Question

Fior real number $x$, if the minimum value of $f(x)=x^{2}+2 b x+2 c^{2}$ is greater than the maximum value of $g(x)=-x^{2}-2 c x+b^{2}$, then

• A. $c^{2}>2 b^{2}$
• B. $c^{2}<2 b^{2}$
• C. $b^{2}=2 c^{2}$
• D. $c^{2}=2 b^{2}$

Answer 1

Answer: (A)

We have,
$f(x) =x^{2}+2 b x+2 c^{2}$
$=(x+b)^{2}+2 c^{2}-b^{2}$
$\therefore $ Minimum value of
$f(x)=2 c^{2}-b^{2} $
Again, $g(x)=-x^{2}-2 c x+b^{2}$
$=-\left[x^{2}+2 c x-b^{2}\right] =-\left[(x+c)^{2}-b^{2}-c^{2}\right]$
$=-(x+c)^{2}+b^{2}+c^{2}$
$\therefore $ Maximum value of
$g(x)=b^{2}+c^{2}$
Now, according to the question.
$2 c^{2}-b^{2}>b^{2}+c^{2}$
$ \Rightarrow \, c^{2}>\,2 b^{2}$