| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Standard +0.3 This question requires algebraic manipulation to convert a logarithmic-exponential equation into quadratic form, then solving it. The steps are systematic: multiply by e^x, use log laws to eliminate ln, and apply the quadratic formula. While it involves multiple techniques (exponential manipulation, logarithms, quadratic solving), the path is relatively standard for P3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Remove logarithms correctly and state \(1 + e^{-x} = e^{-2x}\) | B1 | or equivalent |
| Show equation is \(u^2 + u - 1 = 0\), where \(u = e^x\) | B1 | or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solve a 3-term quadratic for \(u\) | M1 | |
| Obtain root \(\frac{1}{2}(-1+\sqrt{5})\) | A1 | or decimal in [0.61, 0.62] |
| Use correct method for finding \(x\) from a positive root | M1 | |
| Obtain answer \(x = -0.481\) only | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Remove logarithms correctly and state $1 + e^{-x} = e^{-2x}$ | B1 | or equivalent |
| Show equation is $u^2 + u - 1 = 0$, where $u = e^x$ | B1 | or equivalent |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve a 3-term quadratic for $u$ | M1 | |
| Obtain root $\frac{1}{2}(-1+\sqrt{5})$ | A1 | or decimal in [0.61, 0.62] |
| Use correct method for finding $x$ from a positive root | M1 | |
| Obtain answer $x = -0.481$ only | A1 | |
---3
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$\ln \left( 1 + \mathrm { e } ^ { - x } \right) + 2 x = 0$$
can be expressed as a quadratic equation in $\mathrm { e } ^ { x }$.
\item Hence solve the equation $\ln \left( 1 + \mathrm { e } ^ { - x } \right) + 2 x = 0$, giving your answer correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q3 [6]}}