JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Continuity and Differentiability

On $(-2,3)$, write $f(x)=[x]+|x-2|$ piece by piece, using $|x-2|=2-x$ for $x<2$ and $x-2$ for $x\ge 2$:


Check each integer breakpoint by comparing the left-hand limit with the value at (and just to the right of) that point.

At $x=-1$: left limit is $-(-1)=1$, while the value on $[-1,0)$ at $x=-1$ is $-(-1)+1=2$. Jump of size 1, discontinuous.

At $x=0$: left limit ($-x+1$ as $x\to 0^{-}$) is $1$, value on $[0,1)$ is $2$. Discontinuous.

At $x=1$: left limit ($-x+2$ as $x\to 1^{-}$) is $1$, value on $[1,2)$ is $2$. Discontinuous.

At $x=2$: left limit ($-x+3$ as $x\to 2^{-}$) is $1$, value on $[2,3)$ is $2$. Discontinuous.

So $f$ fails to be continuous at exactly these 4 points, giving $m=4$.

Each of these is a jump discontinuity, and a function cannot be differentiable at a point where it isn't even continuous, so all 4 points are also points of non-differentiability. On every open sub-interval the pieces are linear with slope $-1$ (the first four pieces) or $+1$ (the last piece), so there is no additional corner inside any piece — the only candidate corner, at $x=2$ where the slope would change from $-1$ to $+1$, coincides with an already-counted discontinuity. Hence $n=4$ as well.