Question 8 - A-Level Maths

OCR MEI Paper 1 2021 November — Question 8 12 marks

Exam Board	OCR MEI
Module	Paper 1 (Paper 1)
Year	2021
Session	November
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Newton-Raphson method
Type	Show root in interval
Difficulty	Standard +0.3 This is a multi-part question on Newton-Raphson method with standard components: commenting on sign change argument, deriving the iterative formula, performing iterations, and explaining failure cases. While it requires understanding of the method and graph interpretation, all parts are routine applications of textbook techniques with no novel problem-solving required. The graphical analysis of failure is a common exam scenario.
Spec	1.09a Sign change methods: locate roots 1.09d Newton-Raphson method 1.09e Iterative method failure: convergence conditions

8 Kareem wants to solve the equation \(\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0\). He uses his calculator to create the following table of values for \(\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75\).

\(x\)	0	1	2	3	4	5	6
\(\mathrm { f } ( x )\)	1.750	0.361	1.875	0.263	0.480	1.670	- 0.153

He argues that because \(\mathrm { f } ( 6 )\) is the first negative value in the table, there is a root of the equation between 5 and 6 .

Comment on the validity of his argument. The diagram shows the graph of \(y = \sin 4 x + e ^ { - x } + 0.75\). \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
Explain why Kareem failed to find other roots between 0 and 6 . Kareem decides to use the Newton-Raphson method to find the root close to 3 .
1. Determine the iterative formula he should use for this equation.
2. Use the Newton-Raphson method with \(x _ { 0 } = 3\) to find a root of the equation \(\mathrm { f } ( x ) = 0\). Show three iterations and give your answer to a suitable degree of accuracy. Kareem uses the Newton-Raphson method with \(x _ { 0 } = 5\) and also with \(x _ { 0 } = 6\) to try to find the root which lies between 5 and 6 . He produces the following tables.
  \(x _ { 0 }\) 5
  \(x _ { 1 }\) 3.97288
  \(x _ { 2 }\) 4.12125
  
  \(x _ { 0 }\) 6
  \(x _ { 1 }\) 6.09036
  \(x _ { 2 }\) 6.07110
1. For the iteration beginning with \(x _ { 0 } = 5\), represent the process on the graph in the Printed Answer Booklet.
2. Explain why the method has failed to find the root which lies between 5 and 6 .
3. Explain how Kareem can adapt his method to find the root between 5 and 6 .

Show mark scheme Show mark scheme source

Question 8(a):

Answer	Marks	Guidance
Answer	Marks	Guidance
Karim has a valid argument that there is a root between 5 and 6 because there is a change of sign on his table	E1 [1]	Argues from change of sign that this argument is valid. Allow argument is not valid as he does not state that the function is continuous

Question 8(b):

Answer	Marks	Guidance
Answer	Marks	Guidance
There are two roots between 2 and 3 (and/or between 4 and 5) so there is no change of sign in the table	E1 [1]	Allow for a comment that implies changes of sign are missed

Question 8(c)(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(f'(x) = 4\cos 4x - e^{-x}\)	M1	Attempt to differentiate
So N-R formula is \(x_{n+1} = x_n - \dfrac{\sin 4x_n + e^{-x_n} + 0.75}{4\cos 4x_n - e^{-x_n}}\)	A1 [2]	cao, but condone missing subscripts in the fraction

Question 8(c)(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(x_0 = 3\); \(x_1 = 2.920853\ldots\); \(x_2 = 2.908274\ldots\); \(x_3 = 2.907846\ldots\)	M1	Produces at least two iterations
A1	Three iterations with correct values either rounded or truncated to at least 3 d.p. Root is 2.907845109 to 10 sf
The root is 2.908 to 4 s.f.	A1 [3]	Correct to at least 3 s.f. FT their values if their sequence seems to converge

Question 8(d)(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
[Tangent drawn at \(x=5\) reaching \(x\)-axis]	B1	Attempt to draw a tangent at \(x=5\) as far as the \(x\)-axis
[Second tangent drawn approximately at point where \(x = 3.97\)]	B1 [2]	Drawing the second tangent approximately at the point where \(x = 3.97\) as far as the \(x\)-axis

Question 8(d)(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
The start value is close to a stationary point [so the gradient is very small] and the tangent meets the \(x\)-axis far away from the required root	B1	Conveys the idea that the stationary point or the value of the gradient causes the problem
The sequence converges to a root, but not the required root	B1 [2]	Conveys the idea that the wrong root is found

