October 2021 Paper 1 Q8

OCR MEICurrent spec12 marksNumerical Methods

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

\(x\)	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.

(a) Comment on the validity of his argument. [1]

The diagram shows the graph of \(y = \sin 4x + \mathrm{e}^{-x} + 0.75\).

Graph of y = sin 4x + e^(−x) + 0.75 for x from about −0.7 to 6.5: it oscillates, touching or dipping just below the x-axis near x = 1.3, between 2 and 3, between 4 and 5 and between 5 and 6

(b) Explain why Kareem failed to find other roots between 0 and 6. [1]

Kareem decides to use the Newton-Raphson method to find the root close to 3.

(c)

(i) Determine the iterative formula he should use for this equation. [2]

(ii) 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. [3]

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

(d)

(i) For the iteration beginning with \(x_0 = 5\), represent the process on the graph in the Printed Answer Booklet. [2]

The graph in the Printed Answer Booklet:

Printed Answer Booklet graph of y = sin 4x + e^(−x) + 0.75 for x from 3 to 5.5 on a grid with gridlines every 0.5; the curve has maxima near x = 3.55 and x = 5.1 and dips below the x-axis between about 4.1 and 4.6

(ii) Explain why the method has failed to find the root which lies between 5 and 6. [2]

(iii) Explain how Kareem can adapt his method to find the root between 5 and 6. [1]

Mark scheme (a)

Scheme	Marks	AO
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	2.3
	[1]

Notes

E1: 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

Mark scheme (b)

Scheme	Marks	AO
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	2.3
	[1]

Notes

E1: Allow for a comment that implies changes of sign are missed

Mark scheme (c)

Scheme

Marks

(i) \(\mathrm{f}^\prime(x) = 4\cos 4x - \mathrm{e}^{-x}\)

1.1b

So N-R formula is \(x_{n+1} = x_n - \dfrac{\sin 4x_n + \mathrm{e}^{-x_n} + 0.75}{4\cos 4x_n - \mathrm{e}^{-x_n}}\)

1.1b

[2]

(ii)

\(x_0\)	3
\(x_1\)	2.920853…
\(x_2\)	2.908274…
\(x_3\)	2.907846…

The root is 2.908 to 4 s.f.

M1
A1
A1

1.1b
1.1b
2.2a

[3]

Notes

(i) M1: Attempt to differentiate

A1: cao, but condone missing subscripts in the fraction

(ii) M1: Produces at least two iterations

A1: Three iterations with correct values either rounded or truncated to at least 3 decimal places

A1: Correct to at least 3 s.f. FT their values if their sequence seems to converge

(root is 2.907845109 to 10 sf)

Graph with the Newton-Raphson process from x0 = 5: a vertical line from x = 5 up to the curve, the tangent there drawn down to the x-axis at about 3.97, then a vertical line up to the curve and the second tangent drawn to the x-axis — Mark scheme (d)