| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring understanding of complex roots, Vieta's formulas, graphical interpretation for determining parameter ranges, and polynomial transformations. Part (b) requires connecting the graph to the condition that two roots are complex (requiring p outside [-2,2]), and part (c)(i) involves algebraic manipulation of transformed roots. While systematic, it demands synthesis of several A-level concepts beyond routine application. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks |
|---|---|
| 14(a) | Gives a full and correct explanation. |
| Answer | Marks | Guidance |
|---|---|---|
| justification. | 2.4 | E1 |
| Answer | Marks |
|---|---|
| 14(b) | πΌπΌ+π½π½+πΎπΎ = 0 πΌπΌ+π½π½ = βπΎπΎ |
| Answer | Marks | Guidance |
|---|---|---|
| π₯π₯ β3π₯π₯ | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS scores 2/2 | 2.2a | A1 |
| Answer | Marks |
|---|---|
| 14(c)(i) | Correctly expands |
| Answer | Marks | Guidance |
|---|---|---|
| π¦π¦ = π₯π₯ β3π₯π₯+ππ π₯π₯+1 π₯π₯β1 | 3.1a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| π₯π₯β1 π₯π₯ π₯π₯ β3π₯π₯+ππ | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS scores 0/3 | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 14(c)(ii) | Gives the correct x-intercepts and no others. | 2.2a |
| Total | 7 | |
| PAPER TOTAL | 80 |
Question 14:
--- 14(a) ---
14(a) | Gives a full and correct explanation.
Accept followed by without further
justification. | 2.4 | E1 | The coefficient of is zero
2
π₯π₯
--- 14(b) ---
14(b) | πΌπΌ+π½π½+πΎπΎ = 0 πΌπΌ+π½π½ = βπΎπΎ
Substitutes β1 or 1 into
Possibly implied by or3
π₯π₯ β3π₯π₯ | 3.1a | M1 | β΄ πΌπΌ+π½π½+πΎπΎ = 0
max poπΌπΌin+t π½π½ = βπΎπΎ and
min point
= (β1,2)
= (1,β2)
,
ππ < β2 ππ > 2
Finds the correct setβ o2f val2ues for p.
Condone βandβ.
NMS scores 2/2 | 2.2a | A1
--- 14(c)(i) ---
14(c)(i) | Correctly expands
Or recognises that is a horizontal translation of
(πΌπΌ+1)(π½π½+1)(πΎπΎ+1)
(possibly implied by the sight of or )
π¦π¦ = f(π₯π₯)
Accep3t any sensible alternative for .
π¦π¦ = π₯π₯ β3π₯π₯+ππ π₯π₯+1 π₯π₯β1 | 3.1a | B1 | Let
π€π€ = π₯π₯+1
π₯π₯ = π€π€β1
3
(π€π€β1) β3(π€π€β1)+ππ = 0
3 2
π€π€ β3π€π€ +3π€π€β1β3π€π€+3+ππ = 0
co3nstant 2term
π€π€ β3π€π€ +ππ+2 = 0
π₯π₯
Substitutes 0 for , and for , and for
Or substitutes for in
πΌπΌ+π½π½+πΎπΎ Β±3 πΌπΌπ½π½+π½π½πΎπΎ+πΎπΎπΌπΌ Β±ππ πΌπΌπ½π½πΎπΎ
Accept any sensible alternativ3e for .
π₯π₯β1 π₯π₯ π₯π₯ β3π₯π₯+ππ | 1.1a | M1
π₯π₯
Shows the required result.
NMS scores 0/3 | 2.1 | R1
--- 14(c)(ii) ---
14(c)(ii) | Gives the correct x-intercepts and no others. | 2.2a | B1 | 0 and 2= ππ+2
Total | 7
PAPER TOTAL | 80The graph of $y = x^3 - 3x$ is shown below.
\includegraphics{figure_14}
The two stationary points have $x$-coordinates of $-1$ and $1$
The cubic equation
$$x^3 - 3x + p = 0$$
where $p$ is a real constant, has the roots $\alpha$, $\beta$ and $\gamma$.
The roots $\alpha$ and $\beta$ are not real.
\begin{enumerate}[label=(\alph*)]
\item Explain why $\alpha + \beta = -\gamma$ [1 mark]
\item Find the set of possible values for the real constant $p$. [2 marks]
\item $f(x) = 0$ is a cubic equation with roots $\alpha + 1$, $\beta + 1$ and $\gamma + 1$
\begin{enumerate}[label=(\roman*)]
\item Show that the constant term of $f(x)$ is $p + 2$ [3 marks]
\item Write down the $x$-coordinates of the stationary points of $y = f(x)$ [1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q14 [7]}}