A-LevelMathematicsNumerical methods9709/32 · May/June 2021 · Q1010 Marks

The diagram shows a trapezium ABCD in which AD = BC = r and AB = 2r. The acute angles BAD and ABC are both equal to x radians. Circular arcs of radius r with centres A and B meet at M, the midpoint of AB. [Figure X.X]

(a)Given that the sum of the areas of the shaded sectors is 90% of the area of the trapezium, show that x satisfies the equation x = 0.9(2 – cos x) sin x.[3]
(b)Verify by calculation that x lies between 0.5 and 0.7.[2]
(c)Show that if a sequence of values in the interval 0 < x < π/2 given by the iterative formula x_(n+1) = cos⁻¹(2 - x_n / (0.9 sin x_n)) converges, then it converges to the root of the equation in part (a).[2]
(d)Use this iterative formula to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B1] State or imply CD = 2r – 2r cos x
[M1] Using correct formulae for area of sector and trapezium, or equivalent, form an equation in r and x
[A1] Obtain x = 0.9(2−cos x) sin x
[M1] Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 0.7
[A1] Complete the argument correctly with correct values
[B1] State a suitable equation, e.g. cos x = 2 − x/(0.9 sin x)
[B1] Rearrange this as x = 0.9 sin x(2−cos x)
[M1] Use the iterative process correctly at least once
[A1] Obtain answer 0.62
[A1] Show sufficient iterations to at least 4 d.p.to justify 0.62 to 2 d.p., or show there is a sign change in the interval (0.615, 0.625)

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