Question 14 - A-Level Maths

AQA Paper 1 2024 June — Question 14 10 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Fixed Point Iteration
Type	Staircase/cobweb diagram
Difficulty	Standard +0.3 This is a standard A-level iteration question covering routine techniques: change of sign to locate a root, algebraic rearrangement, applying an iterative formula with a calculator, and drawing a convergence diagram. All parts follow textbook procedures with no novel problem-solving required. The algebraic manipulation in part (b) is straightforward, and the conceptual understanding needed (why iteration fails at x=0) is basic. Slightly easier than average due to its highly procedural nature.
Spec	1.06g Equations with exponentials: solve a^x = b 1.09a Sign change methods: locate roots 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams 1.09d Newton-Raphson method

The equation $$x^3 = e^{6-2x}$$ has a single solution, $x = \alpha$ By considering a suitable change of sign, show that $\alpha$ lies between 0 and 4 [2 marks]
Show that the equation $x^3 = e^{6-2x}$ can be rearranged to give $$x = 3 - \frac{3}{2}\ln x$$ [3 marks]
1. Use the iterative formula $$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$ with $x_1 = 4$, to find $x_2$, $x_3$ and $x_4$ Give your answers to three decimal places. [2 marks]
2. Figure 1 below shows a sketch of parts of the graphs of $$y = 3 - \frac{3}{2}\ln x \quad \text{and} \quad y = x$$ On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place. Label, on the $x$-axis, the positions of $x_2$, $x_3$ and $x_4$ [2 marks]
3. Explain why the iterative formula $$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$ fails to converge to $\alpha$ when the starting value is $x_1 = 0$ [1 mark]

Show mark scheme Show mark scheme source

Question 14:

Answer	Marks
14(a)	Rearranges the given equation

to equal zero and evaluates

their non-zero expression in the

interval [0,4] at least once.

Answer	Marks	Guidance
Must have equated to zero.	1.1a	M1

Let f (x)= x3 −e6−2x

f ( 0 ) = − 4 0 3  0

f ( 4 ) = 6 3 .8 6  0

Hence lies between 0 and 4

Completes argument with two

correct evaluations of their

correct expression in the interval

[0,4] either side of the solution,

with comparison to zero or a

comment about change of sign.

AND concludes that the solution

lies between 0 and 4

Evaluations must be correct to

at least two significant figures

rounded or truncated.

Answer	Marks	Guidance
Accept exact evaluation at x = 0.	2.1	R1
Subtotal	2
Q	Marking instructions	AO

Answer	Marks	Guidance
14(b)	Uses natural logs to correctly
remove exponential.	1.1a	M1

lnx3 =6−2x

3lnx=6−2x

2x=6−3lnx

3 x=3− lnx

2 Uses log rule or cube roots to

Answer	Marks	Guidance
remove power of 3.	1.1a	M1

Completes reasoned argument

3 to show x = 3 − l n x

2

Answer	Marks	Guidance
AG	2.1	R1
Subtotal	3
Q	Marking instructions	AO

Answer	Marks
14(c)(i)	Obtains any correct value to at

least 3 decimal places, ignoring

Answer	Marks	Guidance
labels.	1.1a	M1

2 x = 3 .1 2 4

3 x = 1 .2 9 1

4 Obtains x , x and x correct to

2 3 4

at least 3 decimal places

If no labels only accept the

three correct answers in the

correct order with no extras

seen beyond x

4 x =0.92055...

2 x =3.12416...

3 x =1.29125...

