Let x= θ ^3 θ and y= ^3 θ θ , then:

Options

1. Aif $0 < \theta < \frac{\pi}{4}$, then $x+y>0$
2. Bif $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, then $x < y$
3. Cif $\frac{\pi}{2} < \theta < \frac{3 \pi}{4}$, then $x+y>0$
4. Dif $\frac{3 \pi}{4} < \theta < \pi$, then $x < y$