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: (A), (C)

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

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

$f\left(x\right)=\cos 2x({\cos }^{2}x+{\sin }^{2}x)-\cos 2x(-{\cos }^{2}x+{\sin }^{2}x)+\sin 2x(-\sin x\cos x-\sin x\cos x)$

$f\left(x\right)=\cos 2x+{\cos }^{2}2x-{\sin }^{2}2x$

$f\left(x\right)=\cos 2x+\cos 4x$

$f\left(x\right)=\frac{-\sin 2x}{2}-\frac{\sin 4x}{4}$

In $(-\pi ,\pi )f\left(x\right)=o$ at $x=o,\frac{\pi }{2},\frac{-\pi }{2}$

total $3$ points

$f\left(x\right)=\frac{-\cos 2x}{24}-\frac{\cos 4x}{16}=-ive$ at $x=0$

So $f\left(x\right)$ is maximum when $x=0$

$A,C$ are correct.