$f(x) = k³x+ k³-2$ cuts the curve $g(x) = \ln(|x|)$ at exactly one point then $'k'$ may lie ...
My attempt : $f(x)=g(x)$ Then $\quad K^3=\frac{\ln(|x|)+2)}{(x+1)}$
Then plot the graph.
And by seeing the graph for $k^3=(1,\infty)$ you will get corresponding one value of $x$. (Use the calculator to plot the graph)
But answer is $\left(\frac{1}{\sqrt e},e\right)$, but here $0<\frac{1}{\sqrt e}<1$.
So, where did I make a mistake here?