Question

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

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

Answer 1

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

Given : $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2$

Differentiating $f\left(x\right)$ and then equating with $0$, we get

$2x+2b=0$

$x=-b$

so, now, $f(-b)$ is the minimum $f\left(x\right)$

now, differentiating $g\left(x\right)$ and then equating with $0$, we get

$-2x-2c=0$

$x=-c$

so, now $g(-c)$ is the maximum $g\left(x\right)$

now, according to question

minimum $f\left(x\right)>maximumg\left(x\right)$

$f(-b)>g(-c)$

$(-b{)}^2+2b(-b)+2c^2>-(-c{)}^2-2c(-c)+b^2$

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

$c^2>2b^2$

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