Let f(x) = min\ 1, 1 + xsinx\ , 0 ≤ x ≤ 2π . If m is the number of...

Given,

$f\left(x\right) = min\left\{1, 1 + xsinx\right\}x \in \left[0, 2\pi\right]</p><p>$f\left(x\right) = \left\{$$\begin{array}{ll}1, 0 \le x \le \pi \\ 1 + xsinx, \pi \le x \le 2\pi\end{array}$$\right

Now, at $x = \pi, \lim_{x \to {\pi}^{-}}f\left(x\right) = 1 = \lim_{x \to {\pi}^{+}}f\left(x\right)</p><p>$∴ f \left(x\right)$is continuous in \left[0, 2\pi\right]$

So, number of points of discontinuity is zero or $n = 0$

Now, at $x = \pi$

L.H.D $= \lim_{h \to 0}\dfrac{f\left(\pi - h\right) - f\left(\pi\right)}{- h} = 0$

R.H.D $= \lim_{h \to 0}\dfrac{f\left(\pi + h\right) - f\left(\pi\right)}{h} = \lim_{h \to 0}\dfrac{1 - \left(\pi + h\right)sinh - 1}{h}$

$= - \pi$

$∴ f\left(x\right)$ is not differentiable at $x = \pi$ so, $m = 1$

$∴$\left(m, n\right) = $\left(1, 0\right)</p>$