JEE Main 2025 Jan 24 Shift 2, Mathematics Q7: Continuity of Function

Both $[x]$ and $|x-2|$ contribute the only possible "bad" points:

• $[x]$ has integer jump discontinuities at every integer.
• $|x-2|$ is continuous everywhere but is non-differentiable at $x=2$.

On the open interval $(-2,3)$, the integers are $-1,0,1,2$. At each of these, $[x]$ jumps, so $f(x)=[x]+|x-2|$ is discontinuous and (consequently) non-differentiable.

At $x=2$, $[x]$ also jumps (the integer point), so it is already counted among the four discontinuity points. The corner of $|x-2|$ at $x=2$ adds no new non-differentiability beyond what discontinuity already implies.

Therefore:

Number of discontinuity points: $m=4$ (at $x=-1,0,1,2$).

Number of non-differentiability points: $n=4$ (same four points).

$m+n=4+4=8$

Correct option: (3)