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Mathematics · Application of Derivatives

JEE Main 2025 — 3 April, Evening Shift — Question 26

Let f:R→Rf: R \rightarrow R be a function defined by f(x)=∥x+2∣f(x)=\| x+2 \mid −2∣x∥-2 \mid x \|. If mm is the number of points of local minima and n is the number of points of local maxima of ff, then m+nm+n is :

  1. Option A:

    3

    Correct
  2. Option B:

    2

  3. Option C:

    4

  4. Option D:

    5

Answer: A

Step-by-step solution

f(x)={∣−x−2+2x∣x≤−2∣x+2+2x∣−2≤x≤0∣x+2−2x∣x≥0={∣x−2∣x≤−2∣3x+2∣−2≤x≤0∣2−x∣x≥0f(x) = \begin{cases} |-x - 2 + 2x| & x \leq -2 \\ |x + 2 + 2x| & -2 \leq x \leq 0 \\ |x + 2 - 2x| & x \geq 0 \end{cases} = \begin{cases} |x - 2| & x \leq -2 \\ |3x + 2| & -2 \leq x \leq 0 \\ |2 - x| & x \geq 0 \end{cases} f(x)={x−2x≤−2−3x−2−2≤x≤−233x+2−23<x≤02−x0<x<2x−2x≥2f(x) = \begin{cases} x - 2 & x \leq -2 \\ -3x - 2 & -2 \leq x \leq -\frac{2}{3} \\ 3x + 2 & -\frac{2}{3} < x \leq 0 \\ 2 - x & 0 < x < 2 \\ x - 2 & x \geq 2 \end{cases}

figure

No. of maxima=1\text{No. of maxima} = 1 No. of minima=2\text{No. of minima} = 2 m=2m = 2 n=1n = 1 m+n=3m + n = 3

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Maxima and Minima

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