| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with derivative given or simple |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on standard numerical methods (sign change, interval bisection, Newton-Raphson) with routine calculations. Part (a) requires simple substitution to verify sign change, part (b) is mechanical bisection following a prescribed algorithm, and part (c) is direct application of Newton-Raphson formula. While it requires careful arithmetic and knowledge of multiple techniques, there's no conceptual challenge or problem-solving insight needed—it's below average difficulty for Further Maths FP1. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Let \(f(x) = 24x^3 + 36x^2 + 18x - 5\) | |
| 7(a) | \(f(0.1) = -2.816, f(0.2) = 0.232\) | M1 |
| 7(a) | Change of sign so α lies between 0.1 and 0.2 | A1 |
| 7(b) | \(f(0.15) = -1.409 (< 0 \text{ so root } > 0.15)\) | M1 |
| 7(b) | \(f(0.175) \approx -0.619 (< 0 \text{ so root } > 0.175)\) | A1 |
| 7(b) | α lies between 0.175 and 0.2 | A1 |
| 7(c) | \(f'(x) = 72x^2 + 72x + 18\) \((x_2=)\) | B1 |
| 7(c) | \(0.2 - \frac{24(0.2)^3 + 36(0.2)^2 + 18(0.2) - 5}{72(0.2)^2 + 72(0.2) + 18}\) | B1 |
| 7(c) | B1 | |
| 7(c) | \(= 0.1934\) (to 4dp) | B1 |
7(a) | Let $f(x) = 24x^3 + 36x^2 + 18x - 5$ | |
7(a) | $f(0.1) = -2.816, f(0.2) = 0.232$ | M1 | Both attempted and at least one evaluated correctly to at least 1sf rounded or truncated OE fraction
7(a) | Change of sign so α lies between 0.1 and 0.2 | A1 | Need both evaluations correct to above degree of accuracy and 'change of sign OE' and relevant reference to 0.1 and 0.2
7(b) | $f(0.15) = -1.409 (< 0 \text{ so root } > 0.15)$ | M1 | $f(0.15)$ considered first
7(b) | $f(0.175) \approx -0.619 (< 0 \text{ so root } > 0.175)$ | A1 | $f(0.15)$ then $f(0.175)$ both evaluated correctly to at least 1sf OE fractions. Dependent on both previous marks gained and no other additional evaluations other than that at 0.15 and 0.175
7(b) | α lies between 0.175 and 0.2 | A1 |
7(c) | $f'(x) = 72x^2 + 72x + 18$ $(x_2=)$ | B1 | PI
7(c) | $0.2 - \frac{24(0.2)^3 + 36(0.2)^2 + 18(0.2) - 5}{72(0.2)^2 + 72(0.2) + 18}$ | B1 | B1 for numerator in correct formula
7(c) | | B1 | B1 for denominator in correct formula
7(c) | $= 0.1934$ (to 4dp) | B1 | CAO Must be 0.1934. Do not apply ISW. NMS scores 0/47 The equation
$$24 x ^ { 3 } + 36 x ^ { 2 } + 18 x - 5 = 0$$
has one real root, $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies in the interval $0.1 < x < 0.2$.
\item Starting from the interval $0.1 < x < 0.2$, use interval bisection twice to obtain an interval of width 0.025 within which $\alpha$ must lie.
\item Taking $x _ { 1 } = 0.2$ as a first approximation to $\alpha$, use the Newton-Raphson method to find a second approximation, $x _ { 2 }$, to $\alpha$. Give your answer to four decimal places.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q7 [9]}}