A-LevelMathematicsNumerical methods9709/31 · Oct/Nov 2025 · Q98 Marks

By sketching a suitable pair of graphs, show that the equation sec 2x = -e^(-x) has only one root in the interval 0 < x < 1/2π.

(a)By sketching a suitable pair of graphs, show that the equation sec 2x = -e^(-x) has only one root in the interval 0 < x < 1/2π.[2]
(b)Show by calculation that this root lies between 0.9 and 1.[2]
(c)Show that if a sequence of values given by the iterative formula x_(n+1) = 1/2 cos⁻¹(-e^(-x_n)) converges, then it converges to the root of the equation in part (a).[1]
(d)Use the iterative formula given in part (c) to calculate x correct to 3 decimal places. Give the result of each iteration to 5 decimal places.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

[B1] Sketch a relevant graph, e.g. y = sec 2x
[B1] Sketch a second relevant graph, e.g. y = –e^(-x) and justify the given statement
[M1] Calculate the values of a relevant expression or pair of expressions at x = 0.9 and x = 1
[A1] Complete the argument correctly with correct calculated values
[B1] State the equation x = 1/2 cos⁻¹ (-e^(-x)) and rearrange to obtain sec 2x = −e^(-x) Or rearrange sec 2x = −e^(-x) as x = 1/2 cos⁻¹ (-e^(-x)) and state x_n+1 = 1/2 cos⁻¹ (-e^(-x_n))
[M1] Use the iterative process correctly at least once
[A1] Obtain final answer 0.978
[A1] Show sufficient iterations to at least 5 d.p. to justify 0.978 to 3 d.p. or show there is a sign change in the interval (0.9775, 0.9785)

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