| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a straightforward Further Pure 1 question testing routine Newton-Raphson application (substitute into formula) and showing a root lies in an interval (evaluate at endpoints). Both parts require only direct substitution with no problem-solving insight, making it slightly easier than average despite being Further Maths content. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.09a Sign change methods: locate roots1.09d Newton-Raphson method4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(39) = 2(39)^3 + 5(39)^2 + 3(39) - 132000 = -1110 < 0\) | M1 | Evaluate \(f\) at both endpoints |
| \(f(40) = 2(40)^3 + 5(40)^2 + 3(40) - 132000 = 1252 > 0\) | A1 | Sign change and conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = 6x^2 + 10x + 3\) | M1 | Correct differentiation |
| \(x_2 = 40 - \frac{f(40)}{f'(40)} = 40 - \frac{1252}{10003}\) | M1 | Correct Newton-Raphson application |
| \(x_2 = 39.87\) | A1 | To 2 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{n} 2r(3r+2) = \sum_{r=1}^{n}(6r^2 + 4r)\) | M1 | Expanding |
| \(= 6 \cdot \frac{n(n+1)(2n+1)}{6} + 4 \cdot \frac{n(n+1)}{2}\) | M1 | Using both standard formulae |
| \(= n(n+1)(2n+1) + 2n(n+1)\) | A1 | Correct substitution |
| \(= n(n+1)(2n+1+2) = n(n+1)(2n+3)\) | M1 | Factorising |
| \(= n(n+p)(2n+q)\) where \(p=1\), \(q=3\) | A1 | Both values stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_8 4^r = r\log_8 4 = \frac{2r}{3}\) | B1 | \(\lambda = \frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=k+1}^{60}(3r+2)\cdot\frac{2r}{3} = \frac{2}{3}\sum_{r=k+1}^{60}r(3r+2)\) | M1 | Using result from (c)(i) and (b) |
| \(\frac{1}{3}[60(61)(123) - k(k+1)(2k+3)] > 106060\) | M1 | Applying summation formula |
| \(k(k+1)(2k+3) < 150270\) | A1 | Correct inequality |
| \(k = 39\) | A1 | Greatest value stated |
# Question 7:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(39) = 2(39)^3 + 5(39)^2 + 3(39) - 132000 = -1110 < 0$ | M1 | Evaluate $f$ at both endpoints |
| $f(40) = 2(40)^3 + 5(40)^2 + 3(40) - 132000 = 1252 > 0$ | A1 | Sign change and conclusion stated |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = 6x^2 + 10x + 3$ | M1 | Correct differentiation |
| $x_2 = 40 - \frac{f(40)}{f'(40)} = 40 - \frac{1252}{10003}$ | M1 | Correct Newton-Raphson application |
| $x_2 = 39.87$ | A1 | To 2 d.p. |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n} 2r(3r+2) = \sum_{r=1}^{n}(6r^2 + 4r)$ | M1 | Expanding |
| $= 6 \cdot \frac{n(n+1)(2n+1)}{6} + 4 \cdot \frac{n(n+1)}{2}$ | M1 | Using both standard formulae |
| $= n(n+1)(2n+1) + 2n(n+1)$ | A1 | Correct substitution |
| $= n(n+1)(2n+1+2) = n(n+1)(2n+3)$ | M1 | Factorising |
| $= n(n+p)(2n+q)$ where $p=1$, $q=3$ | A1 | Both values stated |
## Part (c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_8 4^r = r\log_8 4 = \frac{2r}{3}$ | B1 | $\lambda = \frac{2}{3}$ |
## Part (c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=k+1}^{60}(3r+2)\cdot\frac{2r}{3} = \frac{2}{3}\sum_{r=k+1}^{60}r(3r+2)$ | M1 | Using result from (c)(i) and (b) |
| $\frac{1}{3}[60(61)(123) - k(k+1)(2k+3)] > 106060$ | M1 | Applying summation formula |
| $k(k+1)(2k+3) < 150270$ | A1 | Correct inequality |
| $k = 39$ | A1 | Greatest value stated |
These pages appear to be blank answer spaces (pages 17-19 for question 7, and pages 20-21 for question 8). The actual mark scheme content is not visible in these images — they only show lined answer spaces for students to write in.
Page 20 does show the **question text** for Question 8:
$$y = \frac{x(x-3)}{x^2+3}$$
- **(a)** State the equation of the asymptote of C. [1 mark]
- **(b)** The line $y = k$ intersects the curve C. Show that $4k^2 - 4k - 3 \leq 0$. [5 marks]
- **(c)** Hence find the coordinates of the stationary points of the curve C. (No credit will be given for solutions based on differentiation.) [5 marks]
To extract a **mark scheme**, I would need images of the actual mark scheme document rather than the student answer booklet pages. Could you share the mark scheme pages instead?
The images you've shared show only **blank answer spaces** (pages 22, 23, and 24) from an AQA exam paper (P/Jun15/MFP1). These pages contain:
- Lined answer spaces for Question 8
- "END OF QUESTIONS" notice
- A blank page with "DO NOT WRITE ON THIS PAGE"
**There is no mark scheme content visible in these images.** These are pages from the student question paper, not the mark scheme.
To find the mark scheme for this paper (AQA MFP1, June 2015), I'd recommend checking:
- **AQA's website** (aqa.org.uk) under past papers
- **Physics & Maths Tutor** (physicsandmathstutor.com)
Would you like help with a specific question from this paper instead?7
\begin{enumerate}[label=(\alph*)]
\item The equation $2 x ^ { 3 } + 5 x ^ { 2 } + 3 x - 132000 = 0$ has exactly one real root $\alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies in the interval $39 < \alpha < 40$.
\item Taking $x _ { 1 } = 40$ as a first approximation to $\alpha$, use the Newton-Raphson method to find a second approximation, $x _ { 2 }$, to $\alpha$. Give your answer to two decimal places.
\end{enumerate}\item Use the formulae for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that
$$\sum _ { r = 1 } ^ { n } 2 r ( 3 r + 2 ) = n ( n + p ) ( 2 n + q )$$
where $p$ and $q$ are integers.
\item \begin{enumerate}[label=(\roman*)]
\item Express $\log _ { 8 } 4 ^ { r }$ in the form $\lambda r$, where $\lambda$ is a rational number.
\item By first finding a suitable cubic inequality for $k$, find the greatest value of $k$ for which $\sum _ { r = k + 1 } ^ { 60 } ( 3 r + 2 ) \log _ { 8 } 4 ^ { r }$ is greater than 106060.\\[0pt]
[4 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2015 Q7 [15]}}