Question

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

• A. no real value of b & c
• B. $0<c<b\sqrt{2}$
• C. $|c|<|b|\sqrt{2}$
• D. $|c|>|b|\sqrt{2}$

Answer 1

Answer: (D)

$|c|>|b|\sqrt{2}$

$f\left(x\right)=x^2+2bx+2c^2$

$f^{\prime }\left(x\right)=2x+2b$

$\Rightarrow 0=2x+2b$

$\begin{array}{cc}\Rightarrow x& =-b\\ f^{\prime \prime }\left(x\right)& =2>0\end{array}$

minima occurs atx=-b$

$f(-b)=b^2-2b^2+2c^2$

$=-b^2+2c^2$

$=2c^2-b^2$

Now,

$g\left(x\right)=-x^2-2cx+b^2$

$g^{\prime }\left(x\right)=-2x-2c$

$\Rightarrow 0=-2x-2c$

$\Rightarrow x=-c$

$g^{\prime \prime }\left(x\right)=-2<0$

Moxima occurs at $x=-c$

$g(-c)=-c^2+2c^2+b^2$

$=b^2+c^2$

Now,

$minf\left(x\right)>maxg\left(x\right)$

$\Rightarrow 2c^2-b^2>b^2+c^2$

$\Rightarrow c^2>2b^2$

$\Rightarrow |c|>|b|\sqrt{2}$