$\begin{aligned}
& \left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \
\sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \
\sin ^2 x & \cos ^2 x & 1+4 \sin 4 x
\end{array}\right|, x \in R \
& R_2 \rightarrow R_2-R_1 & R_3 \rightarrow R_3 \rightarrow R_1 \
& f(x)\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \
-1 & 1 & 0 \
-1 & 0 & 1
\end{array}\right|
\end{aligned}Expandabout\mathrm{R}_1,usegetf(x)=2+4 \sin 4 x\therefore \mathrm{M}=maxvalueof\mathrm{f}(\mathrm{x})=6m=\minvalueoff(x)=-2\therefore \mathrm{m}^4-\mathrm{M}^4=1280$
Let M and m respectively be the maximum and the minimum values of f(x)= | … |, x ∈ R Then M^4-m^4 is equal to :
JEE Main 2025 — Mathematics Algebra
Let M and m respectively be the maximum and the minimum values of f(x)=1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin4x4sin4x1+4sin4x,x∈R Then M4−m4 is equal to :
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
1280
- B
1295
- C
1215
- D
1040
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