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

<p>$f(x) = \min \{ 1,\,1 + x\sin x\}$, $0 \le x \le x$</p> <p>$$f(x) = \left\{ {\matrix{ {1,} & {0 \le x < \pi } \cr {1 + x\sin x,} & {\pi \le x \le 2\pi } \cr } } \right.$$</p> <p>Now at $$x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)$$</p> <p>$\therefore$ f(x) is continuous in [0, 2$\pi$]</p> <p>Now, at x = $\pi$ $$L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$$</p> <p>$$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $$</p> <p>$\therefore$ f(x) is not differentiable at x = $\pi$</p> <p>$\therefore$ (m, n) = (1, 0)</p>