The sum of all local minimum values of the f(x) = begin cases 1-2x, & x < -1 \\ 1/3(7+2 x ), & -1 le x le 2 \\ 11/18(x-4)(x-5), & x 2 end cases

Mathematics · Application of Derivatives

JEE Main 2025 — 28 January, Morning Shift — Question 14

The sum of all local minimum values of the $f(x) = \begin{cases} 1-2x, & x < -1 \\ \frac{1}{3}(7+2|x|), & -1 \le x \le 2 \\ \frac{11}{18}(x-4)(x-5), & x > 2 \end{cases}$

Option A:
$\frac{171}{72}$
Option B:
$\frac{131}{72}$
Option C:
$\frac{157}{72}$
Correct
Option D:
$\frac{167}{72}$

Answer: C

Step-by-step solution

f(x) = \begin{cases} 1-2x, & x < -1 \\ \frac{1}{3}(7-2x), & -1 \le x < 0 \\ \frac{1}{3}(7+2x), & 0 \le x \le 2 \\ \frac{11}{18}(x-4)(x-5), & x > 2 \end{cases}

$\therefore$ Local minimum values at A & B

\frac{7}{3} - \frac{11}{72}

\Rightarrow \frac{168-11}{72} \Rightarrow \frac{157}{72}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Local, Global extremum