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

The correct options are:
$p\left(\frac{3+\sqrt{2}}{4}\right)<0$
and
$p\left(\frac{5\sqrt{2}-1}{4}\right)>0$

Step 1: Simplify the function f(θ)

The given function is defined as:
f(θ) = ¹/₂ · |A| + |B|

Let us first evaluate the determinant of the first matrix A:
$\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

Expanding the determinant along the first row:
|A| = 1 · (1 - (-sin²θ)) - sin θ · (-sin θ - (-sin θ)) + 1 · (sin²θ - (-1))
|A| = (1 + sin²θ) - 0 + (1 + sin²θ)
|A| = 2(1 + sin²θ)

Thus, the first part of f(θ) is:
¹/₂ · |A| = 1 + sin²θ

Now, let us analyze the second matrix B:
$\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta +\frac{\pi }{4}\right)& \tan \left(\theta -\frac{\pi }{4}\right)\\ \sin \left(\theta -\frac{\pi }{4}\right)& -\cos \frac{\pi }{2}& {\log }_e\left(\frac{4}{\pi }\right)\\ \cot \left(\theta +\frac{\pi }{4}\right)& {\log }_e\left(\frac{\pi }{4}\right)& \tan \pi \end{array}\right|$

Notice the diagonal elements:
sin π = 0, -cos(π/2) = 0, and tan π = 0.

Also, notice the off-diagonal pairs:
• sin(θ - π/4) = -cos(θ + π/4)
• cot(θ + π/4) = -tan(θ - π/4)
• log_e(π/4) = -log_e(4/π)

Since B_ij = -B_ji for all i, j and all diagonal entries are 0, matrix B is a skew-symmetric matrix of odd order 3.
The determinant of any odd-order skew-symmetric matrix is always 0. Therefore, |B| = 0.

Hence, we have:
f(θ) = 1 + sin²θ

Step 2: Find the expression for g(θ)

The function g(θ) is defined for θ ∈ [0, π/2] as:
$g\left(\theta \right)=\sqrt{f\left(\theta \right)-1}+\sqrt{f\left(\frac{\pi }{2}-\theta \right)-1}$

Substituting f(θ) = 1 + sin²θ:
$f\left(\theta \right)-1={\sin }^2\theta$
$f\left(\frac{\pi }{2}-\theta \right)-1={\sin }^2\left(\frac{\pi }{2}-\theta \right)={\cos }^2\theta$

Since θ ∈ [0, π/2], both sin θ ≥ 0 and cos θ ≥ 0:
g(θ) = sin θ + cos θ

Step 3: Determine the maximum and minimum values of g(θ)

We can rewrite g(θ) as:
g(θ) = √2 · sin(θ + π/4)

For θ ∈ [0, π/2]:
• At θ = 0 or θ = π/2, g(θ) attains its minimum value: 1
• At θ = π/4, g(θ) attains its maximum value: √2

Thus, the roots of the quadratic polynomial p(x) are 1 and √2.

Step 4: Find the quadratic polynomial p(x)

Since 1 and √2 are the roots of p(x), we can express p(x) as:
p(x) = k(x - 1)(x - √2)

Given that p(2) = 2 - √2:
p(2) = k(2 - 1)(2 - √2) = k(2 - √2) = 2 - √2 ⇒ k = 1

Therefore, the polynomial is:
p(x) = (x - 1)(x - √2)

Since the coefficient of x² is positive, the sign of p(x) behaves as follows:
• p(x) < 0 for x ∈ (1, √2)
• p(x) > 0 for x < 1 or x > √2

Step 5: Check the given options

1. Evaluate at x = (3 + √2)/4:
Since √2 ≈ 1.414, we have (3 + 1.414)/4 = 4.414/4 = 1.1035.
Since 1 < 1.1035 < √2, the value lies strictly between the roots.
Hence, $p\left(\frac{3+\sqrt{2}}{4}\right)<0$ is TRUE.

2. Evaluate at x = (1 + 3√2)/4:
(1 + 3(1.414))/4 = 5.242/4 = 1.3105.
Since 1 < 1.3105 < √2, p(x) must be negative. Thus, option 2 is FALSE.

3. Evaluate at x = (5√2 - 1)/4:
(5(1.414) - 1)/4 = 6.07/4 = 1.5175.
Since 1.5175 > √2, this value is greater than the larger root.
Hence, $p\left(\frac{5\sqrt{2}-1}{4}\right)>0$ is TRUE.

4. Evaluate at x = (5 - √2)/4:
(5 - 1.414)/4 = 3.586/4 = 0.8965.
Since 0.8965 < 1, p(x) must be positive. Thus, option 4 is FALSE.