Question

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

• 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 1

Answer: (B), (C)

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