$\Rightarrow f (\theta)=3\left(\cos ^4 \theta+\sin ^4 \theta\right)-2 \cos ^2 2 \theta$
$\Rightarrow f (\theta)=3\left(1-\frac{1}{2} \sin ^2 2 \theta\right)-2 \cos ^2 2 \theta$ $\Rightarrow f (\theta)=3-\frac{3}{2} \sin ^2 2 \theta-2 \cos ^2 \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^2 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$f^{\prime}(\theta)=\sin 4 \theta$
$\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}$
$\Rightarrow 4 \theta=n \pi+(-1)^n \frac{\pi}{3}$
$\Rightarrow \theta=\frac{ n \pi}{4}+(-1)^{ n } \frac{\pi}{12}$
$\Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right)$
$\Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2}$ $\Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f (\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4}$
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Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where
$a_{i j}=J_{6+i, 3}-J_{i+3,3}, \quad i \leq j$
$\quad\quad\quad\quad\quad\quad0 , \quad\quad\quad i>j$.
Then $\left|\operatorname{adj} A^{-1}\right|$ is :