Verified Solution • 2025–26 Session moderate

QJEEM24Ma7ba7: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7ba7 Solution: lim x → π 2 ⁡ ( 1 ( x − π 2 ) 2 ∫ x 3 ( π 2 ) 3 cos ⁡ ( t 1 3 ) …

limx→π2⁡(1(x−π2)2∫x3(π2)3cos⁡(t13)dt) is equal to

Multiple Choice Options

  • A 3π24
  • B 3π28
  • C 3π4
  • D 3π8

Step-by-Step Verified Solution

Using L'hospital rule

=limx→π−20−cos⁡x×3x22(x−π2)=limx→π−2sin⁡(x−π2)2(x−π2)×3π24=3π28

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