If `f(x)={{:(mx+1,xle(pi)/(2)),(sinx+n,xgt(pi)/(2)):}` is continuous at `x=(pi)/(2)`. Then which one of the following is correct?

Options

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

Answer

To determine the correct relationship between \( m \) and \( n \) such that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), we follow these steps:

### Step 1: Understand the function definition
The function \( f(x) \) is defined as:
- \( f(x) = mx + 1 \) for \( x \leq \frac{\pi}{2} \)
- \( f(x) = \sin x + n \) for \( x > \frac{\pi}{2} \)

### Step 2: Set up the continuity condition
For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), the left-hand limit as \( x \) approaches \( \frac{\pi}{2} \) must equal the right-hand limit at \( x = \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) \]

### Step 3: Calculate the left-hand limit
For \( x \) approaching \( \frac{\pi}{2} \) from the left:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = m\left(\frac{\pi}{2}\right) + 1 = \frac{m\pi}{2} + 1 \]

### Step 4: Calculate the right-hand limit
For \( x \) approaching \( \frac{\pi}{2} \) from the right:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \sin\left(\frac{\pi}{2}\right) + n = 1 + n \]

### Step 5: Set the limits equal to each other
Setting the left-hand limit equal to the right-hand limit gives:
\[ \frac{m\pi}{2} + 1 = 1 + n \]

### Step 6: Simplify the equation
Subtracting 1 from both sides:
\[ \frac{m\pi}{2} = n \]

### Conclusion
Thus, the relationship we found is:
\[ n = \frac{m\pi}{2} \]

### Step 7: Identify the correct option
From the options provided, the correct option is:
\[ n = m \cdot \frac{\pi}{2} \]