Key Concepts and Formulas

• Monotonicity of a function: A function $f(x)$ is increasing on an interval if $f'(x) > 0$ for all $x$ in the interval, and decreasing if $f'(x) < 0$ for all $x$ in the interval.
• Concavity of a function: A function $f(x)$ is concave up on an interval if $f''(x) > 0$ for all $x$ in the interval, and concave down if $f''(x) < 0$ for all $x$ in the interval. This also implies $f'(x)$ is increasing or decreasing respectively.
• Differentiation Rules: We will use the standard rules of differentiation for trigonometric and polynomial functions.

Step-by-Step Solution

Step 1: Find the first derivative, $f'(x)$

We need to determine the intervals where $f(x)$ is increasing or decreasing, so we first find its derivative. $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$ $f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$ $f'(x) = \cos x + 3 - \frac{4x}{\pi} - \frac{2}{\pi}$

Step 2: Analyze the sign of $f'(x)$ on the interval $(0, \frac{\pi}{2})$

To determine if $f(x)$ is increasing in $(0, \frac{\pi}{2})$, we need to show that $f'(x) > 0$ in this interval. Let's analyze $f'(x)$: $f'(x) = \cos x + 3 - \frac{4x}{\pi} - \frac{2}{\pi}$

Consider $x = 0$: $f'(0) = \cos 0 + 3 - \frac{4(0)}{\pi} - \frac{2}{\pi} = 1 + 3 - 0 - \frac{2}{\pi} = 4 - \frac{2}{\pi} > 0$ Consider $x = \frac{\pi}{2}$: $f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} + 3 - \frac{4(\frac{\pi}{2})}{\pi} - \frac{2}{\pi} = 0 + 3 - 2 - \frac{2}{\pi} = 1 - \frac{2}{\pi} > 0$ Since $\cos x$ is decreasing from 1 to 0 on $[0, \frac{\pi}{2}]$ and $-\frac{4x}{\pi}$ is decreasing from 0 to -2 on $[0, \frac{\pi}{2}]$, it's not immediately obvious that $f'(x) > 0$ for all $x$ in $(0, \frac{\pi}{2})$. Let's examine the second derivative.

Step 3: Find the second derivative, $f''(x)$

To determine if $f'(x)$ is increasing or decreasing, we find the second derivative: $f''(x) = -\sin x - \frac{4}{\pi}$

Step 4: Analyze the sign of $f''(x)$ on the interval $(0, \frac{\pi}{2})$

Since $\sin x > 0$ for $x \in (0, \frac{\pi}{2})$, we have $-\sin x < 0$. Also, $-\frac{4}{\pi} < 0$. Therefore, $f''(x) = -\sin x - \frac{4}{\pi} < 0$ for all $x \in (0, \frac{\pi}{2})$. This means that $f'(x)$ is decreasing in $(0, \frac{\pi}{2})$.

Step 5: Revisit the analysis of $f'(x)$

Since $f'(x)$ is decreasing and $f'(0) > 0$ and $f'(\frac{\pi}{2}) > 0$, it's plausible that $f'(x) > 0$ for all $x \in (0, \frac{\pi}{2})$. To prove this, we need to show that the minimum value of $f'(x)$ on $[0, \frac{\pi}{2}]$ is positive.

Let's find where $f''(x) = 0$. Since $f''(x) = -\sin x - \frac{4}{\pi} < 0$ for all $x \in (0, \frac{\pi}{2})$, $f''(x)$ is never equal to zero in the open interval. Therefore, $f'(x)$ has no minimum in the interval $(0, \frac{\pi}{2})$. Since $f'(x)$ is decreasing, its minimum value occurs at $x = \frac{\pi}{2}$. As we calculated before, $f'(\frac{\pi}{2}) = 1 - \frac{2}{\pi} > 0$. Thus, $f'(x) > 0$ for all $x \in (0, \frac{\pi}{2})$. Therefore, $f(x)$ is increasing in $(0, \frac{\pi}{2})$.

Step 6: Evaluate the truth of the given statements

Statement (I): $f$ is increasing in $(0, \frac{\pi}{2})$. This is true since $f'(x) > 0$ in $(0, \frac{\pi}{2})$. Statement (II): $f'$ is decreasing in $(0, \frac{\pi}{2})$. This is true since $f''(x) < 0$ in $(0, \frac{\pi}{2})$.

Step 7: Identify the correct option Since only statement (I) is true and statement (II) is false, option (A) is the correct answer.

Common Mistakes & Tips

• Sign Errors: Be careful with the signs when computing derivatives, especially with trigonometric functions.
• Interval Endpoints: Remember to consider the endpoints of the interval when analyzing the sign of the derivative.
• Second Derivative Test: The second derivative test can help determine local maxima and minima, but it's not always necessary to find the exact critical points to determine monotonicity.

Summary

We analyzed the monotonicity of the given function $f(x)$ by examining the sign of its first derivative $f'(x)$. We found that $f'(x) > 0$ for all $x$ in the interval $(0, \frac{\pi}{2})$, implying that $f(x)$ is increasing in this interval. We also found that $f''(x) < 0$ for all $x$ in the interval $(0, \frac{\pi}{2})$, which means that $f'(x)$ is decreasing in the same interval. Thus, statement (I) is true and statement (II) is false.

The final answer is \boxed{A}.