Question

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

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

Answer 1

Answer: (A)

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