Verified Solution • 2025–26 Session moderate

QJEEM21Maf1c0: JEE Main 2021 Mathematics - Class 11 JEEM21Maf1c0 Solution: The range of a R for which the function f(x) = (4a 3)(x + log e …

The range of a ∈\in R for which the

function f(x) = (4a −- 3)(x + loge 5) + 2(a −- 7) cot (x2)\left( {{x \over 2}} \right) sin2 (x2)\left( {{x \over 2}} \right) , x ≠\ne 2n π\pi , n ∈\in N has critical points, is :

Multiple Choice Options

  • A [1, ∞\infty )
  • B ( −- 3, 1)
  • C [−43,2]\left[ { - {4 \over 3},2} \right]
  • D ( −- ∞\infty , −- 1]

Step-by-Step Verified Solution

f(x)=(4a−3)(x+ln⁡5)+2(a−7)(cos⁡x2sin⁡x2.sin⁡2x2)f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)

f(x)=(4a−3)(x+ln⁡5)+(a−7)sin⁡xf(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x

f′(x)=(4a−3)+(a−7)cos⁡x=0f'(x) = (4a - 3) + (a - 7)\cos x = 0

cos⁡x=−(4a−3)a−7\cos x = {{ - (4a - 3)} \over {a - 7}}

−1≤−4a−3a−7≤1- 1 \le - {{4a - 3} \over {a - 7}} \le 1

−1≤4a−3a−7≤1- 1 \le {{4a - 3} \over {a - 7}} \le 1

4a−3a−7−1≤0{{4a - 3} \over {a - 7}} - 1 \le 0 and 4a−3a−7+1≥0{{4a - 3} \over {a - 7}} + 1 \ge 0

⇒−43≤a≤2\Rightarrow {{ - 4} \over 3} \le a \le 2

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