JEE Main 2025 — Determinants Question with Solution

$\begin{aligned} & \left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\right|, x \in R \\ & R_2 \rightarrow R_2-R_1 \& R_3 \rightarrow R_3 \rightarrow R_1 \\ & f(x)\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right| \end{aligned}$ Expand about ${\mathrm{R}}_1$, use get $f(x)=2+4\sin 4x$ $\therefore \mathrm{M}=$ max value of $\mathrm{f}(\mathrm{x})=6$ $m=\min$ value of $f(x)=-2$ $\therefore {\mathrm{m}}^4-{\mathrm{M}}^4=1280$