JEE Main · 2006 · 30 Apr 2006 · Offline · Straight Lines

For the point $(a,a^2)$ to lie inside the angle bounded by $y=\frac{x}{2}$ and $y=3x$ in the first quadrant, it must sit above the gentler line and below the steeper one. Being above $y=\frac{x}{2}$ requires $a^2>\frac{a}{2}$, i.e. $a\left(a-\frac{1}{2}\right)>0$, which gives $a<0$ or $a>\frac{1}{2}$. Being below $y=3x$ requires $a^2<3a$, i.e. $a(a-3)<0$, which gives $0<a<3$. The lines are specified for $x>0$, so $a>0$, and the two conditions together give $\frac{1}{2}<a<3$.