Verified Solution • 2025–26 Session moderate

QJEEM25Ma4e2b: JEE Main 2025 Mathematics - Class 12 JEEM25Ma4e2b Solution: Let M and m respectively be the maximum and the minimum values…

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

f(x)=∣1+sin⁡2xamp;cos⁡2xamp;4sin⁡4xsin⁡2xamp;1+cos⁡2xamp;4sin⁡4xsin⁡2xamp;cos⁡2xamp;1+4sin⁡4x∣,x∈Rf(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 M4−m4M^4 - m^4 is equal to :

Multiple Choice Options

  • A 1280
  • B 1040
  • C 1215
  • D 1295

Step-by-Step Verified Solution

amp;∣1+sin⁡2xamp;cos⁡2xamp;4sin⁡4xsin⁡2xamp;1+cos⁡2xamp;4sin⁡4xsin⁡2xamp;cos⁡2xamp;1+4sin⁡4x∣,x∈Ramp;R2→R2−R1&R3→R3−R1amp;f(x)∣1+sin⁡2xamp;cos⁡2xamp;4sin⁡4x−1amp;1amp;0−1amp;0amp;1∣\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-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 , we get

f(x)=2+4sin⁡4xf(x)=2+4 \sin 4 x

∴M=max⁡\therefore M=\max value of f(x)=6f(x)=6

m=min\mathrm{m}=\mathrm{min} value of f(x)=−2\mathrm{f}(\mathrm{x})=-2

∴M4−m4=1280\therefore \mathrm{M}^4-\mathrm{m}^4=1280

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