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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Determinants

Let M and m respectively be the maximum and the minimum values of...

Let M and m respectively be the maximum and the minimum values of (f(x)=\left|1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x\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 M4−m4M^4-m^4 is equal to :

Options

  1. A1280
  2. B1295
  3. C1215
  4. D1040

Correct answer

A. 1280

Step-by-step solution

∣1+sin⁡2xcos⁡2x4sin⁡4xsin⁡2x1+cos⁡2x4sin⁡4xsin⁡2xcos⁡2x1+4sin⁡4x∣,x∈RR2→R2−R1R3→R3→R1f(x)∣1+sin⁡2xcos⁡2x4sin⁡4x−110−101∣\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 R1\mathrm{R}_1, use get f(x)=2+4sin⁡4x∴M=f(x)=2+4 \sin 4 x \therefore \mathrm{M}= max value of f(x)=6m=min⁡\mathrm{f}(\mathrm{x})=6 m=\min value of f(x)=−2∴m4−M4=1280f(x)=-2 \therefore \mathrm{m}^4-\mathrm{M}^4=1280
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