Question

If non zero real numbers $b$ and $c$ are such that min $f\left(x\right)>$max $g\left(x\right)$ , where $f\left(x\right)=x^2+2bx+2c^2$ and $g\left(x\right)=-x^2-2cx+b^2(x\in R)$, then $\left|\frac{c}{b}\right|$ lies in the interval :

• A. $(0,\frac{1}{2})$
• B. $[\frac{1}{2},\frac{1}{\sqrt{2}}]$
• C. $[\frac{1}{\sqrt{2}},\sqrt{2}]$
• D. $(\sqrt{2},\infty )$

Answer 1

The correct option is D $(\sqrt{2},\infty )$

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

$=x^2+2\times b\times x+b^2-b^2+2c^2$...... (Add and subtract $b^2$)

$=(x+b{)}^2-b^2+2c^2$

From the above expression, minimum value of $f\left(x\right)$ is $-b^2+2c^2$ $....\left(1\right)(\because$ minimum value of square term is zero. $)$

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

$=-(x^2+2cx-b^2)$

$=-(x^2+2cx+c^2-c^2-b^2)$

$=-\left(\right(x+c{)}^2-c^2-b^2)$

$=-(x+c{)}^2+b^2+c^2$

From the above expression, maximum value of $g\left(x\right)$ is $b^2+c^2....\left(2\right)$

It is given in the question that $minf\left(x\right)>maxg\left(x\right)$

Using $\left(1\right)$ and $\left(2\right)$ we get,

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

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

$\Rightarrow \frac{c^2}{b^2}>2$

$\Rightarrow \left|\frac{c}{b}\right|>\sqrt{2}$