| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with derivative given or simple |
| Difficulty | Standard +0.3 This is a straightforward application of Newton-Raphson with a simple cubic and given starting value, combined with routine summation formula manipulation and verification. Part (a) requires only one iteration of a standard formula with easy arithmetic; part (b) is algebraic manipulation of standard summation formulas; part (c) connects the results but requires minimal insight since the cubic from part (a) directly gives the answer. Slightly above average difficulty due to the multi-part structure and connection between parts, but all techniques are standard FP1 material with no novel problem-solving required. |
| Spec | 1.04g Sigma notation: for sums of series1.09d Newton-Raphson method4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) \(x_2 = 50 - \frac{50^3 + 2(50)^2 + 50 - 100\,000}{3(50)^2 + 4(50) + 1}\) | B1 | |
| B1 | For denominator (PI by value 7701) | |
| \(x_2 = 46.1\) | B1 | 3 marks |
| 8(b)(i) \(\sum(3r + 1) = 3\sum r^2 + \sum r\) | M1 | |
| \(\ldots = 3\left(\frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)\right)\) | m1 | |
| \(\ldots = \frac{1}{2}n(n+1)(2n+1+1)\) | m1m1 | |
| \(\ldots = n(n+1)^2\) convincingly shown | A1 | 5 marks |
| 8(b)(ii) Correct expansion of \(n(n+1)^2\) | B1 | 1 mark |
| 8(c) Attempt at value of \(S_{46}\) | M1 | 3 marks |
| Attempt at value of \(S_{45}\) | m1 | |
| \(S_{45} < 100000 < S_{46}\), so \(N = 46\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Root of equation in (a) is 45.8 | ||
| So lowest integer value is 46 | (B3) |
**8(a)** $x_2 = 50 - \frac{50^3 + 2(50)^2 + 50 - 100\,000}{3(50)^2 + 4(50) + 1}$ | B1 | | For numerator (PI by value 30050)
| B1 | | For denominator (PI by value 7701)
$x_2 = 46.1$ | B1 | 3 marks | Allow AWRT 46.1
**8(b)(i)** $\sum(3r + 1) = 3\sum r^2 + \sum r$ | M1 | | correct formulae substituted
$\ldots = 3\left(\frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)\right)$ | m1 | | m1 for each factor ($n$ and $n+1$)
$\ldots = \frac{1}{2}n(n+1)(2n+1+1)$ | m1m1 |
$\ldots = n(n+1)^2$ convincingly shown | A1 | 5 marks | AG
**8(b)(ii)** Correct expansion of $n(n+1)^2$ | B1 | 1 mark | and conclusion drawn (AG)
**8(c)** Attempt at value of $S_{46}$ | M1 | 3 marks
Attempt at value of $S_{45}$ | m1 |
$S_{45} < 100000 < S_{46}$, so $N = 46$ | A1 |
**Alternative method**
Root of equation in (a) is 45.8 | | | Allow AWRT 45.7 or 45.8
So lowest integer value is 46 | (B3) |
**Total: 12 marks**
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## **GRAND TOTAL: 75 marks**8
\begin{enumerate}[label=(\alph*)]
\item The equation
$$x ^ { 3 } + 2 x ^ { 2 } + x - 100000 = 0$$
has one real root. Taking $x _ { 1 } = 50$ as a first approximation to this root, use the Newton-Raphson method to find a second approximation, $x _ { 2 }$, to the root.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $S _ { n } = \sum _ { r = 1 } ^ { n } r ( 3 r + 1 )$, use the formulae for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that
$$S _ { n } = n ( n + 1 ) ^ { 2 }$$
\item The lowest integer $n$ for which $S _ { n } > 100000$ is denoted by $N$.
Show that
$$N ^ { 3 } + 2 N ^ { 2 } + N - 100000 > 0$$
\end{enumerate}\item Find the value of $N$, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2011 Q8 [12]}}