Question

If $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$ are such that $f(x{)}_{min}>g(x{)}_{max},$ then the relation between $b$ and $c$ is $($where $b,c\in R)$

• A. Not possible
• B. $0<c<\frac{b}{2}$
• C. $|c|<|b|\sqrt{2}$
• D. $|c|>|b|\sqrt{2}$

Answer 1

The correct option is D $|c|>|b|\sqrt{2}$

Given,$f\left(x\right)=x^2+2bx+2c^2$
$\Rightarrow f^{\prime }\left(x\right)=2x+2b\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow 2x+2b=0$
$\Rightarrow x=-b$
Now, $f^{\prime \prime }\left(x\right)=2>0$
$\Rightarrow$ at $x=-b,f\left(x\right)$ is minimum
$\Rightarrow \text{minimum}{f\left(x\right)|}_{x=-b}=b^2-2b^2+2c^2$
$\Rightarrow f(x{)}_{min}=-b^2+2c^2$
And $g\left(x\right)=-x^2-2cx+b^2$
$\Rightarrow g^{\prime }\left(x\right)=-2x-2c$
For $g^{\prime }\left(x\right)=0$
$x=-c$
$g^{\prime \prime }\left(x\right)=-2<0$
$\Rightarrow$ at $x=-c,g\left(x\right)$ is maximum
$\Rightarrow \text{max}{g\left(x\right)|}_{x=-c}=-c^2+2c^2+b^2$
$\Rightarrow g(x{)}_{max}=(b^2+c^2)$
Since $f(x{)}_{min}>g(x{)}_{max}$
$\Rightarrow (2c^2-b^2)>(c^2+b^2)$
$\Rightarrow c^2>2b^2$
$\Rightarrow |c|>|b|\sqrt{2}$