Question

Let $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$, then $min\left\{f\right(x\left)\right\}>max\left\{g\right(x\left)\right\}$ holds for

• A. $0<c<b\sqrt{2}$
• B. $|c|>|b|\sqrt{2}$
• C. $|c|<|b|\sqrt{2}$
• D. no real value of $b$ and $c$

Answer 1

The correct option is B $|c|>|b|\sqrt{2}$

The minimum value of $f\left(x\right)$ occurs at $x=-b$ . The maximum value of $g\left(x\right)$ occurs at $x=-c$ .
$f(-b)=2c^2-b^2$
$g(-c)=c^2+b^2$
Given that $f\left(min\right)>g\left(max\right)$, we have
$2c^2-b^2>c^2+b^2$
$\to c^2>2b^2$
Hence, option B is correct .