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. $(-2,6)$
• D. $(-2,0)$

Answer 1

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

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

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

Also, $D>0$
$\Rightarrow \left[2\right(m-3){]}^2-4\cdot 1\cdot 9>0$
$\Rightarrow m^2-6m>0$
$\Rightarrow m(m-6)>0$
$\Rightarrow m\in (-\infty ,0)\cup (6,\infty )\cdots \left(ii\right)$

Taking intersection of $\left(i\right)$ and $\left(ii\right)$ we get,
$m\in (-2,0)$