Question

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

• A. $f'(x)=0$ at exactly three points in $(-\pi, \pi)$
• B. $f'(x)=0$ 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 1

Answer: (B), (C)

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