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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?

jayant98
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  • Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 17 '18 at 20:15

1 Answers1

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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

$$p=\frac m e$$

user
  • 154,566
  • Sorry but it is not the correct answer. Correct option is (A). Also you have just considered the case where $x>\frac{1}{m}$. – jayant98 Aug 25 '18 at 21:52
  • @jayant98 Ok I revise that! Thanks – user Aug 25 '18 at 21:53
  • @jayant98 Yes of course, the line y=px always intercepts f(x) for x<1/m and to intercept at one point f(x) for x>1/m we need tha f(x) and px are tangent that is p=m/e. – user Aug 25 '18 at 21:56
  • Yes. That's what I had given in my attempt. So, now, I think my solution is correct. :)) – jayant98 Aug 25 '18 at 21:57
  • @jayant98 Yes and we can obtain that by the tangency condition. – user Aug 25 '18 at 21:59