| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Rationalize denominator simple |
| Difficulty | Moderate -0.3 This is a straightforward rationalization problem requiring students to find a common denominator and simplify. Part (a) involves standard algebraic manipulation with difference of squares (the denominators multiply to 9 - 25n), and part (b) is a trivial deduction once part (a) is complete. While it requires careful algebra, it's a routine textbook exercise with no novel insight needed, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{7}{3+5\sqrt{n}} - \frac{7}{5\sqrt{n}-3}\) → multiplies numerator and denominator of at least one fraction by appropriate conjugate; or obtains common denominator with numerators simplifying to \(35\sqrt{n}-21-(21+35\sqrt{n})\) | M1 | |
| \(= \frac{7(5\sqrt{n}-3)}{(3+5\sqrt{n})(5\sqrt{n}-3)} - \frac{7(3+5\sqrt{n})}{(3+5\sqrt{n})(5\sqrt{n}-3)} = \frac{35\sqrt{n}-21-(21+35\sqrt{n})}{(3+5\sqrt{n})(5\sqrt{n}-3)}\) | A1 | Obtains correct single unsimplified fraction |
| \(= -\frac{42}{25n-9}\) | A1 | Obtains correct simplified fraction OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since 42 and \(25n-9\) are both integers, the expression is rational | E1F | Explains that numerator and denominator are both integers, or rational, and concludes it is rational |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{7}{3+5\sqrt{n}} - \frac{7}{5\sqrt{n}-3}$ → multiplies numerator and denominator of at least one fraction by appropriate conjugate; or obtains common denominator with numerators simplifying to $35\sqrt{n}-21-(21+35\sqrt{n})$ | M1 | |
| $= \frac{7(5\sqrt{n}-3)}{(3+5\sqrt{n})(5\sqrt{n}-3)} - \frac{7(3+5\sqrt{n})}{(3+5\sqrt{n})(5\sqrt{n}-3)} = \frac{35\sqrt{n}-21-(21+35\sqrt{n})}{(3+5\sqrt{n})(5\sqrt{n}-3)}$ | A1 | Obtains correct single unsimplified fraction |
| $= -\frac{42}{25n-9}$ | A1 | Obtains correct simplified fraction OE |
## Question 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since 42 and $25n-9$ are both integers, the expression is rational | E1F | Explains that numerator and denominator are both integers, or rational, and concludes it is rational |
---7
\begin{enumerate}[label=(\alph*)]
\item Given that $n$ is a positive integer, express
$$\frac { 7 } { 3 + 5 \sqrt { n } } - \frac { 7 } { 5 \sqrt { n } - 3 }$$
as a single fraction not involving surds.\\
7
\item Hence, deduce that
$$\frac { 7 } { 3 + 5 \sqrt { n } } - \frac { 7 } { 5 \sqrt { n } - 3 }$$
is a rational number for all positive integer values of $n$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2023 Q7 [4]}}