Let $M$ and $m$ respectively be the maximum and the minimum values of $f(x)=\begin{vmatrix}1+\sin^{2}x & \cos^{2}x & 4\sin 4x\\ \sin^{2}x & 1+\cos^{2}x & 4\sin 4x\\ \sin^{2}x & \cos^{2}x & 1+4\sin 4x\end{vmatrix},\,x\in\mathbb{R}.$ Then $M^{4}-m^{4}$ is equal to:
Step-by-Step Solution
Key Concept: $R_{2}\to R_{2}-R_{1},\,R_{3}\to R_{3}-R_{1}$ turns the lower two rows into $(-1,1,0)$ and $(-1,0,1)$, dramatically simplifying the expansion. Then use $\sin^{2}x+\cos^{2}x=1.$
$R_{2}\to R_{2}-R_{1}$ and $R_{3}\to R_{3}-R_{1}$:
$$f(x)=\begin{vmatrix}1+\sin^{2}x & \cos^{2}x & 4\sin 4x\\ -1 & 1 & 0\\ -1 & 0 & 1\end{vmatrix}.$$
Expand along $R_{1}$:
$f(x)=(1+\sin^{2}x)(1)-\cos^{2}x(-1)+4\sin 4x(1)=1+\sin^{2}x+\cos^{2}x+4\sin 4x=2+4\sin 4x.$
$\sin 4x\in[-1,1]\Rightarrow f(x)\in[-2,6].$ So $M=6,\,m=-2.$
$M^{4}-m^{4}=1296-16=1280.$
Correct Answer: 1