Question

If non-zero real numbers b and c are such that min.(f(x)) > max.(g(x)), where $f\left(x\right)=x^2+2bx+2c^2$,where $g\left(x\right)=-x^2-2cx+b^2$, $\left(xϵR\right)$, then $\left|\frac{c}{b}\right|$ belongs to

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

Answer 1

The correct option is C $(\sqrt{2},\infty )$

Given min(f(x)) > max(g(x))
$\Rightarrow \frac{8c^2-4b^2}{4}>\frac{-4b^2-4c^2}{-4}\Rightarrow c^2>2b^2\Rightarrow \frac{c^2}{b^2}>2\Rightarrow \left|\frac{c}{b}\right|>\sqrt{2}\Rightarrow \left|\frac{c}{b}\right|ϵ(\sqrt{2},\infty )$