Question

Let $|M|$ denote the determinant of a square matrix $M$ Let $g:\left[0, \frac{\pi}{2}\right] \rightarrow R$ 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?

• A. $p \left(\frac{3+\sqrt{2}}{4}\right)<0$
• B. $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
• C. $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
• D. $p \left(\frac{5-\sqrt{2}}{4}\right)<0$

Answer 1

Answer: (A), (C)

$p \left(\frac{3+\sqrt{2}}{4}\right)<0$

Answer 2

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

As second determinant is skew symmetric hence its value is 0.
$⇒ f(\theta) = (1 + sin^2 \theta)$
$⇒ g(\theta)=|sin\theta|+|cos\theta| ∈ [1,\sqrt2]$
⇒$$p(x) = a(x - 1) (x-\sqrt{2})\ as\ p(2) = 2 - \sqrt{2} ⇒a = 1
Hence $p \left(\frac{3+\sqrt{2}}{4}\right)<0$ and $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$