For the function

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

consider the following two statements :

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

(II) $f^{\prime}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.

Between the above two statements,

<p>$$\begin{aligned} & f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right] \\ & f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1) \\ & =\cos x-\frac{4 x}{\pi}-\frac{2}{\pi}+3 \\ & \text { as } x \in\left[0, \frac{\pi}{2}\right] \\ & \frac{4 x}{\pi} \in[0,2] \end{aligned}$$</p> <p>$\Rightarrow 3-\frac{2}{\pi}-\frac{4 x}{\pi}>0$</p> <p>and also $\cos x>0$ when $x \in\left[0, \frac{\pi}{2}\right]$</p> <p>$\Rightarrow f^{\prime}(x)>0$</p> <p>$\Rightarrow f(x)$ is increasing</p> <p>Now, $$f^{\prime \prime}(x)=-\sin x-\frac{4}{\pi}<0 \forall x \in\left[0, \frac{\pi}{2}\right]$$</p> <p>Hence, $f^{\prime}(x)$ is decreasing</p> <p>$\therefore \quad$ Both statements (I) and (II) are true</p>