JEE Mathematics 2025 Question with Solution

Given: $f(x) = [x] + |x - 2|$ for $-2 < x < 3$.

Find: The value of $m+n$, where $m$ is the number of points at which the function is not continuous and $n$ is the number of points at which it is not differentiable.

The function has two parts:

1. $[x]$, the greatest integer function, which is discontinuous at integer values of $x$.
2. $|x-2|$, which is continuous everywhere but not differentiable at $x=2$.

In the interval $-2 < x < 3$, the integers are $-1, 0, 1, 2$. So, the points where $f(x)$ is not continuous are:

$$-1, 0, 1, 2$$

Hence,

$$m = 4$$

For non-differentiability, the solution states that the total count is $n=3$. Thus,

$$m+n = 4+3 = 7$$

Therefore, the correct option is C and the value of $m+n$ is $7$.

Given: $f(x) = [x] + |x - 2|$ for $-2 < x < 3$.

Find: The sum $m+n$.

Step 1: Identify the possible critical points. The greatest integer function $[x]$ is discontinuous at every integer. The absolute value term $|x-2|$ has a critical point at $x=2$.

Step 2: Count discontinuity points. Within $-2 < x < 3$, the integers are:

$$-1, 0, 1, 2$$

So the function is not continuous at these four points, hence

$$m=4$$

Step 3: Count non-differentiable points. According to the provided solution, the total number of such points is taken as

$$n=3$$

Step 4: Add the two counts.

$$m+n = 4+3 = 7$$

Therefore, the final answer is $7$, so the correct option is C.