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The angle between the curve $y=e^{mx}$ and $y=e^{-mx}$ for $m>1 $ is

\[\begin{array}{1 1}(1)\tan^{-1}\bigg(\frac{2m}{m^{2}-1}\bigg)&(2)\tan^{-1}\bigg(\frac{2m}{1-m^{2}}\bigg)\\(3)\tan^{-1}\bigg(\frac{-2m}{1+m^{2}}\bigg)&(4)\tan^{-1}\bigg(\frac{2m}{m^{2}+1}\bigg)\end{array}\]

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$y=e^{-mx}$
$y=e^{mx}$
$\large \frac{dy}{dx}$$=-me^{-mx}$
$\large \frac{dy}{dx}$$=me^{mx}$
$m_2=-me^{-mx}$
$m_1=me^{mx}$
Also $e^{-mx}=e^{mx}=> \large\frac{1}{e^{mx}}$$=e^{mx}$
$e^{2mx}=1=> (e^{mx})^2=1$
$=> e^{mx}=1 => \large\frac{1}{e^{mx}}=\frac{1}{1}$
$\qquad= e^{mx}=1$
Angle between the curves
$\tan \theta= \bigg| \large\frac{m_1 -m_2}{1+m_1m_2} \bigg|$
Point of contact $x=0$
$\qquad= \large\frac{me^{mx}+me^{-mx}}{1+me^{mx}(-me^{-mx})}$
$\qquad= \large\frac{m.1+m.1}{1-m^2}$
$\qquad=\bigg| \large\frac{2m}{1-m^2}\bigg|$
$\qquad= \large\frac{2m}{m^2-1} $$ \quad m >1$
$=>\tan^{-1}\bigg(\frac{2m}{m^{2}-1}\bigg)$
Hence 1 is the correct answer
answered May 19, 2014 by meena.p
 

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