JEE Main 2025 Jan 28 Shift 1, Mathematics Q24: Continuity of Function

Resolve the middle piece by intervals where $[x]$ is constant.

On $0\le x<1$: $[x]=0$, so $\min (1+x,x)=x$.
On $1\le x<2$: $[x]=1$, so $\min (2+x,x+2)=x+2$.
At $x=2$: $[x]=2$, so $\min (3+x,x+4)=3+x=5$.

Hence $f$ simplifies to a piecewise-linear function:

$f(x)=\{\begin{array}{l}3x,x<0\\ x,0\le x<1\\ x+2,1\le x<2\\ 5,x\ge 2\end{array}$

Continuity. Check the junctions:
At $x=0$: ${\lim }_{x\to 0^{-}}3x=0$, $f(0)=0$, continuous.
At $x=1$: ${\lim }_{x\to 1^{-}}x=1$, $f(1)=3$, jump $\Rightarrow$ discontinuous.
At $x=2$: ${\lim }_{x\to 2^{-}}(x+2)=4$, $f(2)=5$, jump $\Rightarrow$ discontinuous.

Number of discontinuities: $\alpha =2$.

Differentiability. At $x=0$: left slope 3, right slope 1, kink $\Rightarrow$ not differentiable. At $x=1$ and $x=2$: not even continuous, so not differentiable.

Number of non-differentiability points: $\beta =3$.

$\therefore \alpha +\beta =2+3=5$.