Answer	Marks	Guidance
4	1.1b	A1
Subtotal	2

Obtains x , x and x correct to

2 3 4

at least 3 decimal places

If no labels only accept the

three correct answers in the

correct order with no extras

seen beyond x

4

Answer	Marks	Guidance
Q	Marking instructions	AO

Answer	Marks
14(c)(ii)	Draws correct cobweb diagram

Condone missing vertical line at

Answer	Marks	Guidance
x = 4	1.1a	M1

Shows positions of x , x and x

2 3 4

on the x-axis

Accept correct values in place of

x AWRT 0.92, 3.12 and 1.29

n

Do not accept labels on y = x

Answer	Marks	Guidance
without indication on x-axis	1.1b	A1
Subtotal	2

Shows positions of x , x and x

2 3 4

on the x-axis

Accept correct values in place of

x AWRT 0.92, 3.12 and 1.29

n

Do not accept labels on y = x

Answer	Marks	Guidance
Q	Marking instructions	AO

Answer	Marks
14(c)(iii)	Explains that (it is not possible

to evaluate x as) ln 0 has no

2

Answer	Marks	Guidance
value or that y is undefined OE	2.4	E1

2 ln 0 has no value

Answer	Marks	Guidance
Subtotal	1
Question 14 Total	10
Q	Marking instructions	AO

Question 14:
--- 14(a) ---
14(a) | Rearranges the given equation
to equal zero and evaluates
their non-zero expression in the
interval [0,4] at least once.
Must have equated to zero. | 1.1a | M1 | x 3 = e 6 − 2 x  x 3 − e 6 − 2 x = 0
Let f (x)= x3 −e6−2x
f ( 0 ) = − 4 0 3  0
f ( 4 ) = 6 3 .8 6  0
Hence lies between 0 and 4
Completes argument with two
correct evaluations of their
correct expression in the interval
[0,4] either side of the solution,
with comparison to zero or a
comment about change of sign.
AND concludes that the solution
lies between 0 and 4
Evaluations must be correct to
at least two significant figures
rounded or truncated.
Accept exact evaluation at x = 0. | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b) ---
14(b) | Uses natural logs to correctly
remove exponential. | 1.1a | M1 | x3 =e6−2x
lnx3 =6−2x
3lnx=6−2x
2x=6−3lnx
3
x=3− lnx
2
Uses log rule or cube roots to
remove power of 3. | 1.1a | M1
Completes reasoned argument
3
to show x = 3 − l n x
2
AG | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c)(i) ---
14(c)(i) | Obtains any correct value to at
least 3 decimal places, ignoring
labels. | 1.1a | M1 | x = 0 .9 2 1
2
x = 3 .1 2 4
3
x = 1 .2 9 1
4
Obtains x , x and x correct to
2 3 4
at least 3 decimal places
If no labels only accept the
three correct answers in the
correct order with no extras
seen beyond x
4
x =0.92055...
2
x =3.12416...
3
x =1.29125...
4 | 1.1b | A1
Subtotal | 2
Obtains x , x and x correct to
2 3 4
at least 3 decimal places
If no labels only accept the
three correct answers in the
correct order with no extras
seen beyond x
4
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c)(ii) ---
14(c)(ii) | Draws correct cobweb diagram
Condone missing vertical line at
x = 4 | 1.1a | M1
Shows positions of x , x and x
2 3 4
on the x-axis
Accept correct values in place of
x AWRT 0.92, 3.12 and 1.29
n
Do not accept labels on y = x
without indication on x-axis | 1.1b | A1
Subtotal | 2
Shows positions of x , x and x
2 3 4
on the x-axis
Accept correct values in place of
x AWRT 0.92, 3.12 and 1.29
n
Do not accept labels on y = x
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c)(iii) ---
14(c)(iii) | Explains that (it is not possible
to evaluate x as) ln 0 has no
2
value or that y is undefined OE | 2.4 | E1 | It is not possible to evaluate x as
2
ln 0 has no value
Subtotal | 1
Question 14 Total | 10
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item The equation
$$x^3 = e^{6-2x}$$
has a single solution, $x = \alpha$

By considering a suitable change of sign, show that $\alpha$ lies between 0 and 4
[2 marks]

\item Show that the equation $x^3 = e^{6-2x}$ can be rearranged to give
$$x = 3 - \frac{3}{2}\ln x$$
[3 marks]

\item \begin{enumerate}[label=(\roman*)]
\item Use the iterative formula
$$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$
with $x_1 = 4$, to find $x_2$, $x_3$ and $x_4$

Give your answers to three decimal places.
[2 marks]

\item Figure 1 below shows a sketch of parts of the graphs of
$$y = 3 - \frac{3}{2}\ln x \quad \text{and} \quad y = x$$

On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.

Label, on the $x$-axis, the positions of $x_2$, $x_3$ and $x_4$
[2 marks]

\begin{tikzpicture}[scale=1.5, >=stealth]
    
    % Draw axes
    \draw[->] (-0.6, 0) -- (4.8, 0) node[below] {$x$};
    \draw[->] (0, -0.6) -- (0, 4.8) node[left] {$y$};
    
    % Origin label
    \node[below left] at (0,0) {$O$};
    
    % Tick mark for x = 4
    \draw (4, 0.06) -- (4, -0.06) node[below=2pt] {$4$};
    
    % Draw the line y = x
    \draw[domain=-0.4:4.4, smooth, variable=\x] plot ({\x}, {\x});
    
    % Draw the curve y = 3 - 1.5 * ln(x)
    \draw[domain=0.35:4.6, samples=100, smooth, variable=\x] plot ({\x}, {3 - 1.5*ln(\x)});
    
\end{tikzpicture}

\item Explain why the iterative formula
$$x_{n+1} = 3 - \frac{3}{2}\ln x_n$$
fails to converge to $\alpha$ when the starting value is $x_1 = 0$
[1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2024 Q14 [10]}}

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