Mathematics - Determinant Question with Solution | TestHub

${\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1$

$f\left(\theta \right)=\left|\begin{array}{ccc}-{\sin }^2\theta & -1& 1\\ -{\cos }^2\theta & -1& 1\\ 12& -2& -2\end{array}\right|=4\left({\cos }^2\theta -{\sin }^2\theta \right)=4\left(\cos 2\theta \right)$

Since $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

Again, $\frac{\mathrm{\pi }}{4}\le \theta \le \frac{\mathrm{\pi }}{2}\Rightarrow 2\left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right)\le 2\theta \le 2\left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}\right)$

$\Rightarrow \frac{\mathrm{\pi }}{2}\le \theta \le \pi$$\therefore \theta \in \left[\frac{\pi }{4},\frac{\pi }{2}\right]\Rightarrow 2\theta \in \left[\frac{\pi }{2},\mathrm{\pi }\right]$ i.e., Second Quadrant.

$\Rightarrow -1\le \cos 2\theta \le 0$

$f\left(\theta \right){}_{\max }=M=0$

$f{\left(\theta \right)}_{min} =m=-4$