JEE Mathematics 2025 Question with Solution

Given: $f(x) = ||x+2| - 2|x||$

Find: the total number of points of local maxima and local minima, that is $m+n$.

Break the function at the points where the inner absolute value expressions change sign, namely at $x=-2$ and $x=0$.

For $x \ge 0$,

For $-2 \le x < 0$,

For $x < -2$,

Now simplify each part further:

From these linear pieces, the graph changes direction at three points:

• At $x=-\frac{2}{3}$, the function changes from decreasing to increasing, so this is a local minimum.
• At $x=0$, the function changes from increasing to decreasing, so this is a local maximum.
• At $x=2$, the function changes from decreasing to increasing, so this is a local minimum.

At $x=-2$, the slope is negative on both sides, so there is no local extremum there.

Hence, number of local maxima is $m=1$ and number of local minima is $n=2$.

Therefore,

The correct option is B.

Given: $f(x) = ||x+2| - 2|x||$

Find: the number of local extrema points.

The critical points mentioned in the extracted solution are $x=-2, 0, -\frac{2}{3}$ and the graph also shows the corner at $x=2$ where the outer absolute value changes branch.

Reading the graph:

• there is a valley at $x=-\frac{2}{3}$,
• a peak at $x=0$,
• another valley at $x=2$. So the graph has one local maximum and two local minima.

Therefore, $m+n=3$ and the correct option is B.