| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Solve p(exponential) = 0 |
| Difficulty | Moderate -0.3 This is a structured multi-part question with clear signposting through each step. Part (a) is routine factor theorem application, (b) is straightforward polynomial division, (c) requires checking discriminant of the quadratic (standard technique), and (d) connects to part (c) via substitution e^x = y. While it spans multiple techniques, each step is standard and the 'hence' structure guides students through the solution path, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Substitutes x = 2 into function | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| is a factor | 2.1 | R1 |
| Subtotal | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | Obtains correct factor | 1.1b |
| Subtotal | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(c) | Calculates discriminant for their | |
| quadratic OE | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| solutions from the quadratic | 2.1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| solution coming from factor (x – 2) | 2.2a | B1 |
| Subtotal | 3 |
| Answer | Marks |
|---|---|
| 6(d) | Expresses equation as a cubic in a |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑥𝑥 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains ln 2. ISW | 1.1b | A1 |
| Subtotal | 3 | |
| Question Total | 9 | |
| Q | Marking Instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Substitutes x = 2 into function | 1.1a | M1 | f (2) = 23 – 22 + 2 – 6
f(2) = 0
which shows that (x – 2) is a factor
Completes reasoned argument to
explain that f(2) = 0 shows (x – 2)
is a factor | 2.1 | R1
Subtotal | 2
--- 6(b) ---
6(b) | Obtains correct factor | 1.1b | B1 | x2 + x + 3
Subtotal | 1
--- 6(c) ---
6(c) | Calculates discriminant for their
quadratic OE | 3.1a | M1 | x2 + x + 3 = 0
Discriminant = 12 – 4 × 1 × 3
= –11 < 0
So no real solutions to the quadratic
Therefore x = 2 is the only solution
States that there are no real
solutions from the quadratic | 2.1 | A1
Deduces that there is only one
solution coming from factor (x – 2) | 2.2a | B1
Subtotal | 3
--- 6(d) ---
6(d) | Expresses equation as a cubic in a
single different variable or in terms
of
𝑥𝑥
𝑒𝑒
𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 | 3.1a | M1 | 3 whe2re y =
𝑦𝑦 − 𝑦𝑦 + 𝑦𝑦−6 = 0
𝑥𝑥
Solution y =𝑒𝑒 2
= 2
𝑥𝑥
x𝑒𝑒 = ln 2
(𝑒𝑒 ) − (𝑒𝑒 ) + (𝑒𝑒 )−6 = 0
Obtains solution
𝑥𝑥 | 1.1b | A1
𝑒𝑒 = 2
Obtains ln 2. ISW | 1.1b | A1
Subtotal | 3
Question Total | 9
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item It is given that
$$f(x) = x^3 - x^2 + x - 6$$
Use the factor theorem to show that $(x - 2)$ is a factor of $f(x)$. [2 marks]
\item Find the quadratic factor of $f(x)$. [1 mark]
\item Hence, show that there is only one real solution to $f(x) = 0$ [3 marks]
\item Find the exact value of $x$ that solves
$$e^{3x} - e^{2x} + e^x - 6 = 0$$ [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2020 Q6 [9]}}