JEE Main 2014 — Mathematics Calculus

Given $f(x)={x}^{2}+2bx+2{c}^{2}$f(x)=x2+2bx+2c2 and $g(x)=-{x}^{2}-2cx+{b}^{2}$g(x)=−x2−2cx+b2

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

$${f}^{'}(x)=2x+2b=0$$f′(x)=2x+2b=0

$$\Rightarrow x=-b$$⇒x=−b

Also, ${f}^{"}(x)=2$f"(x)=2

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

So, $f(x)$f(x) has minimum value at $x=-b$x=−b

Hence, ${f}_{min}={b}^{2}-2{b}^{2}+2{c}^{2}$fmin​=b2−2b2+2c2

$$\Rightarrow {f}_{min}=2{c}^{2}-{b}^{2}$$⇒fmin​=2c2−b2

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

$${g}^{'}(x)=-2x-2c=0$$g′(x)=−2x−2c=0

$$\Rightarrow x=-c$$⇒x=−c

Also, ${g}^{"}(x)=-2$g"(x)=−2

$${g}^{"}(-c)=-2<0$$g"(−c)=−2<0

So, $g(x)$g(x) has maximum value at $x=-c$x=−c.

Hence, ${g}_{\mathrm{max}}=-{c}^{2}+2{c}^{2}+{b}^{2}$gmax​=−c2+2c2+b2

$$\Rightarrow {g}_{\mathrm{max}}={c}^{2}+{b}^{2}$$⇒gmax​=c2+b2

Given ${f}_{min}>{g}_{max}$fmin​>gmax​$\Rightarrow 2{c}^{2}-{b}^{2}>{c}^{2}+{b}^{2}$⇒2c2−b2>c2+b2$\text{⇒}({c}^{2}-2{b}^{2})>0$⇒(c2−2b2)>0 $\text{⇒}(c-\sqrt{2}b)(c+\sqrt{2}b)>0$⇒(c−2​b)(c+2​b)>0

$$\Rightarrow (\frac{c}{b}-\sqrt{2})(\frac{c}{b}+\sqrt{2})>0$$⇒(bc​−2​)(bc​+2​)>0

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

$$\therefore |\frac{c}{b}|\in (\sqrt{2},\infty ).$$∴∣bc​∣∈(2​,∞).