| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| 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 (explicitly given), simplification using index laws, then solving a quadratic. The substitution is provided, making it more routine than if students had to identify it themselves. The algebraic manipulation is straightforward for A-level students, making this 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(9^x = p^2\), \(3^{x+2} = 9p\) or \(3^{x-1} = \frac{p}{3}\) | B1 | Uses index law and states/implies any of these or equivalent forms e.g. \(9^x = p \times p\), \(3^{x+2} = 3^2 \times p\), \(3^{x-1} = p \times 3^{-1}\). If awarding first term must be from correct work so \(3 \times 9^x = 3 \times 3^2 \times 3^x = 3p^2\) is fine |
| \(3 \times 9^x + 3^{x+2} = 1 + 3^{x-1} \Rightarrow 3p^2 + 3^2 \times p = 1 + \frac{p}{3}\) | M1 | Look for \(3 \times 9^x + 3^{x+2} = 1 + 3^{x-1} \Rightarrow 3p^2 \pm kp = 1 \pm \frac{p}{3}\) obtained from correct work, with \(k = 6\) or \(9\) |
| \(9p^2 + 26p - 3 = 0\) via \(3p^2 + 9p = 1 + \frac{p}{3}\) * | A1* | Proceeds to given answer \(9p^2 + 26p - 3 = 0\) with no errors or omissions. Intermediate line of \(3p^2 + 9p = 1 + \frac{p}{3}\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(9p^2 + 26p - 3 = 0 \Rightarrow (9p-1)(p+3) = 0\) | M1 | Valid non-calculator attempt at solving \(9p^2 + 26p - 3 = 0\) |
| \(3^x = \frac{1}{9}\) | A1, M1 in EPEN | For \(3^x = \frac{1}{9}\) seen. Must be clear it is a value for \(3^x\) not \(p\) or \(x\). May be implied by e.g. \(p = \frac{1}{9} \Rightarrow x = -2\) |
| \(x = -2\) | A1 | \(x = -2\) only |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $9^x = p^2$, $3^{x+2} = 9p$ or $3^{x-1} = \frac{p}{3}$ | B1 | Uses index law and states/implies any of these or equivalent forms e.g. $9^x = p \times p$, $3^{x+2} = 3^2 \times p$, $3^{x-1} = p \times 3^{-1}$. If awarding first term must be from correct work so $3 \times 9^x = 3 \times 3^2 \times 3^x = 3p^2$ is fine |
| $3 \times 9^x + 3^{x+2} = 1 + 3^{x-1} \Rightarrow 3p^2 + 3^2 \times p = 1 + \frac{p}{3}$ | M1 | Look for $3 \times 9^x + 3^{x+2} = 1 + 3^{x-1} \Rightarrow 3p^2 \pm kp = 1 \pm \frac{p}{3}$ obtained from correct work, with $k = 6$ or $9$ |
| $9p^2 + 26p - 3 = 0$ via $3p^2 + 9p = 1 + \frac{p}{3}$ * | A1* | Proceeds to given answer $9p^2 + 26p - 3 = 0$ with no errors or omissions. Intermediate line of $3p^2 + 9p = 1 + \frac{p}{3}$ must be seen |
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## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $9p^2 + 26p - 3 = 0 \Rightarrow (9p-1)(p+3) = 0$ | M1 | Valid non-calculator attempt at solving $9p^2 + 26p - 3 = 0$ |
| $3^x = \frac{1}{9}$ | A1, M1 in EPEN | For $3^x = \frac{1}{9}$ seen. Must be clear it is a value for $3^x$ not $p$ or $x$. May be implied by e.g. $p = \frac{1}{9} \Rightarrow x = -2$ |
| $x = -2$ | A1 | $x = -2$ only |
---\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 $p = 3 ^ { x }$, show that the equation
$$3 \times 9 ^ { x } + 3 ^ { x + 2 } = 1 + 3 ^ { x - 1 }$$
can be rewritten in the form
$$9 p ^ { 2 } + 26 p - 3 = 0$$
(b) Hence solve
$$3 \times 9 ^ { x } + 3 ^ { x + 2 } = 1 + 3 ^ { x - 1 }$$
\hfill \mbox{\textit{Edexcel P1 2023 Q5 [6]}}