Question

If $f\left(x\right)=\frac{x}{1+\left({\log }_{e}x\right)\left({\log }_{e}x\right)\cdots \infty }\forall x\in [1,3]$ is non-differentiable at $x=k,$ then the value of $\left[k^2\right],$ is (where $[.]$ denotes the greatest integer function)

Answer 1

Let $g\left(x\right)=\left({\log }_{e}x\right)\left({\log }_{e}x\right)\cdots \infty$
$\Rightarrow g\left(x\right)=\{\begin{array}{cc}1,& x=e\\ 0,& 1\le x<e\\ \infty ,& x>e\end{array}$

$f\left(x\right)=\frac{x}{1+\left({\log }_{e}x\right)\left({\log }_{e}x\right)\cdots \infty }$
$=\frac{x}{1+g\left(x\right)}$
$f\left(x\right)=\{\begin{array}{cc}x,& 1\le x<e\\ \frac{x}{2},& x=e\\ 0,& e<x\le 3\end{array}$
Clearly $f\left(e^{+}\right)=0$ and $f\left(e^{-}\right)=e$
$\therefore f\left(x\right)$ is not continuous and hence not differentiable at $x=e=k$
$\therefore \left[k^2\right]=7$