Question

If $(a,a^2)$ falls inside the triangle made by the lines $y=\frac{x}{2},x>0$ and $y=3x,x>0$, then $a$ belongs to the interval

• A. $(0,\frac{1}{2})$
• B. $(3,\infty )$
• C. $\left(\frac{1}{2},3\right)$
• D. $(-3,-\frac{1}{2})$

Answer 1

The correct option is C $\left(\frac{1}{2},3\right)$

Clearly, the sides of the triangle are on the opposite sides of the point $(a,a^2)$.

Substituting the points in the equation we get:

$\frac{a}{2}-a^2$ and $3a-a^2$ must have opposite sign

$\therefore \left(\frac{a}{2}-a^2\right)(3a-a^2)<0$

$\Rightarrow a^2(1-2a)(3-a)<0$

$\Rightarrow (2a-1)(a-3)<0$

$\Rightarrow \frac{1}{2}<a<3$

$\Rightarrow a\in \left(\frac{1}{2},3\right)$