For the function f(x)=sin x+3 x-frac{2}{pi}le | vedclass

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(D) Given $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$.Step $1$: Analyze statement $(I)$.$f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$.For $x \in (0,

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Both $(I)$ and $(II)$ are true.

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(D) Given $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$.Step $1$: Analyze statement $(I)$.$f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$.For $x \in (0, \pi/2)$,$\cos x \in (0, 1)$ and $2x+1 \in (1, \pi+1)$.$f'(x) = \cos x + 3 - \frac{4x}{\pi} - \frac{2}{\pi}$.At $x=0$,$f'(0) = 1 + 3 - 2/\pi = 4 - 2/\pi > 0$.At $x=\pi/2$,$f'(\pi/2) = 0 + 3 - \frac{2}{\pi}(\pi + 1) = 3 - 2 - 2/\pi = 1 - 2/\pi > 0$.Since $f''(x) = -\sin x - 4/\pi  0$. Thus,$f'(x) > 0$ for all $x \in (0, \pi/2)$,so $f$ is increasing.Step $2$: Analyze statement $(II)$.$f''(x) = -\sin x - 4/\pi$.Since $\sin x > 0$ for $x \in (0, \pi/2)$,$f''(x) = -(\sin x + 4/\pi) Since $f''(x) Conclusion: Both $(I)$ and $(II)$ are true.

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