Question

Let $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2,b,c\ne 0$. If the minimum of $f\left(x\right)$ is greater than maximum value of $g\left(x\right)$, then the range of $\frac{b}{c}$ is

• A. $(0,\frac{1}{\sqrt{2}})$
• B. $(-\frac{1}{\sqrt{2}},0)$
• C. $(\frac{1}{\sqrt{2}},\infty )$
• D. $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Answer 1

The correct option is D $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Given:
$f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2,b,c\ne 0$
As $a>0$ for $f\left(x\right)$, so the minimum
$-\frac{D}{4a}=-\left(\frac{4b^2-8c^2}{4}\right)=2c^2-b^2$

As $a<0$ for $g\left(x\right)$, so the maximum
$-\frac{D}{4a}=\left(\frac{4c^2+4b^2}{4}\right)=b^2+c^2$

As minimum of $f\left(x\right)$ is greater than maximum of $g\left(x\right)$, so
$2c^2-b^2>b^2+c^2\Rightarrow c^2>2b^2\Rightarrow \frac{b^2}{c^2}<\frac{1}{2}\therefore \frac{b}{c}\in (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$