Question

Find the values of $m$ such that exactly one root of the quadratic equation $x^2-(m-3)x+m=0;m\in R$ lie in the interval $(1,2).$

• A. $(1,9)$
• B. $(-\infty ,1)$
• C. $(10,\infty )$
• D. $(9,\infty )$

Answer 1

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

Given: $x^2-(m-3)x+m=0;m\in R$
On comparing with standard quadratic expression $f\left(x\right)=a{x}^2+bx+c,$ we have $a=1,b=-(m-3),c=m.$

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

When exactly one root of $f\left(x\right)=0$ lies in the interval $(1,2)$


Condition :
$\left(i\right)D>0$
$\left(ii\right)f\left(1\right).f\left(2\right)<0$

Now, on solving it,
$\left(i\right)D>0$
$\Rightarrow (m-3{)}^2-4m>0$
$\Rightarrow m^2-6m+9-4m>0$
$\Rightarrow m^2-10m+9>0$
$\Rightarrow (m-1)(m-9)>0$
$\Rightarrow m\in (-\infty ,1)\cup (9,\infty )$

$\left(ii\right)f\left(1\right).f\left(2\right)<0$
$\Rightarrow (1^2-(m-3).1+m).(2^2-(m-3).2+m)<0$
$\Rightarrow (1-m+3+m).(4-2m+6+m)<0$
$\Rightarrow 4(10-m)<0$
$\Rightarrow m>10$
$\Rightarrow m\in (10,\infty )$

Now, taking intersection of both the conditions, we get:
$m\in \left\{\right(-\infty ,1)\cup (9,\infty \left)\right\}\cap \left\{\right(10,\infty \left)\right\}$
$m\in (10,\infty )$