O-LevelAdditional MathematicsFunctions4037/12 · May/June 2025 · Q98 Marks

It is given that f(x) = 2ln(3x-4), for x > a, and that f⁻¹ exists.

(a)Find the least possible value of a.[1]
(b)For your value of a, find the range of f.[1]
(c)For your value of a, find an expression for f⁻¹(x).[2]
(d)It is given that the equation f(x) = f⁻¹(x) has two roots. For your value of a, sketch the graphs of y = f(x) and y = f⁻¹(x) on the axes. Label each graph. State the intercepts of each graph with the axes. State the equations of any asymptotes.[4]

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Mark scheme

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

[B1] a = 4/3 (Award B1 when a correct answer is seen)
[B1] ℝ oe (May be in terms of f or y but not x)
[M1] f⁻¹(x) = (4+e^(x/2))/3 oe (for a complete valid method, allow a sign error)
[A1] Must be using the correct notation
[B1] Sketch y = f(x) in the 1st and 4th quadrants and appropriate asymptotic behaviour
[B1] Sketch y = f⁻¹(x) intersecting twice with y = f(x), in the 1st and 2nd quadrants and appropriate asymptotic behaviour (dep on previous B mark)
[B1] Intercepts: f(x) x-intercept at x = 5/3 (no y-intercept); f⁻¹(x) y-intercept at y = 5/3 (no x-intercept). (5/3 marked correctly on each axis or stated, and no other intercepts) (dep on first B1)
[B1] Asymptotes: f(x) vertical asymptote x = 4/3; f⁻¹(x) horizontal asymptote y = 4/3 (independent, either on the graph or at the side.)

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