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3 and 4 .Determinants and Matrices
hard
Let $M$ and $m$ respectively be the maximum and the minimum values of $f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R$ Then $M ^4- m ^4$ is equal to :____
A$1280$
B$1295$
C$1040$
D$1215$
(JEE MAIN-2025)
Solution
$\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R$
$R _2 \rightarrow R _2- R _1 \& R _3 \rightarrow R _3 \rightarrow R _1$
$f ( x )\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$
Expand about $R _1$, use get
$f(x)=2+4 \sin 4 x$
$\therefore M=\max$ value of $f(x)=6$
$M=\min$ value of $f(x)=-2$
$\therefore M^4-M^4=1280$
$R _2 \rightarrow R _2- R _1 \& R _3 \rightarrow R _3 \rightarrow R _1$
$f ( x )\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$
Expand about $R _1$, use get
$f(x)=2+4 \sin 4 x$
$\therefore M=\max$ value of $f(x)=6$
$M=\min$ value of $f(x)=-2$
$\therefore M^4-M^4=1280$
Standard 12
Mathematics
