Let $\left|M\right|$ denote the determinant of a square matrix $M$. Let $g:\left[0,\frac{\pi }{2}\right]\to \mathrm{ℝ}$ be the function defined by $g\left(\theta \right)=\sqrt{f\left(\theta \right)-1}+\sqrt{f\left(\frac{\pi }{2}-\theta \right)-1}$ where $f\left(\theta \right)=\frac{1}{2}\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta +\frac{\pi }{4}\right)& \tan \left(\theta -\frac{\pi }{4}\right)\\ \sin \left(\theta -\frac{\pi }{4}\right)& -\cos \frac{\pi }{2}& {\log }_e\left(\frac{4}{\pi }\right)\\ \cot \left(\theta +\frac{\pi }{4}\right)& {\log }_e\left(\frac{\pi }{4}\right)& \tan \pi \end{array}\right|$
Let $p\left(x\right)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g\left(\theta \right)$, and $p\left(2\right)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?

Given,

$f\left(\theta \right)=\frac{1}{2}\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta +\frac{\pi }{4}\right)& \tan \left(\theta -\frac{\pi }{4}\right)\\ \sin \left(\theta -\frac{\pi }{4}\right)& -\cos \frac{\pi }{2}& {\log }_e\left(\frac{4}{\pi }\right)\\ \cot \left(\theta +\frac{\pi }{4}\right)& {\log }_e\left(\frac{\pi }{4}\right)& \mathrm{tan\pi }\end{array}\right|$

$\Rightarrow f\left(\theta \right)=\frac{1}{2}\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}0& \cos \left(\theta +\frac{\pi }{4}\right)& \tan \left(\theta -\frac{\pi }{4}\right)\\ \sin \left(\theta -\frac{\pi }{4}\right)& 0& {\log }_e\left(\frac{4}{\pi }\right)\\ -\tan \left(\theta -\frac{\pi }{4}\right)& -{\log }_e\left(\frac{4}{\pi }\right)& 0\end{array}\right|$

Here we used $\cos \left(\theta +\frac{\pi }{4}\right)=-\sin \left(\theta -\frac{\pi }{4}\right)$

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 $\sin \pi =-\cos \frac{\pi }{2}=\tan \pi =0$

So, $\Rightarrow f\left(\theta \right)=\frac{1}{2}\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|+\left(\text{skew symmetric}\right)$

$\Rightarrow f\left(\theta \right)=1+{\sin }^2\theta$

So, $g\left(\theta \right)=\left|\sin \theta \right|+\left|\cos \theta \right|$

Now maximum and minimum values are $\sqrt{2}$ and $1$ respectively.

Now quadratic polynomial will be $P\left(x\right)=a\left(x-\sqrt{2}\right)\left(x-1\right)$, where $a\in R-\left\{0\right\}$,

But given $P\left(2\right)=2-\sqrt{2}$ , so $a=1$.

$\therefore P\left(x\right)=\left(x-\sqrt{2}\right)\left(x-1\right)$

Now solving all options we get,

$P\left(\frac{3+\sqrt{2}}{4}\right)=\left(\frac{3-3\sqrt{2}}{4}\right)·\left(\frac{\sqrt{2}-1}{4}\right)<0$

$P\left(\frac{1+3\sqrt{2}}{4}\right)=\left(\frac{1-\sqrt{2}}{4}\right)·\left(\frac{3\sqrt{2}-3}{4}\right)<0$

$P\left(\frac{5\sqrt{2}-1}{4}\right)=\left(\frac{\sqrt{2}-1}{4}\right)·\left(\frac{5\sqrt{2}-5}{4}\right)>0$

$P\left(\frac{5-\sqrt{2}}{4}\right)=\left(\frac{5-5\sqrt{2}}{4}\right)\left(\frac{1-\sqrt{2}}{4}\right)>0$