Let $$|M|$$ denote the determinant of a square matrix $$M$$. Let $$g: \left[0, \frac{\pi}{2}\right] \to \mathbb{R}$$ be the function defined by

$$g(\theta) = \sqrt{f(\theta) - 1} + \sqrt{f\left(\frac{\pi}{2} - \theta\right) - 1}$$

where

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

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?

Write $$s=\sin\theta$$ for brevity.

The first determinant in $$f(\theta)$$ is

$$ \begin{vmatrix} 1 & s & 1\\ -s & 1 & s\\ -1 & -s & 1 \end{vmatrix}. $$

Expanding along the first row,

$$ |A| =1\!\left| \begin{matrix} 1&s\\ -s&1 \end{matrix}\right| -\;s\! \left| \begin{matrix} -s&s\\ -1&1 \end{matrix}\right| +\;1\! \left| \begin{matrix} -s&1\\ -1&-s \end{matrix}\right| =1\,(1+s^{2})-s\,(0)+1\,(1+s^{2}) =2(1+s^{2}). $$

Hence

$$\frac{1}{2}|A|=1+\sin^{2}\theta.$$ So the first part of $$f(\theta)$$ is $$1+\sin^{2}\theta.$$

Now examine the second determinant. Note the numerical values

$$\sin\pi=0,\qquad\cos\frac{\pi}{2}=0,\qquad\tan\pi=0,$$ $$\ln\!\left(\frac{\pi}{4}\right)=-\ln\!\left(\frac{4}{\pi}\right).$$

Thus the matrix becomes

$$ B= \begin{pmatrix} 0 & \cos\!\left(\theta+\frac{\pi}{4}\right) & \tan\!\left(\theta-\frac{\pi}{4}\right)\\[2pt] \sin\!\left(\theta-\frac{\pi}{4}\right) & 0 & \ln\!\left(\frac{4}{\pi}\right)\\[2pt] \cot\!\left(\theta+\frac{\pi}{4}\right) & -\ln\!\left(\frac{4}{\pi}\right) & 0 \end{pmatrix}. $$

Denote

$$ \begin{aligned} c &=\cos\!\left(\theta+\frac{\pi}{4}\right),& k &=\cot\!\left(\theta+\frac{\pi}{4}\right),\\ s_1&=\sin\!\left(\theta-\frac{\pi}{4}\right),& t &=\tan\!\left(\theta-\frac{\pi}{4}\right),\\ L &=\ln\!\left(\frac{4}{\pi}\right). \end{aligned} $$

Expanding $$|B|$$ along the first row,

$$ |B| =-c\! \begin{vmatrix} s_1 & L\\[2pt] k & 0 \end{vmatrix} +t\! \begin{vmatrix} s_1 & 0\\[2pt] k & -L \end{vmatrix} =L\,(c\,k-t\,s_1). $$

Put $$A=\theta+\dfrac{\pi}{4},\;B=\theta-\dfrac{\pi}{4}.$$ Then $$A-B=\dfrac{\pi}{2}\implies \sin A=\cos B,\quad \cos A=-\sin B.$$ Hence

$$c=\cos A=-\sin B=-s_1,\qquad k=\cot A=\frac{\cos A}{\sin A}=\frac{-s_1}{\cos B}=-t.$$

Therefore $$c\,k=t\,s_1,$$ giving $$|B|=0$$ for every $$\theta.$$

Consequently

$$f(\theta)=1+\sin^{2}\theta,\qquad f\!\left(\frac{\pi}{2}-\theta\right)=1+\sin^{2}\!\left(\frac{\pi}{2}-\theta\right)=1+\cos^{2}\theta.$$ So

$$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\!\left(\frac{\pi}{2}-\theta\right)-1} =\sqrt{\sin^{2}\theta}+\sqrt{\cos^{2}\theta} =\sin\theta+\cos\theta\qquad\left(0\le\theta\le\frac{\pi}{2}\right).$$

The derivative $$g'(\theta)=\cos\theta-\sin\theta$$ vanishes only at $$\theta=\frac{\pi}{4},$$ which is a maximum because $$g''(\theta)=-\sin\theta-\cos\theta\lt 0.$$ Hence

Maximum value $$=g\!\left(\frac{\pi}{4}\right)=\sqrt{2},\qquad \text{Minimum value}=g(0)=g\!\left(\frac{\pi}{2}\right)=1.$$

Let $$p(x)$$ be a quadratic whose roots are $$1$$ and $$\sqrt{2}.$$ Then $$p(x)=k(x-1)(x-\sqrt{2}).$$ Given $$p(2)=2-\sqrt{2},$$

$$k(2-1)(2-\sqrt{2})=2-\sqrt{2}\;\Longrightarrow\;k=1.$$ Thus

$$p(x)=(x-1)(x-\sqrt{2})=x^{2}-(1+\sqrt{2})x+\sqrt{2}.$$ Because the coefficient of $$x^{2}$$ is positive,

• $$p(x)\lt0$$ when $$1\lt x\lt \sqrt{2},$$
• $$p(x)\gt0$$ when $$x\lt1$$ or $$x\gt\sqrt{2}.$$

Evaluate the four numbers:

$$ \begin{aligned} x_A&=\frac{3+\sqrt{2}}{4},& x_B&=\frac{1+3\sqrt{2}}{4},\\[3pt] x_C&=\frac{5\sqrt{2}-1}{4},& x_D&=\frac{5-\sqrt{2}}{4}. \end{aligned} $$

Check their positions relative to $$1$$ and $$\sqrt{2}$$ (using $$\sqrt{2}\approx1.414$$):

$$ \begin{aligned} 1\lt x_A=1.1035\lt \sqrt{2}&\;\Longrightarrow\;p(x_A)\lt0,\\ 1\lt x_B=1.3106\lt \sqrt{2}&\;\Longrightarrow\;p(x_B)\lt0,\\ x_C=1.5178\gt \sqrt{2}&\;\Longrightarrow\;p(x_C)\gt0,\\ x_D=0.8964\lt 1&\;\Longrightarrow\;p(x_D)\gt0. \end{aligned} $$

Hence

Case A: $$p\!\left(\dfrac{3+\sqrt{2}}{4}\right)\lt0$$  is TRUE.
Case B: $$p\!\left(\dfrac{1+3\sqrt{2}}{4}\right)\gt0$$  is FALSE.
Case C: $$p\!\left(\dfrac{5\sqrt{2}-1}{4}\right)\gt0$$  is TRUE.
Case D: $$p\!\left(\dfrac{5-\sqrt{2}}{4}\right)\lt0$$  is FALSE.

Therefore the correct options are:
Option A and Option C.