Start free

Mathematics · Application of Derivatives

JEE Main 2025 — 7 April, Morning Shift — Question 25

Let x=−1x=-1 and x=2x=2 be the critical points of the function f(x)=x3+ax2+blog⁡e∣x∣+1,x≠0f(x)=x^{3}+a x^{2}+b \log _{e}|x|+1, x \neq 0. Let mm and MM respectively be the absolute minimum and the absolute maximum values of ff in the interval [−2,−12]\left[-2,-\frac{1}{2}\right]. Then ∣M+m∣|M+m| is equal to (Take log⁡2=\log 2= 0.7)0.7) :

  1. Option A:

    21.1

    Correct
  2. Option B:

    20.9

  3. Option C:

    19.8

  4. Option D:

    22.1

Answer: A

Step-by-step solution

f′(x)=3x2+2ax+bxf'(x)=3 x^{2}+2 a x+\frac{b}{x} f′(−1)=3−2a−b=0,f′(−2)=12+4a+b2=0f'(-1)=3-2 a-b=0 , f'(-2)=12+4 a+\frac{b}{2}=0 a=−92b=12\begin{aligned} & a=\frac{-9}{2} & b=12\end{aligned} f′(x)=3x2−9x+12x=0f'(x)=3 x^{2} - 9x+\frac{12}{x}=0 ⇒x=−1,−2,−2\Rightarrow x = -1, -2, -2 ∴f(x)=x3−92x2+12ln⁡∣x∣+1\therefore f(x)=x^{3}-\frac{9}{2} x^{2}+12 \ln |x|+1 f(−1)=−1−92+1=−92=−4.5f(-1)=-1-\frac{9}{2}+1=-\frac{9}{2}=-4.5 f(−2)=−8−18+12ln⁡2+1f(-2)=-8-18+12 \ln 2+1 =−25+12ln⁡2=−16.6=-25+12 \ln 2=-16.6 f(−12)=−18−98+12ln⁡(12)+1=−8.5f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12 \ln \left(\frac{1}{2}\right)+1=-8.5

∣M+m∣=∣−16.6−4.5∣=21.1|M+m|=|-16.6-4.5|=21.1

Answer key and solution verified before publishing.

Practise Application of Derivatives

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
7 April, Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Maxima and Minima
Let x=-1 and x=2 be the critical points of the function f(x)=x 3 +a x… | JEE Main 2025 PYQ with Solution · DhiX AI