Question

Let $f\left(x\right)=4e^x{\cos }^{2}x-9e^{-x}{\cot }^{2}x,g\left(x\right)=4e^x+9e^{-x},$ then the minimum value of $g\left(x\right)-f\left(x\right)$ is

• A. $6$
• B. $12$
• C. $36$
• D. does not exist.

Answer 1

The correct option is B $12$

Given : $f\left(x\right)=4e^x{\cos }^{2}x-9e^{-x}{\cot }^{2}x$
$\Rightarrow f\left(x\right)=4e^x(1-{\sin }^{2}x)-9e^{-x}({\text{cosec}}^{2}x-1)\Rightarrow f\left(x\right)=(4e^x+9e^{-x})-(4e^x{\sin }^{2}x+9e^{-x}{\text{cosec}}^{2}x)$ and $g\left(x\right)=4e^x+9e^{-x}$
$\Rightarrow g\left(x\right)-f\left(x\right)=4e^x{\sin }^{2}x+9e^{-x}{\text{cosec}}^{2}x$
As all terms are non-negative : using $A.M.\ge G.M.$

$\Rightarrow \frac{4e^x{\sin }^{2}x+9e^{-x}{\text{cosec}}^{2}x}{2}\ge \sqrt{4e^x{\sin }^{2}x\cdot 9e^{-x}{\text{cosec}}^{2}x}\Rightarrow 4e^x{\sin }^{2}x+9e^{-x}{\text{cosec}}^{2}x\ge 2\times 6\Rightarrow g\left(x\right)-f\left(x\right)\ge 12$