The equation $$|\ln (mx)|=px$$ Where m is a positive constant has exactly two roots for
$(A) p=\frac{m} {e} $
$(B) p=\frac{e} {m} $
$(C) \frac{e} {m} \geqslant p >0$
$(D) \frac{m} {e} \geqslant p>0$
My attempt:
Drew the graph for $|\ln(mx)|$ and then found that for a single root $y=px$ would have $p>\frac{m} {e} $ And hence, if $p$ is less than that value, then $y=px$ would have two roots until $p$ becomes less than$ \frac{m} {e}$ . So $p$ has only one value for two roots and that is $\frac {m} {e}$. Is my attempt correct?
We have that
$$f(x)=|\ln (mx)|$$
has a minimum at $mx=1$ that is $f(1/m)=0$ then to intercept $f(x)$ we need that $p\ge 0$.
For $p=0$ we have only one solution, then assume $p>0$.
For $x<\frac1m$ the line $y=px$ always intercepts $f(x)$ at one point.
For $x>\frac1m$, in order to have exactly one intersection point, we need that $y=px$ is tangent to $f(x)$ that is $$f'(x)=\frac m {mx}=p \implies x=\frac 1p \implies y=1 \implies mx=e \quad x=\frac e m\implies p=\frac m e$$
therefore to have exactly $2$ solution we need
$$p=\frac m e$$