Question 2 - A-Level Maths

Edexcel PURE 2024 October — Question 2

Exam Board	Edexcel
Module	PURE
Year	2024
Session	October
Paper	Download PDF ↗
Topic	Indices and Surds
Type	Rationalize denominator simple
Difficulty	Easy -1.2 Part (i) is routine index law manipulation requiring only basic algebraic simplification. Part (ii) involves rationalizing a denominator and simplifying surds—standard techniques with no problem-solving insight needed. Both are textbook exercises testing fundamental skills, making this easier than average for A-level.
Spec	1.02a Indices: laws of indices for rational exponents 1.02b Surds: manipulation and rationalising denominators

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Simplify fully $$\frac { 3 y ^ { 3 } \left( 2 x ^ { 4 } \right) ^ { 3 } } { 4 x ^ { 2 } y ^ { 4 } }$$
Find the exact value of $a$ such that $$\frac { 16 } { \sqrt { 3 } + 1 } = a \sqrt { 27 } + 4$$ Write your answer in the form $p \sqrt { 3 } + q$ where $p$ and $q$ are fully simplified rational constants.

Show mark scheme Show mark scheme source

(i)

Answer	Marks	Guidance
$\frac{3y^3 \left[2x^{-1}\right]^{10}}{4x^{-2}y^4} = \frac{6x^{10}}{y}$	B1, B1, B1 (3)	For any one of 6, $x^{10}$ or $\frac{1}{y}$ Allow $y^{-1}$ for $\frac{1}{y}$ - may be seen separately, and allow for one correctly simplified term even if a spurious extra term in the same variable is present.

(ii)

Answer	Marks	Guidance
$\frac{16}{\sqrt{3}+1} = a\sqrt{27}+4$	B1	States or uses $\sqrt{27} = 3\sqrt{3}$ . May be implied by working as long as it is seen before the final answer.
$\text{Correct attempt at rationalising seen e.g. } \frac{16}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$	M1	Correct attempt at rationalisation seen anywhere. This can be on any expression with a two term surd expression in a denominator during their working. The denominator may be simplified without the multiplication being seen (and need not be correct), but the evidence of the process must be clear.
$\text{Correct attempt to make "a" the subject and make progress towards the form } p\sqrt{3}+q \text{ e.g.} \\ 8\sqrt{3}-8 = 3a\sqrt{3}+4 \Rightarrow 3a\sqrt{3} = 8\sqrt{3}-12 \Rightarrow a = \frac{8\sqrt{3}}{3\sqrt{3}}-\frac{12}{3\sqrt{3}}$	M1	Attempts to make "a" the subject and make progress towards the form $p\sqrt{3}+q$ ($p, q \neq 0$) - must reach a two term expression for a (but may be over common denominator), though the $\sqrt{27}$ may not have been simplified. E.g. accept for $a = \frac{8\sqrt{3}-12}{\sqrt{27}}$ . Can be scored if a calculator was used to rationalise the surd expression but there must be an intermediate step before their final answer to show the method.
$a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$	A1cso (4)	$a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$ but isw after a correct expression. Must have score all previous marks, showing the relevant steps as outline above before the final answer is given but condone if "invisible brackets" are recovered. Accept with terms in the other order, and allow $\frac{1}{3}(8-4\sqrt{3})$

Total: 7 marks

Guidance for (i):

- B1: For any one of 6, $x^{10}$ or $\frac{1}{y}$ Allow $y^{-1}$ for $\frac{1}{y}$ - may be seen separately, and allow for one correctly simplified term even if a spurious extra term in the same variable is present.

- B1: For any two of the above as the only term in that variable within a single expression.

- B1: For $\frac{6x^{10}}{y}$ or $6x^{10}y^{-1}$ - must be one expression. Allow with $y^1$ in the denominator.

Guidance for (ii):

- B1: States or uses $\sqrt{27} = 3\sqrt{3}$ . May be implied by working as long as it is seen before the final answer.

