JEE Advance - Mathematics (2022 - Paper 1 Online - No. 14)
Let $$|M|$$ denote the determinant of a square matrix $$M$$. Let $$g:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$$ be the function defined by
$$ g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1} $$
where
$$ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right| . $$
Let $$p(x)$$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $$g(\theta)$$, and $$p(2)=2-\sqrt{2}$$. Then, which of the following is/are TRUE ?
Explanation
$$ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right| $$
Here $\cos \left(\theta+\frac{\pi}{4}\right)=-\sin \left(\theta-\frac{\pi}{4}\right)$
and $\tan \left(\theta-\frac{\pi}{4}\right)=-\cot \left(\theta+\frac{\pi}{4}\right)$
and $\log _e\left(\frac{4}{\pi}\right)=-\log _e\left(\frac{\pi}{4}\right)$
Also $\sin \pi=-\cos \frac{\pi}{2}=\tan \pi=0$
$$ \Rightarrow $$ $$ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 2 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \\ 0 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} 0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{\mathrm{e}}\left(\frac{4}{\pi}\right) \\ -\tan \left(\theta-\frac{\pi}{4}\right) & -\log _{\mathrm{e}}\left(\frac{4}{\pi}\right) & 0 \end{array}\right| $$
$$ \begin{aligned} & f(\theta)=\left(1+\sin ^2 \theta\right)+0 \text { (skew symmetric) } \\\\ & g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1} \\\\ & =|\sin \theta|+|\cos \theta| \quad \\\\ & g(\theta) \in[1, \sqrt{2}] \text { for } \theta \in\left[0, \frac{\pi}{2}\right] \end{aligned} $$
Again let $\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}-\sqrt{2})(\mathrm{x}-1)$
$$
\begin{aligned}
& 2-\sqrt{2}=\mathrm{k}(2-\sqrt{2})(2-1) \\\\
& \Rightarrow \mathrm{k}=1 \quad(\mathrm{P}(2)=2-\sqrt{2} \text { given }) \\\\
& \therefore \mathrm{P}(\mathrm{x})=(\mathrm{x}-\sqrt{2})(\mathrm{x}-1)
\end{aligned}
$$
For option (A) $P\left(\frac{3+\sqrt{2}}{4}\right)<0$ Correct.
For option (B) $P\left(\frac{1+3 \sqrt{2}}{4}\right)<0$ Incorrect.
For option (C) $P\left(\frac{5 \sqrt{2}-1}{4}\right)>0$ Correct.
For option (D) $P\left(\frac{5-\sqrt{2}}{4}\right)>0$ Incorrect.
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