If `f(x)=[{:(mx+1,if x le (pi)/(2)),(sinx+n,ifxgt(pi)/(2)):}` is continuous at `x = (pi)/(2)`, then find the relation between m and n.

Options

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

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). The function is defined piecewise as follows:

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

### Step 1: Set up the continuity condition
For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), we need the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the left to equal the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \) from the right. This can be mathematically expressed as:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) \]

### Step 2: Calculate the left-hand limit
For \( x \) approaching \( \frac{\pi}{2} \) from the left, we use the first piece of the function:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} (mx + 1) = m \cdot \frac{\pi}{2} + 1 \]

### Step 3: Calculate the right-hand limit
For \( x \) approaching \( \frac{\pi}{2} \) from the right, we use the second piece of the function:

\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} (\sin x + n) = \sin\left(\frac{\pi}{2}\right) + n = 1 + n \]

### Step 4: Set the limits equal to each other
Now, we set the two limits equal to establish the continuity condition:

\[ m \cdot \frac{\pi}{2} + 1 = 1 + n \]

### Step 5: Simplify the equation
We can simplify this equation by subtracting 1 from both sides:

\[ m \cdot \frac{\pi}{2} = n \]

### Conclusion
Thus, the relation between \( m \) and \( n \) is:

\[ n = m \cdot \frac{\pi}{2} \]