JEE Advanced 2022 — Paper 1 — Question 14

$f(x)=\left|\begin{array}{ccc}\cos 2 x & \cos 2 x & \sin 2 x \\ -\cos x & \cos x & -\sin x \\sin x & \sin x & \cos x\end{array}\right|=\cos 4 x+\cos 2 x$

Now $f^{\prime}(x)=-2 \sin 2 x-4 \sin 4 x=0$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \sin 2 \mathrm{x}(1+4 \cos 2 \mathrm{x})=0$

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.

$\mathrm{f}^{\prime}(\mathrm{x})=0$ has more than three solutions.

Again $f^{\prime \prime}(x)=-(4 \cos 2 x+16 \cos 4 x)$

$\Rightarrow \quad \mathrm{f}^{\prime \prime}(0)<0$