Question

Let $\rm|M|$ denote the determinant of a square matrix M. Let $\rm g : \bigg[0, \dfrac{\pi}{2} \bigg] \to \mathbb{R}$ be the function defined by $\rm g( \theta) = \sqrt{f( \theta) - 1} + \sqrt{f \bigg( \dfrac{\pi}{2} - \theta\bigg) - 1}$ where
$\rm f( \theta) = \dfrac{1}{2} \left| \begin{matrix} 1& \sin( \theta) &1 \\ - \sin( \theta) &1& \sin( \theta) \\ - 1& - \sin( \theta)&1 \end{matrix} \right | + \left| \begin{matrix} \sin(\pi) & \cos( \theta + \frac{\pi}{4} ) & \tan( \theta - \frac{\pi}{4} ) \\ \sin( \theta - \frac{\pi}{4} ) & - \cos( \frac{\pi}2 ) & \log_{e} ( \frac{4}\pi ) \\ \cot( \theta + \frac{\pi}{4} ) & \log_{e} ( \frac{\pi}4 )& \tan(\pi) \end{matrix} \right |$
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 True?

$\rm(1) \: p \bigg( \frac{3 + \sqrt{2} }{4} \bigg) < 0 \\ \rm(2) \: p \bigg( \frac{1 + 3 \sqrt{2} }{4} \bigg) > 0 \\ \rm(3) \: p \bigg( \frac{5\sqrt{2} - 1 }{4} \bigg) > 0 \\ \rm(4) \: p \bigg( \frac{5 - \sqrt{2} }{4} \bigg) < 0$

​

Answer 1

The second matrix in the definition of $f$ is singular, since

$\sin(\pi) = -\cos\left(\dfrac\pi2\right) = \tan(\pi) = 0$

$\cos\left(\theta+\dfrac\pi4\right) = \sin\left(\dfrac\pi2 - \left(\theta+\dfrac\pi4\right)\right) = \sin\left(\dfrac\pi4-\theta\right)=-\sin\left(\theta-\dfrac\pi4\right)$

$\cot\left(\theta+\dfrac\pi4\right) = \tan\left(\dfrac\pi2 - \left(\theta+\dfrac\pi4\right)\right) = \tan\left(\dfrac\pi4 - \theta\right) = -\tan\left(\theta-\dfrac\pi4\right)$

$\ln\left(\dfrac4\pi\right) = -\ln\left(\dfrac\pi4\right)$

In other words, it's antisymmetric; $A^\top=-A$. It's easy to show that $\det(A)=0$ if $A$ is 3x3 and antisymmetric.

The other determinant reduces to

$\begin{vmatrix}1 & \sin(\theta) & 1 \\ - \sin(\theta) & 1 & \sin(\theta) \\ -1 & -\sin(\theta) & 1 \end{vmatrix} = 2 + 2\sin^2(\theta)$

Hence

$f(\theta) = 1 + \sin^2(\theta) \implies f\left(\dfrac\pi2\right) = 1 + \cos^2(\theta)$

With $g$ defined on $\left[0,\frac\pi2\right]$, both $\sin(\theta)$ and $\cos(\theta)$ are non-negative. So

$g(\theta) = \sqrt{f(\theta)-1} + \sqrt{f\left(\dfrac\pi2-\theta\right)-1} \\\\ ~~~~ = \sqrt{\sin^2(\theta)} + \sqrt{\cos^2(\theta)} \\\\ ~~~~ = |\sin(\theta)| + |\cos(\theta)| \\\\ ~~~~ = \sin(\theta) + \cos(\theta) \\\\ ~~~~ = \sqrt2\,\sin\left(\theta + \dfrac\pi4\right)$

which is maximized at $t=\frac\pi4$ with a value of $\sqrt2\,\sin\left(\frac\pi2\right)=\sqrt2$, and minimized at $t=0$ and $t=\frac\pi2$ with a value of $\sqrt2\,\sin\left(\frac{3\pi}4\right)=1$.

Edit: The rest of my answer wouldn't fit. Continued in attachment.