| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Express in terms of substitution |
| Difficulty | Easy -1.2 This is a straightforward P1 question testing basic index laws and surd manipulation. Part (i) requires simple substitution using m=2^n with standard index rules (2^{n+3}=8m, 16^{3n}=m^12). Part (ii) is a routine linear equation with surds requiring collection of like terms and rationalizing the denominator—standard textbook exercises with no problem-solving insight needed. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2^{n+3} = 2^n \times 2^3 = 8m\) | B1 | For \(8m\) (condone \(8M\)). Do not allow \(2^3m\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16^{3n} = \left(2^4\right)^{3n}\) | M1 | For writing \(16^{3n}\) correctly as an expression involving a power of 2. Examples: \(\left(2^4\right)^{3n}\), \(\left(2^2\right)^{6n}\), \(\left(2^{3n}\right)^4\), \(\left(2^n\right)^{12}\), \(\left(2^{12}\right)^n\), \(2^{12n}\) |
| \(= 2^{12n} = \left(2^n\right)^{12} = m^{12}\) | A1 | For \(m^{12}\) (condone \(M^{12}\)). Correct answer only scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\sqrt{3} - 3 = x + \sqrt{3} \Rightarrow x\sqrt{3} - x = 3 + \sqrt{3} \Rightarrow x(\sqrt{3}-1) = 3 + \sqrt{3} \Rightarrow x = \ldots\) | M1 | Collects \(x\) terms to one side, factorises and makes \(x\) the subject. Reaching \(x = \frac{\alpha + \beta\sqrt{3}}{\gamma + \delta\sqrt{3}}\), \(\alpha,\beta,\gamma,\delta \neq 0\) |
| \(= \frac{3+\sqrt{3}}{\sqrt{3}-1} \times \frac{\pm(\sqrt{3}+1)}{\pm(\sqrt{3}+1)}\) | M1 | Correct attempt to rationalise denominator. Requires \(\frac{\ldots}{\gamma + \delta\sqrt{3}} = \frac{\ldots}{\gamma + \delta\sqrt{3}} \times \frac{\gamma - \delta\sqrt{3}}{\gamma - \delta\sqrt{3}}\) or equivalent |
| \(= \frac{\pm(4\sqrt{3}+6)}{\pm(3-1)} = 3 + 2\sqrt{3}\) | A1 | For \(3 + 2\sqrt{3}\) with at least one intermediate step. Allow \(2\sqrt{3}+3\). A1 depends on both M marks. |
## Question 2(i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2^{n+3} = 2^n \times 2^3 = 8m$ | B1 | For $8m$ (condone $8M$). Do **not** allow $2^3m$ |
---
## Question 2(i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16^{3n} = \left(2^4\right)^{3n}$ | M1 | For writing $16^{3n}$ correctly as an expression involving a power of 2. Examples: $\left(2^4\right)^{3n}$, $\left(2^2\right)^{6n}$, $\left(2^{3n}\right)^4$, $\left(2^n\right)^{12}$, $\left(2^{12}\right)^n$, $2^{12n}$ |
| $= 2^{12n} = \left(2^n\right)^{12} = m^{12}$ | A1 | For $m^{12}$ (condone $M^{12}$). Correct answer only scores both marks. |
---
## Question 2(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\sqrt{3} - 3 = x + \sqrt{3} \Rightarrow x\sqrt{3} - x = 3 + \sqrt{3} \Rightarrow x(\sqrt{3}-1) = 3 + \sqrt{3} \Rightarrow x = \ldots$ | M1 | Collects $x$ terms to one side, factorises and makes $x$ the subject. Reaching $x = \frac{\alpha + \beta\sqrt{3}}{\gamma + \delta\sqrt{3}}$, $\alpha,\beta,\gamma,\delta \neq 0$ |
| $= \frac{3+\sqrt{3}}{\sqrt{3}-1} \times \frac{\pm(\sqrt{3}+1)}{\pm(\sqrt{3}+1)}$ | M1 | Correct attempt to rationalise denominator. Requires $\frac{\ldots}{\gamma + \delta\sqrt{3}} = \frac{\ldots}{\gamma + \delta\sqrt{3}} \times \frac{\gamma - \delta\sqrt{3}}{\gamma - \delta\sqrt{3}}$ or equivalent |
| $= \frac{\pm(4\sqrt{3}+6)}{\pm(3-1)} = 3 + 2\sqrt{3}$ | A1 | For $3 + 2\sqrt{3}$ with at least one intermediate step. Allow $2\sqrt{3}+3$. A1 depends on both M marks. |
---\begin{enumerate}
\item (i) Given that $m = 2 ^ { n }$, express each of the following in simplest form in terms of $m$.\\
(a) $2 ^ { n + 3 }$\\
(b) $16 ^ { 3 n }$\\
(ii) In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.
Solve the equation
$$x \sqrt { 3 } - 3 = x + \sqrt { 3 }$$
giving your answer in the form $p + q \sqrt { 3 }$ where $p$ and $q$ are integers.
\hfill \mbox{\textit{Edexcel P1 2024 Q2 [6]}}