If \(f(x) = \begin{cases} mx + 1, & x \ge \pi/2 \\ \sin x + n, & x \le \pi/2 \end{cases}\) is continuous at \(x = \pi/2\), where \(m \in \mathrm{Z}\) (set of integers), then \(\sin 2n =\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
For continuity at \(x = \frac{\pi}{2}\), the left-hand limit, the right-hand limit and \(f\left(\frac{\pi}{2}\right)\) must be equal.
\(\displaystyle \lim_{x \to \frac{\pi}{2}^-} (\sin x + n) = 1 + n\)
\(\displaystyle \lim_{x \to \frac{\pi}{2}^+} (mx + 1) = \frac{m\pi}{2} + 1\)
Equating them:
\(\displaystyle 1 + n = \frac{m\pi}{2} + 1\)
\(\displaystyle n = \frac{m\pi}{2}\)
\(2n = m\pi\)
Since m is an integer, \(\sin m\pi = 0\):
\(\sin 2n = \sin m\pi = 0\)