JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Continuity and Differentiability

For $0\le x<1$, $[x]=0$, so $\min \{1+x,x\}=x$ (since $x<1+x$).

For $1\le x<2$, $[x]=1$, so $\min \{2+x,x+2\}=x+2$ (the two expressions are equal).

At $x=2$, $[x]=2$, giving $\min \{5,6\}=5$, matching the value $5$ used for $x>2$.

So the function simplifies to:

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

Continuity:

At $x=0$: left value $\to 0$, right value $\to 0$, $f(0)=0$ — continuous.

At $x=1$: left limit $\to 1$, but $f(1)=1+2=3$ — discontinuous.

At $x=2$: from the $1\le x<2$ side, the limit is $2+2=4$, but $f(2)=5$ (matching the $x>2$ side) — discontinuous.

So discontinuities occur at $x=1,2$: $\alpha =2$.

Differentiability:

At $x=0$: left derivative (of $3x$) is $3$, right derivative (of $x$) is $1$ — not differentiable, even though continuous.

At $x=1,2$: already discontinuous, hence not differentiable.

So non-differentiable points are $x=0,1,2$: $\beta =3$.

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