JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Continuity and Differentiability

The function $f(x)=\min \{1,1+x\sin x\}$ switches between the two expressions where $1=1+x\sin x$, i.e., where $x\sin x=0$.

In $[0,2\pi ]$, $x\sin x=0$ at $x=0,\pi ,2\pi$.

At $x=0$ and $x=2\pi$, the function equals 1 from both sides (boundary behavior).

At $x=\pi$, the function transitions from $1+x\sin x$ to 1, creating a corner where the derivative does not exist.

The function is continuous everywhere (both expressions are continuous and equal at transition points).

Therefore, $m=1$ (non-differentiable at $x=\pi$) and $n=0$ (continuous everywhere), giving $(m,n)=(1,0)$.