If $f(x)=\left|\begin{array}{rrr}\cos 2 x & \cos 2 x & \sin 2 x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x\end{array}\right|$, then
(a) $f^{\prime}(x)=0$ at exactly three points in $(-\pi, \pi)$
(b) $f^{\prime}(x)=0$ at more than three points in $(-\pi, \pi)$
(c) $f(x)$ attains its maximum at $x=0$
(d) $f(x)$ attains its minimum at $x=0$
Answer & explanation
Correct answer: option 2
We have,
$f(x) =\left|\begin{array}{rrr} \cos 2 x & \cos 2 x & \sin 2 x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{array}\right|$
$\Rightarrow f(x) =\left|\begin{array}{rrr} \cos 2 x & 0 & \sin 2 x \\ -\cos x & 2 \cos x & -\sin x \\ \sin x & 0 & \cos x \end{array}\right|$ [Applying $C_2 \rightarrow C_2-C_1$]
$\Rightarrow f(x) =2 \cos x\left|\begin{array}{rr} \cos 2 x & \sin 2 x \\ \sin x & \cos x
\end{array}\right|=2 \cos x \cos 3 x$
$\Rightarrow f(x) =\cos 4 x+\cos 2 x$
$\Rightarrow f'(x)=-4 \sin 4 x-2 \sin 2 x=-2 \sin 2 x(1+4 \cos 2 x)$
∴ $f'(x)=0$
$\Rightarrow -2 \sin 2 x(1+4 \cos 2 x)=0 $
$\Rightarrow \sin 2 x=0 \text { or } \cos 2 x=-\frac{1}{4}$
Now, $\sin 2 x=0 \Rightarrow 2 x=0,-\pi, \pi \Rightarrow x=0,-\frac{\pi}{2}, \frac{\pi}{2}$
$\cos 2 x=-\frac{1}{4}$ has four solutions in $(-\pi, \pi)$ as $y=-\frac{1}{4}$ intersects $y=\cos 2 x$ at four points in $(-\pi, \pi)$.
Thus, $f'(x)=0$ at more than three points in $(-\pi, \pi)$.
Now, $f'(x)=-4 \sin 4 x-2 \sin 2 x$
$\Rightarrow f''(x)=-16 \cos 4 x-4 \cos 2 x$
$\Rightarrow f''(0)=-20<0$
Thus, f(x) attains maximum at x = 0.