Determinant Function Values and Quadratic Polynomial Roots

To determine which options are true, we start by analyzing the function $f(\theta)$ given by the sum of two determinants:

$f(\theta) = \frac{1}{2} D_1 + D_2$

where $D_1 = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix}$

and $D_2 = \begin{vmatrix} \sin \pi & \cos\left(\theta + \frac{\pi}{4}\right) & \tan\left(\theta - \frac{\pi}{4}\right) \\ \sin\left(\theta - \frac{\pi}{4}\right) & -\cos\frac{\pi}{2} & \log_e\left(\frac{4}{\pi}\right) \\ \cot\left(\theta + \frac{\pi}{4}\right) & \log_e\left(\frac{\pi}{4}\right) & \tan \pi \end{vmatrix}$

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Step 1: Evaluating the first determinant $D_1$

Expanding $D_1$ along the first row: $D_1 = 1 \cdot (1 + \sin^2 \theta) - \sin\theta \cdot (-\sin\theta + \sin\theta) + 1 \cdot (\sin^2 \theta + 1)$ $D_1 = (1 + \sin^2 \theta) - 0 + (1 + \sin^2 \theta) = 2(1 + \sin^2 \theta)$

Thus: $\frac{1}{2} D_1 = 1 + \sin^2 \theta$

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Step 2: Evaluating the second determinant $D_2$

Notice the values of the diagonal elements:

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

Now, let us inspect the off-diagonal elements of $D_2$:

1. $a_{21} = \sin\left(\theta - \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4} - \theta\right) = -\cos\left(\frac{\pi}{2} - \left(\frac{\pi}{4} - \theta\right)\right) = -\cos\left(\theta + \frac{\pi}{4}\right) = -a_{12}$
2. $a_{31} = \cot\left(\theta + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right)\right) = \tan\left(\frac{\pi}{4} - \theta\right) = -\tan\left(\theta - \frac{\pi}{4}\right) = -a_{13}$
3. $a_{32} = \log_e\left(\frac{\pi}{4}\right) = -\log_e\left(\frac{4}{\pi}\right) = -a_{23}$

Since $a_{ij} = -a_{ji}$ for all $i, j$ and all diagonal terms are $0$, the matrix corresponding to $D_2$ is a skew-symmetric matrix of order $3$ (an odd order).

Since the determinant of any odd-order skew-symmetric matrix is zero, we have: $D_2 = 0$

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

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Step 3: Determining the function $g(\theta)$

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

Substituting $f(\theta) = 1 + \sin^2 \theta$: $f(\theta) - 1 = \sin^2 \theta$ $f\left(\frac{\pi}{2} - \theta\right) - 1 = 1 + \sin^2\left(\frac{\pi}{2} - \theta\right) - 1 = \cos^2 \theta$

For $\theta \in \left[0, \frac{\pi}{2}\right]$, both $\sin\theta \ge 0$ and $\cos\theta \ge 0$, so: $g(\theta) = \sqrt{\sin^2 \theta} + \sqrt{\cos^2 \theta} = \sin \theta + \cos \theta$

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Step 4: Finding the Maximum and Minimum Values of $g(\theta)$

We can rewrite $g(\theta)$ as: $g(\theta) = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right)$

For $\theta \in \left[0, \frac{\pi}{2}\right]$, the angle $\left(\theta + \frac{\pi}{4}\right) \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$.

• Maximum value: Occurs at $\theta = \frac{\pi}{4}$, giving: $g_{\max} = \sqrt{2} \sin\left(\frac{\pi}{2}\right) = \sqrt{2}$

• Minimum value: Occurs at the endpoints $\theta = 0$ or $\theta = \frac{\pi}{2}$, giving: $g_{\min} = \sin(0) + \cos(0) = 1$

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Step 5: Constructing the Quadratic Polynomial $p(x)$

The roots of the quadratic polynomial $p(x)$ are $1$ and $\sqrt{2}$. Therefore, $p(x)$ can be written in the form: $p(x) = k(x - 1)(x - \sqrt{2})$

Given that $p(2) = 2 - \sqrt{2}$: $p(2) = k(2 - 1)(2 - \sqrt{2}) = k(2 - \sqrt{2})$

Comparing with $p(2) = 2 - \sqrt{2}$, we get: $k = 1$

Thus, the polynomial is: $p(x) = (x - 1)(x - \sqrt{2})$

Since the leading coefficient $k = 1 > 0$, $p(x)$ is an upward-opening parabola such that:

• $p(x) < 0$ for $x \in (1, \sqrt{2})$
• $p(x) > 0$ for $x \in (-\infty, 1) \cup (\sqrt{2}, \infty)$

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Step 6: Checking the Options

1. Option A: $x = \frac{3 + \sqrt{2}}{4}$ $\frac{3 + \sqrt{2}}{4} - 1 = \frac{\sqrt{2} - 1}{4} > 0 \implies \frac{3 + \sqrt{2}}{4} > 1$ $\sqrt{2} - \frac{3 + \sqrt{2}}{4} = \frac{3\sqrt{2} - 3}{4} = \frac{3(\sqrt{2} - 1)}{4} > 0 \implies \frac{3 + \sqrt{2}}{4} < \sqrt{2}$ Since $\frac{3 + \sqrt{2}}{4} \in (1, \sqrt{2})$, we have $p\left(\frac{3 + \sqrt{2}}{4}\right) < 0$. Option A is TRUE.

2. Option B: $x = \frac{1 + 3\sqrt{2}}{4}$ $\frac{1 + 3\sqrt{2}}{4} - 1 = \frac{3\sqrt{2} - 3}{4} > 0 \implies \frac{1 + 3\sqrt{2}}{4} > 1$ $\sqrt{2} - \frac{1 + 3\sqrt{2}}{4} = \frac{\sqrt{2} - 1}{4} > 0 \implies \frac{1 + 3\sqrt{2}}{4} < \sqrt{2}$ Since $\frac{1 + 3\sqrt{2}}{4} \in (1, \sqrt{2})$, we have $p\left(\frac{1 + 3\sqrt{2}}{4}\right) < 0$. Option B is FALSE.

3. Option C: $x = \frac{5\sqrt{2} - 1}{4}$ $\frac{5\sqrt{2} - 1}{4} - \sqrt{2} = \frac{\sqrt{2} - 1}{4} > 0 \implies \frac{5\sqrt{2} - 1}{4} > \sqrt{2}$ Since $\frac{5\sqrt{2} - 1}{4} > \sqrt{2}$, we have $p\left(\frac{5\sqrt{2} - 1}{4}\right) > 0$. Option C is TRUE.

4. Option D: $x = \frac{5 - \sqrt{2}}{4}$ $1 - \frac{5 - \sqrt{2}}{4} = \frac{\sqrt{2} - 1}{4} > 0 \implies \frac{5 - \sqrt{2}}{4} < 1$ Since $\frac{5 - \sqrt{2}}{4} < 1$, we have $p\left(\frac{5 - \sqrt{2}}{4}\right) > 0$. Option D is FALSE.

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Conclusion

The correct options are A and C.