Verified Solution • 2025–26 Session moderate QJEEM25Ma09ff: JEE Main 2025 Mathematics - Class 12 JEEM25Ma09ff Solution: The sum of all local minimum values of the function is The sum of all local minimum values of the function f(x)={1−2x,amp;xlt;−113(7+2∣x∣),amp;−1≤x≤21118(x−4)(x−5),amp;xgt;2\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.f(x)=⎩⎨⎧1−2x,31(7+2∣x∣),1811(x−4)(x−5),amp;xamp;−1≤x≤2amp;xlt;−1gt;2 is Multiple Choice Options A 16772\frac{167}{72}72167 B 15772\frac{157}{72}72157 C 17172\frac{171}{72}72171 D 13172\frac{131}{72}72131 Step-by-Step Verified Solution f(x)={1−2x,amp;xlt;−113(7−2x),amp;−1≤x≤213(7+2x)amp;0≤xlt;21118(x−4)(x−5),amp;xgt;2f(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.f(x)=⎩⎨⎧1−2x,31(7−2x),31(7+2x)1811(x−4)(x−5),amp;xamp;−1≤x≤2amp;0≤xamp;xlt;−1lt;2gt;2 amp;∴ Local minimum values at A& Bamp;amp;73−1172amp;⇒168−1172⇒15772\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}amp;∴ Local minimum values at A& Bamp;amp;37−7211amp;⇒72168−11⇒72157 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
