Let $f(x)=\min\{1,\,1+x\sin x\}, \qquad 0\le x\le 2\pi.$

We need:

1. the number of points where $f$ is not differentiable,
2. the number of points where $f$ is not continuous.

1. Rewrite the function piecewise

We compare the two quantities inside the minimum: $1 \quad \text{and} \quad 1+x\sin x.$

Now, $1+x\sin x \le 1 \iff x\sin x \le 0.$

Since $x\in[0,2\pi]$, we have $x\ge 0$. So the sign of $x\sin x$ is determined by $\sin x$ (except at $x=0$, where $x\sin x=0$).

Thus:

• when $\sin x>0$, we get $1+x\sin x>1$, so $f(x)=1;$
• when $\sin x<0$, we get $1+x\sin x<1$, so $f(x)=1+x\sin x;$
• when $\sin x=0$, both are equal, so $f(x)=1.$

On $[0,2\pi]$:

• $\sin x>0$ for $x\in(0,\pi)$,
• $\sin x<0$ for $x\in(\pi,2\pi)$,
• $\sin x=0$ at $x=0,\pi,2\pi$.

Hence the piecewise form is

(At $x=\pi$, both formulas give $1$.)

2. Check continuity

Each branch is continuous on its interval:

• $1$ is continuous,
• $1+x\sin x$ is continuous.

So only the joining point $x=\pi$ needs checking.

At $x=\pi$

Left-hand value: $f(\pi^-)=1.$ Right-hand value: $f(\pi^+)=1+\pi\sin\pi=1.$ And $f(\pi)=1.$

So $f$ is continuous at $x=\pi$.

At the endpoints $x=0$ and $x=2\pi$, the function values are well-defined and the one-sided continuity also holds.

Therefore, there are no points of discontinuity.

So, $n=0.$

3. Check differentiability

Possible issue points are where the formula changes, i.e. at $x=\pi$.

For $0<x<\pi$

$f(x)=1 \implies f'(x)=0.$

For $\pi<x<2\pi$

$f(x)=1+x\sin x$ so $f'(x)=\sin x + x\cos x.$

At $x=\pi$

Left derivative: $f'_-(\pi)=0.$

Right derivative: $f'_+(\pi)=\sin\pi+\pi\cos\pi=0+\pi(-1)=-\pi.$

Since $f'_-(\pi)\ne f'_+(\pi),$ $f$ is not differentiable at $x=\pi$.

At $x=0$ and $x=2\pi$, differentiability is usually considered only for interior points of the interval; in any case, the only corner/cusp in the domain occurs at $x=\pi$.

Thus, $m=1.$

4. Final ordered pair

$(m,n)=(1,0).$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B, i.e. $(m,n)=(1,0).$