Question

If \( f(x) = \left\{ \begin{array}{ll} mx + 1, & \text{for } x \leq \frac{\pi}{2}, \\ \sin x + n, & \text{for } x>\frac{\pi}{2}, \end{array} \right. \) is continuous at \( x = \frac{\pi}{2} \), then:

• A. \( m = 1, n = 0 \)
• B. \( m = \frac{\pi}{2} + 1 \)
• C. \( n = m \frac{\pi}{2} \)
• D. \( m = n = \frac{\pi}{2} \)

Hint:

For piecewise functions to be continuous at a point, ensure that the left-hand and right-hand limits at that point are equal and also equal to the function value at that point.

Answer 1

The Correct Option is C

The function \( f(x) \) is defined piecewise: 1. \( f(x) = mx + 1 \) for \( x \leq \frac{\pi}{2} \), 2. \( f(x) = \sin x + n \) for \( x>\frac{\pi}{2} \). We are asked to find the values of \( m \) and \( n \) such that \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). Step 1: Continuity condition at \( x = \frac{\pi}{2} \). For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), the following condition must hold: \[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = f\left( \frac{\pi}{2} \right). \] This means the left-hand limit (as \( x \to \frac{\pi}{2}^- \)) must equal the right-hand limit (as \( x \to \frac{\pi}{2}^+ \)), and both must equal the value of the function at \( x = \frac{\pi}{2} \). Step 2: Left-hand limit. For \( x \leq \frac{\pi}{2} \), \( f(x) = mx + 1 \). So, the left-hand limit is: \[ \lim_{x \to \frac{\pi}{2}^-} f(x) = m\left( \frac{\pi}{2} \right) + 1. \] Step 3: Right-hand limit. For \( x>\frac{\pi}{2} \), \( f(x) = \sin x + n \). So, the right-hand limit is: \[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \sin\left( \frac{\pi}{2} \right) + n = 1 + n. \] Step 4: Equating the limits. For continuity, the left-hand and right-hand limits must be equal. So, we have: \[ m\left( \frac{\pi}{2} \right) + 1 = 1 + n. \] Simplifying: \[ m \frac{\pi}{2} + 1 = 1 + n. \] Subtracting 1 from both sides: \[ m \frac{\pi}{2} = n. \] Thus, the relationship between \( m \) and \( n \) is: \[ n = m \frac{\pi}{2}. \] Step 5: Conclusion. Therefore, the correct answer is (c) \( n = m \frac{\pi}{2} \).