JEE Mathematics 2025 Question with Solution

Given: The piecewise function is

$$f(x)=\left\{\begin{array}{ll}1-2x & \text{if } x<-1\\ \frac{1}{3}(7+2|x|) & \text{if } -1 \leq x \leq 2\\ \frac{11}{18}(x-4)(x-5) & \text{if } x>2\end{array}\right.$$

Find: The sum of all local minimum values of $f(x)$.

For $x<-1$,

$$f(x)=1-2x$$

This is a decreasing linear function, so this part has no local minimum.

For $-1 \leq x \leq 2$,

$$f(x)=\frac{1}{3}(7+2|x|)$$

Now split at $x=0$:

$$f(x)=\frac{1}{3}(7-2x), \quad -1 \leq x \leq 0$$

which is decreasing, and

$$f(x)=\frac{1}{3}(7+2x), \quad 0 \leq x \leq 2$$

which is increasing. Hence, there is a local minimum at $x=0$ with value

$$f(0)=\frac{7}{3}$$

For $x>2$,

$$f(x)=\frac{11}{18}(x-4)(x-5)$$

This is an upward-opening quadratic, so its vertex gives a local minimum. The vertex lies at

$$x=\frac{4+5}{2}=\frac{9}{2}$$

Then

$$f\left(\frac{9}{2}\right)=\frac{11}{18}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)=-\frac{11}{72}$$

The middle part $\frac{1}{3}(7+2|x|)$ is a V-shaped graph, so its minimum occurs at the vertex $x=0$. The right part $\frac{11}{18}(x-4)(x-5)$ is an upward-opening parabola, so its minimum occurs at the midpoint of roots $4$ and $5$, namely $\frac{9}{2}$. Therefore, the local minimum values are

$$\frac{7}{3} \quad \text{and} \quad -\frac{11}{72}$$

Their sum is

$$\frac{7}{3}-\frac{11}{72}=\frac{168}{72}-\frac{11}{72}=\frac{157}{72}$$

Therefore, the correct option is D. The solution states option A, but the working gives $\frac{157}{72}$, which matches option D.