Statement I : The maximum value of  $\frac{\log x}{x}$  is equal to $\frac{1}{e}$
Statement II : If $f\left(x\right)=\frac{1}{4x^2+2x+1},$ then its maximum value is $\frac{4}{3}$: which of the below statements are true 

Let  $y=\frac{\log x}{x}$
$\therefore \frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x}{x^2}$ 
$\Rightarrow \frac{dy}{dx}=\frac{1-\log x}{x^2}$ 
Put $\frac{dy}{dx}=0$, for maxima or minima
$\therefore x=e$
Now,  $\frac{d^{2}y}{d{x}^2}=\frac{x^2(-\frac{1}{x})-(1-\log x)2x}{x^4}=\frac{-[3-2\log x]}{x^3}$
And ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=e}=-\frac{1}{e^3}<0$ 
$\therefore$ Function is maximum at  $x=e$.
$\therefore y_{\max }=\frac{\log e}{e}=\frac{1}{e}$ .
Given $f\left(x\right)=\frac{1}{4x^2+2x+1}$   
On differentiating, we get  $f\text{'}\left(x\right)=\frac{-(8x+2)}{{(4x^2+2x+1)}^2}$
For maxima or minima put  $f\text{'}\left(x\right)=0$
$\Rightarrow 8x+2=0\Rightarrow x=-\frac{1}{4}$ 
Again differentiating 
$f"\left(x\right)=-\frac{[{(4x^2+2x+1)}^2\left(8\right)-(8x+2)2\times (4x^2+2x+1)(8x+2)]}{{(4x^2+2x+1)}^4}$ 
At  $x=-\frac{1}{4},f"(-\frac{1}{4})=-ve$ 
$f\left(x\right)$  is maximum at $x=-\frac{1}{4}$ .
$\therefore$ Maximum value is  $f{(-\frac{1}{4})}_{\max }=\frac{1}{4\times \frac{1}{16}-2\times \frac{1}{4}+1}=\frac{1}{\frac{1}{4}-\frac{2}{4}+1}$
$=\frac{4}{1-2+4}=\frac{4}{3}$ .