CAIE P3 2019 June — Question 4

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
SessionJune
TopicDifferentiating Transcendental Functions

4 The equation of a curve is \(y = \frac { 1 + \mathrm { e } ^ { - x } } { 1 - \mathrm { e } ^ { - x } }\), for \(x > 0\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is always negative.
  2. The gradient of the curve is equal to - 1 when \(x = a\). Show that \(a\) satisfies the equation \(\mathrm { e } ^ { 2 a } - 4 \mathrm { e } ^ { a } + 1 = 0\). Hence find the exact value of \(a\).