- M1: Correct attempt at rationalisation seen anywhere. This can be on any expression with a two term surd expression in a denominator during their working. The denominator may be simplified without the multiplication being seen (and need not be correct), but the evidence of the process must be clear.

- M1: Attempts to make "a" the subject and make progress towards the form $p\sqrt{3}+q$ ($p, q \neq 0$) - must reach a two term expression for a (but may be over common denominator), though the $\sqrt{27}$ may not have been simplified. E.g. accept for $a = \frac{8\sqrt{3}-12}{\sqrt{27}}$ . Can be scored if a calculator was used to rationalise the surd expression but there must be an intermediate step before their final answer to show the method.

- A1cso: $a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$ but isw after a correct expression. Must have score all previous marks, showing the relevant steps as outline above before the final answer is given but condone if "invisible brackets" are recovered. Accept with terms in the other order, and allow $\frac{1}{3}(8-4\sqrt{3})$

**(i)**

| $\frac{3y^3 \left[2x^{-1}\right]^{10}}{4x^{-2}y^4} = \frac{6x^{10}}{y}$ | B1, B1, B1 (3) | For any one of 6, $x^{10}$ or $\frac{1}{y}$ Allow $y^{-1}$ for $\frac{1}{y}$ - may be seen separately, and allow for one correctly simplified term even if a spurious extra term in the same variable is present. |

**(ii)**

| $\frac{16}{\sqrt{3}+1} = a\sqrt{27}+4$ | B1 | States or uses $\sqrt{27} = 3\sqrt{3}$ . May be implied by working as long as it is seen before the final answer. |
| $\text{Correct attempt at rationalising seen e.g. } \frac{16}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$ | M1 | Correct attempt at rationalisation seen anywhere. This can be on any expression with a two term surd expression in a denominator during their working. The denominator may be simplified without the multiplication being seen (and need not be correct), but the evidence of the process must be clear. |
| $\text{Correct attempt to make "a" the subject and make progress towards the form } p\sqrt{3}+q \text{ e.g.} \\ 8\sqrt{3}-8 = 3a\sqrt{3}+4 \Rightarrow 3a\sqrt{3} = 8\sqrt{3}-12 \Rightarrow a = \frac{8\sqrt{3}}{3\sqrt{3}}-\frac{12}{3\sqrt{3}}$ | M1 | Attempts to make "a" the subject and make progress towards the form $p\sqrt{3}+q$ ($p, q \neq 0$) - must reach a two term expression for a (but may be over common denominator), though the $\sqrt{27}$ may not have been simplified. E.g. accept for $a = \frac{8\sqrt{3}-12}{\sqrt{27}}$ . Can be scored if a calculator was used to rationalise the surd expression but there must be an intermediate step before their final answer to show the method. |
| $a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$ | A1cso (4) | $a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$ but isw after a correct expression. Must have score all previous marks, showing the relevant steps as outline above before the final answer is given but condone if "invisible brackets" are recovered. Accept with terms in the other order, and allow $\frac{1}{3}(8-4\sqrt{3})$ |

**Total: 7 marks**

**Guidance for (i):**
- B1: For any one of 6, $x^{10}$ or $\frac{1}{y}$ Allow $y^{-1}$ for $\frac{1}{y}$ - may be seen separately, and allow for one correctly simplified term even if a spurious extra term in the same variable is present.
- B1: For any two of the above as the only term in that variable within a single expression.
- B1: For $\frac{6x^{10}}{y}$ or $6x^{10}y^{-1}$ - must be one expression. Allow with $y^1$ in the denominator.

