If

$f(x) = \begin{cases} mx + 1,\ x \geq \pi/2 \\sin x + n, x \leq \pi/2, & \end{cases}$

is continuous at $x = \pi/2$, where $m \in \mathbb{Z}$ (set of integers), then $\sin 2n =$

For $f(x)$ to be continuous at $x = \pi/2$, the left-hand limit must equal the right-hand limit.

From the left $(x \leq \pi/2)$:

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

$= 1 + n$

From the right $(x \geq \pi/2)$:

$\lim_{x \to \pi/2^+} f(x) = m \cdot \frac{\pi}{2} + 1$

$= \frac{m\pi}{2} + 1$

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Setting the two limits equal for continuity:

$1 + n = \frac{m\pi}{2} + 1$

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

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Substituting into $\sin 2n$:

$\sin(2n) = \sin\left(2 \cdot \frac{m\pi}{2}\right)$

$= \sin(m\pi)$

Since $m \in \mathbb{Z}$, we have $\sin(m\pi) = 0$ for any integer $m$.

Therefore, $\sin 2n = 0$