QJEEM22Maeb37: JEE Main 2022 Mathematics - Class 12 JEEM22Maeb37 Solution: Let f(x) = min {1, 1 + x sin x}, 0 x 2 . If m is the number of…

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

$$f(x)=\{\begin{array}{cc}1,& 0\le x<\pi \\ 1+x\sin x,& \pi \le x\le 2\pi \end{array}$$

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)