Verified Solution • 2025–26 Session moderate

QJEEM21Maba10: JEE Main 2021 Mathematics - Class 12 JEEM21Maba10 Solution: Let , . Then, f is :

Let f(x)=3sin⁡4x+10sin⁡3x+6sin⁡2x−3f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3 , x∈[−π6,π2]x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right] . Then, f is :

Multiple Choice Options

  • A increasing in (−π6,π2)\left( { - {\pi \over 6},{\pi \over 2}} \right)
  • B decreasing in (0,π2)\left( {0,{\pi \over 2}} \right)
  • C increasing in (−π6,0)\left( { - {\pi \over 6},0} \right)
  • D decreasing in (−π6,0)\left( { - {\pi \over 6},0} \right)

Step-by-Step Verified Solution

f(x)=3sin⁡4x+10sin⁡3x+6sin⁡2x−3,x∈[−π6,π2]f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3,x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]

f′(x)=12sin⁡3xcos⁡x+30sin⁡2xcos⁡x+12sin⁡xcos⁡xf'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x

=6sin⁡xcos⁡x(2sin⁡2x+5sin⁡x+2)= 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)

=6sin⁡xcos⁡x(2sin⁡x+1)(sin⁡+2)= 6\sin x\cos x(2\sin x + 1)(\sin + 2)

JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Application of Derivatives Question 94 English Explanation
Decreasing in (−π6,0)\left( { - {\pi \over 6},0} \right)

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