If the function \(f(x) = \begin{cases} mx + 1, & x \le \displaystyle\frac{\pi}{2}\\ \sin x + n, & x \gt \displaystyle\frac{\pi}{2}\end{cases}\) is continuous at \(x = \displaystyle\frac{\pi}{2}\), then the relation between m and n is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(n = \displaystyle\frac{m\pi}{2}\)
For continuity at \(\frac{\pi}{2}\), the left-hand limit, the right-hand limit and \(f\left(\frac{\pi}{2}\right)\) must be equal.
\( \displaystyle \lim_{x \to \frac{\pi}{2}^-} f(x) = f\left(\frac{\pi}{2}\right) = \frac{m\pi}{2} + 1 \)
\( \displaystyle \lim_{x \to \frac{\pi}{2}^+} f(x) = \sin\frac{\pi}{2} + n = 1 + n \)
Equating the two:
\( \displaystyle \frac{m\pi}{2} + 1 = 1 + n \)
\( \displaystyle n = \frac{m\pi}{2} \)