JEE Advanced 2022 Paper 1, Mathematics Section 2 Q6: Algebra Of Determinants

Expanding the first determinant: it equals $1+{\sin }^2\theta$ after using cofactor expansion along the first row. Specifically, $\frac{1}{2}\left|\begin{array}{c}1\sin \theta 1\\ -\sin \theta 1\sin \theta \\ -1-\sin \theta 1\end{array}\right|=\frac{1}{2}(2+2{\sin }^2\theta )=1+{\sin }^2\theta$.

For the second determinant, observe the entries: $\sin \pi =0$, $\cos \frac{\pi }{2}=0$, $\tan \pi =0$, $\cos \left(\theta +\frac{\pi }{4}\right)=-\sin \left(\theta -\frac{\pi }{4}\right)$, $\tan \left(\theta -\frac{\pi }{4}\right)=-\cot \left(\theta +\frac{\pi }{4}\right)$, and ${\log }_e\left(\frac{4}{\pi }\right)=-{\log }_e\left(\frac{\pi }{4}\right)$. This makes the matrix skew-symmetric (with antisymmetric corresponding pairs across the diagonal of zeros), so its determinant is $0$.

Hence $f(\theta )=1+{\sin }^2\theta$.

Then $f(\theta )-1={\sin }^2\theta$ and $f\left(\frac{\pi }{2}-\theta \right)-1={\cos }^2\theta$, so

$g(\theta )=|\sin \theta |+|\cos \theta |.$

On $\left[0,\frac{\pi }{2}\right]$, the minimum of $|\sin \theta |+|\cos \theta |$ is $1$ (at the endpoints) and the maximum is $\sqrt{2}$ (at $\theta =\frac{\pi }{4}$).

Therefore $p(x)=a(x-1)(x-\sqrt{2})$. Using $p(2)=a(1)(2-\sqrt{2})=2-\sqrt{2}$, we get $a=1$.

So $p(x)=(x-1)(x-\sqrt{2})$.

Sign of $p(x)$: positive for $x<1$ or $x>\sqrt{2}$, negative for $1<x<\sqrt{2}$.

(A) $x=\frac{3+\sqrt{2}}{4}\approx \frac{4.414}{4}\approx 1.10$, which lies in $(1,\sqrt{2})$. So $p(x)<0$. TRUE.

(B) $x=\frac{1+3\sqrt{2}}{4}\approx \frac{5.243}{4}\approx 1.31$, which lies in $(1,\sqrt{2})$. So $p(x)<0$, not $>0$. FALSE.

(C) $x=\frac{5\sqrt{2}-1}{4}\approx \frac{6.071}{4}\approx 1.52$, which is $>\sqrt{2}\approx 1.414$. So $p(x)>0$. TRUE.

(D) $x=\frac{5-\sqrt{2}}{4}\approx \frac{3.586}{4}\approx 0.90$, which is $<1$. So $p(x)>0$, not $<0$. FALSE.

$\therefore$ The correct options are (A) and (C).