If `f(x)={{:(mx+1",",xle(pi)/(2)),(sinx+n",",xge (pi)/(2)):}` is continuous at `x=(pi)/(2)` , then

Options

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

Answer

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

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

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

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) is given by the expression for \( f(x) \) when \( x \leq \frac{\pi}{2} \):

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

### Step 2: Calculate the Right-Hand Limit (RHL)

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

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

### Step 3: Calculate the Function Value at \( x = \frac{\pi}{2} \)

The function value at \( x = \frac{\pi}{2} \) is:

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

### Step 4: Set the Limits Equal for Continuity

For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), we need:

\[ \text{LHL} = \text{RHL} = f\left(\frac{\pi}{2}\right) \]

This gives us the equation:

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

### Step 5: Simplify the Equation

Subtracting 1 from both sides, we have:

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

### Conclusion

Thus, we find that:

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

### Final Answer

The condition for continuity at \( x = \frac{\pi}{2} \) is:

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