JEE MAIN - Mathematics (2025 - 28th January Morning Shift - No. 15)
The sum of all local minimum values of the function
$$\mathrm{f}(x)=\left\{\begin{array}{lr} 1-2 x, & x<-1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x>2 \end{array}\right.$$
is
$\frac{167}{72}$
$\frac{157}{72}$
$\frac{171}{72}$
$\frac{131}{72}$
Explanation
$f(x)=\left\{\begin{array}{cc}1-2 x, & x<-1 \\ \frac{1}{3}(7-2 x), & -1 \leq x \leq 2 \\ \frac{1}{3}(7+2 x) & 0 \leq x<2 \\ \frac{11}{18}(x-4)(x-5), & x>2\end{array}\right.$
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$$\begin{aligned} &\therefore \text { Local minimum values at } \mathrm{A} \& \mathrm{~B}\\ &\begin{aligned} & \frac{7}{3}-\frac{11}{72} \\ & \Rightarrow \frac{168-11}{72} \Rightarrow \frac{157}{72} \end{aligned} \end{aligned}$$
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