**Guidance for (ii):**
- B1: States or uses $\sqrt{27} = 3\sqrt{3}$ . May be implied by working as long as it is seen before the final answer.
- M1: Correct attempt at rationalisation seen anywhere. This can be on any expression with a two term surd expression in a denominator during their working. The denominator may be simplified without the multiplication being seen (and need not be correct), but the evidence of the process must be clear.
- M1: Attempts to make "a" the subject and make progress towards the form $p\sqrt{3}+q$ ($p, q \neq 0$) - must reach a two term expression for a (but may be over common denominator), though the $\sqrt{27}$ may not have been simplified. E.g. accept for $a = \frac{8\sqrt{3}-12}{\sqrt{27}}$ . Can be scored if a calculator was used to rationalise the surd expression but there must be an intermediate step before their final answer to show the method.
- A1cso: $a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}$ but isw after a correct expression. Must have score all previous marks, showing the relevant steps as outline above before the final answer is given but condone if "invisible brackets" are recovered. Accept with terms in the other order, and allow $\frac{1}{3}(8-4\sqrt{3})$

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Show LaTeX source

\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying on calculator technology are not acceptable.\\
(i) Simplify fully

$$\frac { 3 y ^ { 3 } \left( 2 x ^ { 4 } \right) ^ { 3 } } { 4 x ^ { 2 } y ^ { 4 } }$$

(ii) Find the exact value of $a$ such that

$$\frac { 16 } { \sqrt { 3 } + 1 } = a \sqrt { 27 } + 4$$

Write your answer in the form $p \sqrt { 3 } + q$ where $p$ and $q$ are fully simplified rational constants.

\hfill \mbox{\textit{Edexcel PURE 2024 Q2}}

This paper (9 questions)

View full paper

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Edexcel PURE 2024 October — Question 2

(i)

(ii)

Total: 7 marks

Guidance for (i):

- B1: For any one of 6, \(x^{10}\) or \(\frac{1}{y}\) Allow \(y^{-1}\) for \(\frac{1}{y}\) - may be seen separately, and allow for one correctly simplified term even if a spurious extra term in the same variable is present.

- B1: For any two of the above as the only term in that variable within a single expression.

- B1: For \(\frac{6x^{10}}{y}\) or \(6x^{10}y^{-1}\) - must be one expression. Allow with \(y^1\) in the denominator.

Guidance for (ii):

- B1: States or uses \(\sqrt{27} = 3\sqrt{3}\) . May be implied by working as long as it is seen before the final answer.

This paper (9 questions)

Answer	Marks	Guidance
\(\frac{16}{\sqrt{3}+1} = a\sqrt{27}+4\)	B1	States or uses \(\sqrt{27} = 3\sqrt{3}\) . May be implied by working as long as it is seen before the final answer.
\(\text{Correct attempt at rationalising seen e.g. } \frac{16}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}\)	M1	Correct attempt at rationalisation seen anywhere. This can be on any expression with a two term surd expression in a denominator during their working. The denominator may be simplified without the multiplication being seen (and need not be correct), but the evidence of the process must be clear.
\(\text{Correct attempt to make "a" the subject and make progress towards the form } p\sqrt{3}+q \text{ e.g.} \\ 8\sqrt{3}-8 = 3a\sqrt{3}+4 \Rightarrow 3a\sqrt{3} = 8\sqrt{3}-12 \Rightarrow a = \frac{8\sqrt{3}}{3\sqrt{3}}-\frac{12}{3\sqrt{3}}\)	M1	Attempts to make "a" the subject and make progress towards the form \(p\sqrt{3}+q\) (\(p, q \neq 0\)) - must reach a two term expression for a (but may be over common denominator), though the \(\sqrt{27}\) may not have been simplified. E.g. accept for \(a = \frac{8\sqrt{3}-12}{\sqrt{27}}\) . Can be scored if a calculator was used to rationalise the surd expression but there must be an intermediate step before their final answer to show the method.
\(a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}\)	A1cso (4)	\(a = -\frac{4}{3}\sqrt{3}+\frac{8}{3}\) but isw after a correct expression. Must have score all previous marks, showing the relevant steps as outline above before the final answer is given but condone if "invisible brackets" are recovered. Accept with terms in the other order, and allow \(\frac{1}{3}(8-4\sqrt{3})\)