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)

$f(\theta)=\frac{1}{2}\begin{vmatrix} & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{vmatrix}+\begin{vmatrix}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _c\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _c \frac{\pi}{4} & \tan \pi\end{vmatrix}$
$f(\theta)=\frac{1}{2}\begin{vmatrix}2 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \\ 0 & -\sin \theta & 1\end{vmatrix}+\begin{vmatrix}0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{ c }\left(\frac{4}{\pi}\right) \\ -\tan \left(\theta-\frac{\pi}{4}\right) & -\log _{ c }\left(\frac{4}{\pi}\right) & 0\end{vmatrix}$
$ f (\theta)=\left(1+\sin ^2 \theta\right)+0 \text { (skew symmetric) } $
$ g (\theta)=\sqrt{ f (\theta)-1}+\sqrt{ f \left(\frac{\pi}{2}-\theta\right)-1}$
$=|\sin \theta|+|\cos \theta| \text { for } \theta \in\left[0, \frac{\pi}{2}\right] $
$ g (\theta) \in[1, \sqrt{2}] $
$ \text { Again let } P ( x )= k ( x -\sqrt{2})( x -1) $
$ 2-\sqrt{2}= k (2-\sqrt{2})(2-1) $
$ \Rightarrow k =1 ( P (2)=2-\sqrt{2} \text { given) } $
$ \therefore P ( x )=( x -\sqrt{2})( x -1) $
$\text { for option (A) } P \left(\frac{3+\sqrt{2}}{4}\right)<0 \text { correct }$ $ \text { option (B) } P \left(\frac{1+3 \sqrt{2}}{4}\right)<0 \text { incorrect }$
option (C) $P \left(\frac{5 \sqrt{2}-1}{4}\right) >0$ correct
option (D) $P \left(\frac{5-\sqrt{2}}{4}\right) >0$ incorrect