| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2004 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Substitution to solve disguised quadratic |
| Difficulty | Moderate -0.3 This is a straightforward substitution question requiring students to recognize that 2^(-x) = 1/y, form the quadratic y - 1/y = 1, then solve it using standard methods. While it involves exponentials, the actual work is routine algebraic manipulation with no conceptual challenges beyond the initial substitution insight, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) State or imply \(2^{-x} = \frac{1}{y}\) | B1 | |
| Obtain 3-term quadratic e.g. \(y^2 - y - 1 = 0\) | B1 | Total: 2 |
| (ii) Solve a 3-term quadratic, obtaining 1 or 2 roots | M1 | |
| Obtain answer \(y = (1+\sqrt{5})/2\), or equivalent | A1 | |
| Carry out correct method for solving equation of form \(2^x = a\), where \(a > 0\), reaching a ratio of logarithms | M1 | |
| Obtain answer \(x = 0.694\) only | A1 | Total: 4 |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** State or imply $2^{-x} = \frac{1}{y}$ | B1 | |
| Obtain 3-term quadratic e.g. $y^2 - y - 1 = 0$ | B1 | **Total: 2** |
| **(ii)** Solve a 3-term quadratic, obtaining 1 or 2 roots | M1 | |
| Obtain answer $y = (1+\sqrt{5})/2$, or equivalent | A1 | |
| Carry out correct method for solving equation of form $2^x = a$, where $a > 0$, reaching a ratio of logarithms | M1 | |
| Obtain answer $x = 0.694$ only | A1 | **Total: 4** |
---4 (i) Show that if $y = 2 ^ { x }$, then the equation
$$2 ^ { x } - 2 ^ { - x } = 1$$
can be written as a quadratic equation in $y$.\\
(ii) Hence solve the equation
$$2 ^ { x } - 2 ^ { - x } = 1$$
\hfill \mbox{\textit{CAIE P3 2004 Q4 [6]}}