For the function \(\rm f(x)=\sin x+3x-\frac{2}{\pi}(x^2+x)\) where x ∈ \(\rm \left[0, \frac{\pi}{2}\right]\) consider the following two statements : 

(I) f is increasing in \(\rm \left(0,\frac{\pi}{2}\right)\)

(II) f' is decreasing in \(\rm \left(0,\frac{\pi}{2}\right)\)

Between the above two statements, 

Calculation:

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

⇒ \(f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1)>0 \quad f(x) \uparrow\)

⇒ \(f^{\prime}(x)=-\sin x+0-\frac{\pi}{2}(2)\)

= \(-\sin x-\frac{4}{\pi}<0\ {f}^{\prime}({x}) \downarrow\)

\(0<{x}<\frac{\pi}{2}\)

⇒ \(-\frac{2}{\pi}(\underset{+1}{0}<\underset{+1}{2 x}<\underset{+1}{\pi})\) 

∴ f is increasing in \(\rm \left(0,\frac{\pi}{2}\right)\).

The correct answer is Option 1.