1. Given function

$f(x)=\sin x+3x-\frac{2}{\pi}(x^2+x), \qquad x\in\left[0,\frac{\pi}{2}\right]$

We need to check:

• (I) whether $f$ is increasing on $\left(0,\frac{\pi}{2}\right)$
• (II) whether $f'$ is decreasing on $\left(0,\frac{\pi}{2}\right)$

2. Compute the first derivative

Differentiate term by term:

$f'(x)=\cos x+3-\frac{2}{\pi}(2x+1)$

So,

$f'(x)=\cos x+3-\frac{4x+2}{\pi}$

To check statement (I), we must see whether $f'(x)>0$ for all $x\in\left(0,\frac{\pi}{2}\right)$.

Now,

• for $x\in\left(0,\frac{\pi}{2}\right)$, we have $\cos x\ge 0$
• also, since $x\le \frac{\pi}{2}$, $4x+2\le 2\pi+2$ hence $\frac{4x+2}{\pi}\le 2+\frac{2}{\pi}$

Therefore,

$f'(x)\ge 0+3-\left(2+\frac{2}{\pi}\right)=1-\frac{2}{\pi}$

Since

$1-\frac{2}{\pi}>0,$

we get

$f'(x)>0 \quad \text{for all } x\in\left(0,\frac{\pi}{2}\right).$

Hence, $f$ is increasing on $\left(0,\frac{\pi}{2}\right)$.

So, Statement (I) is true.

3. Compute the second derivative

Differentiate $f'(x)$:

$f''(x)=-\sin x-\frac{4}{\pi}$

Now for $x\in\left(0,\frac{\pi}{2}\right)$,

$\sin x>0$

So,

$f''(x)=-\sin x-\frac{4}{\pi}<0$

for all $x\in\left(0,\frac{\pi}{2}\right)$.

Since $f''(x)<0$ throughout the interval, $f'(x)$ is decreasing on $\left(0,\frac{\pi}{2}\right)$.

Thus, Statement (II) is true.

4. Conclusion

• (I) is true
• (II) is true

Therefore, the correct option is:

$\boxed{\text{B: both (I) and (II) are true}}$

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.