Question

If atleast one of the root of the equation $x^2-(a-3)x+a=0$ is greater than $2$, then $a$ lies in the interval

• A. $[7,9]$
• B. $[7,\infty )$
• C. $[9,\infty )$
• D. $[7,9)$

Answer 1

The correct option is C $[9,\infty )$

Let $f\left(x\right)=x^2-(a-3)x+a$

Case $1:$ Both roots are greater than $2$,


Required conditions are
$\left(i\right)$$D\ge 0$
$\Rightarrow$$D=(a-3{)}^2-4a\ge 0$
$\Rightarrow a^2-10a+9\ge 0\Rightarrow (a-1)(a-9)\ge 0\Rightarrow a\in (-\infty ,1]\cup [9,\infty )\cdots \left(1\right)$

$f\left(2\right)>0\Rightarrow 4-(a-3)2+a>0\Rightarrow a<10\cdots \left(2\right)$
Now, $x$ coordinate of the vertex will be greater than $2.$
$\therefore \frac{a-3}{2}>2\Rightarrow a>7\cdots \left(3\right)$
From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$\Rightarrow a\in [9,10)\cdots \left(4\right)$

Case $2:$ One root is greater than $2$ and other root is less than $2$

$\because$ $2$ lies in between the roots
$\Rightarrow f\left(2\right)<0$
$\Rightarrow 4-(a-3)2+a<0$
$\Rightarrow a>10\cdots \left(5\right)$

Case $3:$ One root is greater than $2$ and other root (lowest root) $=2$

$\because$ Lowest root is $2.$
Required conditions are $\left(i\right)f\left(2\right)=0$ and $\left(ii\right)\frac{-b}{2a}>0$
$f\left(2\right)=0\Rightarrow a=10...\left(6\right)$
$\frac{-b}{2a}>0\Rightarrow a>7...\left(7\right)$
From $\left(6\right)$ and $\left(7\right)$
$a=10...\left(8\right)$

$\therefore$ From $\left(4\right),\left(5\right)$ and $\left(8\right)$
$a\in [9,\infty )$