| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Solve equations with surds |
| Difficulty | Moderate -0.8 This is a straightforward two-part question testing basic manipulation of surds and indices. Part (i) requires simplifying √18 = 3√2 then collecting like terms—a routine algebraic exercise. Part (ii) involves expressing both sides as powers of 2 and equating exponents—a standard technique. Both parts are textbook exercises with no problem-solving or insight required, making this easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\sqrt{2} - \sqrt{18} = x \Rightarrow x(\sqrt{2}-1) = \sqrt{18} \Rightarrow x = \frac{\sqrt{18}}{\sqrt{2}-1}\) | M1 | Combines terms in \(x\), factorises and divides to find \(x\). Condone sign slips and ignore attempts to simplify \(\sqrt{18}\). Alt: squares both sides \(x\sqrt{2}-\sqrt{18}=x \Rightarrow 2x^2-12x+18=x^2\) |
| \(x = \frac{\sqrt{18}}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\) | dM1 | Complete method to find \(x\): making \(x\) subject then multiplying by \(\frac{\sqrt{2}+1}{\sqrt{2}+1}\). In alt method: squaring both sides to produce 3TQ then factorising |
| \(x = \frac{\sqrt{18}(\sqrt{2}+1)}{1} = 6+3\sqrt{2}\) | A1 | Only following correct intermediate line. Allow \(\frac{6+3\sqrt{2}}{1}\) as intermediate. In alt method \(6-3\sqrt{2}\) must be discarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4^{3x-2} = \frac{1}{2\sqrt{2}} \Rightarrow 2^{6x-4} = 2^{-\frac{3}{2}}\) | M1 | Uses correct notation attempting to set both sides as powers of 2 or 4. Alt: uses logs (base 2 or 4) to get linear equation in \(x\) |
| \(6x-4 = -\frac{3}{2} \Rightarrow x = \ldots\) | dM1 | Setting indices of 2 or 4 equal and solving for \(x\). Must attempt both sides. Condone bracketing errors e.g. \(4^{3x-2}=2^{2\times3x-2}\) or \(\frac{1}{2\sqrt{2}}=2^{-1+\frac{1}{2}}\) |
| \(x = \frac{5}{12}\) | A1 | With correct intermediate work |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\sqrt{2} - \sqrt{18} = x \Rightarrow x(\sqrt{2}-1) = \sqrt{18} \Rightarrow x = \frac{\sqrt{18}}{\sqrt{2}-1}$ | M1 | Combines terms in $x$, factorises and divides to find $x$. Condone sign slips and ignore attempts to simplify $\sqrt{18}$. Alt: squares both sides $x\sqrt{2}-\sqrt{18}=x \Rightarrow 2x^2-12x+18=x^2$ |
| $x = \frac{\sqrt{18}}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$ | dM1 | Complete method to find $x$: making $x$ subject then multiplying by $\frac{\sqrt{2}+1}{\sqrt{2}+1}$. In alt method: squaring both sides to produce 3TQ then factorising |
| $x = \frac{\sqrt{18}(\sqrt{2}+1)}{1} = 6+3\sqrt{2}$ | A1 | Only following correct intermediate line. Allow $\frac{6+3\sqrt{2}}{1}$ as intermediate. In alt method $6-3\sqrt{2}$ must be discarded |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4^{3x-2} = \frac{1}{2\sqrt{2}} \Rightarrow 2^{6x-4} = 2^{-\frac{3}{2}}$ | M1 | Uses correct notation attempting to set both sides as powers of 2 or 4. Alt: uses logs (base 2 or 4) to get linear equation in $x$ |
| $6x-4 = -\frac{3}{2} \Rightarrow x = \ldots$ | dM1 | Setting indices of 2 or 4 equal and solving for $x$. Must attempt both sides. Condone bracketing errors e.g. $4^{3x-2}=2^{2\times3x-2}$ or $\frac{1}{2\sqrt{2}}=2^{-1+\frac{1}{2}}$ |
| $x = \frac{5}{12}$ | A1 | With correct intermediate work |
---\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) Solve the equation
$$x \sqrt { 2 } - \sqrt { 18 } = x$$
writing the answer as a surd in simplest form.\\
(ii) Solve the equation
$$4 ^ { 3 x - 2 } = \frac { 1 } { 2 \sqrt { 2 } }$$
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q3 [6]}}