| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show gradient condition leads to equation |
| Difficulty | Standard +0.3 This is a straightforward quotient rule differentiation followed by algebraic manipulation. Part (i) requires applying the quotient rule to an exponential function and showing the result is negative (routine sign analysis). Part (ii) involves setting the derivative equal to -1 and algebraic rearrangement to form a quadratic in e^a, then solving using the quadratic formula—all standard techniques with no novel insight required. Slightly easier than average due to the mechanical nature of the steps. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the quotient or product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Reduce to \(-\dfrac{2e^{-x}}{\left(1-e^{-x}\right)^2}\), or equivalent, and explain why this is always negative | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate derivative to \(-1\) and obtain the given equation | B1 | |
| State or imply \(u^2 - 4u + 1 = 0\), or equivalent in \(e^a\) | B1 | |
| Solve for \(a\) | M1 | |
| Obtain answer \(a = \ln(2+\sqrt{3})\) and no other | A1 | |
| Total: 4 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the quotient or product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Reduce to $-\dfrac{2e^{-x}}{\left(1-e^{-x}\right)^2}$, or equivalent, and explain why this is always negative | A1 | |
| **Total: 3** | | |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate derivative to $-1$ and obtain the given equation | B1 | |
| State or imply $u^2 - 4u + 1 = 0$, or equivalent in $e^a$ | B1 | |
| Solve for $a$ | M1 | |
| Obtain answer $a = \ln(2+\sqrt{3})$ and no other | A1 | |
| **Total: 4** | | |4 The equation of a curve is $y = \frac { 1 + \mathrm { e } ^ { - x } } { 1 - \mathrm { e } ^ { - x } }$, for $x > 0$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ is always negative.\\
(ii) 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 $a$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q4 [7]}}