Question

The abscissas of points $P$ and $Q$ on the curve $y=e^x+e^{-x}$ such that tangents at $P$ and $Q$ make ${60}^{\circ }$ with the $x$-axis are

• A. $\ln \left(\frac{\sqrt{3}+\sqrt{7}}{7}\right)$ and $\ln \left(\frac{\sqrt{3}+\sqrt{5}}{2}\right)$
• B. $\ln \left(\frac{\sqrt{3}+\sqrt{7}}{2}\right)$
• C. $\ln \left(\frac{\sqrt{7}-\sqrt{3}}{7}\right)$
• D. $\pm \ln \left(\frac{\sqrt{3}+\sqrt{7}}{2}\right)$

Answer 1

The correct option is B $\ln \left(\frac{\sqrt{3}+\sqrt{7}}{2}\right)$

$y=e^x+e^{-x}$
$\frac{dy}{dx}=e^x-e^{-x}=\frac{e^{2x}-1}{e^x}$
Let P$(x_1,y_1)$ and Q$(x_2,y_2)$ be the points on the given curve.
Slope of tangent at P is $m_1=\frac{e^{2x_1}-1}{e^{x_1}}$
$\Rightarrow \sqrt{3}=\frac{e^{2x_1}-1}{e^{x_1}}$
$\Rightarrow e^{2x_1}-\sqrt{3}{e}^x-1=0$
$\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2}{)}^2=\frac{7}{4}$
$\Rightarrow e^{x_1}=\pm \frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$
$\Rightarrow x_1=ln(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})$ as logarithm of negative numbers is not defined.
Similarly, $\Rightarrow x_2=ln(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})$ as it makes same angle,${60}^0$ with x-axis.