Verified Solution • 2025–26 Session moderate

QJEEM22Ma207e: JEE Main 2022 Mathematics - Class 12 JEEM22Ma207e Solution: If the absolute maximum value of the function in the interval…

If the absolute maximum value of the function f(x)=(x2−2x+7)e(4x3−12x2−180x+31)f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)} in the interval [−3,0][-3,0] is f(α)f(\alpha) , then :

Multiple Choice Options

  • A α=0\alpha=0
  • B α=−3\alpha=-3
  • C α∈(−1,0)\alpha \in(-1,0)
  • D α∈(−3,−1]\alpha \in(-3,-1]

Step-by-Step Verified Solution

Given, f(x)=(x2−2x+7)⏟f1(x)e(4x3−12x2−180x+31)⏟f2(x)f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}

f1(x)=x2−2x+7f_{1}(x)=x^{2}-2 x+7

f1′(x)=2x−2f_{1}^{\prime}(x)=2 x-2 , so f(x)f(x) is decreasing in [−3,0][-3,0] and positive also

f2(x)=e4x3−12x2−180x+31f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}

f2′(x)=e4x3−12x2−180x+31⋅12x2−24x−180f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180

=12(x−5)(x+3)e4x3−12x2−180x+31=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}

So, f2(x)f_{2}(x) is also decreasing and positive in {−3,0}\{-3,0\}

∴\therefore absolute maximum value of f(x)f(x) occurs at x=−3x=-3

∴α=−3\therefore \quad \alpha=-3

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring