| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Rational curve analysis with turning points and range restrictions |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on partial fractions and curve sketching. Parts (a)-(c) involve routine identification of asymptotes and partial fraction decomposition, while part (d) requires basic algebraic reasoning without calculus. The multi-part structure and Further Maths context place it slightly above average, but the techniques are standard and well-practiced. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian |
| Hence, or otherwise, write down the coordinates of the \(y\)-intercept of \(C\) | ||
| Without using calculus, show that the line \(y = - 1\) does not intersect \(C\) |
| Answer | Marks |
|---|---|
| \(y = 3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Denominator is \((x+4)(x+1) = x^2 + 5x + 4\); \(\therefore m = 4\) and \(p = 5\) | M1, A1 | M1: identifies correct factors of denominator |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 0\), \(y = \frac{p}{m} = \frac{5}{4}\); \(\therefore\) \(y\)-intercept is \(\left(0, \frac{5}{4}\right)\) | B1F | Follow through their \(\frac{p}{m}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -1\) intersects \(C\) when \(\frac{3x^2 + 4x + 5}{x^2 + 5x + 4} = -1\); \(\Rightarrow 3x^2 + 4x + 5 = -(x^2 + 5x + 4)\); \(\Rightarrow 4x^2 + 9x + 9 = 0\); \(b^2 - 4ac = 9^2 - 4 \times 4 \times 9 = -63 < 0\); \(\therefore\) there are no real roots; \(\therefore y = -1\) does not intersect \(C\) | M1, M1, A1, M1, R1 | Could equate to letter \(k\) instead of \(-1\); allow one arithmetic error; correct quadratic could be in terms of \(k\): \((k-3)x^2 + (5k-4)x + 4k - 5 = 0\); or consider sign of discriminant in terms of \(k\): \(\Delta = 9k^2 + 28k - 44\) |
## Question 10:
### Part 10(a):
$y = 3$ | B1 |
### Part 10(b):
Denominator is $(x+4)(x+1) = x^2 + 5x + 4$; $\therefore m = 4$ and $p = 5$ | M1, A1 | M1: identifies correct factors of denominator
### Part 10(c):
When $x = 0$, $y = \frac{p}{m} = \frac{5}{4}$; $\therefore$ $y$-intercept is $\left(0, \frac{5}{4}\right)$ | B1F | Follow through their $\frac{p}{m}$
### Part 10(d):
$y = -1$ intersects $C$ when $\frac{3x^2 + 4x + 5}{x^2 + 5x + 4} = -1$; $\Rightarrow 3x^2 + 4x + 5 = -(x^2 + 5x + 4)$; $\Rightarrow 4x^2 + 9x + 9 = 0$; $b^2 - 4ac = 9^2 - 4 \times 4 \times 9 = -63 < 0$; $\therefore$ there are no real roots; $\therefore y = -1$ does not intersect $C$ | M1, M1, A1, M1, R1 | Could equate to letter $k$ instead of $-1$; allow one arithmetic error; correct quadratic could be in terms of $k$: $(k-3)x^2 + (5k-4)x + 4k - 5 = 0$; or consider sign of discriminant in terms of $k$: $\Delta = 9k^2 + 28k - 44$10
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of the horizontal asymptote of $C$
10
\item Find the value of $m$ and the value of $p$\\
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10
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\end{tabular} & Hence, or otherwise, write down the coordinates of the $y$-intercept of $C$ \\
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& Without using calculus, show that the line $y = - 1$ does not intersect $C$ \\
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\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q10 [9]}}