| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2006 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.8 This is a straightforward substitution problem requiring recognition that 4^x = (2^2)^x = (2^x)^2 = y^2, leading to a simple quadratic equation. The technique is standard and commonly practiced, making it easier than average, though it does require multiple steps including solving the quadratic and taking logarithms. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply that \(4^x = y^2\) \((=2^{2x})\) | B1 | 1 mark |
| (ii) Carry out recognizable solution method for a quadratic equation in \(y\) | M1 | |
| Obtain \(y = 3\) and \(y = \frac{1}{3}\) from \(3y^2 - 10y + 3 = 0\) | A1 | |
| Use logarithmic method to solve an equation of the form \(2^x = k\), where \(k > 0\) | M1 | |
| State answer \(1.58\) | A1 | |
| State answer \(-1.58\) | A1 \((A1 \checkmark \text{ if } 1.59)\) | 5 marks |
(i) State or imply that $4^x = y^2$ $(=2^{2x})$ | B1 | 1 mark |
(ii) Carry out recognizable solution method for a quadratic equation in $y$ | M1 |
Obtain $y = 3$ and $y = \frac{1}{3}$ from $3y^2 - 10y + 3 = 0$ | A1 |
Use logarithmic method to solve an equation of the form $2^x = k$, where $k > 0$ | M1 |
State answer $1.58$ | A1 |
State answer $-1.58$ | A1 $(A1 \checkmark \text{ if } 1.59)$ | 5 marks |2 (i) Express $4 ^ { x }$ in terms of $y$, where $y = 2 ^ { x }$.\\
(ii) Hence find the values of $x$ that satisfy the equation
$$3 \left( 4 ^ { x } \right) - 10 \left( 2 ^ { x } \right) + 3 = 0 ,$$
giving your answers correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2006 Q2 [6]}}