The abscissas of points P and Q on the curve y=ex+e-x such that tangents at P and Q make 60° with the x - axis are

Question

The abscissas of points P and Q on the curve y=ex+e-x such that tangents at P and Q make 60° with the x - axis are (A)  ln ((√3+√7/7)) and ln ((√3+√5/2)) (B)  ln ((√3+√7/2)) (C)  ln ((√7-√3/2)) (D) ± ln ((√3+√7/2))

Tardigrade Admin · Accepted Answer

y=ex+e-x or (dy/dx)=ex-e-x= tan θ, where θ is the angle of the tangent with the x -axis. For θ=60°, we have tan 60°=ex-e-x or e2 x-√3 ex-1=0 or ex=(√3 ± √7/2) or x= log e((√3+√7/2))

Q. The abscissas of points $P$ and $Q$ on the curve $y = e^{x} + e^{- x}$ such that tangents at $P$ and $Q$ make $6 0^{\circ}$ with the $x$ - axis are

$ln (\frac{3 + 7}{7})$ and $ln (\frac{3 + 5}{2})$

$ln (\frac{3 + 7}{2})$

$ln (\frac{7 - 3}{2})$

$\pm ln (\frac{3 + 7}{2})$

$y = e^{x} + e^{- x}$ or $\frac{d y}{d x} = e^{x} - e^{- x} = tan θ$ ,
where $θ$ is the angle of the tangent with the $x$ -axis.
For $θ = 6 0^{\circ}$ , we have
$tan 6 0^{\circ} = e^{x} - e^{- x}$
or $e^{2 x} - 3 e^{x} - 1 = 0$
or $e^{x} = \frac{3 \pm 7}{2}$ or $x = lo g_{e} (\frac{3 + 7}{2})$

Q. The abscissas of points P and Q on the curve y=ex+e−x such that tangents at P and Q make 60∘ with the x - axis are

ln(73​+7​​) and ln(23​+5​​)

ln(23​+7​​)

ln(27​−3​​)

±ln(23​+7​​)