Question:

Step 1: To determine the validity of the given statements about the function $f(x)=\sin x+3x-\dfrac{2}{\pi}(x^2+x)$, we first need to find its first and second derivatives, as these will help us understand the behavior of the function and its rate of change. Step 2: The first derivative of the function $f(x)$ with respect to $x$ is given by $f'(x) = \cos x + 3 - \dfrac{2}{\pi}(2x + 1)$. This derivative represents the rate of change of the function with respect to $x$, and it will help us determine if the function is increasing or decreasing in the interval $\left(0, \dfrac{\pi}{2}\right)$. Step 3: To assess if $f$ is increasing in $\left(0, \dfrac{\pi}{2}\right)$, we need to check the sign of $f'(x)$ within this interval. If $f'(x) > 0$ for all $x$ in $\left(0, \dfrac{\pi}{2}\right)$, then the function is indeed increasing. Step 4: Next, to evaluate the truth of statement (II), which claims that $f'$ is decreasing in $\left(0, \dfrac{\pi}{2}\right)$, we must find the second derivative of the function, $f''(x)$. The second derivative is given by $f''(x) = -\sin x - \dfrac{4}{\pi}$. If $f''(x) < 0$ for all $x$ in $\left(0, \dfrac{\pi}{2}\right)$, then $f'$ is decreasing, validating statement (II). Step 5: We calculate $f''(x)$ to check its sign in the given interval. Given $f''(x) = -\sin x - \dfrac{4}{\pi}$, since $\sin x$ is positive in the first quadrant and $-\dfrac{4}{\pi}$ is a negative constant, $f''(x)$ will be negative, indicating that $f'$ is indeed decreasing in $\left(0, \dfrac{\pi}{2}\right)$, thus statement (II) is true. Step 6: Now, let's examine the sign of $f'(x)$ in the interval $\left(0, \dfrac{\pi}{2}\right)$. Given $f'(x) = \cos x + 3 - \dfrac{2}{\pi}(2x + 1)$, we observe that $\cos x$ is positive in the first quadrant, and the term $3 - \dfrac{2}{\pi}(2x + 1)$ needs to be evaluated. However, since we've established that $f''(x) < 0$, implying $f'(x)$ is decreasing, and given that $f'(0) = 1 + 3 - \dfrac{2}{\pi} > 0$, it suggests that $f'(x)$ remains positive throughout the interval $\left(0, \dfrac{\pi}{2}\right)$ because the decrease is from a positive value and $f''(x)$ being negative does not immediately imply $f'(x)$ becomes negative within the interval. Step 7: Considering the above steps, both statements (I) and (II) are true. Statement (I) is true because $f'(x) > 0$ in the interval $\left(0, \dfrac{\pi}{2}\right)$, indicating the function is increasing. Statement (II) is true because $f''(x) < 0$ in the interval, meaning $f'$ is decreasing. Therefore, the correct answer is that both (I) and (II) are true, which corresponds to Option 4. The final answer is: $\boxed{4}$