JEE Main 2025 Jan 28 Shift 1, Mathematics Q14: Maxima & Minima

Examine each piece for local minima.

Piece 1 ($x<-1$): $f=1-2x$ is strictly decreasing, no local minimum in the open interval.

Piece 2 ($-1\le x\le 2$): $f=\frac{1}{3}(7+2|x|)$ is a V-shape with vertex at $x=0$, value $f(0)=\frac{7}{3}$. Check the boundary $x=-1$: as we move left from $-1$ (into piece 1), $f$ rises; moving right into piece 2, $f$ first decreases toward 0 then rises. So $x=-1$ is a not minimum. Hence the only minimum from piece 2 is at $x=0$ with $f(0)=\frac{7}{3}$.

Piece 3 ($x>2$): $f=\frac{11}{18}(x-4)(x-5)$. The quadratic opens upward, with vertex at $x=\frac{9}{2}$, value $f(9/2)=\frac{11}{18}\cdot \frac{1}{2}\cdot \left(-\frac{1}{2}\right)=-\frac{11}{72}$. This vertex lies in the domain ($\frac{9}{2}>2$), so it is a local minimum.

Sum of local minimum values $=\frac{7}{3}+\left(-\frac{11}{72}\right)=\frac{168-11}{72}=\frac{157}{72}$.