Question

If at least one of the roots of the equation $x^2-(a-3)x+a=0$ lies in the interval $(1,2),$ then $a$ lies in the interval

• A. $[9,\infty )$
• B. $(10,\infty )$
• C. $[9,10)$
• D. $(5,7)\cup (10,\infty )$

Answer 1

The correct option is B $(10,\infty )$

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

Case $1:$ Both the roots lie in the interval $(1,2)$
$\left(i\right)D\ge 0\Rightarrow (a-3{)}^2-4a\ge 0$
$\Rightarrow a^2-10a+9\ge 0\Rightarrow (a-1)(a-9)\ge 0\Rightarrow a\in (-\infty ,1]\cup [9,\infty )\cdots \left(1\right)$

$\left(ii\right)1<-\frac{b}{2a}<2\Rightarrow 1<\frac{a-3}{2}<2\Rightarrow 5<a<7\cdots \left(2\right)$

$\left(iii\right)f\left(1\right)>0f\left(2\right)>0\Rightarrow 4>010-a>0\Rightarrow a<10\cdots \left(3\right)$
$\therefore \left(1\right)\cap \left(2\right)\cap \left(3\right)\Rightarrow a\in \varphi \cdots \left(4\right)$

Case $2:$ Exactly one root lies in the interval $(1,2)$
$f\left(1\right)f\left(2\right)<0$
$\Rightarrow 4(10-a)<0$
$\Rightarrow a\in (10,\infty )\cdots \left(5\right)$

Checking for boundary condition, when $a=10$, then
$f\left(x\right)=x^2-7x+10=(x-2)(x-5)f\left(x\right)=0\Rightarrow x=2,5\notin (1,2)$

Hence, from equation $\left(4\right)$ and $\left(5\right),$ we get
$x\in (10,\infty )$