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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Application of Derivatives

Let x=-1 and x=2 be the critical points of the function...

Let x=−1andx=2x=-1 and x=2 be the critical points of the function f(x)=x3+ax2+blog⁡c∣x∣+1,x≠0\mathrm{f}(\mathrm{x})=\mathrm{x}^3+\mathrm{ax}^2+\mathrm{b} \log _{\mathrm{c}}|\mathrm{x}|+1, \mathrm{x} \neq 0. Let mandMm and M respectively be the absolute minimum and the absolute maximum values of ff in the interval \left[-2,-\frac{1}{2}\right].Then. Then|\mathrm{M}+m|isequalto(Takeis equal to (Take\log _{\mathrm{c}} 2=0.7$ ):

Options

  1. A21.1
  2. B19.8
  3. C22.1
  4. D20.9

Correct answer

A. 21.1

Step-by-step solution

<br/>f(x)=x2+ax2+bℓn∣x∣+1,x≠0f′(x)=3x2+2ax+bxf′(−1)=3−2a−b=0f′(−2)=12+4a−b2=0a=−92, b=12f′(x)=3x2−9x+12x=3(x+1)(x+2)2x<br/>\begin{aligned}<br/>& \mathrm{f}(\mathrm{x})=\mathrm{x}^2+\mathrm{ax}^2+\mathrm{b} \ell \mathrm{n}|\mathrm{x}|+1, \quad \mathrm{x} \neq 0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+2 \mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \\ & \mathrm{f}^{\prime}(-1)=3-2 \mathrm{a}-\mathrm{b}=0 \\ & \mathrm{f}^{\prime}(-2)=12+4 \mathrm{a}-\frac{\mathrm{b}}{2}=0 \\ & \mathrm{a}=\frac{-9}{2}, \mathrm{~b}=12 \\ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2-9 \mathrm{x}+\frac{12}{\mathrm{x}}=\frac{3(\mathrm{x}+1)(\mathrm{x}+2)^2}{\mathrm{x}}<br/>\end{aligned}
Max. at n=−1\mathrm{n}=-1
f(x)=x2−92x2+12ln⁡∣x∣+1f(x)=x^2-\frac{9}{2} x^2+12 \ln |x|+1
<br/>f(−1)=−1−92+1=−92M=−4.5<br/>\begin{aligned}<br/>& \mathrm{f}(-1)=-1-\frac{9}{2}+1=-\frac{9}{2} \\ & \mathrm{M}=-4.5<br/>\end{aligned}
Min. value at x=−2\mathrm{x}=-2
<br/>f(−2)=−8−18+12ℓn2+1 m=−25+12ℓn2=−16.6∣M+m∣=21.1<br/>\begin{aligned}<br/>& \mathrm{f}(-2)=-8-18+12 \ell \mathrm{n} 2+1 \\ & \mathrm{~m}=-25+12 \ell \mathrm{n} 2=-16.6 \\ & |\mathrm{M}+\mathrm{m}|=21.1<br/>\end{aligned}
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