The problem defines a function $g(\theta)$ in terms of a determinant $f(\theta)$. We are given $g(\theta) = \sqrt{f(\theta)-1} + \sqrt{f(\pi/2 - \theta) - 1}$, where $f(\theta)$ is a determinant of a $3 \times 3$ matrix. We are also given that $p(x)$ is a quadratic polynomial whose roots are the maximum and minimum values of $g(\theta)$, and $p(2) = 2 - \sqrt{2}$. We need to determine which of the given inequalities involving $p(x)$ are true.

1. Problem Description

The problem defines a function $g(\theta)$ in terms of a determinant $f(\theta)$. We are given $g(\theta) = \sqrt{f(\theta)-1} + \sqrt{f(\pi/2 - \theta) - 1}$, where $f(\theta)$ is a determinant of a $3 \times 3$ matrix. We are also given that $p(x)$ is a quadratic polynomial whose roots are the maximum and minimum values of $g(\theta)$, and $p(2) = 2 - \sqrt{2}$. We need to determine which of the given inequalities involving $p(x)$ are true.

2. Solution Steps

First, let's compute $f(\theta)$:

$f(\theta) = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix}$

$f(\theta) = 1(1+\sin^2 \theta) - \sin \theta (-\sin \theta + \sin \theta) + 1(\sin^2 \theta + 1)$

$f(\theta) = 1 + \sin^2 \theta + 0 + \sin^2 \theta + 1 = 2 + 2 \sin^2 \theta$

Now, let's compute $f(\pi/2 - \theta)$:

$f(\pi/2 - \theta) = 2 + 2 \sin^2(\pi/2 - \theta) = 2 + 2 \cos^2 \theta$

Therefore, $g(\theta) = \sqrt{2 + 2 \sin^2 \theta - 1} + \sqrt{2 + 2 \cos^2 \theta - 1}$

$g(\theta) = \sqrt{1 + 2 \sin^2 \theta} + \sqrt{1 + 2 \cos^2 \theta}$

We need to find the maximum and minimum values of $g(\theta)$.

Note that $g(\pi/2 - \theta) = \sqrt{1 + 2 \cos^2 \theta} + \sqrt{1 + 2 \sin^2 \theta} = g(\theta)$. We can also calculate the derivative:

$g'(\theta) = \frac{4 \sin \theta \cos \theta}{2 \sqrt{1 + 2 \sin^2 \theta}} - \frac{4 \sin \theta \cos \theta}{2 \sqrt{1 + 2 \cos^2 \theta}} = 2 \sin \theta \cos \theta (\frac{1}{\sqrt{1+2\sin^2\theta}} - \frac{1}{\sqrt{1+2\cos^2\theta}})$

Setting $g'(\theta) = 0$, either $\sin \theta \cos \theta = 0$, or $\sqrt{1+2\sin^2\theta} = \sqrt{1+2\cos^2\theta}$

If $\sin \theta \cos \theta = 0$, then $\sin \theta = 0$ or $\cos \theta = 0$, so $\theta = 0$ or $\theta = \pi/2$.

If $\sqrt{1+2\sin^2\theta} = \sqrt{1+2\cos^2\theta}$, then $\sin^2 \theta = \cos^2 \theta$, so $\tan^2 \theta = 1$, meaning $\tan \theta = \pm 1$.

Since $\theta \in [0, \pi/2]$, $\theta = \pi/4$.

When $\theta = 0$, $g(0) = \sqrt{1} + \sqrt{1+2} = 1 + \sqrt{3}$.

When $\theta = \pi/2$, $g(\pi/2) = \sqrt{1+2} + \sqrt{1} = \sqrt{3} + 1$.

When $\theta = \pi/4$, $g(\pi/4) = \sqrt{1 + 2(1/2)} + \sqrt{1 + 2(1/2)} = \sqrt{2} + \sqrt{2} = 2 \sqrt{2}$.

Since $1+\sqrt{3} \approx 1 + 1.732 = 2.732$ and $2\sqrt{2} \approx 2(1.414) = 2.828$, the maximum value is $2\sqrt{2}$ when $\theta = \pi/4$, and the minimum value is $1+\sqrt{3}$ when $\theta=0$ or $\theta = \pi/2$. This is incorrect because $2\sqrt{2}$ is the min and $1 + \sqrt{3}$ is the max.

When $\theta = 0$, $g(0) = 1 + \sqrt{3}$. When $\theta = \frac{\pi}{4}$, $g(\frac{\pi}{4}) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$. When $\theta = \frac{\pi}{2}$, $g(\frac{\pi}{2}) = \sqrt{3} + 1$.

Thus, minimum = $2\sqrt{2}$ and maximum is $1 + \sqrt{3}$.

Since the roots of $p(x)$ are $2\sqrt{2}$ and $1+\sqrt{3}$, $p(x) = a(x-2\sqrt{2})(x-(1+\sqrt{3}))$ for some constant $a$.

We are given $p(2) = 2 - \sqrt{2}$. Thus, $a(2-2\sqrt{2})(2 - (1+\sqrt{3})) = 2 - \sqrt{2}$

$a(2-2\sqrt{2})(1-\sqrt{3}) = 2-\sqrt{2}$

$a(2(1-\sqrt{2}))(1-\sqrt{3}) = 2-\sqrt{2}$

$a(1-\sqrt{2})(1-\sqrt{3}) = 1 - \sqrt{2}/2$

$2a(1-\sqrt{2})(1-\sqrt{3}) = 2-\sqrt{2}$

Since $a(1-\sqrt{2})(1-\sqrt{3}) = \frac{2-\sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$, $a = \frac{1-\frac{\sqrt{2}}{2}}{(1-\sqrt{2})(1-\sqrt{3})} = \frac{\frac{2-\sqrt{2}}{2}}{(1-\sqrt{2})(1-\sqrt{3})}$.

It turns out the roots are minimum $2\sqrt{2}$ and maximum $1+\sqrt{3}$ so the shape of the graph is upside down and a is negative. Thus $p(x)=a(x-(1+\sqrt{3}))(x-2\sqrt{2})$.

$p(2)=2-\sqrt{2}$ so $a(2-1-\sqrt{3})(2-2\sqrt{2})=2-\sqrt{2}$

$a(1-\sqrt{3})(2-2\sqrt{2})=2-\sqrt{2}$

$a(1-\sqrt{3})2(1-\sqrt{2})=2-\sqrt{2}$

$a=\frac{2-\sqrt{2}}{2(1-\sqrt{3})(1-\sqrt{2})}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(1-\sqrt{3})(1-\sqrt{2})}=-\frac{\sqrt{2}}{2(1-\sqrt{3})}=\frac{\sqrt{2}(\sqrt{3}+1)}{2(-2)}=-\frac{\sqrt{2}(\sqrt{3}+1)}{4}$

Since $a < 0$, $p(x)$ is a downward-facing parabola. The roots are $2\sqrt{2} \approx 2.828$ and $1+\sqrt{3} \approx 2.732$.

$p(x) = a(x-2\sqrt{2})(x-1-\sqrt{3})$.

A. $p(\frac{3+\sqrt{2}}{4}) < 0$. $\frac{3+\sqrt{2}}{4} \approx \frac{3+1.414}{4} \approx \frac{4.414}{4} \approx 1.1$. Since $\frac{3+\sqrt{2}}{4}$ is less than the roots, we expect $p(x)$ is negative.

B. $p(\frac{1+3\sqrt{2}}{4}) > 0$. $\frac{1+3\sqrt{2}}{4} \approx \frac{1+3(1.414)}{4} \approx \frac{1+4.242}{4} \approx \frac{5.242}{4} \approx 1.3$.

C. $p(\frac{5\sqrt{2}-1}{4}) > 0$. $\frac{5\sqrt{2}-1}{4} \approx \frac{5(1.414)-1}{4} \approx \frac{7.07-1}{4} \approx \frac{6.07}{4} \approx 1.5$.

D. $p(\frac{5-\sqrt{2}}{4}) < 0$. $\frac{5-\sqrt{2}}{4} \approx \frac{5-1.414}{4} \approx \frac{3.586}{4} \approx 0.8965$.

Since $p(2\sqrt{2})=0$ and $p(1+\sqrt{3})=0$, and knowing a is negative ($a<0$) then values outside of this range make $p(x)<0$. Values inside this range make $p(x)>0$.

$p(\frac{3+\sqrt{2}}{4})<0$ since $\frac{3+\sqrt{2}}{4} \approx 1.1$ which is less than $2.732$ and $2.828$.

$p(\frac{5-\sqrt{2}}{4})<0$ since $\frac{5-\sqrt{2}}{4} \approx 0.9$ which is less than $2.732$ and $2.828$.

Options A and D are correct.

3. Final Answer

A, D