A-LevelMathematicsNumerical methods9709/22 · Oct/Nov 2023 · Q610 Marks

The diagram shows the curve with parametric equations x = 3 ln(2t – 3), y = 4t ln t. The curve crosses the y-axis at the point A. At the point B, the gradient of the curve is 12. [Figure 6.1]

(a)Find the exact gradient of the curve at A.[5]
(b)Show that the value of the parameter t at B satisfies the equation t = 9/(1 + ln t) + 3/2.[2]
(c)Use an iterative formula, based on the equation in (b), to find the value of t at B, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[B1] (a) Obtain dx/dt = 6/(2t-3)
[M1] (a) Use product rule to find dy/dt Allow unsimplified.
[A1] (a) Obtain dy/dt = 4lnt + 4 OE
[M1] (a) Attempt to find t corresponding to point A using a complete and correct method
[A1] (a) Obtain t = 2 and hence gradient is 2/3 ln 2 + 2/3 Or exact equivalent.
[M1] (b) Equate dy/dx to 12 and attempt rearrangement to 2t-3 = k/(4lnt+4)
[A1] (b) Confirm given result t = 9/(1+lnt) + 3/2 with sufficient detail AG
[M1] (c) Use iteration process correctly at least once Need to see 4.9626.
[A1] (c) Obtain final answer 4.96 Answer required to exactly 3 s.f.
[A1] (c) Show sufficient iterations to 5 sf to justify answer or show sign change in interval [4.955, 4.965]

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