For 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…
For 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
- $c^2>2 b^2$
- $c^2 < 2 b^2$
- $b^2=2 c^2$
- $c^2=2 b^2$
Solution
We have,
$
\begin{aligned}
f(x) & =x^2+2 b x+2 x^2 \\
& =(x+b)^2+2 x^2-b^2
\end{aligned}
$
$\therefore$ Minimum value of
$
\begin{aligned}
f(x) & =2 c^2-b^2 \\
g(x) & =-x^2-2 c x+b^2 \\
\left.x-b^2\right] & =-\left[(x+c)^2-b^2-c^2\right] \\
& =-(x+c)^2+b^2+c^2
\end{aligned}
$
Again,
$
\begin{aligned}
=-\left[x^2+2 x-b^2\right] & =-\left[(x+c)^2-b^2-c^2\right] \\
& =-(x+c)^2+b^2+c^2
\end{aligned}
$
$\therefore$ 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
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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