If `f(x)=[{:(mx+1,if x le (pi)/(2)),(sinx+n,ifxgt(pi)/(2)):}` is continuous at `x = (pi)/(2)`, then

Options

A. `m = 1, n = 0`
B. `m = (n pi)/(2)+1`
C. `n = (mpi)/(2)`
D. `m = n = (pi)/(2)`

Explanation

To determine the values of \( m \) and \( n \) such that the function

\[ f(x) = \begin{cases} mx + 1 & \text{if } x \leq \frac{\pi}{2} \\ \sin x + n & \text{if } x > \frac{\pi}{2} \end{cases} \]

is continuous at \( x = \frac{\pi}{2} \), we need to ensure that the left-hand limit and the right-hand limit at \( x = \frac{\pi}{2} \) are equal.

### Step-by-step Solution:

1. **Identify the left-hand limit**:
The left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) is given by the expression for \( f(x) \) when \( x \leq \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right)^- = m\left(\frac{\pi}{2}\right) + 1 \]

2. **Identify the right-hand limit**:
The right-hand limit as \( x \) approaches \( \frac{\pi}{2} \) is given by the expression for \( f(x) \) when \( x > \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right)^+ = \sin\left(\frac{\pi}{2}\right) + n = 1 + n \]

3. **Set the left-hand limit equal to the right-hand limit**:
For the function to be continuous at \( x = \frac{\pi}{2} \), we must have:
\[ m\left(\frac{\pi}{2}\right) + 1 = 1 + n \]

4. **Simplify the equation**:
By simplifying the equation, we can isolate \( n \):
\[ m\left(\frac{\pi}{2}\right) + 1 - 1 = n \]
\[ n = m\left(\frac{\pi}{2}\right) \]

5. **Final relation**:
Thus, we have established the relationship between \( m \) and \( n \):
\[ n = \frac{m\pi}{2} \]

### Summary:
The function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \) if and only if the relationship \( n = \frac{m\pi}{2} \) holds.