A-LevelMathematicsNumerical methods9709/23 · May/June 2021 · Q711 Marks

The diagram shows the curve with parametric equations x = 4t + e^(2t), y = 6t sin 2t, for 0 ≤ t ≤ 1. The point P on the curve has parameter p and y-coordinate 3. [Figure X.X]

(a)Show that p = 1 / (2 sin 2p).[1]
(b)Show by calculation that the value of p lies between 0.5 and 0.6.[2]
(c)Use an iterative formula, based on the equation in part (a), to find the value of p correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.[3]
(d)Find the gradient of the curve at P.[5]

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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 AG] Equate y to 3 and confirm p = 1/(2 sin 2p)
[M1] Consider sign of p - 1/(2 sin 2p) or equivalent for 0.5 and 0.6
[A1 AG] Obtain -0.09... and 0.06... or equivalents and justify conclusion
[M1] Use iteration process correctly at least once
[A1] Obtain final answer 0.557 only
[A1] Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [0.5565, 0.5575]
[B1] Obtain dx/dt = 4 + 2e^(2t)
[M1] Use product rule to find dy/dt
[A1] Obtain 6sin 2t + 12t cos 2t
[DM1] Divide to obtain dy/dx using their dy/dt and dx/dt correctly
[A1] Obtain 0.826

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