AQA C3 2010 January — Question 2 11 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2010
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeSolve trigonometric equation via iteration
DifficultyStandard +0.3 This is a structured multi-part question on inverse trig functions and iteration that guides students through each step. Part (a) requires sketching arcsin and drawing a line (routine C3 skills), part (b) is straightforward substitution to verify an inequality, and part (c) involves standard iteration with calculator work and drawing a cobweb diagram. While it covers several techniques, each part is scaffolded and uses familiar C3 methods without requiring novel insight or complex problem-solving.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams
2 [Figure 1, printed on the insert, is provided for use in this question.]
    1. Sketch the graph of \(y = \sin ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
    2. By drawing a suitable straight line on your sketch, show that the equation $$\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$$ has only one solution.
  2. The root of the equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) is \(\alpha\). Show that \(0.5 < \alpha < 1\).
  3. The equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) can be rewritten as \(x = \sin \left( \frac { 1 } { 4 } x + 1 \right)\).
    1. Use the iteration \(x _ { n + 1 } = \sin \left( \frac { 1 } { 4 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
    2. The sketch on Figure 1 shows parts of the graphs of \(y = \sin \left( \frac { 1 } { 4 } x + 1 \right)\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
Question 2:
Part (a)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct shape graph passing through origin, stopping at \(A\) and \(B\)B1 Correct shape passing through origin and stopping at \(A\) and \(B\)
\(A\!\left(1,\ \dfrac{\pi}{2}\right)\)B1
\(B\!\left(-1,\ -\dfrac{\pi}{2}\right)\)B1 SC \(A(1, 90)\) and \(B(-1,-90)\) scores B1
Part (a)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Line intersecting the curve (positive gradient, positive \(y\)-intercept)M1 One solution only, stated or indicated on sketch — must be in the first quadrant (curve intersects line once). Must have scored B1 for graph in (a)(i)
Correct statementA1
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
LHS\((0.5)=0.5\), RHS\((0.5)=1.1\); LHS\((1)=1.6\), RHS\((1)=1.3\)M1 Allow \(f(0.5)<0\), \(f(1)>0\)
At \(0.5\): LHS \(<\) RHS; At \(1\): LHS \(>\) RHS \(\therefore 0.5 < \alpha < 1\)A1 CSO
Alternative 1: \(f(x) = \sin^{-1}(x) - \dfrac{1}{4}x - 1\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(0.5) = -0.6\), \(f(1) = 0.3\) (AWRT)(M1) \(f(x)\) must be defined
Change of sign \(\Rightarrow 0.5 < \alpha < 1\)(A1)
Alternative 2: \(f(x) = \sin\!\left(\dfrac{1}{4}x+1\right) - x\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(0.5)=0.4\), \(f(1)=-0.1\)(M1) \(f(x)\) must be defined
Change of sign \(\Rightarrow 0.5 < \alpha < 1\)(A1)
Alternative 3: \(f(x) = 4\sin^{-1}x - x - 4\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(0.5)=-2.4\), \(f(1)=1.3\)(M1) \(f(x)\) must be defined
Change of sign \(\Rightarrow 0.5 < \alpha < 1\)(A1)
Part (c)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x_2 = 0.902\)M1 Sight of AWRT \(0.902\) or AWRT \(0.941\)
\(x_3 = 0.941\)A1 These values only
Part (c)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Staircase diagram shownM1 Staircase (vertical line) from \(x_1\) to curve, horizontal to line, vertical to curve
\(x_2, x_3\) at approximately correct positions on \(x\)-axisA1 \(x_2, x_3\) approx correct position on \(x\)-axis 
# Question 2:

## Part (a)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape graph passing through origin, stopping at $A$ and $B$ | B1 | Correct shape passing through origin and stopping at $A$ and $B$ |
| $A\!\left(1,\ \dfrac{\pi}{2}\right)$ | B1 | |
| $B\!\left(-1,\ -\dfrac{\pi}{2}\right)$ | B1 | SC $A(1, 90)$ and $B(-1,-90)$ scores B1 |

## Part (a)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Line intersecting the curve (positive gradient, positive $y$-intercept) | M1 | One solution only, stated or indicated on sketch — must be in the first quadrant (curve intersects line once). Must have scored B1 for graph in (a)(i) |
| Correct statement | A1 | |

## Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| LHS$(0.5)=0.5$, RHS$(0.5)=1.1$; LHS$(1)=1.6$, RHS$(1)=1.3$ | M1 | Allow $f(0.5)<0$, $f(1)>0$ |
| At $0.5$: LHS $<$ RHS; At $1$: LHS $>$ RHS $\therefore 0.5 < \alpha < 1$ | A1 | CSO |

**Alternative 1:** $f(x) = \sin^{-1}(x) - \dfrac{1}{4}x - 1$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.5) = -0.6$, $f(1) = 0.3$ (AWRT) | (M1) | $f(x)$ must be defined |
| Change of sign $\Rightarrow 0.5 < \alpha < 1$ | (A1) | |

**Alternative 2:** $f(x) = \sin\!\left(\dfrac{1}{4}x+1\right) - x$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.5)=0.4$, $f(1)=-0.1$ | (M1) | $f(x)$ must be defined |
| Change of sign $\Rightarrow 0.5 < \alpha < 1$ | (A1) | |

**Alternative 3:** $f(x) = 4\sin^{-1}x - x - 4$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.5)=-2.4$, $f(1)=1.3$ | (M1) | $f(x)$ must be defined |
| Change of sign $\Rightarrow 0.5 < \alpha < 1$ | (A1) | |

## Part (c)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_2 = 0.902$ | M1 | Sight of AWRT $0.902$ or AWRT $0.941$ |
| $x_3 = 0.941$ | A1 | These values only |

## Part (c)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Staircase diagram shown | M1 | Staircase (vertical line) from $x_1$ to curve, horizontal to line, vertical to curve |
| $x_2, x_3$ at approximately correct positions on $x$-axis | A1 | $x_2, x_3$ approx correct position on $x$-axis |
2 [Figure 1, printed on the insert, is provided for use in this question.]
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = \sin ^ { - 1 } x$, where $y$ is in radians. State the coordinates of the end points of the graph.
\item By drawing a suitable straight line on your sketch, show that the equation

$$\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$$

has only one solution.
\end{enumerate}\item The root of the equation $\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$ is $\alpha$. Show that $0.5 < \alpha < 1$.
\item The equation $\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$ can be rewritten as $x = \sin \left( \frac { 1 } { 4 } x + 1 \right)$.
\begin{enumerate}[label=(\roman*)]
\item Use the iteration $x _ { n + 1 } = \sin \left( \frac { 1 } { 4 } x _ { n } + 1 \right)$ with $x _ { 1 } = 0.5$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.
\item The sketch on Figure 1 shows parts of the graphs of $y = \sin \left( \frac { 1 } { 4 } x + 1 \right)$ and $y = x$, and the position of $x _ { 1 }$.

On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x _ { 2 }$ and $x _ { 3 }$ on the $x$-axis.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C3 2010 Q2 [11]}}
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