Let M and m respectively be the maximum and the minimum values of
$f(x)=\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x
\end{array}\right|, x \in \mathrm{R}$
Then $M^4-m^4$ is equal to :

Question

Aicharya · Accepted Answer

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

Let M and m respectively be the maximum and the minimum values of \(f(x)=\left|\begin{array}{ccc} 1+\sin ^2…

Solution