A-LevelMathematicsNumerical methods9709/32 · Feb/Mar 2018 · Q79 Marks

By sketching suitable graphs, show that the equation e²ˣ = 6 + e⁻ˣ has exactly one real root.

(i)By sketching suitable graphs, show that the equation e²ˣ = 6 + e⁻ˣ has exactly one real root.[2]
(ii)Verify by calculation that this root lies between 0.5 and 1.[2]
(iii)Show that if a sequence of values given by the iterative formula xₙ₊₁ = ½ ln(1 + 6e⁻ˣⁿ) converges, then it converges to the root of the equation in part (i).[2]
(iv)Use this iterative formula to calculate the root 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 (Feb/Mar 2018). The mark or mark type is shown in brackets.

[B1] Sketch a relevant graph, e.g. y = e²ˣ
[B1] Sketch a second relevant graph, e.g. y = 6 + e⁻ˣ, and justify the given statement
[M1] Calculate the value of a relevant expression or values of a pair of relevant expressions at x = 0.5 and x = 1
[A1] Complete the argument correctly with correct calculated values
[B1] State a suitable equation, e.g. x = ½ ln(1 + 6e⁻ˣ)
[B1] Rearrange this as e²ˣ = 6 + e⁻ˣ, or commence working vice versa
[M1] Use the iterative formula correctly at least once
[A1] Obtain final answer 0.928
[A1] Show sufficient iterations to 5 d.p. to justify 0.928 to 3 d.p., or show there is a sign change in the interval (0.9275, 0.9285)

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