June 2023 Paper 1 Q13

AQACurrent spec9 marksNumerical Methods Trigonometry

13 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \arccos x \quad \text{for } 0 \leqslant x \leqslant a\]

The curve with equation \(y = \mathrm{f}(x)\) is shown below.

Graph of y = arccos x, a decreasing curve from (0, π/2) on the y-axis to (a, 0) on the x-axis, with π/2 also marked on the x-axis to the right of a

(a) State the value of \(a\) [1 mark]

(b)

(i) On the diagram above, sketch the curve with equation\[y = \cos x \quad \text{for } 0 \leqslant x \leqslant \frac{\pi}{2}\]and

sketch the line with equation

\[y = x \quad \text{for } 0 \leqslant x \leqslant \frac{\pi}{2}\] [4 marks]

(ii) Explain why the solution to the equation\[x - \cos x = 0\]must also be a solution to the equation\[\cos x = \arccos x\] [1 mark]

(c) 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. [3 marks]

Scheme	Marks	AO
States 1	B1	1.2
	(1)

The curve y = arccos x from (0, π/2) to (1, 0), the curve y = cos x from (0, 1) to (π/2, 0) and the line y = x through O, all three meeting at one point — Mark scheme (b)

Scheme	Marks	AO
(i) Draws a concave arc for \(0 \leqslant x \leqslant \dfrac{\pi}{2}\) Must intersect \(y\)-axis below \(\dfrac{\pi}{2}\) Condone dotted section	M1	1.1a
Labels the \(y\)-intercept of their concave arc 1 or \(a\).	A1	1.1b
Draws straight line through \(O\) at approximately \(45^\circ\) crossing the given curve \(y = \arccos x\)	M1	1.1b
Shows all three graphs intersecting at a common point with the maximum of the cosine graph in the correct position and \(y = x\) shown as a straight line through \(O\).	A1	2.2a
	(4)
(ii) Explains that \(y = \cos x\) and \(y = \arccos x\) are reflections in \(y = x\) Accept \(y = x\) is a line of symmetry. Accept all three graphs meet at the same point. Or Starts with \(x = \cos x\) and obtains \(\arccos x = x\) Accept \(\cos^{-1} x\) for \(\arccos x\) throughout.	E1	2.4
	(1)

Mark scheme (c)

Scheme	Marks	AO
Obtains \(1 + \sin x\) PI by \(x_2 = 0.75036\ldots\) AWRT 0.75	B1	1.1b
Obtains \(x_n - \dfrac{x_n - \cos x_n}{1 \pm \sin x_n}\) Ignore subscripts, condone ANS for \(x_n\) PI by \(x_2 = 0.75036\ldots\) AWRT 0.75	M1	1.1a
Obtains AWRT \(x_3 = 0.7391\) condone missing label provided this is their final answer. Must have scored M1.	A1	1.1b
	(3)
	(9 marks)

Typical solution

\[x_{n+1} = x_n - \frac{x_n - \cos x_n}{1 + \sin x_n}\]\[x_3 = 0.7391\]