Question

Let a quadratic function $f\left(x\right)=x^2+bx+c$ have two distinct roots $\alpha ,\beta$ and $\alpha <\beta .$ Then the maximum value of $g\left(x\right)=2f\left(x\right)+\frac{18}{f\left(x\right)}$ in $(\alpha ,\beta ),$ is

• A. $36$
• B. $-6$
• C. $12$
• D. $-12$

Answer 1

The correct option is D $-12$

Given : $f\left(x\right)=x^2+bx+c$ have two distinct roots
Sign scheme for $f\left(x\right):$

$\Rightarrow f\left(x\right)<0$ in $(\alpha ,\beta )$
Putting, $h\left(x\right)=-f\left(x\right)>0$ in $(\alpha ,\beta )$
Now,
$g\left(x\right)=2f\left(x\right)+\frac{18}{f\left(x\right)}g\left(x\right)=-[-2f(x)+\frac{18}{-f\left(x\right)}]g\left(x\right)=-\left[2h\right(x)+\frac{18}{h\left(x\right)}]$

Using $A.M.\ge G.M.:h\left(x\right)>0$
$\frac{1}{2}\left[2h\right(x)+\frac{18}{h\left(x\right)}]\ge \sqrt{2h\left(x\right)\cdot \frac{18}{h\left(x\right)}}\Rightarrow 2h\left(x\right)+\frac{18}{h\left(x\right)}\ge 2\times 6\Rightarrow -\left[2h\right(x)+\frac{18}{h\left(x\right)}]\le -12\Rightarrow g\left(x\right)\le -12$
$\therefore$ maximum value of $g\left(x\right)$ is $-12.$