Question

Let $x_1,x_2(x_1\ne x_2)$ be the roots of the equation $x^2+2(m-3)x+9=0$. If $-6<x_1,x_2<1,$ then ${}^{\prime }{m}^{\prime }$ lies in the interval

• A. $(6,\frac{27}{4}]$
• B. $(\frac{27}{4},9)$
• C. $(2,\frac{27}{4})$
• D. $(6,\frac{27}{4})$

Answer 1

The correct option is D $(6,\frac{27}{4})$

Let $f\left(x\right)=x^2+2(m-3)x+9$,
$-6<x_1,x_2<1$


Conditions:
$\left(i\right)\Delta >0\Rightarrow 4(m-3{)}^2-36>0$
$\Rightarrow m^2-6m>0\Rightarrow m(m-6)>0\Rightarrow m\in (-\infty ,0)\cup (6,\infty )\dots \left(1\right)$

$\left(ii\right)f(-6)>0\Rightarrow 12m-81<0\Rightarrow m<\frac{27}{4}\dots \left(2\right)$

$\left(iii\right)f\left(1\right)>0\Rightarrow 2m+4>0\Rightarrow m>-2\dots \left(3\right)$

$\left(iv\right)-6<-\frac{b}{2a}<1\Rightarrow -6<3-m<1\Rightarrow 2<m<9\dots \left(4\right)$

From $\left(1\right),\left(2\right),\left(3\right)\text{and}\left(4\right)$, we get
$m\in (6,\frac{27}{4})$