2 x + 16 \cos 4 x)$
$\Rightarrow f'' (0) < 0$