AQA FP1 2005 January — Question 8

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionJanuary
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton-Raphson method
TypeNewton-Raphson with derivative given or simple
DifficultyModerate -0.3 This is a straightforward application of the Newton-Raphson formula with a simple cubic function. Parts (a) and (d) require direct substitution into the standard formula, part (b) is a basic sketch, and part (c) is simple evaluation. The function is elementary (cubic polynomial) and the method is routine for FP1, though slightly above average difficulty due to being Further Maths content.
Spec1.09d Newton-Raphson method

8 [Figure 2, printed on the insert, is provided for use in this question.]
The diagram shows a part of the graph of \(y = \mathrm { f } ( x )\), where $$f ( x ) = x ^ { 3 } - 2 x - 1$$ The point \(P\) has coordinates \(( 1 , - 2 )\). \includegraphics[max width=\textwidth, alt={}, center]{a77cc9c3-5ff6-4abc-931e-e811740267f2-05_606_565_717_740}
  1. Taking \(x _ { 1 } = 1\) as a first approximation to a root of the equation \(\mathrm { f } ( x ) = 0\), use the NewtonRaphson method to find a second approximation, \(x _ { 2 }\), to the root.
  2. On Figure 2, draw a straight line to illustrate the Newton-Raphson method as used in part (a). Mark \(x _ { 1 }\) and \(x _ { 2 }\) on Figure 2
  3. By considering \(f ( 2 )\), show that the second approximation found in part (a) is not as good as the first approximation.
  4. Taking \(x _ { 1 } = 1.6\) as a first approximation to the root, use the Newton-Raphson method to find a second approximation to the root. Give your answer to three decimal places.
    (2 marks)

Question 8:
Part (a)
AnswerMarks
\(f'(x) = 3x^2 - 2\)B1
\(x_2 = 1 - \frac{-2}{1} = 3\)M1A1 (3 marks)
Part (b)
AnswerMarks
Tangent at \(P\) drawnB1
\(x_1\) and \(x_2\) shown correctlyB1 (2 marks)
Part (c)
AnswerMarks Guidance
\(f(2) = 3 > 0\), so root \(< 2\)E2,1 (2 marks) E1 for incomplete explanation
Part (d)
AnswerMarks
\(x_2 = 1.6 - \frac{-0.104}{5.68} \approx 1.618\)M1A1 (2 marks)
## Question 8:

**Part (a)**
$f'(x) = 3x^2 - 2$ | B1 |
$x_2 = 1 - \frac{-2}{1} = 3$ | M1A1 (3 marks) |

**Part (b)**
Tangent at $P$ drawn | B1 |
$x_1$ and $x_2$ shown correctly | B1 (2 marks) |

**Part (c)**
$f(2) = 3 > 0$, so root $< 2$ | E2,1 (2 marks) | E1 for incomplete explanation

**Part (d)**
$x_2 = 1.6 - \frac{-0.104}{5.68} \approx 1.618$ | M1A1 (2 marks) |

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8 [Figure 2, printed on the insert, is provided for use in this question.]\\
The diagram shows a part of the graph of $y = \mathrm { f } ( x )$, where

$$f ( x ) = x ^ { 3 } - 2 x - 1$$

The point $P$ has coordinates $( 1 , - 2 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a77cc9c3-5ff6-4abc-931e-e811740267f2-05_606_565_717_740}
\begin{enumerate}[label=(\alph*)]
\item Taking $x _ { 1 } = 1$ as a first approximation to a root of the equation $\mathrm { f } ( x ) = 0$, use the NewtonRaphson method to find a second approximation, $x _ { 2 }$, to the root.
\item On Figure 2, draw a straight line to illustrate the Newton-Raphson method as used in part (a).

Mark $x _ { 1 }$ and $x _ { 2 }$ on Figure 2
\item By considering $f ( 2 )$, show that the second approximation found in part (a) is not as good as the first approximation.
\item Taking $x _ { 1 } = 1.6$ as a first approximation to the root, use the Newton-Raphson method to find a second approximation to the root. Give your answer to three decimal places.\\
(2 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2005 Q8}}