Explanation

(d)
Given that, $f(x) = x^{2} + 2bx + 2c^{2}\$
and $g(x) = - x^{2} - 2cx + b^{2}$
min $f(x) = - \frac{D}{4a} = - \frac{4b^{2} - 8c^{2}}{4}$
$= - (b^{2} - 2c^{2})$ (upward parabola)
max $g(x) = - \frac{D}{4a} = \frac{4c^{2} + 4b^{2}}{4}$
$= b^{2} + c^{2}$ (downward parabola)
Now, $2c^{2} - b^{2} > b^{2} + c^{2}$
$\Rightarrow c^{2} > 2b^{2} \Rightarrow |c| > |b|\sqrt{2}$