Question

If $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$ are such that the minimum value of $f\left(x\right)$ always exceeds maximum value of $g\left(x\right),$ then which of the following is/are correct?

• A. $b=1,c=0$
• B. $b=2,c=3$
• C. $b=3,c=5$
• D. $b=4,c=5$

Answer 1

Answer: (B), (C)

The correct options are
B $b=2,c=3$
C $b=3,c=5$

$f\left(x\right)=x^2+2bx+2c^2$
$=(x+b{)}^2+2c^2-b^2$
$g\left(x\right)=-x^2-2cx+b^2$
$=-(x+c{)}^2+b^2+c^2$

Given that
$min\left[f\right(x\left)\right]>max\left[g\right(x\left)\right]$
$\Rightarrow 2c^2-b^2>b^2+c^2$
$\Rightarrow c^2>2b^2\dots \left(1\right)$
Clearly, option $\left(b\right)$ and $\left(c\right)$ are satisfying equation $\left(1\right)$