Let M and m respectively be the maximum and the minimum values of f(x)= 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 xend array , x in R Then M 4 -m 4 is equal to

Mathematics · Determinants

JEE Main 2025 — 29 January, Morning Shift — Question 46

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

Option A:
1280
Correct
Option B:
1295
Option C:
1040
Option D:
1215

Answer: A

Step-by-step solution

$\left|\begin{array}{ccc}1+\sin ^{2} x & \cos ^{2} x & 4 \sin 4 x \\ \sin ^{2} x & 1+\cos ^{2} x & 4 \sin 4 x \\ \sin ^{2} x & \cos ^{2} x & 1+4 \sin 4 x\end{array}\right|, x \in R$

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \& \mathrm{R}_{3} \rightarrow \mathrm{R}_{3} - \mathrm{R}_{1}$

$f(x)\left|\begin{array}{ccc}1+\sin ^{2} x & \cos ^{2} x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$

Expand about $\mathrm{R}_{1}$ , use get $f(x)=2+4 \sin 4 x$

$\therefore \mathrm{M}=$ max value of $\mathrm{f}(\mathrm{x})=6$

$M=\min$ value of $f(x)=-2$

$\therefore \mathrm{M}^{4}-\mathrm{m}^{4}=1280$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Mathematics
Chapter: Determinants
Topic: Determinants