JEE Main 2025 — Application of Derivatives Question with Solution
JEE Main 2025 (7 Apr Shift 1)
Question
Let and be the critical points of the function . Let and respectively be the absolute minimum and the absolute maximum values of in the interval . Then is equal to (Take ):
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Show full solutionCorrect option: A
Correct answer
A21.1
Step-by-step explanation
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\mathrm{x}^2+\mathrm{ax}^2+\mathrm{b} \ell \mathrm{n}|\mathrm{x}|+1, \quad \mathrm{x} \neq 0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+2 \mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \\ & \mathrm{f}^{\prime}(-1)=3-2 \mathrm{a}-\mathrm{b}=0 \\ & \mathrm{f}^{\prime}(-2)=12+4 \mathrm{a}-\frac{\mathrm{b}}{2}=0 \\ & \mathrm{a}=\frac{-9}{2}, \mathrm{~b}=12 \\ & \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2-9 \mathrm{x}+\frac{12}{\mathrm{x}}=\frac{3(\mathrm{x}+1)(\mathrm{x}+2)^2}{\mathrm{x}}
\end{aligned}\mathrm{n}=-1f(x)=x^2-\frac{9}{2} x^2+12 \ln |x|+1\begin{aligned}
& \mathrm{f}(-1)=-1-\frac{9}{2}+1=-\frac{9}{2} \\ & \mathrm{M}=-4.5
\end{aligned}\mathrm{x}=-2\begin{aligned}
& \mathrm{f}(-2)=-8-18+12 \ell \mathrm{n} 2+1 \\ & \mathrm{~m}=-25+12 \ell \mathrm{n} 2=-16.6 \\ & |\mathrm{M}+\mathrm{m}|=21.1
\end{aligned}$
& \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}}
\end{aligned}\mathrm{n}=-1f(x)=x^2-\frac{9}{2} x^2+12 \ln |x|+1\begin{aligned}
& \mathrm{f}(-1)=-1-\frac{9}{2}+1=-\frac{9}{2} \\ & \mathrm{M}=-4.5
\end{aligned}\mathrm{x}=-2\begin{aligned}
& \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
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.