For the function f(x)=sin x+3 x- 2 π (x^2+x ), where x ∈ [0, π 2 ], consider the following two statements : (I) f is increasing in (0, π 2 ). (II)…

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\sin \mathrm{x}+3 \mathrm{x}-\frac{2}{\pi}\left(\mathrm{x}^2+\mathrm{x}\right) \quad \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ & \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}+3-\frac{2}{\pi}(2 \mathrm{x}+1)>0 \mathrm{f}(\mathrm{x}) \uparrow \\ & \mathrm{f}^{\prime}(\mathrm{x})=-\sin \mathrm{x}+0-\frac{\pi}{2}(2) \\ & =-\sin \mathrm{x}-\frac{4}{\pi} < 0 \quad \mathrm{f}^{\prime}(\mathrm{x}) \downarrow \\ & 0 < \mathrm{x} < \frac{\pi}{2}\end{aligned}$​f(x)=sinx+3x−π2​(x2+x)x∈[0,2π​]f′(x)=cosx+3−π2​(2x+1)>0f(x)↑f′(x)=−sinx+0−2π​(2)=−sinx−π4​<0f′(x)↓0<x<2π​​ $\Rightarrow-\frac{2}{\pi}(\underset{+1}{0}{ < \underset{+1}2 \mathrm{x}} < \underset{+1}\pi)$⇒−π2​(+10​<+12​x<+1π​) $\begin{aligned} & \underset{+3}{-\frac{2}{\pi}}>\frac{-2}{\pi}(\underset{+3}2 \mathrm{x}+1)>-\frac{2}{\pi}(\underset{+3}\pi+1) \\ & \underset{+ve} 3-\frac{2}{\pi}>3-\frac{2}{\pi}(2 x+1)>3-\frac{2}{\pi}(\underset{+ve}\pi+1) \\ & \end{aligned}$​+3−π2​​>π−2​(+32​x+1)>−π2​(+3π​+1)+ve3​−π2​>3−π2​(2x+1)>3−π2​(+veπ​+1)​