| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational functions with parameters: analysis depending on parameter sign/range |
| Difficulty | Challenging +1.2 This is a multi-part Further AS question on rational functions requiring understanding of asymptotes, discriminants for determining vertical asymptote existence, and optimization via inequalities. Part (a) is routine, (b) requires discriminant reasoning (moderately challenging), (c) is standard, and (d) involves algebraic manipulation to derive and solve a quadratic inequality. While requiring several techniques and careful algebra, the problem-solving is relatively guided and uses standard Further Maths methods without requiring deep novel insight. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 1\) | B1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Indicates denominator cannot equal zero; PI by use of discriminant of denominator | M1 | AO 3.1a |
| \(x^2 + px + 7 \neq 0\), therefore \(p^2 - 4 \times 1 \times 7 < 0\) | M1 | AO 1.1a |
| \(p^2 < 28\) | A1 | AO 1.1b |
| \(-2\sqrt{7} < p < 2\sqrt{7}\) (accept \(-\sqrt{28} < p < \sqrt{28}\); condone \(-5.29 < p < 5.29\) or better) | A1 | AO 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=0 \Rightarrow y = \frac{0-3}{0+0+7} = -\frac{3}{7}\) | B1 | AO 1.1b |
| Solves \(x^2 - 3 = 0\) | M1 | AO 1.1a |
| \(\left(0, -\frac{3}{7}\right)\), \(\left(\sqrt{3}, 0\right)\), \(\left(-\sqrt{3}, 0\right)\) — must be written as coordinates | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Multiplies by denominator and forms a quadratic in \(x\) | M1 | AO 1.1a |
| \(k = \frac{x^2-3}{x^2-3x+7}\) leads to \((k-1)x^2 - 3kx + 7k + 3 = 0\); correct quadratic in form \(ax^2+bx+c=0\); PI by correct discriminant | A1 | AO 1.1b |
| Substitutes \(k\) for \(y\) and uses discriminant to form inequality in \(k\): \((-3k)^2 - 4(k-1)(7k+3) \geq 0\) | M1 | AO 3.1a |
| \(-19k^2 + 16k + 12 \geq 0\), i.e. \(19k^2 - 16k - 12 \leq 0\) | A1 | AO 1.1b |
| Completes rigorous proof that \(19k^2 - 16k - 12 \leq 0\) | R1 | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Selects method to find \(y\)-coordinate of minimum point; obtains at least one correct root of given quadratic; PI by \(-0.48\) or \(1.32\) or better | M1 | AO 3.1a; \(k = \frac{8 \pm 2\sqrt{73}}{19}\) |
| \(y = \frac{8 - 2\sqrt{73}}{19}\) | A1 | AO 1.1b |
## Question 14(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 1$ | B1 | AO 1.1b |
## Question 14(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Indicates denominator cannot equal zero; PI by use of discriminant of denominator | M1 | AO 3.1a |
| $x^2 + px + 7 \neq 0$, therefore $p^2 - 4 \times 1 \times 7 < 0$ | M1 | AO 1.1a |
| $p^2 < 28$ | A1 | AO 1.1b |
| $-2\sqrt{7} < p < 2\sqrt{7}$ (accept $-\sqrt{28} < p < \sqrt{28}$; condone $-5.29 < p < 5.29$ or better) | A1 | AO 3.2a |
## Question 14(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=0 \Rightarrow y = \frac{0-3}{0+0+7} = -\frac{3}{7}$ | B1 | AO 1.1b |
| Solves $x^2 - 3 = 0$ | M1 | AO 1.1a |
| $\left(0, -\frac{3}{7}\right)$, $\left(\sqrt{3}, 0\right)$, $\left(-\sqrt{3}, 0\right)$ — must be written as coordinates | A1 | AO 1.1b |
## Question 14(d)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Multiplies by denominator and forms a quadratic in $x$ | M1 | AO 1.1a |
| $k = \frac{x^2-3}{x^2-3x+7}$ leads to $(k-1)x^2 - 3kx + 7k + 3 = 0$; correct quadratic in form $ax^2+bx+c=0$; PI by correct discriminant | A1 | AO 1.1b |
| Substitutes $k$ for $y$ and uses discriminant to form inequality in $k$: $(-3k)^2 - 4(k-1)(7k+3) \geq 0$ | M1 | AO 3.1a |
| $-19k^2 + 16k + 12 \geq 0$, i.e. $19k^2 - 16k - 12 \leq 0$ | A1 | AO 1.1b |
| Completes rigorous proof that $19k^2 - 16k - 12 \leq 0$ | R1 | AO 2.1 |
## Question 14(d)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Selects method to find $y$-coordinate of minimum point; obtains at least one correct root of given quadratic; PI by $-0.48$ or $1.32$ or better | M1 | AO 3.1a; $k = \frac{8 \pm 2\sqrt{73}}{19}$ |
| $y = \frac{8 - 2\sqrt{73}}{19}$ | A1 | AO 1.1b |14 The function f is defined by
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 3 } { x ^ { 2 } + p x + 7 } \quad x \in \mathbb { R }$$
where $p$ is a constant.\\
The graph of $y = \mathrm { f } ( x )$ has only one asymptote.\\
14
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of the asymptote.\\
14
\item Find the set of possible values of $p$\\
□\\
14
\item Find the coordinates of the points at which the graph of $y = \mathrm { f } ( x )$ intersects the axes.
\section*{Question 14 continues on the next page}
14
\item $\quad A$ curve $C$ has equation
$$y = \frac { x ^ { 2 } - 3 } { x ^ { 2 } - 3 x + 7 }$$
The curve $C$ has a local minimum at the point $M$ as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{fd9715c4-9ce1-4608-aed6-f3d4f71208b5-24_371_835_587_605}
The line $y = k$ intersects curve $C$\\
14 (d) (i) Show that
$$19 k ^ { 2 } - 16 k - 12 \leq 0$$
14 (d) (ii) Hence, find the $y$-coordinate of point $M$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q14 [15]}}