Question

If $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$ are such that min $f\left(x\right)>$max $g\left(x\right)$, then the relation between $b$ and $c$ is

• A. no relation
• B. $0<c<b/2$
• C. $\left|c\right|<\left|b\right|\sqrt{2}$
• D. $\left|c\right|>\left|b\right|\sqrt{2}$

Answer 1

Answer: (D)

$\left|c\right|>\left|b\right|\sqrt{2}$

We have
$f\left(x\right)=x^2+2bx+2c^2={(x+b)}^2+2c^2-b^2$
$\Rightarrow minf\left(x\right)=2c^2-b^2$
Also $g\left(x\right)=-x^2-2cx+b^2$
$=b^2+c^2-{(x+c)}^2$.
So $maxg\left(x\right)=b^2+c^2$,
Since $minf\left(x\right)>maxg\left(x\right)$
so$2c^2-b^2>b^2+c^2$
$\Rightarrow c^2>2b^2$
$\Rightarrow \left|c\right|>\sqrt{2}\left|b\right|$
Ans: D