June 2022 Paper 1 Q10

OCR ACurrent spec12 marksNumerical Methods Radians

Sector OAB with centre O and angle AOB of theta radians at O; M is the mid-point of OA, marked by equal-length tick marks; the line MB is drawn

The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(OA\). The angle \(AOB\) is \(\theta\) radians. \(M\) is the mid-point of \(OA\). The ratio of areas \(OMB : MAB\) is 2:3.

(a) Show that \(\theta = 1.25\sin\theta\). [4]

The equation \(\theta = 1.25\sin\theta\) has only one root for \(\theta \gt 0\).

(b) This root can be found by using the iterative formula \(\theta_{n+1} = 1.25\sin\theta_n\) with a starting value of \(\theta_1 = 0.5\).

Write down the values of \(\theta_2\), \(\theta_3\) and \(\theta_4\).
Hence find the value of this root correct to 3 significant figures. [3]

(c) The diagram in the Printed Answer Booklet shows the graph of \(y = 1.25\sin\theta\), for \(0 \leqslant \theta \leqslant \pi\).

Use this diagram to show how the iterative process used in (b) converges to this root.
State the type of convergence. [3]

(d) Draw a suitable diagram to show why using an iterative process with the formula \(\theta_{n+1} = \sin^{-1}(0.8\theta_n)\) does not converge to the root found in (b). [2]

Mark scheme (a)

Scheme	Marks	AO
area \(OMB = \tfrac{1}{2}\left(\tfrac{1}{2}r\right)r\sin\theta\)	B1	1.1
\(2\left(\tfrac{1}{2}r^2\theta - \tfrac{1}{4}r^2\sin\theta\right) = 3\left(\tfrac{1}{4}r^2\sin\theta\right)\) OR \(2\left(\tfrac{1}{2}r^2\theta\right) = 5\left(\tfrac{1}{4}r^2\sin\theta\right)\) OR \(3\left(\tfrac{1}{2}r^2\theta\right) = 5\left(\tfrac{1}{2}r^2\theta - \tfrac{1}{4}r^2\sin\theta\right)\)	M1	3.1a
Correct equation, in two variables (ie \(\theta\) and their \(r\))	A1	2.1
\(\theta - \tfrac{1}{2}\sin\theta = \tfrac{3}{4}\sin\theta\) \(\theta = 1.25\sin\theta\) A.G.	A1	2.1
	[4]

Notes

B1: Correct (possibly unsimplified) area of \(OMB\)
Could use other than \(r\) for the radius
Could set their variable equal to \(OM\), giving a radius that is double this
eg \(OM = x\) so area \(= x^2\sin\theta\)

M1: Attempt to use ratio on two correct areas
Using two of \(OMB\) \(\left(\tfrac{1}{4}r^2\sin\theta\right)\), \(MAB\) \(\left(\tfrac{1}{2}r^2\theta - \tfrac{1}{4}r^2\sin\theta\right)\) and \(OAB\) \(\left(\tfrac{1}{2}r^2\theta\right)\) oe with their variable
Must be two correct areas
Must be using the correct ratio for their two areas ie 2:3 if using \(OMB\) and \(MAB\), 2:5 if using \(OMB\) and \(OAB\) or 3:5 if using \(MAB\) and \(OAB\)
Allow ratio to be used the wrong way around eg \(2OMB = 3MAB\)

A1: Any correct statement linking the two areas
Could use other than \(r\) for the radius
Or \(2x^2\theta - x^2\sin\theta\)

A1: Simplify to given answer
At least one line of working once ratio used

Mark scheme (b)

Scheme	Marks	AO
0.599	B1	1.1a
0.705, 0.810	M1	1.1a
root \(= 1.13\)	A1	1.1
	[3]

Notes

B1: Obtain correct first iterate
3sf or better – more accurate answer is 0.599281923...
Condone truncating if more sig fig given

M1: Attempt correct iterative process to find at least 2 more values
M1 is for the correct process for finding \(\theta_3\) and \(\theta_4\), but these may be incorrect
M0 if working in degrees

A1: Obtain 1.13
Possibly following B0 if first iterate is wrong but process then self corrects
Must follow M1 ie a clear attempt to use the correct iterative process
Must be 3sf
Once M1 is awarded, allow A1 for 1.13 even if an incorrect iterate seen, as process will recover

Graph of y = 1.25 sin theta with the line y = theta from the origin crossing it; a staircase starts from a vertical line at the starting value and steps up between the line and the curve to the point of intersection — Mark scheme (c)

Graph of y = inverse sine of 0.8 theta starting at the origin with increasing gradient, and the line y = theta crossing it; staircases on both sides of the intersection move away from the root — Mark scheme (d)

Scheme	Marks	AO
Draw graph of \(y = \sin^{-1}0.8\theta\), for \(\theta \geqslant 0\)	B1*	3.1a
Draw \(y = \theta\), and show staircase divergence from the root found in (b), on at least one side of the root	B1dep*	3.2a
	[2]