Verified Solution • 2025–26 Session moderate

QJEEM24Ma737d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma737d Solution: If then is equal to :

If f(x)=∣2cos⁡4xamp;2sin⁡4xamp;3+sin⁡22x3+2cos⁡4xamp;2sin⁡4xamp;sin⁡22x2cos⁡4xamp;3+2sin⁡4xamp;sin⁡22x∣,f(x)=\left|\begin{array}{ccc}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x\end{array}\right|, then 15f′(0)=\frac{1}{5} f^{\prime}(0)= is equal to :

Multiple Choice Options

  • A 2
  • B 1
  • C 0
  • D 6

Step-by-Step Verified Solution

amp;∣2cos⁡4xamp;2sin⁡4xamp;3+sin⁡22x3+2cos⁡4xamp;2sin⁡4xamp;sin⁡22x2cos⁡4xamp;3+2sin⁡24xamp;sin⁡22x∣amp;R2→R2−R1,R3→R3−R1amp;∣2cos⁡4xamp;2sin⁡4xamp;3+sin⁡22x3amp;0amp;−30amp;3amp;−3∣amp;f(x)=45amp;f′(x)=0amp;\begin{aligned}& \left|\begin{array}{ccc}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x\end{array}\right| \\& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\& \left|\begin{array}{ccc}2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\3 & 0 & -3 \\0 & 3 & -3\end{array}\right| \\& \mathrm{f}(\mathrm{x})=45 \\& \mathrm{f}^{\prime}(\mathrm{x})=0 \\&\end{aligned}

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