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 where $g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$ $f(\theta)=\frac{1}{2}\begin{vmatrix}1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1\end{vmatrix}+\begin{vmatrix}\sin\pi & \cos\left(\theta+\frac{\pi}{4}\right) & \tan\left(\theta-\frac{\pi}{4}\right) \\ \sin\left(\theta-\frac{\pi}{4}\right) & -\cos\frac{\pi}{2} & \log_e\left(\frac{4}{\pi}\right) \\ \cot\left(\theta+\frac{\pi}{4}\right) & \log_e\left(\frac{\pi}{4}\right) & \tan\pi\end{vmatrix}$ 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 ?

$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\dfrac{\pi}{2}-\theta\right)-1}$ $f(\theta)=\dfrac{1}{2}\det(A)+\det(B),\quad \theta\in\left[0,\dfrac{\pi}{2}\right]$ $p(x):\ \text{roots } = \max g(\theta),\ \min g(\theta);\quad p(2)=2-\sqrt{2}$ $\text{Which of (1)--(4) are TRUE?}$ $\text{Plan: } 1.\ \det A,\quad 2.\ \det B,\quad 3.\ f=1+\sin^2\theta$ $4.\ g=\sin\theta+\cos\theta,\quad 5.\ p \text{ from its roots}$ $A=\begin{vmatrix}1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1\end{vmatrix}$ $\det A=(1+\sin^2\theta)-\sin\theta(0)+(1+\sin^2\theta)=2(1+\sin^2\theta)$ $\sin\!\left(\theta-\tfrac\pi4\right)=-\cos\!\left(\theta+\tfrac\pi4\right)$ $\tan\!\left(\theta-\tfrac\pi4\right)=-\cot\!\left(\theta+\tfrac\pi4\right),\quad \ln\tfrac\pi4=-\ln\tfrac4\pi$ $B=\begin{pmatrix}0 & c & -d \\ -c & 0 & L \\ d & -L & 0\end{pmatrix},\quad B^{T}=-B$ $\text{odd-order skew-symmetric} \Rightarrow \det B=0$ $f(\theta)=\tfrac12\det A+\det B=1+\sin^2\theta$ $g(\theta)=\sqrt{\sin^2\theta}+\sqrt{\sin^2\!\left(\tfrac\pi2-\theta\right)}=|\sin\theta|+|\cos\theta|$ $\theta\in\left[0,\tfrac\pi2\right]\Rightarrow g(\theta)=\sin\theta+\cos\theta$ $g(\theta)=\sqrt2\,\sin\!\left(\theta+\tfrac\pi4\right)$ $\max g=\sqrt2,\qquad \min g=1$ $p(x)=k(x-1)(x-\sqrt2)$ $p(2)=k(1)(2-\sqrt2)=2-\sqrt2 \Rightarrow k=1$ $p(x)=(x-1)(x-\sqrt2)=x^2-(1+\sqrt2)x+\sqrt2$ $p(x)<0 \iff 1<x<\sqrt2$ $(1)\ x=\dfrac{3+\sqrt2}{4}\in(1,\sqrt2)\Rightarrow p<0\ \textbf{TRUE}$ $(2)\ x=\dfrac{1+3\sqrt2}{4}\in(1,\sqrt2)\Rightarrow p<0\ \textbf{FALSE}$ $(3)\ x=\dfrac{5\sqrt2-1}{4}>\sqrt2\Rightarrow p>0\ \textbf{TRUE}$ $(4)\ x=\dfrac{5-\sqrt2}{4}<1\Rightarrow p>0\ \textbf{FALSE}$ $\text{TRUE statements: }(1)\ \text{and}\ (3)$