If $f(x) =\left\{\begin{matrix}mx + 1,&x ≥π/2\\\sin x+n,&x≤π/2\end{matrix}\right.$ is continuous at $x = π/2$, where $m ∈ z$ (set of integers), then $\sin 2n =$

0

The correct answer is Option (1) → 0

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

For continuity at $x = \frac{\pi}{2}$:

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

From the left-hand limit:

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

From the right-hand limit:

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

Equating:

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

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

As $m \in \mathbb{Z}$, for $n$ to be an integer, $\frac{m \pi}{2}$ must also be an integer. This is true when $m = 0$, leading to $n = 0$.

$\sin 2n = \sin 0 = 0$