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

$$f(x) = \left\{ {\matrix{ {1,} & {0 \le x < \pi } \cr {1 + x\sin x,} & {\pi \le x \le 2\pi } \cr } } \right.$$

Now at $$x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)$$

$$\therefore$$ f(x) is continuous in [0, 2$$\pi$$]

Now, at x = $$\pi$$ $$L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$$

$$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $$

$$\therefore$$ f(x) is not differentiable at x = $$\pi$$

$$\therefore$$ (m, n) = (1, 0)