Let $$f(x) = \min\{1, 1 + x\sin x\}, 0 \leq x \leq 2\pi$$. If $$m$$ is the number of points, where $$f$$ is not differentiable and $$n$$ is the number of points, where $$f$$ is not continuous, then the ordered pair $$(m, n)$$ is equal to

Given,

$$f(x)=\min\{1,1+x\sin x\},\qquad 0\le x\le2\pi$$

Now,

$$f(x)=1$$

when

$$1\le1+x\sin x$$

$$x\sin x\ge0$$

Since

$$x\ge0$$

for $$0\le x\le2\pi$$ the sign depends on $$\sin x$$.

Thus,

$$f(x)=1\quad \text{for}\quad 0\le x\le\pi$$

and

$$f(x)=1+x\sin x\quad \text{for}\quad \pi<x<2\pi$$

Also,

$$f(2\pi)=1$$

Now, at $$x=\pi$$

$$\lim_{x\to\pi^-}f(x)=1$$

and

$$\lim_{x\to\pi^+}f(x)=1+\pi\sin\pi=1$$

Hence, $$f$$ is continuous at $$x=\pi$$.

Also, $$f$$ is continuous at $$x=2\pi$$.

Therefore, $$n=0$$

Now, checking differentiability at $$x=\pi$$,

Left derivative: $$f'_-(\pi)=0$$

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

Since,

$$f'_-(\pi)\ne f'_+(\pi)$$

the function is not differentiable at $$x=\pi$$.

Hence, $$m=1$$

Therefore, the ordered pair is $$\boxed{(1,0)}$$.