JEE Main 2014 — Application of Derivatives Question with Solution

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

For finding the minimum value of $f\left(x\right),$

$f^{\text{'}}\left(x\right)=2x+2b=0$

$\Rightarrow x=-b$

Also, $f^{"}\left(x\right)=2$

$\Rightarrow f^{"}\left(-b\right)=2>0$

So, $f\left(x\right)$ has minimum value at $x=-b$

Hence, $f_{min}=b^2-2b^2+2c^2$

      $\Rightarrow f_{min}=2c^2-b^2$

Similarly, for finding the maximum value of $g\left(x\right)$

$g^{\text{'}}\left(x\right)=-2x-2c=0$

                        $\Rightarrow x=-c$

Also, $g^{"}\left(x\right)=-2$

$g^{"}\left(-c\right)=-2<0$

So, $g\left(x\right)$ has maximum value at $x=-c$.

Hence, $g_{\max }=-c^2+2c^2+b^2$

      $\Rightarrow g_{\max }=c^2+b^2$

Given $f_{min}>g_{max}$$\Rightarrow 2c^2-b^2>c^2+b^2$$\text{⇒}\left(c^2-2b^2\right)>0$ $\text{⇒}\left(c-\sqrt{2}b\right)\left(c+\sqrt{2}b\right)>0$

$\Rightarrow \left(\frac{c}{b}-\sqrt{2}\right)\left(\frac{c}{b}+\sqrt{2}\right)>0$

Using wavy curve method, we get $\frac{c}{b}<-\sqrt{2}$ or $\frac{c}{b}>\sqrt{2}$ 

$\therefore \left|\frac{c}{b}\right|\in \left(\sqrt{2}, \infty \right).$