If `sinx+xge|k|x^(2), AA x in [0,(pi)/(2)]`, then the greatest value of k is

Options

A. `(-2(2+pi))/(pi^(2))`
B. `(2(2+pi))/(pi^(2))`
C. can't be determined finitely
D. zero

Explanation

To solve the problem, we need to find the greatest value of \( k \) such that the inequality \( \sin x + x \geq |k| x^2 \) holds for all \( x \) in the interval \( [0, \frac{\pi}{2}] \).

### Step-by-Step Solution:

1. **Define the Functions**:
Let \( f(x) = \sin x + x \) and \( g(x) = |k| x^2 \).

2. **Analyze \( f(x) \)**:
- Find the first derivative:
\[ f'(x) = \cos x + 1 \]
- Since \( \cos x \geq 0 \) for \( x \in [0, \frac{\pi}{2}] \), we have \( f'(x) > 0 \). Thus, \( f(x) \) is an increasing function on the interval.

3. **Analyze the Second Derivative**:
- Find the second derivative:
\[ f''(x) = -\sin x \]
- Since \( -\sin x \leq 0 \) for \( x \in [0, \frac{\pi}{2}] \), it implies that \( f(x) \) is concave down on this interval.

4. **Evaluate \( f(x) \) at the Endpoints**:
- Calculate \( f(0) \):
\[ f(0) = \sin(0) + 0 = 0 \]
- Calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} = 1 + \frac{\pi}{2} \]

5. **Set up the Inequality**:
- We need \( f(x) \geq g(x) \) for all \( x \in [0, \frac{\pi}{2}] \).
- Specifically, at \( x = \frac{\pi}{2} \):
\[ 1 + \frac{\pi}{2} \geq |k| \left(\frac{\pi}{2}\right)^2 \]
- This simplifies to:
\[ 1 + \frac{\pi}{2} \geq |k| \cdot \frac{\pi^2}{4} \]

6. **Solve for \( |k| \)**:
- Rearranging gives:
\[ |k| \leq \frac{4(1 + \frac{\pi}{2})}{\pi^2} \]
- Simplifying further:
\[ |k| \leq \frac{4 + 2\pi}{\pi^2} \]

7. **Determine the Greatest Value of \( k \)**:
- The greatest value of \( k \) is:
\[ k = \frac{4 + 2\pi}{\pi^2} \]

### Final Answer:
The greatest value of \( k \) is:
\[ k = \frac{4 + 2\pi}{\pi^2} \]