AQA Paper 1 2023 June — Question 13 9 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply iteration to find root (pure fixed point)
DifficultyStandard +0.3 This is a straightforward multi-part question on inverse functions and Newton-Raphson iteration. Part (a) requires simple recall that arccos has domain [0,1]. Part (b) involves basic sketching and recognizing that if x = cos x, then arccos(cos x) = x = arccos x (simple function composition). Part (c) is a routine application of the Newton-Raphson formula with f(x) = x - cos x, requiring three iterations from x₀ = 0—a standard textbook exercise with no conceptual challenges.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09d Newton-Raphson method

13 The function f is defined by $$\mathrm { f } ( x ) = \arccos x \text { for } 0 \leq x \leq a$$ The curve with equation \(y = \mathrm { f } ( x )\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-18_842_837_550_603} 13
  1. State the value of \(a\) 13
  2. (i) On the diagram above, sketch the curve with equation $$y = \cos x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ and
    sketch the line with equation $$y = x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ 13 (b) (ii) Explain why the solution to the equation $$x - \cos x = 0$$ must also be a solution to the equation $$\cos x = \arccos x$$ Question 13 continues on the next page 13
  3. Use the Newton-Raphson method with \(x _ { 0 } = 0\) to find an approximate solution, \(x _ { 3 }\), to the equation $$x - \cos x = 0$$ Give your answer to four decimal places. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-21_2491_1716_219_153}

Question 13(a):
AnswerMarks Guidance
\(1\)B1 States 1
Question 13(b)(i):
AnswerMarks Guidance
Concave arc for \(0 \leq x \leq \frac{\pi}{2}\), intersecting \(y\)-axis below \(\frac{\pi}{2}\)M1 Draws concave arc; condone dotted section
\(y\)-intercept labelled 1 or \(a\)A1 Labels \(y\)-intercept of concave arc
Straight line through \(O\) at approximately \(45°\) crossing \(y = \arccos x\)M1 Draws straight line through \(O\) at approximately \(45°\) crossing given curve
All three graphs intersecting at common point, maximum of cosine graph in correct position, \(y = x\) as straight line through \(O\)A1 Shows all three graphs intersecting at a common point with correct relative positions
Question 13(b)(ii):
AnswerMarks Guidance
All three graphs intersect at the same pointE1 \(y = \cos x\) and \(y = \arccos x\) are reflections in \(y = x\); accept \(y = x\) is line of symmetry; accept all three graphs meet at same point; or starts with \(x = \cos x\) and obtains \(\arccos x = x\)
Question 13(c):
AnswerMarks Guidance
\(1 + \sin x\)B1 Obtains \(1 + \sin x\); PI by \(x_2 = 0.75036\ldots\), AWRT \(0.75\)
\(x_{n+1} = x_n - \dfrac{x_n - \cos x_n}{1 + \sin x_n}\)M1 Obtains \(x_n - \dfrac{x_n - \cos x_n}{1 \pm \sin x_n}\); ignore subscripts, condone ANS for \(x_n\)
\(x_3 = 0.7391\)A1 Obtains AWRT \(x_3 = 0.7391\); condone missing label provided this is final answer; must have scored M1
## Question 13(a):

$1$ | B1 | States 1

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## Question 13(b)(i):

Concave arc for $0 \leq x \leq \frac{\pi}{2}$, intersecting $y$-axis below $\frac{\pi}{2}$ | M1 | Draws concave arc; condone dotted section

$y$-intercept labelled 1 or $a$ | A1 | Labels $y$-intercept of concave arc

Straight line through $O$ at approximately $45°$ crossing $y = \arccos x$ | M1 | Draws straight line through $O$ at approximately $45°$ crossing given curve

All three graphs intersecting at common point, maximum of cosine graph in correct position, $y = x$ as straight line through $O$ | A1 | Shows all three graphs intersecting at a common point with correct relative positions

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## Question 13(b)(ii):

All three graphs intersect at the same point | E1 | $y = \cos x$ and $y = \arccos x$ are reflections in $y = x$; accept $y = x$ is line of symmetry; accept all three graphs meet at same point; or starts with $x = \cos x$ and obtains $\arccos x = x$

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## Question 13(c):

$1 + \sin x$ | B1 | Obtains $1 + \sin x$; PI by $x_2 = 0.75036\ldots$, AWRT $0.75$

$x_{n+1} = x_n - \dfrac{x_n - \cos x_n}{1 + \sin x_n}$ | M1 | Obtains $x_n - \dfrac{x_n - \cos x_n}{1 \pm \sin x_n}$; ignore subscripts, condone ANS for $x_n$

$x_3 = 0.7391$ | A1 | Obtains AWRT $x_3 = 0.7391$; condone missing label provided this is final answer; must have scored M1

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13 The function f is defined by

$$\mathrm { f } ( x ) = \arccos x \text { for } 0 \leq x \leq a$$

The curve with equation $y = \mathrm { f } ( x )$ is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-18_842_837_550_603}

13
\begin{enumerate}[label=(\alph*)]
\item State the value of $a$

13
\item (i) On the diagram above, sketch the curve with equation

$$y = \cos x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$

and\\
sketch the line with equation

$$y = x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$

13 (b) (ii) Explain why the solution to the equation

$$x - \cos x = 0$$

must also be a solution to the equation

$$\cos x = \arccos x$$

Question 13 continues on the next page

13
\item Use the Newton-Raphson method with $x _ { 0 } = 0$ to find an approximate solution, $x _ { 3 }$, to the equation

$$x - \cos x = 0$$

Give your answer to four decimal places.\\

\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-21_2491_1716_219_153}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2023 Q13 [9]}}