Question

The quadratic expression $f\left(x\right)=x^2+mx+m^2+6m$ is less than zero for all real values of $x$ satisfying the condition $1<x<2$, then

• A. $m\in (-\frac{7+3\sqrt{5}}{2},-4+2\sqrt{3})$
• B. $m\in (-\frac{7+3\sqrt{5}}{2},\frac{4+2\sqrt{3}}{3})$
• C. $m\in (\frac{3\sqrt{5}-7}{2},\infty )$
• D. None of these

Answer 1

The correct option is A $m\in (-\frac{7+3\sqrt{5}}{2},-4+2\sqrt{3})$

Since coefficient of $x^2>0$, so the parabola opens upwards.
Hence, to satisfy the given condition,
$f\left(1\right)<0$ and $f\left(2\right)<0$
$f\left(1\right)<0$
$m^2+7m+1<0$
$m\in (-\frac{7+3\sqrt{5}}{2},-\frac{7-3\sqrt{5}}{2})$
Now, $f\left(2\right)<0$
$m^2+8m+4<0$
$m\in (4-2\sqrt{3},-4+2\sqrt{3})$
.
From both the conditions above,
$m\in (-\frac{7+3\sqrt{5}}{2},-4+2\sqrt{3})$