Question 8(d)(iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Use \(x_0\) any value [between 5.28 and 5.85] which is nearer to the required root	E1 [1]	Allow for 'a starting value between 5 and 6' oe

## Question 8(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Karim has a valid argument that there is a root between 5 and 6 because there is a change of sign on his table | E1 [1] | Argues from change of sign that this argument is valid. Allow argument is not valid as he does not state that the function is continuous |

---

## Question 8(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| There are two roots between 2 and 3 (and/or between 4 and 5) so there is no change of sign in the table | E1 [1] | Allow for a comment that implies changes of sign are missed |

---

## Question 8(c)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = 4\cos 4x - e^{-x}$ | M1 | Attempt to differentiate |
| So N-R formula is $x_{n+1} = x_n - \dfrac{\sin 4x_n + e^{-x_n} + 0.75}{4\cos 4x_n - e^{-x_n}}$ | A1 [2] | cao, but condone missing subscripts in the fraction |

---

## Question 8(c)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_0 = 3$; $x_1 = 2.920853\ldots$; $x_2 = 2.908274\ldots$; $x_3 = 2.907846\ldots$ | M1 | Produces at least two iterations |
| | A1 | Three iterations with correct values either rounded or truncated to at least 3 d.p. Root is 2.907845109 to 10 sf |
| The root is 2.908 to 4 s.f. | A1 [3] | Correct to at least 3 s.f. FT their values if their sequence seems to converge |

---

## Question 8(d)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| [Tangent drawn at $x=5$ reaching $x$-axis] | B1 | Attempt to draw a tangent at $x=5$ as far as the $x$-axis |
| [Second tangent drawn approximately at point where $x = 3.97$] | B1 [2] | Drawing the second tangent approximately at the point where $x = 3.97$ as far as the $x$-axis |

---

## Question 8(d)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| The start value is close to a stationary point [so the gradient is very small] and the tangent meets the $x$-axis far away from the required root | B1 | Conveys the idea that the stationary point or the value of the gradient causes the problem |
| The sequence converges to a root, but not the required root | B1 [2] | Conveys the idea that the wrong root is found |

---

## Question 8(d)(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $x_0$ any value [between 5.28 and 5.85] which is nearer to the required root | E1 [1] | Allow for 'a starting value between 5 and 6' oe |

Show LaTeX source

8 Kareem wants to solve the equation $\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0$. He uses his calculator to create the following table of values for $\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75$.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { f } ( x )$ & 1.750 & 0.361 & 1.875 & 0.263 & 0.480 & 1.670 & - 0.153 \\
\hline
\end{tabular}
\end{center}

He argues that because $\mathrm { f } ( 6 )$ is the first negative value in the table, there is a root of the equation between 5 and 6 .
\begin{enumerate}[label=(\alph*)]
\item Comment on the validity of his argument.

The diagram shows the graph of $y = \sin 4 x + e ^ { - x } + 0.75$.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
\item Explain why Kareem failed to find other roots between 0 and 6 .

Kareem decides to use the Newton-Raphson method to find the root close to 3 .
\item \begin{enumerate}[label=(\roman*)]
\item Determine the iterative formula he should use for this equation.
\item Use the Newton-Raphson method with $x _ { 0 } = 3$ to find a root of the equation $\mathrm { f } ( x ) = 0$. Show three iterations and give your answer to a suitable degree of accuracy.

Kareem uses the Newton-Raphson method with $x _ { 0 } = 5$ and also with $x _ { 0 } = 6$ to try to find the root which lies between 5 and 6 . He produces the following tables.

\begin{center}
\begin{tabular}{ | l | l | }
\hline
$x _ { 0 }$ & 5 \\
\hline
$x _ { 1 }$ & 3.97288 \\
\hline
$x _ { 2 }$ & 4.12125 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{ | l | l | }
\hline
$x _ { 0 }$ & 6 \\
\hline
$x _ { 1 }$ & 6.09036 \\
\hline
$x _ { 2 }$ & 6.07110 \\
\hline
\end{tabular}
\end{center}
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item For the iteration beginning with $x _ { 0 } = 5$, represent the process on the graph in the Printed Answer Booklet.
\item Explain why the method has failed to find the root which lies between 5 and 6 .
\item Explain how Kareem can adapt his method to find the root between 5 and 6 .
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q8 [12]}}

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