Question

The values of $m$ such that exactly one root of $x^2+2(m-3)x+9=0$ lies between $1$ and $3$, is

• A. $(-\infty ,0)$
• B. $(6,\infty )$
• C. $(-2,0)$
• D. $(-2,6)$

Answer 1

The correct option is C $(-2,0)$

Let $f\left(x\right)=x^2+2(m-3)x+9$
Given that exactly one root lies between $1$ and $3.$


So, $f\left(1\right)\cdot f\left(3\right)<0$
$\Rightarrow (2m+4)\left(6m\right)<0$
$\Rightarrow m(m+2)<0\Rightarrow m\in (-2,0)$

Checking the boundary points
When $m=-2$, we get
$f\left(x\right)=x^2-10x+9\Rightarrow f\left(x\right)=(x-1)(x-9)=0\Rightarrow x=1,9$
When $m=0$, we get
$f\left(x\right)=x^2-6x+9\Rightarrow f\left(x\right)=(x-3{)}^2=0\Rightarrow x=3$

Hence, $x\in (-2,0)$