Question 2 - A-Level Maths

Edexcel FP1 2012 January — Question 2 10 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	January
Marks	10
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 application of standard numerical methods (sign change verification, interval bisection, Newton-Raphson) with clear instructions and simple arithmetic. While it's Further Maths content, the question requires only routine execution of algorithms with no problem-solving insight, making it slightly easier than average overall.
Spec	1.09a Sign change methods: locate roots 1.09b Sign change methods: understand failure cases 1.09d Newton-Raphson method

2. (a) Show that \(\mathrm { f } ( x ) = x ^ { 4 } + x - 1\) has a real root \(\alpha\) in the interval [0.5, 1.0].
[0pt] (b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains \(\alpha\).
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to \(\mathrm { f } ( x )\) to obtain an approximate value of \(\alpha\). Give your answer to 3 decimal places.

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Question 2:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(f(0.5) = -0.4375\ (-\frac{7}{16})\)	M1	Either \(f(0.5) = \text{awrt} -0.4\) or \(f(1) = 1\)
\(f(1) = 1\)
Sign change (positive, negative) and \(f(x)\) is continuous, therefore a root \(\alpha\) is between \(x = 0.5\) and \(x = 1.0\)	A1	\(f(0.5) = \text{awrt} -0.4\) and \(f(1) = 1\), sign change and conclusion

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(f(0.75) = 0.06640625\ (\frac{17}{256})\)	M1	Attempt \(f(0.75)\)
\(f(0.625) = -0.222412109375\ (-\frac{911}{4096})\)	A1	\(f(0.75) = \text{awrt}\ 0.07\) and \(f(0.625) = \text{awrt}\ -0.2\)
\(0.625 < \alpha < 0.75\)	A1	\(0.625 < \alpha < 0.75\) or \([0.625, 0.75]\) or \((0.625, 0.75)\) or equivalent in words

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(f'(x) = 4x^3 + 1\)	B1	Correct derivative (may be implied by e.g. \(4(0.75)^3 + 1\))
\(x_1 = 0.75\)
\(x_2 = 0.75 - \frac{f(0.75)}{f'(0.75)} = 0.75 - \frac{0.06640625}{2.6875(43/16)}\)	M1	Attempt Newton-Raphson
\(x_2 = 0.72529(06976...) = \frac{499}{688}\)	A1	Correct first application: \(0.75 - \frac{17/256}{43/16}\) or awrt \(0.725\) (may be implied)
\(x_3 = 0.724493\left(\frac{499}{688} - \frac{0.002015718978}{2.562146811}\right)\)	A1	Awrt \(0.724\)
\((\alpha) = 0.724\)	A1	cao

> A final answer of \(0.724\) with evidence of NR applied twice with no incorrect work scores 5/5

## Question 2:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0.5) = -0.4375\ (-\frac{7}{16})$ | M1 | Either $f(0.5) = \text{awrt} -0.4$ or $f(1) = 1$ |
| $f(1) = 1$ | | |
| Sign change (positive, negative) and $f(x)$ is continuous, therefore a root $\alpha$ is between $x = 0.5$ and $x = 1.0$ | A1 | $f(0.5) = \text{awrt} -0.4$ **and** $f(1) = 1$, sign change and conclusion |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0.75) = 0.06640625\ (\frac{17}{256})$ | M1 | Attempt $f(0.75)$ |
| $f(0.625) = -0.222412109375\ (-\frac{911}{4096})$ | A1 | $f(0.75) = \text{awrt}\ 0.07$ **and** $f(0.625) = \text{awrt}\ -0.2$ |
| $0.625 < \alpha < 0.75$ | A1 | $0.625 < \alpha < 0.75$ or $[0.625, 0.75]$ or $(0.625, 0.75)$ or equivalent in words |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 4x^3 + 1$ | B1 | Correct derivative (may be implied by e.g. $4(0.75)^3 + 1$) |
| $x_1 = 0.75$ | | |
| $x_2 = 0.75 - \frac{f(0.75)}{f'(0.75)} = 0.75 - \frac{0.06640625}{2.6875(43/16)}$ | M1 | Attempt Newton-Raphson |
| $x_2 = 0.72529(06976...) = \frac{499}{688}$ | A1 | Correct first application: $0.75 - \frac{17/256}{43/16}$ or awrt $0.725$ (may be implied) |
| $x_3 = 0.724493\left(\frac{499}{688} - \frac{0.002015718978}{2.562146811}\right)$ | A1 | Awrt $0.724$ |
| $(\alpha) = 0.724$ | A1 | cao |

> A final answer of $0.724$ with evidence of NR applied twice with no incorrect work scores 5/5

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2. (a) Show that $\mathrm { f } ( x ) = x ^ { 4 } + x - 1$ has a real root $\alpha$ in the interval [0.5, 1.0].\\[0pt]
(b) Starting with the interval [0.5, 1.0], use interval bisection twice to find an interval of width 0.125 which contains $\alpha$.\\
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to $\mathrm { f } ( x )$ to obtain an approximate value of $\alpha$. Give your answer to 3 decimal places.\\

\hfill \mbox{\textit{Edexcel FP1 2012 Q2 [10]}}

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