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: (C)

$p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$

Answer 2

Given :
$f(\theta)=\frac{1}{2}\begin{vmatrix}1 & \sin \theta & 1 \\\ -\sin \theta & 1 & \sin \theta \\\ -1 & -\sin \theta & 1\end{vmatrix}+\begin{vmatrix}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{ c }\left(\frac{4}{\pi}\right) \\\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{ c}\left(\frac{\pi}{4}\right) & \tan \pi\end{vmatrix}$
$f(\theta)=\frac{1}{2}\begin{vmatrix}2 & \sin \theta & 1 \\\ 0 & 1 & \sin \theta \\\ 0 & -\sin \theta & 1\end{vmatrix}+\begin{vmatrix}0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\\ \sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{ c }\left(\frac{4}{\pi}\right) \\\ -\tan \left(\theta-\frac{\pi}{4}\right) & \log _{ c }\left(\frac{\pi}{4}\right) & 0\end{vmatrix}$
$f(\theta)=(1+\sin^2\theta)+0$ ( skew symmetric )
Now,
$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f(\frac{\pi}{2}-\theta)-1}$
$=|\sin\theta|+|\cos\theta|\ \text{for}\ \theta \in[0,\frac{\pi}{2}]$
$g(\theta)\in[1,\sqrt2]$
Now, Again
Let $P(x)=k(x-\sqrt2)(x-1)$
$2-\sqrt2=k(2-\sqrt2(2-1)$
So, k = 1 as $P(2)=2-\sqrt2$ is given.
Hence, $P(x)=(x-\sqrt2)(x-1)$
Now, let's verify the options :
(A) $P(\frac{3+\sqrt2}{4})\lt0$ So, option (A) is correct.
(B) $P(\frac{1+3\sqrt2}{4})\lt0$ So, option (B) is incorrect.
(C) $P(\frac{5\sqrt2-1}{4})\gt0$ So, option (C) is correct.
(D) $P(\frac{5-\sqrt2}{4})\gt0$ So, option (D) is incorrect.
So, the correct options are (A) and (C).