| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.3 This is a standard P1 exponential equation requiring substitution p=2^x, simplifying using index laws (4^x = (2^x)^2, 2^(x+3) = 8ยท2^x, etc.), solving a quadratic, then back-substituting. While it requires careful algebraic manipulation across multiple steps, it follows a well-practiced template with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02f Solve quadratic equations: including in a function of unknown1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses index law, states or implies any of: \(4^x = p^2\), \(2^{x+3} = 8p\), or \(2^{x-1} = \frac{p}{2}\) | B1 | Allow equivalents: \(4^x = p \times p\), \(2^{x+3} = 8 \times p\), \(2^{x-1} = \frac{1}{2}p\) |
| \(2 \times 4^x - 2^{x+3} = 17 \times 2^{x-1} - 4 \Rightarrow 2p^2 - 2^3 \times p = \frac{17p}{2^1} - 4\) | M1 | Attempts to write equation in \(x\) as quadratic in \(p\). All three index laws must be seen but condone slips on signs or \(2^3\) |
| Proceeds to \(4p^2 - 33p + 8 = 0\) via \(2p^2 - 8p = \frac{17p}{2} - 4\) | A1* | CSO. No errors or omissions. Intermediate line \(2p^2 - 8p = \frac{17p}{2} - 4\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4p^2 - 33p + 8 = 0 \Rightarrow (4p-1)(p-8) = 0 \Rightarrow p = \ldots\) | M1 | Valid non-calculator attempt at solving. Award for factorisation, quadratic formula \(\frac{33 \pm \sqrt{(-33)^2 - 4(4)(8)}}{2(4)}\), or completing the square. Roots cannot just appear |
| Sets \(2^x = \frac{1}{4}, 8 \Rightarrow x = \ldots\) | M1 | Valid non-calculator attempt at solving \(2^x = k\), \(k>0\). If \(k\) not a power of 2, score for \(x = \log_2 k\) |
| \(x = -2, 3\) | A1 | Both solutions following correct quadratic. No need to state \(x=\ldots\). If \(x=-2\) is rejected then A0 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses index law, states or implies any of: $4^x = p^2$, $2^{x+3} = 8p$, or $2^{x-1} = \frac{p}{2}$ | B1 | Allow equivalents: $4^x = p \times p$, $2^{x+3} = 8 \times p$, $2^{x-1} = \frac{1}{2}p$ |
| $2 \times 4^x - 2^{x+3} = 17 \times 2^{x-1} - 4 \Rightarrow 2p^2 - 2^3 \times p = \frac{17p}{2^1} - 4$ | M1 | Attempts to write equation in $x$ as quadratic in $p$. All three index laws must be seen but condone slips on signs or $2^3$ |
| Proceeds to $4p^2 - 33p + 8 = 0$ via $2p^2 - 8p = \frac{17p}{2} - 4$ | A1* | CSO. No errors or omissions. Intermediate line $2p^2 - 8p = \frac{17p}{2} - 4$ must be seen |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4p^2 - 33p + 8 = 0 \Rightarrow (4p-1)(p-8) = 0 \Rightarrow p = \ldots$ | M1 | Valid **non-calculator** attempt at solving. Award for factorisation, quadratic formula $\frac{33 \pm \sqrt{(-33)^2 - 4(4)(8)}}{2(4)}$, or completing the square. Roots cannot just appear |
| Sets $2^x = \frac{1}{4}, 8 \Rightarrow x = \ldots$ | M1 | Valid non-calculator attempt at solving $2^x = k$, $k>0$. If $k$ not a power of 2, score for $x = \log_2 k$ |
| $x = -2, 3$ | A1 | Both solutions following correct quadratic. No need to state $x=\ldots$. If $x=-2$ is rejected then A0 |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.\\
(a) By substituting $p = 2 ^ { x }$, show that the equation
$$2 \times 4 ^ { x } - 2 ^ { x + 3 } = 17 \times 2 ^ { x - 1 } - 4$$
can be written in the form
$$4 p ^ { 2 } - 33 p + 8 = 0$$
(b) Hence solve
$$2 \times 4 ^ { x } - 2 ^ { x + 3 } = 17 \times 2 ^ { x - 1 } - 4$$
\hfill \mbox{\textit{Edexcel P1 2024 Q4 [6]}}