If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{m} x+1, & x \leq \frac{\pi}{2} \\ \sin x+\mathrm{n}, & x>\frac{\pi}{2}\end{array}\right.$, is continuous at $x=\frac{\pi}{2},(\mathrm{~m}, \mathrm{n} \in \mathbb{Z})$ then

The function

$f(x)=\begin{cases} m x+1 & x \leq \frac{\pi}{2} \\ \sin x+n & x>\frac{\pi}{2} \end{cases}$

must satisfy the continuity condition at $x = \frac{\pi}{2}$.

The left-hand limit yields

$\lim_{x \to (\frac{\pi}{2})^-} f(x) = m\left(\frac{\pi}{2}\right) + 1$

while the right-hand limit gives

$\lim_{x \to (\frac{\pi}{2})^+} f(x) = \sin\left(\frac{\pi}{2}\right) + n = 1 + n$

Equating these limits with the function value

$f\left(\frac{\pi}{2}\right) = m\left(\frac{\pi}{2}\right) + 1$

produces

$m\left(\frac{\pi}{2}\right) + 1 = 1 + n$

which simplifies to the continuity condition

$n = \frac{m\pi}{2}$

This relationship corresponds exactly with option D.