Question

If $aϵR^{+}$ and the roots of equation $a{x}^2–3x+c=0$ are two consecutive odd positive integers, then

• A. $aϵ(1,\infty )$
• B. $aϵ(1,4)$
• C. $aϵ(0,\frac{3}{4}]$
• D. $aϵ(0,\infty )$

Answer 1

The correct option is C

$aϵ(0,\frac{3}{4}]$

Let $\alpha and\alpha +2$ be two consecutive odd positive integers\\
$\therefore a{\alpha }^2-3\alpha +c=0$ and $a(\alpha +2{)}^2-3(\alpha +2)+c=0$
$\Rightarrow (a{\alpha }^2-3\alpha +c)+4a\alpha +4a-6=0$
$\Rightarrow 4a\alpha +4a-6=0$, since $a{\alpha }^2-3\alpha +c=0$
$\Rightarrow 3=2a(1+\alpha )$ where $\alpha \ge 1$
$\Rightarrow \frac{3}{2a}-1=\alpha \Rightarrow \frac{3}{2a}-1\ge 1\Rightarrow \frac{3}{2a}\ge 2\Rightarrow a\le \frac{3}{4}$