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Suppose $f(x) = \cos(\tan^{-1}x)$ and $g(x) = mx$. Determine range of $m$ so that $f$ and $g$ intersects each other only in one point .

My try:

We know $f(x) = \dfrac{1}{\sqrt{1+x^2}}$, then we should solve $mx = \dfrac{1}{\sqrt{1+x^2}}$.

I don't know where go from here.

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    If $x_0\neq0$, putting $m=\frac{1}{x_0\sqrt{1+x_0^2}}$ guarantees $f(x_0)=g(x_0)$. Now you just need to show that there isn't equality elsewhere. – 2017-02-21
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    @Aweygan Can you explain your answer ? It doesn't make sense now. – 2017-02-21

2 Answers 2

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\begin{align*} mx &= \frac{1}{\sqrt{1+x^2}} > 0 \\ m^2x^2(1+x^2) &= 1 \\ m^2 x^4+m^2 x^2-1 &= 0 \\ x^2 &= \frac{-m^2 \pm \sqrt{m^4+4m^2}}{2m^2} \end{align*}

Rejecting the negative solution, we have

\begin{align*} x^2 &= \frac{\sqrt{m^4+4m^2}-m^2}{2m^2} \\ x &= \frac{\sqrt{\sqrt{m^4+4m^2}-m^2}}{m\sqrt{2}} \tag{$mx > 0$} \end{align*}

which is unique solution for $\fbox{$m\ne 0$}$.

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    In the final step when you have $\begin{align*} x^2 &= \frac{\sqrt{m^4+4m^2}-m^2}{2m^2} \\ \end{align*}$ , then $\begin{align*} x&= \pm\sqrt{\frac{\sqrt{m^4+4m^2}-m^2}{2m^2}} \\ \end{align*}$ – 2017-02-21
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    Why you have considered only positive $x$ ? – 2017-02-21
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    $mx$ is positive instead, $$\frac{\sqrt{\sqrt{m^4+4m^2}-m^2}}{\sqrt{2}}=mx = \frac{1}{\sqrt{1+x^2}} > 0$$ – 2017-02-21
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    okay , we can only say "$m$ and $x$ have same sign ". – 2017-02-21
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    If you explain to me why we accept only positive root , my problem will be solved – 2017-02-21
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    I recap $$\frac{1}{\sqrt{1+x^2}}>0$$ so $$mx>0$$ so ***REJECT*** $$mx=-\sqrt{\frac{\sqrt{m^4+4m^2}-m^2}{2}}$$ – 2017-02-21
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    Okay , thanks for your help! – 2017-02-21
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$$ mx = \frac{1}{\sqrt{1+x^2}} $$

has only one real solution from

$$ x^2 =- \frac12 \pm \sqrt{ \frac14+1/m^2 }$$

when taking positive sign before radical.The domain of parameter $m$ is $ 0

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    Except $m = 0$, where no intersection is obtained. – 2017-02-21
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    for $m=0 $ root occurs at $x \rightarrow + \infty$ – 2017-02-21
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    I think your answer is wrong . $-\infty \lt m \lt \infty$ - $\{0\}$ – 2017-02-21
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    If you count $x \to +\infty$ as a root, you must surely also count $x \to -\infty$ as a root, too. – 2017-02-21
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    @Narasimham I think you are mistaking , real solution is $-\sqrt{\frac{\sqrt{5}}{2}-\frac{1}{2}}$ – 2017-02-21
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    You are right, I am mistaking; $ \,x= \pm \infty$ is ok. – 2017-02-21
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    I have (like Ng Chung Tak) also rejected $<0$ value for $x^2$ but erred in comments. – 2017-02-21
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    Okay , No problem . Thank you for your answers. – 2017-02-21