Question

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

• A. $f^{\prime }\left(x\right)=0$ at exactly three points in $(-\pi ,\pi )$
• B. $f^{\prime }\left(x\right)=0$ at more than three points in $(-\pi ,\pi )$
• C. $f\left(x\right)$ attains its maximum at $x=0$
• D. $f\left(x\right)$ attains its minimum at $x=0$

Answer 1

Answer: (B), (C)

The correct options are
B $f^{\prime }\left(x\right)=0$ at more than three points in $(-\pi ,\pi )$
C $f\left(x\right)$ attains its maximum at $x=0$

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

$\Rightarrow f\left(x\right)=\cos 2x({\cos }^{2}x+{\sin }^{2}x)-\cos 2x(-{\cos }^{2}x+{\sin }^{2}x)+\sin 2x(-\sin x\cos x-\sin x\cos x)$
$\Rightarrow f\left(x\right)=\cos 4x+\cos 2x=2{\cos }^{2}2x+\cos 2x-1$
$\text{Let}\cos 2x=t$ then $t\in [-1,1]$
$\text{and}f\left(t\right)=2t^2+t-1$
$f^{\prime }\left(t\right)=0\Rightarrow t=\frac{-1}{4}$
Also, $f^{\prime \prime }\left(t\right)>0$
Hence, $f\left(x\right)$ attains minimum at $t=\cos 2x=\frac{-1}{4}$
$f\left(x\right)$ attains maximum at $t=1\Rightarrow \cos 2x=1$
$\therefore f\left(x\right)$ attains maximum at $x=0$
$f\left(x\right)=2{\cos }^{2}2x+\cos 2x-1$
$f^{\prime }\left(x\right)=0\Rightarrow \sin 2x(-4\cos 2x-1)=0$


$\sin 2x=0\Rightarrow 2x=0,\pi \Rightarrow x=0,\frac{\pi }{2}\text{in}(-\pi ,\pi )$
Also, $\cos 2x=-\frac{1}{4}$
From the graph, we can conclude that there are four solutions of $\cos 2x=-\frac{1}{4}\text{in}(-\pi ,\pi )$
$\therefore f^{\prime }\left(x\right)=0\text{has}2+4=6\text{solutions in}(-\pi ,\pi ).$