Explanation

(d)
If $f(x) = x^{2} + 2bx + 2c^{2}$ and $g(x) = - x^{2} - 2cx + b^{2}$
Then, $f(x)$ is minimum and $g(x)$ is maximum at
$f(x) = \frac{- D}{4a},\ \ \ \ \left( \because x = - \frac{- b}{a}\ and\ \ f(x) = \frac{- D}{4a}\ \right)$
$\therefore\ \ \min{\left\{ f(x) \right\} = \frac{- (4b^{2} - 8c^{2})}{4} = (2c^{2} - b^{2})}$
And $\max{\left\{ g(x) \right\} = - \frac{(4c^{2} + 4b^{2})}{4( - 1)} = (b^{2} + c^{2})}$
Since, $\min{f(x) > \max{g(x) \Rightarrow 2c^{2} - b^{2} > b^{2} + c^{2}}}$
$\Rightarrow \ \ \ \ c^{2} > 2b^{2}\ \ \ \ \Rightarrow \ \ \ |c| > \sqrt{2}|b|$