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

(B) $f'(x)=0$ at more than three points in $(-\pi, \pi)$; (C) $f(x)$ attains its maximum at $x=0$

(A) $f'(x)=0$ at exactly three points in $(-\pi, \pi)$

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

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Determinants

f (x) = cos (2 x) - cos x sin x cos (2 x) cos x sin x sin (2 x) - sin x cos x

, then

Answer: (B), (C)

Expanding the determinant:

f (x) = cos 4 x + cos 2 x

Differentiating:

f^{'} (x) = - 4 sin 4 x - 2 sin 2 x = - 2 sin 2 x (1 + 4 cos 2 x)

Setting

f^{'} (x) = 0

sin 2 x = 0 or cos 2 x = - \frac{1}{4}

For

sin 2 x = 0

(- π, π)

x = 0, \pm \frac{π}{2}

(3 solutions)

For

cos 2 x = - \frac{1}{4}

(- π, π)

2 x \in (- 2 π, 2 π)

, giving 4 solutions.

Total: more than 3 critical points.

At

x = 0

f (0) = cos 0 + cos 0 = 2

f^{''} (x) = - 8 cos 4 x - 4 cos 2 x

f^{''} (0) = - 8 - 4 = - 12 < 0

f (x)

attains a local maximum at

x = 0

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options