Verified Solution • 2025–26 Session moderate

QJEEM21Ma253c: JEE Main 2021 Mathematics - Class 12 JEEM21Ma253c Solution: The function f(x) = :

The function
f(x) = 4x3−3x26−2sin⁡x+(2x−1)cos⁡x{{4{x^3} - 3{x^2}} \over 6} - 2\sin x + \left( {2x - 1} \right)\cos x :

Multiple Choice Options

  • A increases in (−∞,12]\left( { - \infty ,{1 \over 2}} \right]
  • B decreases in (−∞,12]\left( { - \infty ,{1 \over 2}} \right]
  • C increases in [12,∞)\left[ {{1 \over 2},\infty } \right)
  • D decreases in [12,∞)\left[ {{1 \over 2},\infty } \right)

Step-by-Step Verified Solution

Given, f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡xf(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x

f′(x)=12x2−6x6−2cos⁡x+(2x−1)(−sin⁡x)+cos⁡x(2)f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)

=(2x2−x)−2cos⁡x−2xsin⁡x+sin⁡x+2cos⁡x= (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x

=2x2−x−2xsin⁡x+sin⁡x= 2{x^2} - x - 2x\sin x + \sin x

=2x(x−sin⁡x)−1(x−sin⁡x)= 2x(x - \sin x) - 1(x - \sin x)

f′(x)=(2x−1)(x−sin⁡x)f'(x) = (2x - 1)(x - \sin x)

for x > 0,x - \sin x > 0

x < 0,x - \sin x < 0

for x∈(−∞,0]∪[12,∞),f′(x)≥0x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0

for x∈[0,12],f′(x)≤0x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0

Hence, f(x) increases in [12,∞)\left[ {{1 \over 2},\infty } \right) .

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