Let M and m respectively be the maximum and the minimum values of [f(x)=vmatrix1+^2x & ^2x…

Question

Let $M$ and $m$ respectively be the maximum and the minimum values of
$$f(x)=\begin{vmatrix}1+\sin^2x & \cos^2x & 4\sin4x\ \sin^2x & 1+\cos^2x & 4\sin4x\ \sin^2x & \cos^2x & 1+4\sin4x\end{vmatrix},\ x\in\mathbb{R}.$$
Then $M^4-m^4$ is equal to:

Accepted Answer

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ simplifies the determinant to
$$f(x)=1\cdot\big[(1)(1)-(-1)(0)\big]+\cdots = 2+4\sin4x$$
(after full row-reduction and expansion, as in the source working).

So $f(x)=2+4\sin4x$, giving $M=2+4=6$ and $m=2-4=-2$.

$M^4-m^4=6^4-(-2)^4=1296-16=1280$.

Solutions · 1