Let |M| denote the determinant of a square matrix M . Let g: [0,...

Given,

f\left(\theta\right) = \dfrac{1}{2}\left|\begin{cases}1 & sin\theta & 1 \\ - sin\theta & 1 & sin\theta \\ - 1 & - sin\theta & 1\end{cases}\right| + \left|\begin{cases}sin\pi & \cos \left(\theta + \dfrac{\pi}{4}\right) & \tan \left(\theta - \dfrac{\pi}{4}\right) \\ \sin \left(\theta - \dfrac{\pi}{4}\right) & - cos\dfrac{\pi}{2} & \log_{e}\left(\dfrac{4}{\pi}\right) \\ \cot \left(\theta + \dfrac{\pi}{4}\right) & \log_{e}\left(\dfrac{\pi}{4}\right) & tan \pi\end{cases}\right|

⇒ f\left(\theta\right) = \dfrac{1}{2}\left|\begin{cases}1 & sin\theta & 1 \\ - sin\theta & 1 & sin\theta \\ - 1 & - sin\theta & 1\end{cases}\right| + \left|\begin{cases}0 & \cos \left(\theta + \dfrac{\pi}{4}\right) & \tan \left(\theta - \dfrac{\pi}{4}\right) \\ \sin \left(\theta - \dfrac{\pi}{4}\right) & 0 & \log_{e}\left(\dfrac{4}{\pi}\right) \\ - \tan \left(\theta - \dfrac{\pi}{4}\right) & - \log_{e}\left(\dfrac{4}{\pi}\right) & 0\end{cases}\right|

Here we used $\cos \left(\theta + \dfrac{\pi}{4}\right) = - \sin \left(\theta - \dfrac{\pi}{4}\right)$

And $\tan \left(\theta - \dfrac{\pi}{4}\right) = - \cot \left(\theta + \dfrac{\pi}{4}\right)$

And $\log_{e}\left(\dfrac{4}{\pi}\right) = - \log_{e}\left(\dfrac{\pi}{4}\right)$

Also $sin\pi = - cos\dfrac{\pi}{2} = tan\pi = 0$

So, ⇒ f\left(\theta\right) = \dfrac{1}{2}\left|\begin{cases}1 & sin\theta & 1 \\ - sin\theta & 1 & sin\theta \\ - 1 & - sin\theta & 1\end{cases}\right| + \left(\text{skew symmetric}\right)

$⇒ f\left(\theta\right) = 1 + \sin^{2}\theta$

So, $g\left(\theta\right) = \left|\sin\theta\right| + \left|\cos\theta\right|$

Now maximum and minimum values are $\sqrt{2}$ and $1$ respectively.

Now quadratic polynomial will be $P\left(x\right) = a\left(x - \sqrt{2}\right)\left(x - 1\right)$, where $a \in R - \left\{0\right\}$,

But given $P\left(2\right) = 2 - \sqrt{2}$ , so $a = 1$.

$∴ P\left(x\right) = \left(x - \sqrt{2}\right)\left(x - 1\right)$

Now solving all options we get,

$P\left(\dfrac{3 + \sqrt{2}}{4}\right) = \left(\dfrac{3 - 3\sqrt{2}}{4}\right) \cdot \left(\dfrac{\sqrt{2} - 1}{4}\right) < 0$

$P\left(\dfrac{1 + 3\sqrt{2}}{4}\right) = \left(\dfrac{1 - \sqrt{2}}{4}\right) \cdot \left(\dfrac{3\sqrt{2} - 3}{4}\right) < 0$

$P\left(\dfrac{5\sqrt{2} - 1}{4}\right) = \left(\dfrac{\sqrt{2} - 1}{4}\right) \cdot \left(\dfrac{5\sqrt{2} - 5}{4}\right) > 0$

$P\left(\dfrac{5 - \sqrt{2}}{4}\right) = \left(\dfrac{5 - 5\sqrt{2}}{4}\right)\left(\dfrac{1 - \sqrt{2}}{4}\right) > 0$