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 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: Find the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \)

The left-hand limit is given by the expression for \( f(x) \) when \( x \) is less than or equal to \( \frac{\pi}{2} \):

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

### Step 2: Find the right-hand limit as \( x \) approaches \( \frac{\pi}{2} \)

The right-hand limit is given by the expression for \( f(x) \) when \( x \) is greater than \( \frac{\pi}{2} \):

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

### Step 3: Set the left-hand limit equal to the right-hand limit for continuity

For the function to be continuous at \( x = \frac{\pi}{2} \), the left-hand limit must equal the right-hand limit:

\[ m\left(\frac{\pi}{2}\right) + 1 = 1 + n \]

### Step 4: Simplify the equation

Subtracting 1 from both sides gives:

\[ m\left(\frac{\pi}{2}\right) = n \]

### Conclusion

Thus, we find that:

\[ n = m\left(\frac{\pi}{2}\right) \]

This is the condition for continuity at \( x = \frac{\pi}{2} \).

### Final Answer

The values of \( m \) and \( n \) must satisfy the equation:

\[ n = m\left(\frac{\pi}{2}\right) \]

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