JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumLimit, Continuity and Differentiability420× since 2002The sum of all local minimum values of the function f(x)={1−2x,x<−113(7+2∣x∣),−1≤x≤21118(x−4)(x−5),x>2f(x) = \begin{cases} 1 - 2x, & x < -1 \\ \frac{1}{3}(7 + 2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x - 4)(x - 5), & x > 2 \end{cases}f(x)=⎩⎨⎧1−2x,31(7+2∣x∣),1811(x−4)(x−5),x<−1−1≤x≤2x>2 isAOption A. 12772\frac{127}{72}72127BOption B. 13172\frac{131}{72}72131COption C. 17172\frac{171}{72}72171DOption D. 16772\frac{167}{72}72167CheckStep-by-step hintSkip · show solution
