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JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Application of Derivatives

The sum of all local minimum values of the function f(x)=\ array lr...

The sum of all local minimum values of the function

f(x)=\left\{

\begin{array}{lr} 1-2 x, & x \lt -1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x\gt2 \end{array}

\right.

is $The sum of all local minimum values of the function f(x)=\ array lr 1-2 x, & x -1 (1)/(3)(7+2|x|), & -1 ≤ x ≤ 2$

Options

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

Correct answer

A. $\frac{157}{72}$

Step-by-step solution

f(x)=\left\{

\begin{array}{cl}1-2 x, & x < -1 \\ \frac{1}{3}(7-2 x), & -1 \leq x \leq 2 \\ \frac{1}{3}(7+2 x) & 0 \leq x < 2 \\ \frac{11}{18}(x-4)(x-5), & x>2\end{array}

\right. $The sum of all local minimum values of the function f(x)=\ array lr 1-2 x, & x -1 (1)/(3)(7+2|x|), & -1 ≤ x ≤ 2$

\therefore

Local minimum values at \mathrm{A} \& \mathrm{~B}

\begin{aligned} & \frac{7}{3}-\frac{11}{72} \\ & \Rightarrow \frac{168-11}{72} \Rightarrow \frac{157}{72} \end{aligned}

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